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# Quantum Mechanics PHY 115A

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This 75 page Class Notes was uploaded by Horace McClure on Tuesday September 8, 2015. The Class Notes belongs to PHY 115A at University of California - Davis taught by Andreas Albrecht in Fall. Since its upload, it has received 68 views. For similar materials see /class/191829/phy-115a-university-of-california-davis in Physics 2 at University of California - Davis.

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Notes for Physics 115a Quantum Mechanics 2009 Andreas Albrecht Course web site httpwww nhvsics nodavis edn 39 P115a NB Notes in grey fonts are from previous years and may be updated before use this year Table of Contents Notes for Physics 115a Quantum 1 39 39 1 1 1 Intro 2 2 Probabilities 5 21 Probabilities in everyday experience 5 3 The a 39 quot 39 equation 5 31 Properties ofthe im 5 32 Discrete analogy for continuous variables 6 33 The wavefunction as a vector 7 34 Conservation of probability and Unitarity 7 4 t 10 41 The momentum operator and the uncertainty relation 10 42 Momentum and position eigenstates 12 43 de Broglie 12 44 Ehrenfest s Theorem 13 5 Energy Eigenstates stationary states 15 51 Changing Basis in BraKet notation 15 52 Energy 339 t t 18 53 Real wavefunctions for Energy 339 t t 20 6 The in nite square well 21 61 Stationary states 21 62 More about boundary Jquot 23 63 Discuss HW3 24 64 Grif ths Examples 22 and 2 3 24 7 The harmonic oscillator 26 71 Energy lower bound Grif ths problem 22 26 72 The harmonic oscillator potential 26 73 The ladder operators 27 74 Constructing the harmonic oscillator energy eigenstates 29 75 Nu quot quot 31 76 Working with ladder operators 32 77 Hermite Polvnomial 35 78 Discussion of VIquot x 36 79 Power Series Method 36 8 The Free Particle 36 81 Energy 339 t t 36 Phys 115 Lecture notes 2009 Andreas Albrecht 642009 13210 PM NJI 39 A 83 An example 84 Group velocity 9 The delta function potential 91 Bound states and scattering states 92 Delta Function Well 93 Matching a wavefunction across in nite changes in the potential 94 The bound state 95 Scattering states 96 The delta function peak 10 The nite square well 10 1 Functional form 102 Matching at the boundarie 103 Scattering states 1 1 Review 111 Review of eigenstates for at potential 12 Multiple systems 39 and t 121 Combining two svstems 122 Properties ofthe quot v2 123 Coherence and D 39 1231Nomenclature 1232 The relativity of 39 1233 Pvamp es 124 1241 General Tquot 1242 Description of quantum 1243 Quantum measurements as Hermitian operators 1244 The measurement process 1245 Interpretation 1246 Can we see the many worlds 13 More about measurements and uncertaintie 131 Compatibility of 132 Generalized uncertainty principle 133 The energytime uncertainty principle Stan Mar 31 Lecture Intro HW 1 assigned on Web Readin Assi ed on Web 1 Intro Why I am excited about teaching this course Phys 115 Lecture notes 2009 Andreas Albrecht 642009 13210 PM 38 39 39 40 41 41 43 43 44 47 48 48 49 51 52 52 54 54 56 58 58 59 60 60 60 61 62 63 63 68 71 71 72 73 QM lSt really turned me on to Physics I still think it is amazing how our predecessors just wanted to understand how the atom works and they wound up revolutionizing our understanding of the physical world New course for me not CosmoAstro I will learn from it I ve thought a lot about fundamental questions in QM I m excited to share this rite of passage with you Questions for class what year connections X2 What I eXpect from you Respect this exciting topic Get into it Work hard See me for advice if you are having trouble USE OFFICE HOURS for fun and practical help Don t worry about being a whiz Acquire a strong work ethic The weird stuff 0 Probabilities o InterferenceUncertainty principle Measurement apparatus as link between classical and quantum worlds What is reality Many Worlds Fear FactorSociologyHistorical Baggage Griffiths section 12 QM is less weird than it first looks and less weird than Griffiths claims My position not universally held even among the very best physicists Almost all awkward questions can be addressed putting the measurement apparatus into the Schrodinger equation See my following a collapsing wavefunction paper linked on the course web page PUNCH LINE Learn the technical side well That will put you in the best position to understand what is weird and what is not Thus technical knowledge is basically the sole emphasis of this course Come to office hours to talk about the weird stuff plus brief comments during lectures and a lecture or two at the end if Phys 115 Lecture notes 2009 Andreas Albrecht 642009 13210 PM 3 possible NB Also the technical side is universally agreed among all physicists 9 Review General Info handout linked on web 9 Measurement illustration to introduce next section 9 General comment about probability in everyday life 9 Classical equation for particle of mass m in l dimension dV F quot mx dx 11 9 Schrodinger equation for a particle of mass m moving in a potential V l d 8 1 hz 82 1 zh 2 V P 12 a 2m 8x 39 0 Linear equation only one power of IP 0 IF x 2L is the wavefunction complex 0 139 J 1 34 2 h 1054572 X 10 m kg s J39Sis P1anck s constant technically h 27th is Planck s Constant and h is the Reduced Planck s Constant but few quibble about the words The letter you write says exactly what you mean The probability of finding the particle between positions x1 and x2 is PX1X2lI2Txl11xldl 13 the Born rule 0 IF x 2L can be negative so adding wavefunctions can reduce probability in a given region NB due to linearity adding two solutions to Eqn 12 gives another solution Give an example at the board Phys 115 Lecture notes 2009 Andreas Albrecht 642009 13210 PM 4 NB I replace Griffiths section 12 with Eqn 13 Plus technical discussion of how we match quantum states with our perceptions of the world including what constitutes a measurement 0 NB Don t ask how we can talk about QM in a classical way as in Griffiths section 12 ask how classical world emerges from a fundamentally quantum picture One real number total CM vs one complex number at every point in space QM Adding wavefunctions can reduce probability in a given location NB due to linearity adding two solutions to Eqn 12 gives another solution N Probabilities 21 Probabilities in everyday experience Discuss section 13 from text plus additional related material not in section 13 using Slides httpwwwphvsicsucdaviseduCosmologvP1 15aNotesSl Probppt I Start April 2 Lecture I Continuous parameters Finish Slides Discuss probability density at the board Discuss the units of a wave function when there are varying numbers of continuous coordinates 3 The Schrodinger equation 31 Properties of the wavefunction In class problem to be done in small groups Consider the following particle wavefunctions A lltxltl Plx00 lleI 31 and Phys 115 Lecture notes 2009 Andreas Albrecht 642009 13210 PM 5 B l lt x S 0 I Zx0 B 0ltxltl 32 0 x 21 a Find the values of A and B that normalize the wavefunctions so that the total probability of finding the particle anywhere is unity in each case b Plot the probability density vs x for each wavefunction Now consider a new wavefunction lPSE PI P2 33 c 1s T3 normalized d Plot the probability density vs x for T3 Discuss how adding wavefunctions can be very different from adding probabilities 32 Discrete analogy for continuous variables This section draws on appendiX A Let us consider the discrete analogy of the wavefunction in the continuous variable x This was discussed at the end of the previous lecture but not included in these written notes ll1116196 621 X IIX AC 34 where the sum runs over some discrete approximation to the continuous variable x For all we know real space might be discrete at some very small length scale so there should be no loss of generality by using Eqn 34 for sufficiently fine grids One can absorb the Ax into the wavefunction by defining 12 E 1199 NE 35 which allows one to write ZW xiWxiAxZlgtivgti 36 i Phys 115 Lecture notes 2009 Andreas Albrecht 642009 13210 PM 6 Start April 7 Lecture Collect HW 1 33 The wavefunction as a vector One can think of xix as a vector in a vector space We write such vectors as i or 17 The vector space of interest for quantum mechanics has the inner product or dot product written as 17 17 The quantities 171 are the components of the vector in the basis which correspond to the discrete positions q The inner product expressed in this basis is given by V 9 E lt gtJwlxlxdx 37 This implies that xix 17 This is called the Hermitian conjugate The transpose comes about because you need to multiply a row vector times a column vector to get the sum of components given in Eqn37 By convention 17 is a column vector The 139 W 38 and the probability that the particle can be found at position q is given by Mixilmgkilmz 217571 9 Wxi xidx 39 is the transpose and complex conjugate of components 4 are given by 34 Conservation of probability and Unitarity We want time evolution not to change the fact that the total probability is unity v9 This is sometimes called conservation of probability That means we want WI9Igtv7l019l02v90119013m V9ZV9V9 1 310 Phys 115 Lecture notes 2009 Andreas Albrecht 642009 13210 PM 7 One of the principles of quantum mechanics is that the time evolution from to the state vector at a later time I V I is given by a linear operatorl an initial state vector llgtTl 312 Taking the Hermitian conjugate 1Ilt 90ITTI 313 where the T denotes the Hermitian conjugate complex conjugate and transpose of the operator Equation 311 becomes Wm TTT lt gt0 1 314 TlT 1 315 Eqn 315 is the statement that T is unitary OI Compare With the scalar case T T 1 Th1s 1s true in general 1f 1396 T 6 3 1 6 for any real value of 6 Generalizing to the matrix case What about e161 0 0 0 39 39 39 0 317 0 0 0 eieN This certainly obeys Eqn 316 but one need not look at T in the basis where it is diagonal The full generalization of Eqn 316 is Tt 6100 Where 01 0 0 is Hermitian so that when you do diagonalize T the Q s are real 318 1 In general one can talk about T I2 3 I1 which evolves the system from ll to 2 Phys 115 Lecture notes 2009 Andreas Albrecht 642009 13210 PM 8 Now the Schrodinger equation is about differential amounts of time evolution To talk about those we can perform a Taylor expansion for small values of t given by At T ei m 1i 1AtOAt2 319 where 01 must also be Hermitian This implies that 6T E Z 91 320 or 3 A A Vlgtz 1wgt 321 a Note that 91 has units of inverse time By convention we take the i to the other side giving a l 3t lAgt 1 17 322 It turns out that for quantum mechanical systems with classical analogues 91 is nothing other than the quantum equivalent of the classical energy or Hamiltonian Of course one has to get the units right and this is accomplished by constructing HQ h 1 3 23 which also establishes the sign convention This leads to the standard form of the Schrodinger equation a A A 2h5 w H w 324 As long as His Hermitian time evolution according to Eqn 324 will V9 For the standard 1 dimensional Schrodinger equation given by Eqn 12 the Hamiltonian is h2 62 2m 3x This operator can be shown to be Hermitian once we understand how to preserve the normalization of 325 2 r interpret so this is a special case of Eqn 324 and thus it is not x Phys 115 Lecture notes 2009 Andreas Albrecht 642009 13210 PM 9 surprising that Griffiths verifies conservation of probability in section 14 We have derived a more general result in this section namely that whenever H is Hermitian time evolution according to Eqn 324 will preserve normalization or conserve probability 1n the very common case where H is time independent required for conservation of energy one can integrate Eqn 324 to find z39Ht2 t1 h T12t1 e 326 This uses math facts such as Eqn A91 in Griffiths 4 Momentum 41 The momentum operator and the uncertainty relation The momentum operator is give by h a l ax 41 Some facts about the momentum operator 0 Why Because it works fits data 0 XI 73 Hi This is usually discussed in terms of the commutator xp Exp pxzih 42 Proof operate on a wavefunction Xpl ihxa iy xw if x a x a Z l V l 4 3 8x 8x 39 139 h 1 SO Start April 9 Lecture Remind HWZ is assigned due Tuesday Questions re HWl Phys 115 Lecture notes 2009 Andreas Albrecht 642009 13210 PM 10 Xpih 44 uncertainty principle One can show see eqn 3 62 in text that for any operators A and B 2 l 2 2 a a 2 AB A B 211 1 for momentum and position this leads to h a a 2 x p 2 can t know both x and p with precision 45 46 a What is If you know p has real eigenvalues must have x P 1 47 so 11 a 1 h a 48 1 3x 1 3x thus 11 a 1 h a 49 1 3x 1 3x so 2 I 2 410 3x 3x I a Can relate Eqn 410 to discrete verslon of the a operator CAREFUL 0 h a dx ltpgt 1216 411 I00 dx I 00 1 3x W Phys 115 Lecture notes 2009 Andreas Albrecht 642009 13210 PM 11 42 Momentum and position eigenstates To work on in class Using 1 Aei 412 evaluate p 1 413 p and p2 Note you should just leave A2 a 414 J 00eioc etioch or 2 1 LL e e alx without worrying about how to do the integral 0 What do these results tell us about 1 Discuss how llis an eigenstate of p with eigenvalue hk o How well determined is position in a momentum eigenstate o What is a position eigenstate Need xwa an 416 satisfied by Wu 6x a 417 o How well determined is momentum in a position eigenstate note that the Fourier transform of a delta function is a constant 9 a delta function contains all possible momenta 43 de Broglie wavelength Definition h 27th P P the wavelength of the quantum momentum eigenstate corresponding to classical momentum p 22 418 Phys 115 Lecture notes 2009 Andreas Albrecht 642009 13210 PM 12 i27rh663gtlt1039 m 419 p P kg ms In groups find an example of an object with a mass and a speed which has a de Broglie wavelength of 1m Related to this week s HW problem 24 NOTE The noncommutation ofx andp is one ofthe weird aspects of quantum mechanics in addition working with probabilities and wavefunctions that add to give nonadditive behavior of the probabilities Start April 14 Lecture Collect HWZ Return HW1 solutions on MyUCD Finish discussion of Ehrenfest s thm from last lecture 44 Ehrenfest39s Theorem Worked out at the start of section 15 in Griffiths d x p m lt gt 420 dt Also here is problem 17 from Griffiths worked out algebraically Prove d a p V 421 dt 3x 44pr ltwljplwgtltwlp vgtj 22gt NB use partial derivative when the expression depends on more than one variable to indicate differentiation with respect to only one variable now use the Schrodinger equation a H 51W El W 43923 and its Hermitian conjugate 304 2 ltV 424 at ih Phys 115 Lecture notes 2009 Andreas Albrecht 642009 13210 PM 13 in Eqn 422 to get gum ltwlewgtltwllewgt 1 425 gllH 91 plugging in the Hamiltonian gives 1ltHpgt ilt i V149 if if 2m 426 ltpgt ltVaplgt Since p commutes with itself commutative property of derivatives the first term is zero and we get 29 731K191 ltV3 3Vgt 427 41 which is the desired result Note 1 have used the chain rule to get to the 4th line of Eqn 427 if you find this confusing rewrite this with wavefunctions and integrals over x Basically the upshot is that for quantum systems that correspond to known classical systems the equations for the quantum operators typically Phys 115 Lecture notes 2009 Andreas Albrecht 642009 13210 PM 14 correspond to the classical equations for the variables Then you can just take the expectation value of the operator equations and prove Ehrenfest s theorem namely that the expectation values obey the classical equations But note a couple of caveats o Discus how a symmetrical wavefunction starting at the top of a symmetrical upsidedown potential obeys the trivial Ehrenfest i i Z equations dz p V gt 0 and dz x 0 Thus Ehrenfest s Theorem does not necessarily help you fully understand the emergence of classical behavior in quantum systems 0 There is not always a unique meaning to the equations for the quantum operators typically correspond to the classical equations for the variables For example if the classical equation has a term of the form xzpz should the corresponding quantum expression be szz or Xpo etc These are different expressions when expressed as quantum operators Note that this feature does not represent a fundamental problem with quantum mechanics as some people tried to teach me many years ago There simply are different quantum theories that have the same classical limit We are not able to differentiate right from wrong among these options when only observing classical phenomena but so what 5 Energy Eigenstates stationary states 51 Changing Basis in BraKet notation Here is are a few technical points about linear algebra in the braket notation Just as when we discuss vectors in our 3d space one can express the vectors in different bases or coordinates 3d Example A vector a in a 3d coordinate system One can consider a basis or coordinate system given by the orthonormal mutually orthogonal and normalized basis vectors 51 52 e3 Orthonormal means that Phys 115 Lecture notes 2009 Andreas Albrecht 642009 13210 PM 15 5 5y 51 The coordinates of the vector 5 in the coordinate system given by the basis 51 are given by at 51 i 52 We often give a as an ordered triple of its coordinates alt gta1a2a3 53 The dot product can be written in abstract form a J 54 Or in terms of coordinates a I all1 azb2 a3b3 55 One can also express d in a different basiscoordinate system given by For example if one rotates the X and y axes counterclockwise by an angle 6 the new basis vectors can be written in the original coordinates as 5139 lt gt sin cos 5139 12 625 3 lt gt cos sin60 l 52 3 56 52 lt gt 001 2 a g 22 g Bases in BraKet notation Vectors A vector space can be spanned by a basis which obeys the orthonormality relation ltiJgt6y lt57 A vector can be lll expressed in terms of its components Melt1wgtlt2wgtlt3lwgts 68gt Similar to Eqn 55 the inner product of two vectors can be express in terms of components ltW2W1gtZZltW2igtltiW1gt 59 Phys 115 Lecture notes 2009 Andreas Albrecht 642009 13210 PM 16 One basis commonly used for particles are x1 the eigenstates of the position operator X xlxigtxixigt 510 The standard particle wavefunction I x is the continuum limit of the components of lll in the xigtas discussed in section 32 Bases in BraKet notation Operators A linear operator A is an object which maps a ket lll into a new ket given by yAy 511 One can work in a specific basis in which case A is represented by a matrix Al E ilAl j and the operation of A on vector lll is represented by matrix multiplication on the stated expressed as the components of this is written as ltzIw39gt2A1ltjlwgtzltiIAIjgtltjlwgt am Often one writes J J AWMMUFW1 em and one can see that Eqn 512 is achieved by dotting Eqn 513 by lll from the right and 1quot from the left and using Eqns 511 and 57 Also one can write ltM WM MMMW WWAUWgtltM For any operator A one can define a complete orthonormal space of eigenstates 1411 defined by AAiAA 515 and the orthonormality condition AilAjgt6ij 516 Phys 115 Lecture notes 2009 Andreas Albrecht 642009 13210 PM 17 One can use the Aigtas a basis in which case At takes a special form J AAkltAkAAjgtltAjZAJgtAJltAJI 517 J A1 A5 518 t y In this case Where A1 are the eigenvalues and the matrix Al is diagonal in the eigenvector basis In the diagonal basis ltAgtltwIJgtAltjlwgt W Note that in this section A has different meaning depending on the number of subscripts A1 are the eigenvalues of A and A1 are the matrix elements of A In the special case where the matrix elements are expressed in the eigenbasis the two are related by Eqn 518 Note also that the continuum notation 0 gtxlt A Low xAxtxdx 520 assumes operator A can be expressed in diagonal from Axin the xbasis compare with Eqns 514 and 52 Energy Eigenstates Start April 16 Lecture Clarify from HWZ o Tt is the same for all states in same system in same vector space called for QM a Hilbert space 0 Tt I gtT TITT t Hermitian conjugation In algebraic form the Schrodinger equation is given by 3 17 4E 17 521 a h The hermitian operator H can be diagonalized producing eigenvalues E and the corresponding eigenstates By definition these obey Phys 115 Lecture notes 2009 Andreas Albrecht 642009 13210 PM 18 HEEE 522 since His hermitian the E are real For bound systems namely particles trapped in wells in the potential V x the spectrum of energy eigenstates is discrete but often still countably infinite so we can use the discrete index i on the energy eigenstates and eigenvalues without apologies Assuming as we probably will do for the entire course that H is time independent the Schrodinger equation is trivial to integrate for the energy eigenstates giving IEJIgteXp iEjIhEj0gt 523 note that here the I refers to the time evolution of the vector not the eigenvlaue One can eXpress the energy eigenstates as wavefunctions that is write them in the basis of X eigenstates to get 1 x x E 524 So 1 xz eXp iEjIh l j x0 525 Griffiths uses the convention that lower case l is wj x 2 PJ x0 526 The energy eigenstates are often called stationary states because the probability density does not change with time 1 xt 1 jxt eXp iEj Ejth 1 j x0 1 j x0 HP MOW x30 The energy eigenstates form a complete basis so any wavefunction or state can be expanded in terms of energy eigenstates l l l0gtZlt1j l l0gtyjgtEch1jgt 528 J J and its time evolution can be easily determined l l lgtchexp iEjIhljgt 529 139 So even though the energy eigenstates are stationary the really tell you all about the time evolution of any state in the space Note that what is stationary is the probability density but the sum in Eqn 529 is a sum over 527 Phys 115 Lecture notes 2009 Andreas Albrecht 642009 13210 PM 19 many wavefunctions before squaring and as we have already seen adding wavefunction does not necessarily mean adding probability densities Also note the difference between I llj and is that yEiz0 530 which allows us to express the time evolution explicitly in Eqn 529 or equivalently allows us to not write Eqn 529 using the right hand side of Eqn 530 which would be more clumsy In Class Problem similar to Griffiths Example 21 A quantum system starts out in the following combination of energy eigenstates l l l0gta111gta212 531 For simplicity we take both at s real and we normalize with 0112 a 1 What are the probabilities of finding the system in states IS as a function of time without loss of generality we can assume sl VII are both real see section 53 for specifics Answer lts l tgt a1 exp iElthlts11 a2 exp iEZI hlts12gt532 So lts l lgt2 011011 sly12 012a2 ltsw2gt2 2011012 cos E2 E1IhltSV1gtltSV2gt 533 Note how assuming E1 7 E2 a significant time dependence of the probability has emerged due to the fact that different stationary states were combined 53 Real wavefunctions for Energy Eigenstates Consider the expansion of an energy eigenstate in a particular basis EjltkEjkEZkakJk 534 One can write each expansion coefficient in terms of real and imaginary parts a a j mid 535 Phys 115 Lecture notes 2009 Andreas Albrecht 642009 13210 PM 20 One can plug Eqn 534 into Eqn 522 and note that as with any equation the real and imaginary parts must each separately be equal That means that IEJgtR E gam 536 I 1 Ej akjk 537 are separately energy eigenstates the key here is that the eigenvalues are real so Eqn 522 does not miX the real and imaginary parts of the am In particular when the basis is the position eigenstates one can chose the wavefunctions yj x to be realz Of course once one chooses that convention one must stick to it and not readjust to real wavefunctions at a later time After all the compleX phase contains all the information about the time evolution 6 The infinite square well 61 Stationary states We now consider the case treated in Griffiths section 22 0 0 S x S a V x 61 00 otherw1se Classically this corresponds to a frictionless particle that bounces elastically from each boundary Boundary conditions Discussed further in section 62 o l continuous a 0 at can be discontinuous at the barrier edge o l 0 under the barrier Stationary states find solutions to 2 There still is an ambiguity under 1 gt l One just has to make a choice and stick to it 11 and l represent exactly the same state Phys 115 Lecture notes 2009 Andreas Albrecht 642009 13210 PM 21 1 12 dy E 62 2m a x2 W Or fly k2 63 dxz 1 E w1th k and assuming can be proven E gt 0 Note that not surprisingly we are looking for momentum eigenstates inside 2 P the well H 2 there We ve already discussed that momentum In eigenstates take the form wk 2 A eXp z kx 64 Here we will follow the convention that requires the stationary states to be real which gives eigenstates of the momentum operator llk Ak coskx wk Bk sinkx 65 The subset of these that obeys the boundary conditions specifically ya 0 is llk Bk sinkx k 2 66 a Normalizing appropriately gives the following complete set of stationary states F mz s1n x OSxSa ynx a a 67 0 otherwise With corresponding eigenvalues hk 112 th E 2 2m 21710 n 68 Discussion of Eqns 67 and 68 o The functions are alternatively even or odd with respect to the center of the well corresponding to even or odd 1 0 Number of nodes places where Vx 0 n 1 Phys 115 Lecture notes 2009 Andreas Albrecht 642009 13210 PM 22 o Orthonormal Ixj Iva 611 69 0 Complete fx chyn x 610 111 0 with 6 ltwn fIvxfxdx 611 NB this is only true for f x obeying the same boundary conditions 0 Thus the general solution to the Schrodinger equation is given by l l lgtchexp iEjlh1jgt 612 f 39ust repeating Eqn 529 Stalt April 21 Lecture 0 Hand back HWZ comments from reader I Check units check that answer makes sense He worked hard to nd mistakes will not do so in the future I Learn Bra Ket notation look over HW solutions if you did integrals for this HW Look at solutions and see me if you still don t understand the material For grading issues contact reader first Collect HW3 1 HW 4 Assigned a On April 23 I will discuss the midterm coming up on April 30 62 More about boundary conditions How general is the completeness Eqn 610 Consider a wavefunction which is only nonzero under the bariier All cquot 0 so such wavefunctions are not treated by this complete set Still the stationaIy states we have found are complete for wavefunctions inside the well Phys 115 Lecture notes 2009 Andreas Albrecht 642009 13210 PM 23 Further discussion of boundary conditions using slides 63 Discuss HW3 Prob 31 Supposing a finite universe introduces discreteness in quantities that we are used to think of as continuous but the discreteness appears on such a fine scale that we would never have noticed it so far Prob 32 Here we see that putting in the barrier removes have of the old eigenstates but these get replaced with new states See here httpwwwchemucieduundergradappletsdwelldwellhtm for an illustration using a nite barrier As the barrier is raised you can see the states continuously deform in the direction of the states you found for the infinite central barrier Used these slides httpwwwphvsicsucdaviseduCosmologvPl lSaNotesSZ SquareWellB Csppt 64 Grif ths Examples 22 and 23 Discussion based on text Key points 0 Expansion in energy eigenstates ch eXp iEjIh wj 613 Eqn 529 with c gjokm ngyx lx gjokm xa xazx 0 11 even 614 8JEn7r3 11 odd 0 Energy eXpectation value ltEgticlzEl ml 615 which gives 0 Dominance of lowest energy eigenstate Phys 115 Lecture notes 2009 Andreas Albrecht 642009 13210 PM 24 2 qu 8JZ 21998555m 61a 7139 0 Periodic behavior 1 sz 1 xl Zcck exp iEj EkIh w x wk x617 jk Each term has a period Tjk E 618 IE1 Ekl Maximised for minimum IE j Ekl The energy difference is minimized for large k and j which differ by unity approaches E1 as k and j get large Periodic behavior generalizes to recurrences in any nite system due to finite number of energy eigenvalues and thus finite number of frequencies in Eqn618 optional not on exams Periodic behavior of universe optional not on exams RU lP 2 SU T tPe tPe 2 1025m10 35m 120 45 10 619 tPe 10 e s Discussion based on Matlab program parabolam 0 Plot time evolution of probability density Discuss What is the minimum possible expectation value for the energy of all possible states for a possible particle in an infinite square well E Elci 392 E 2E1 620 using the convention where E Emin Phys 115 Lecture notes 2009 Andreas Albrecht 642009 13210 PM 25 7 The harmonic oscillator 71 Energy lower bound Grif ths problem 22 For all energy eigenstates for a potential V El gtminV 71 Rough proof Rewrite the eigenvalue equation as 2 d 1 2m VxElv 72 dxz 7 12 If El lt minV then Vx E gt 0 everywhere Discuss pictorially how a wavefunction which obeys Eqn 72 cannot be normalized Start April 23 Lecture 0 Discuss MT1 handout also posted on Web 72 The harmonic oscillator potential Classically a particle moving in a potential 1 Vx3kx2 73 is knows as the harmonic oscillator ie a pendulum for small amplitudes or a mass on an ideal spring The oscillation frequency is k a 74 In so one can write Vx mwzx2 75 The eigenvalue equation time independent Schrodinger equation for a particle moving in the potential given in Eqn 75 is 1 12 fly 1 2 2 ma x E 76 2m dxz 2 V V Or H1E1 77 Phys 115 Lecture notes 2009 Andreas Albrecht 642009 13210 PM 26 with p lma2x2 78 2m 2 73 The ladder operators Define f and iby3 1 1 tha 2 i2 79 2p 2 j lt NB I use f and i as an intermediate step to de ning a i Gri iths does not but he gets to the same de nitions of ai To work in class What are the dimensions of f and i and what is if i Compare with Xp if Note the classical versions of f and i are also useful for classical studies of the SH0 Now try and factor H If f and iwere numbers not operators one could use the expression u2v2iuv iuv 710 To factor Eqn 79 Lets try this anyway and see where it takes us Define4 E i 39N N 7 11 di J5 zp X the reason for the peculiar looking sign convention will become clear below To work in class iWhatis 5551 aa1 ii What is al 3 The tilde parameters are defined by separately equating the mom entum and potential terms My INK is the same as Griffith s gused in his section 232 4 This is the same Eqn247 of Griffiths but I have chosen to define the dimensionless p and x variables in order to get there in a way I think is more convenient and commonly used in physics Phys 115 Lecture notes 2009 Andreas Albrecht 642009 13210 PM 27 Now let us see what the analog of Eqn 710 gives M 2215 ix 715 32 1 N N W w Ep2 x2 zxp pX 1 712 152 i2 iif 2 2132 222 1 So H 2 ha 0101 713 In general by definition of the cccommutator BAAB AB 714 So apt a7a aa aft l 715 Which means we can also write Hhw aal ha aaj 716 Thus the time independent Schrodinger or H eigenvalue equation reads szhw aa jyha aa jy2Ey 717 Now to the punch line It turns out that if llE is an eigenstate with eigenvalue E then a lIE is an eigenstate with eigenvalue E 0 and a lIE is an eigenstate with eigenvalue E wh Proof of the case Phys 115 Lecture notes 2009 Andreas Albrecht 642009 13210 PM 28 afla 0101 lll 73918 2 01 H hwI 2 01 E hwI Ehaat In class Prove the case The a i are called the raising and lowering operators BTW This is why the positive and negative signs were assigned to a i the way they were in Eqn 711 74 Constructing the harmonic oscillator energy eigenstates First off it seems we have proved that there can be energy eigenstates of arbitrary low energy just keep operating with a This is not in con ict with Eqn 71 because it turns out as discussed in Section 71 these eigenstates will not be normalizable Note if we have a non normalizable wavefunciton it is impossible to make it normalizable by operating with a 1ifiampgt1 6H 719 J5 J5 ax 39 Multiplying by x or differentiating will not make a nonnormalizable function normalizable There is one way out There is one wavefunction that has the property that Phys 115 Lecture notes 2009 Andreas Albrecht 642009 13210 PM 29 gill0 0 That terminates the chain so that subsequent operations with a don t produce different solutions Find IIO by solving gill0 ZLEEN ijVo 0 J2 ax Or a 2 l0 xVo 3x Integrating gives I M I id W0 Or 2 ln10 7 consl Which gives W0 2 AeXp i2 2 Normalizing and converting to x gives 14 m0 m0 We 0151 In class problem What is EO answer 1 ha HVo ha 0 3 7W0 where we have used Eqn 721 So E 2 1ha 2 0 NB HWO gmon hw1o 720 721 722 723 724 725 726 727 728 729 So the zero comes from the wavefunction being zero not the energy being zero Phys 115 Lecture notes 2009 Andreas Albrecht 642009 13210 PM Have evaluated EO we can use Eqn 718 to construct the higher energy eigenvalue which gives E 2 ha 17 1 730 75 Normalization 1f is the normalized energy eigenstate constructed by n Anaf WU 731 where Aquot is chosen to get the normalization right We can check the normalization of the next state5 01 n2 n aim 732 l ltn wrap 17 1 So a ngtnlnlgt 733 and Am1 Aquot 734 17 1 GiVing l A 2 735 So 5 1n the first step of Eqn 732 we use 01 2 CL and solve Eqn 713 for H Phys 115 Lecture notes 2009 Andreas Albrecht 642009 13210 PM 31 n af wo 736 17 Also6 angtr ltnlafa In H l n n lt 2139 gt 737 l l ltnln3 3Ingt 17 So anZn 1 738 76 Working with ladder operators Ladder operators are useful for evaluating all kinds of expectation values between energy eigenstates Here are some expressions that will be useful i a7 a 5 739 f 2 a 07 739 2 And H aha ha 2 H l aft 740 ha 2 i a7a a 01 ha 2 New we will do Griffiths example 25 6 1n the first step of Eqn 737We use Eli 2 0 and solve Eqn 717 for H Phys 115 Lecture notes 2009 Andreas Albrecht 642009 13210 PM 32 ltnIVIngth ltnIi2Ingt ha Tltna aa 01 n ha Tltna a aa7 a aa n ltnl2lngtltnlaa aa 1ltnn 2gt1 n1n 2 2gtj hwE I 2 11 2 2 741 Stan April 28 Lecture 0 Take all questions re MT1 Also ltningtfltnlw agtngt 2ltnlalngtltn a n 742 Jnn1mltnn1 Now consider 2 0 Phys 115 Lecture notes 2009 Andreas Albrecht 642009 13210 PM 33 1 743 ltlt4a mgtltn a 4 Giving l J nlm l Jn nml lt lt gt lt 744 l lt 6mmil H I l 16mm1gt Note The first line of Eqn 744 is not defined for m 0 However the last line of Eqn 743 is well defined since we know that a7 390 0 Also the last line of Eqn 744 is correct for all allowable m and n In general to avoid concerns about treating the special case of a7 390 0 one can equally well think of a7 operating to the left To understand the meaning a7 operating to the left just take the hermitian na ltnlnl 745 Thus one can replace Eqn 744 with LJmn1mmnm1 J5 1 Ewn 16Mquot1 m 16m1gt which does not involve the special case conjugate of Eqn 733 746 Note that one also can use p Vhwmf 747 and Phys 115 Lecture notes 2009 Andreas Albrecht 642009 13210 PM 34 X X 7 48 1710 to translate between expectation values of the tildeed and untildeed quantities 77 Hermite Polynomials The energy eigenstate wavefunctions of the harmonic oscillator are related to some well know functions called the Hermite polynomials Griffiths discusses these in the context of an alternative analytic calculation7 of the energy eigenstates in his section 23 2 note that Griffiths s variable g is the same as my variable i One can write the harmonic oscillator energy eigenstates as 12 Wquot x 1 Hquot iexp Sc2 2 749 7m 2quotn Where HO 1 H122 H2 24562 2 750 H3 83 12E H4 164 48E2 12 Note that the even and oddness of the H n s is the same as the even or oddness of the indeX The Hermite polynomials have been well studies by mathematicians and there are many know theorems about them which must be equivalent to things you can derive about the VIquot using the algebraic techniques used in these lectures 7 We will not cover the analytic approach now but it will be applied to other systems later in the 115abc course series Phys 115 Lecture notes 2009 Andreas Albrecht 642009 13210 PM 35 78 Discussion of WW x Discuss the properties of the harmonic oscillator eigenstates using these slides SHO Probabilities and SHO momentum http wwwphysicsucdaviseduCosmologyPl lSaNotesSHOpppt http wwwphvsicsucdaviseduCosmologvPl lSaNotesSHOProbppt Points to discuss More nearly classical probability distribution as one goes to larger energies Rough notion of classical momentum as a function of position applies for large energies but note these are stationary solutions with zero expectation value for the momentum something like combining classical solutions with different phases Note that the above discussion is not sufficient to eXplain the emergence of classical behavior in quantum mechanics but it is part of the story vs infinite square well Similar decrease of wavelength with increasing E vs infinite square well Leakage into classically forbidden zone different from infinite square well Increasing E results in more broad wavefunction well gets more broad at high E unlike inf1nite square well 79 Power Series Method Read section 232 of Griffiths Not covered in this class but you should be aware of these methods You will leam and use them in other courses 8 The Free Particle 81 Energy Eigenstates Hamiltonian 2 2 2 h a 2 p 2 81 2m 2m 3x So the eigenvalue equation can be written Phys 115 Lecture notes 2009 Andreas Albrecht 642009 13210 PM 36 HVzEy L 82 dzy a x2 4621 Where k E i ZZZE 83 We ve already discussed in section 42 that the eigenstates of p are wk 2 A eXp 84 Where the momentum eigenvalues which can be positive or negative are given by p hk 85 Since the Hamiltonian depends only on p The p eigenstates given by Eqn 84 are also eigenstates of the free particle H with eigenvalues 2 k 131sz Ek p l 2m 2m This means the time dependence is given by 2 Pk xz AeXp z 87 Which describes waves traveling to the rightleft for positivenegative k 86 Interestingly the propagation speed of the waves is hk p l vquantum vclassical 2m 2m 2 We will see in section 84 that vquamm is not the speed at which a particle actually travels Start May 4 Lecture 0 Return MT 1 0 Return HW 4 o Announce Kolb Lecture Phys 115 Lecture notes 2009 Andreas Albrecht 642009 13210 PM 37 82 Normalization The energy and momentum eigenstates for a free particle given by Eqn 84 are not normalizable LO wgykdx A200 89 Thus the llk are not realistic states for a real particle However thanks to theorems from Fourier transform theory we know that we still can use the llk as a basis Realistic normalizable wavefunctions can be expressed in terms of energy eigenstates by expressions like Pxl j 5kexpikx zdk 810 which is the equivalent of Eqn 529 except here there are a continuum of energy eigenstates so we have in integral instead of a sum A normalizable wavefunction for a free particle is called a wave packet Note that according to Fourier transform theory one can determine k in Eqn 810 by performing the integral 1 w k 1 0 39kx dx 811 ltgt 71 xgtexpz lt gt Note also that the standard orthonormality expression from Fourier transform theory ICC exp iklx exp ikzxdx 5 k1 k2 812 1s equivalent to Eqn 89 except 5 0 is just a more palatable way of writing 00 Also as 1 have discussed before we don t really know if the universe is in a some massively large box or not 0 Once can equally well expand the free particle in the discrete Fourier series corresponding to the energy eigenstates of a square well 0 The differences between the time evolution under the two different descriptions will only be noticeable when the wave packet approaches Phys 115 Lecture notes 2009 Andreas Albrecht 642009 13210 PM 38 the edge of the well A true free particle will just propagate on through but a particle that is really in the well and thus expressed in terms of the finite Fourier series will bounce off the wall and come back the other way I May 6 Special Guest lecture by Prof Rocky Kolb I 83 An example Example 26 from Griffiths Consider q 0 A altxlta 813 x 3 0 otherwise What is A9Answer 1 I l2a As worked out in Griffiths 1 sinka k 814 m k lt gt Leading to 1 sinka hk2 1 ac ex 239 kx z dk 815 EVZaI k p m The time evolution of this wavefunction is exhibited in these slides http wwwphvsicsucdaviseduCosmologvPl l 5aNotesG2p104ppt 84 Group velocity The above example showed a wave packet with no average velocity Now let as assume that k is peaked around some nonzero value k0 Let us write Eqn 810 in the form Pxz j kexpikx az 816 Phys 115 Lecture notes 2009 Andreas Albrecht 642009 13210 PM 39 2 The equivalent of Eqn 816 can appear in other where here a circumstances with different functions a k Now let us expand a k as a Taylor series aka0 a39k k0 817 Where mo ak0 da 818 0 gko Keeping only the first two terms the Taylor series and changing variables to S E k k0 measuring relative to the central value at k0 one can rewrite Eqn 816 as 1 xl E J7r eXp i aol k0 seXp 139 k0 sx w ld s 819 Since the integral is the same at all times if one simply makes the substitution x gt x mm one finds 1 xl eXp i w0zk0agtg 1x agtgz0 820 Which is just the initial packet moving along at speed as Thus one has d0 hk p Vgroup velassical dk k0 m m So velassical Vgroup zvphase Note that there are corrections to Eqn 820 which involve higher terms in the Taylor expansion They typically describe 9 The delta function potential My discussion closely follows Griffiths section 25 Phys 115 Lecture notes 2009 Andreas Albrecht 642009 13210 PM 40 91 Bound states and scattering states Discuss bound states and scattering states in classical physics with drawings on the board We have seen both types of situations in quantum mechanics square well and harmonic oscillator were bound free particle was scattering in a triVial way The following is generally true of energy eigenstates for quantum systems If E lt Vx 00 and E lt Vx 00 the energy eigenstates are bound states with a discrete index If E gt Vx 00 or E gt Vx 00 the energy eigenstates are scattering states with a continuous index In many situations we can take V ioo 0 in which case the sign of the energy eigenvalue dictates whether the state is scattering or bound For discussion in class What do eigenstates look like in semiinfinite potential draw on board Answer Draw images like the ones on these slides httpwwwphysicsucdaViseduCosmologyPl lSaNotesSHOpppt Stan May 7 Lecture 0 Guest lecture by Prof Rocky Kolb on Schrodinger s alarming results Stan May 12 Lecture 0 Guest lecture by Daniel Phillips on Delta function well 92 Delta Function WeII Consider the case where Phys 115 Lecture notes 2009 Andreas Albrecht 642009 13210 PM 41 Vx a6x 91 Except at x 0 the Hamiltonian is 2 62 H zgz hzy 92 In most of space you are solving H l 2 El 1 93 dzll W Kzll E lt 0 012 if 48w E gt 0 Where k E i 94 as before and 2mE K E T 95 Note that Eqns 94 and 95 are just chosen to re ect the overall sign of E in Eqn 93 For E lt 0 the general solution is of the form yx A eXp Kx BeXp Kx 96 And for E gt 0 the general solution is of the form Vx A eXp BeXp ikx 97 The solutions will take these general forms to either side of x 0 in general with different coefficients on either side The remaining task is to match these solutions together at x 0 The results from the next section will help Start May 14 Lecture 0 Take questions Phys 115 Lecture notes 2009 Andreas Albrecht 642009 13210 PM 42 93 Matching a wavefunction across in nite changes in the potential Let us consider the integral of the Schrodinger equation over a small region around x 0 h2 5 5 5 78 dxz dxJ1 VxtxclxEJEIxdx 98 We are specifically interested in the limit as 8 gt 0 Since llis always required to be continuous the right hand side of Eqn 98 goes to zero in this limit giving if g 0121 5 5 alx2 dx ligVx xdx 99 The left side can be trivially integrated to give 011 dy g Ei5 ligVx xdx 910 If Vis continuous in this region the right side of Eqn 910 will go to zero giving the result that ymust be continuous in that case8 However for the case of the delta function potential f Vx1xdx5 gt 2a 10 911 So for this case we must have dry dry dry Zma EA 2 0 912 01x dag de if M 0 We will now use this result to find energy eigenstates for the delta function potential 94 The bound state In the case where E lt 0 one has VxAeXp KxBeXpltx xlt0 913 VxFeXp KxGeXpltx xgt0 I 8 Once can also contemplate equation 910 in the case of the infinite square well and learn that the case we considered involves a special limit where the right side remains finite but nonzero Phys 115 Lecture notes 2009 Andreas Albrecht 642009 13210 PM 43 To achieve normalizability one must set G 0 914 Also continuity requires BeXp 0 FeXp 0 915 B F So our solution is VxBeXpltx xlt0 916 1xBeXp Kx xgt0 I Now applying Eqn 912 Gives Zma 2BK 2 B h 917 mot K Note there is only one bound state Every delta function potential has exactly one bound state with wavefunction Vmot Vx h eXp ma lxlhz 918 And energy maz 9 19 2h2 39 95 Scattering states Now considerE gt 0 For x lt O the Schrodinger equation reads 0 2 2mE dx f hz w k21 920 with 2mE And the general solution is Vx A eXp B eXp ikx 922 Unlike the bound case at this point we have no reason to exclude either term Similarly for x gt 0 we can write Phys 115 Lecture notes 2009 Andreas Albrecht 642009 13210 PM 44 Vx FeXp GeXp ikx 923 The continuity of at x 0 requires FGAB 924 The derivatives are d V ikF G dx 925 0 z39kAB Earlier we derived Eqn 912 city city city Zma EA 2 0 926 01x dxig de 1312 Using Eqn 925 in Eqn 926 gives ikF G AB 2O AB 927 Or F GAl2i Bl 2i 928 Where ma 2 929 8 hzk For each C there are two equations Eqn 928 and Eqn 924 and four unknowns The remaining freedom corresponds to a large variety of possible solutions Here is a convenient way to organize these First recall that eXp ikxis a plane wave moving to the right and eXp 1706 is a plane wave moving to the left Why 1 kxAI AeXp ikx EkAzh 930 E AeXpikx Ax So the value of 1 now has the value it used to have a bit to the left Similarly 1 kxAz A eXp 2 kx EkAzh 931 E A exp i k x So 1 now has the value it used to have a bit to the right Phys 115 Lecture notes 2009 Andreas Albrecht 642009 13210 PM 45 Draw Picture similar to Grif ths figure 215 on board If we set G 0 we have removed the incoming wave from the right the solution we get will correspond to a particle incident from the left with amplitude A SolVing for B and F gives le A T 932 F A 1 m In this picture B the amplitude of the re ected wave and F is the amplitude of the transmitted wave One can consider the wave incoming from the right by setting A 0 In class project Use symmetry only to determine B and F for a particle incident from the right 36 l z F m G l i Going back to the incident from the right case the probability per unit length 933 in the incident wave islAl2 the probability per unit length in the re ected wave is IB 392 and the probability per unit length of the transmitted wave is 2 IFI From these we can construct the relative probabilities that the incoming particle will be re ected RZIBIZ z IAIZ W 934 And that the incoming particle will be transmitted Phys 115 Lecture notes 2009 Andreas Albrecht 642009 13210 PM 46 2 IF I 1 IAIZ 1 z These are called the re ection coefficient and transmission coefficient respectively Note that we must have R T l 936 Which is indeed the case Writing 8 in terms of the original parameters gives 935 R 4 937 12h2Ema2 and T 2 4 1 ma2 2h2E Note that higher E produces stronger transmission a reasonable result 938 Discuss relationship between these time independent states and actual time dependent scattering the discussion on pp 7576 of Griffiths is good 96 The delta function peak One can use similar tools to consider the same potential with the sign ipped Vx a6x 939 You loose the ground state but the logic of the scattering states remains the same and you can just take 06 gt 0 in the above discussion Since m0 8 Eqns 932 and 933 will give different answers for the factors in hzk the wavefunction but since the transmission and refection coefficients are ratios of probabilities wave function squared they wind up depending only on a2 Thus the transmission and re ection coefficients are unchanged when you ip the sign of the potential Discuss how this is very different from the classical case Discuss quantum tunneling Similar to the discussion at the end of section 25 of Griffiths Phys 115 Lecture notes 2009 Andreas Albrecht 642009 13210 PM 47 10 The finite square well My discussion closely follows Griffiths section 26 101 Functional form For convenience we place the well slightly differently on the XaXis V a S x S a Vx 0 This system has both bound and scattering states We ll start with the bound states lxl gt0 101 In the region lxl gt a the situation is similar to the exterior of the delta function well H 1 2 By 1 102 2 61 KZIl E lt 0 With K 10 3 h GiVing Vx AeXp KxBeXpKx xlt a 104 VxFeXp KxGeXpltx xgta I To achieve normalizability we set A G 0 to get Vx BeXp Kx x lt a 10 5 Vx FeXp Kx x gt a In the interior of the Schrodinger equation reads Phys 115 Lecture notes 2009 Andreas Albrecht 642009 13210 PM 48 1 106 012 512 W Where 2m EV EO 107 Remember from section 71 we know that we must have E gt V0 so I is real and positive The general solution is IxCsinlxDcoslx xlta 108 The general plane wave solutions would also work but this fOI H l gets us to our answer faster 102 Matching at the boundaries Now we must match up the solutions in the different regions We ll use continuity of I and since the potential is finite we get to use continuity of 1 as well One can prove see Griffiths problem 21c that for a symmetric potential well the solutions are either even or odd We can use that symmetry ya ilI a to relate the matching at a to that at a so lets focus on a Now we ll consider the even functions so C 0 Continuity of at 01 gives FeXp Ka 2 Boos la 109 and the continuity of 1 gives KF eXp Ka 2 1D sin la 1010 Dividing the above two equations gives Kltanla 1011 Phys 115 Lecture notes 2009 Andreas Albrecht 642009 13210 PM 49 Since both K and I depend on E we can solve Eqn 1011 for E to get the allowed energies It helps to use the variables Z E la 1012 20 2J2mVO 1013 From the definitions of K and I one can get Kaszg zz 1014 And And Eqn 1011 becomes tanz 2022 1 1015 Plot gure 218 from Griffiths and discuss 1 Wide and deep When Z0 gtgt 1 tan Z z 20 Z So 112 th 2m 2002 Which corresponds to half the energy eigenvalues of the infinite square well Remember we are only doing the even functions right now Note that for finite V0 there are only a finite number of bound states EnV0 2 1016 2 Shallow and narrow As Z0 takes on smaller values show how one curve moves to the left the total bound state number falls but there always is one remaining Note 7139 that the lowest odd state drops out for Z0 lt 3 Start May 19 Lecture 0 Take questions 0 Return HW 5 read reader s comments 0 Collect HW 6 Phys 115 Lecture notes 2009 Andreas Albrecht 642009 13210 PM 50 HW 7 assigned due June 2 o HW 8 posted not due but you must know this material for the nal 0 Announce MT 2 review sheet and discussion for Thursday 0 Special of ce hours No of ce hours on memorial day Special of ce hours 101050am Wed May 27 103 Scattering states Starting on the left VxAeXpikxBeXp ikx xlt a 1017 x 2mE h Inside the well Where k E IxCsinlxDcoslx xlta 1018 With 2 E V 15 mh 0 1019 Let s consider a scattering state incoming from the left so we remove the wave incoming from the right leaving Vx FeXp x gt a 1020 Continuity of l and I39 at a gives A eXp ika BeXp ika C sin la Dcos la 1021 And ik A map ika B eXp ika C cos la D sin 101 1022 While both continuities at a give Csin la Dcos la 2 FeXp ika 1023 and C cos la Dsin 1a 2 ikFeXp ika 1024 One can eliminate C and D and solve for B and F to give Phys 115 Lecture notes 2009 Andreas Albrecht 642009 13210 PM 51 sin 2101 B 2 2712 k2F 1025 and eXp 2ikaA F 1026 k2 2 cos 21a i 2 sin 2101 Using T E IFI2 A2and going back to the original variables gives 71 V2 201 T 1 0sm2 2mEV0 1027 4E E it Note that there is perfect transmission T l at the zeros o 39 39 occur at 2 2mEV0 2117 1028 Which gives 2 2 2 En VO 2 Liz 1029 2m 2a Sketch Grif ths gure 219 on the board and discuss Think destructive interference on re ected wave 1 1 Review Take questions re MT2 111 Review of eigenstates for flat potentials The time independent Schrodinger equation can be written as d l 2m V x E W 111 dxz hzt ltgt 1 If Vx const one has 2 d I lzcnx 112 d2 x If E gt Vthen C lt 0 and Eqn 112 is solved by Phys 115 Lecture notes 2009 Andreas Albrecht 642009 13210 PM 52 w Aexpizx 113 Which gives oscillating behavior If E lt Vthen C gt 0 and Eqn 112 is solved by VleeXp ixEx 114 which gives decaying or growing behavior If Vx is piecewise constant then the solution is produced by combining the above solutions in their relevant regions using appropriate boundary conditions BOUNDARY CONDITONS l Vx is continuous everywhere 2 Qty x is continuous everywhere if V does not have an infinite discontiunity Otherwise there may be discontinuities in axll We have seen specific examples of these in the infinite square well see section 61 and for the delta function potential see section 93 3 Vx is normalizable for bound states Comment Note that even when V x is not constant the rough characteristics described above apply For example the harmonic oscillator energy eigenstates have oscillating behavior in regions where E gt V and decaying behavior in regions where E lt V Start May 22 Lecture 9 Go over MT2 info sheet Phys 115 Lecture notes 2009 Andreas Albrecht 642009 13210 PM 53 12 Multiple systems coherence and measurement We have so far only discussed the behavior of single systems The real world is made of many physical objects including a great many harmonic oscillators for example This section introduces some of the tools that we use to handle this multiplicity of systems in quantum mechanics Along the way we will introduce the important concept of quantum coherence and introduce tools that are key to understanding quantum measurements This discussion is quite different from what is found in Griffiths so I will try and make these notes selfcontained 121 Combining two systems We have discussed the behavior of simple systems whose quantum states live in a simple vector space For example we can consider a vector space spanned by the orthonormal basis and write any state in that space in that basis as yZci 121 Now suppose we want to consider a large quantum systems made up of two subsystems designated SA and SB Suppose SA is spanned by the orthonormal basis igtA and SB is spanned by the orthonormal basis B One construct a larger vector space out of SA and SB formally denoted SA gtlt SB One can construct a basis that spans SA gtlt SB formally written igtA gtlt jgtB which contains elements Iii lib JgtB 022 for all the i s and j s that label the respective bases for SA and SB The inner product in SA gtlt SBis given by AxB vikith A ltikgtA B ltjlgtB 5ik5kl 123 Veg important It only takes a difference in one of the subsystems to cause orthognality For example Phys 115 Lecture notes 2009 Andreas Albrecht 642009 13210 PM 54 AXB3539AX3 A33A B59B 533559 lgtlt 0 0 124 The fact that subsystem SA is in the same state in both basis states in Eqn 124 does not prevent the two states from having an inner product or overlap of zero The general state in the combined space can be expanded in the basis to give Ir Iii igtA x JgtB 025 if if the 2nd equality gives an alternate form that will be useful below It is tempting to think of states in the combined space such that each system is in a specific state For example one could have system SA in state A and system SB in state 3915 NOTE from MAY 22 2009 After class I redid this section with a slightly di erent notation which I think is clearer Now A and B appear only outside the brackets and only label which subspace The fact that we are looking at potentially di erent states in each subspace indicated by the di erent Greek letters inside the brackets Let us define the eXpansion coefficients for each subsystems state in the subsystem basis as 125 gtA ZaiigtA 126 and B Zbiigt3 027 Then one could construct the following state for the combined system IIIXi gtAfgt3 ZKZailzijgajljg 128 27 Phys 115 Lecture notes 2009 Andreas Albrecht 642009 13210 PM 55 Note that it does not make sense to write IV gtA lfgtB 029 Each term in Eqn 129 is defined in a different vector space so Eqn 129 does not make sense Note that the state given in Eqn 128is not nearly as general as the general state given in Eqn 125 For example if the sizes9 of the subspaces are N A and N B respectively then Eqn 128 contains only N A N B 1 independent parameters whereas the general state Eqn 125 contains N A X N B 1 independent parameters Note that the l comes from the normalization constraint For subsystems larger than dimension 2 there are many possible states that cannot be written in the product form of Eqn 128 If the state of the combined system cannot be written in the product form the two subsystem are said to be in an entangled state Discuss examples from everyday experience such as a macroscopic pendulum interacting with the air and photons in the room where entanglement is a natural consequence of interactions with the environment 122 Properties of the wavefunction v2 Loss of coherence through correlation An illustration In Section 31 we considered a simple exercise I ve reproduced in here with the correct normalizations put in In class problem to be done in small groups Consider the following particle wavefunctions l 1 lt lt1 I lx0 J5 x 1210 0 x 21 and 9 Here size means the number of orthogonal basis states required to span the space aka the dimension of the space Phys 115 Lecture notes 2009 Andreas Albrecht 642009 13210 PM 56 lltxS0 I Zx0 0ltxltl 1211 0 J llel a Find the values of A and B that normalige the wavefunctions so that the total probability of finding the particle anywhere is unity in each case These normalizations are give this time around b Plot the probability density vs x for each wavefunction Both have constant probability densities for l lt x lt 1 Now consider a new wavefunction I l T2 3 J5 e Is lIgnormalized Yes once the xE is included as it is done in Eqn1212 but not in section 31 f Plot the probability density vs x for T3 1212 We learned that even though both states had equal probably of finding the particle for positive or negative x when the two wavefunctions were added we found that there was zero probability of finding the particle for all positive values of x Now let us consider a variation on this problem Suppose the particle is just one subsystem and is combined with another dimension 2 subsystem The second subsystem is spanned by the orthonormal basis 1 2gtB B Start May 26 Lecture a Any questions re MT2 3 Discuss Q s re reading use MT2 info sheet as a guide some of all reading will be examined but less so of the later assignments Phys 115 Lecture notes 2009 Andreas Albrecht 642009 13210 PM 57 Now consider the state for the whole system given by 1 IwltrogtgtM5ltIWIgtAIIgtBIWZgtAI2gtBgt am Now we can evaluation the probability of finding the particle in position eigenstate xgtA by evaluating le lz ltxW1gtA I1B XIV2h 2gtB2 Blt1W1xgt Blt2ltW2 xgtxV1gtA I1B xlll2h 2B Blt11gtBltW1IXXXIW1gt Blt12BltW1xgtltxlgt 0214 3 lt21gtB W2 IxXxW1gt 32I2B wzlxXlez W1x2 x2 1 The first line of Eqn 1214uses facts about how to define probabilities of subsystems which may not yet be familiar to you 2 The probability of finding the particle at x gt 0 is nonzero in this case Compare this with the original problem 123 Coherence and Decoherence 1231 Nomenclature Here are some of the words that go with the situation discussed in Section 122 Note that you can already understand this simple example just using what we have discussed in section 122 The additional words called out in bullets below just introduce you to terminology that is often used to discuss this type of situation 0 We say that the particle and system B are entangled or correlated because Eqn 1213 cannot be written in the form of a single product of states in the two subsystems 0 We also say the particle is in an incoherent state or is decohered due to its entanglement with system B Phys 115 Lecture notes 2009 Andreas Albrecht 642009 13210 PM 58 o The state lP321111P2 1215 is considered a coherent superposition 0 We attribute the cancellation of the wavefunction or probability amplitude for 0 lt x lt1 in the case where the particle was not entangled with system B to quantum coherence o A common example of such quantum coherence is often discussed in the double slit experiment where the coherent addition of the wavefunctions emerging from each slit leads to interesting patterns of cancellations and enhancements o The entanglement described by Eqn 1213prevents the cancellation in the region 0 lt x lt1 and thus we say the entanglement destroys the coherence of the particle or decoheres the particle 1232 The relativity of Decoherence Instead of Eqn 1213 we could have written IHPO 0 i W1gtAJEW2gtA 3 W1gtAJ iW2gtA 2 1216 AXE J5 Eqn 1216also has an entangled form But note that the states entangled with lgtB and 2gtB are quite different and seem to involve some of the quantum coherence we discussed for the nonentangled case For example the probability amplitude for 0 lt x lt1 cancels to zero in the state correlated with lgtB similarly the region 1 lt x lt 0 has zero probability in the state correlated with 2gtB These cancellations are the result of quantum coherence much like that discussed for the coherence case in Section 31 even though there is entanglement Note however that the probability for finding the particle at point x is the same for the wavefunctions given in Eqn 1216and in Eqn 1213 The upshot There are many interesting effects that come about when quantum systems become entangled These often are associated with the destruction of quantum coherence or decoherence However be careful The presence of decoherence or entanglement does not mean that there are no effects associated with quantum coherence as illustrated in the above Phys 115 Lecture notes 2009 Andreas Albrecht 642009 13210 PM 59 example One just has to carefully crank through the calculation and see how things turn out This section is called the relativity of decoherence because typically the decoherence is best understood in terms of which states are correlated with orthogonal states in the other system That is decoherence can be different depending on which states are entangled for example one can imagine entangling either position or momentum eigenstates of a particle 1233 Examples Discuss how interacting systems are the norm and often the interaction will change the state of at least one of the systems dramatically That will rapidly lead to quantum coherence even if you imagine starting out in a coherence nonentangled state Discuss Large pendulum interacting with air and light Double slit interacting with airlight or measuring which slit BoseEinstein condensation neutrino mixing super uidityconductiVity examples of maintaining coherence despite the environment 124 Measurement 1241 General Discussion Ask class What are important properties of a measurement quantum or otherwise Look for these points Correlate apparatus with state of system being measured Do not change the state of the system being measured Be able to read the apparatus compare with airphoton bouncing off macroscopic pendulum Able to keep reliable record of measurement Phys 115 Lecture notes 2009 Andreas Albrecht 642009 13210 PM 60 1242 Description of quantum measurement A good quantum measurement apparatus will Correlate apparatus with state of system being measured Not change the state of the system being measured as best as possible see below Be readable and enable a permanent record to be made Jp A U Here is a simple model of a quantum measurement S indicates the system being measured and A indicates the apparatus Before the measurement let us assume the system is in pure product state with the apparatus in the zero state IWO 1 M 0gtA 0217 At some time after the measurement the system and apparatus become entangled WI1gtSXA clW1gtS1gtA 02W2gtS2gtA c3lw3gts 3gtA 123918 Note that the system is now in an entangled state which is not generally the same as the initial state A key part of a quantum measurement apparatus is that it is identified with a measurement basis or pointer basis in the system space This basis the llligts s in Eqn1218 define the quantity being measured ie position or momentum The only way a measurement cannot change the state of the system is if the system starts out in a single pointer basis statelo The next best thing one can hope for in terms of the measurement not impacting the system is that the ct s given in Eqn 1218are given by cl S1ixS 1219 where llgtS is the state which appears in Eqn1217 In other words the probabilities of finding the system in the state llllgts are unchanged by the measurement although the probabilities of finding the system in linear 10 This is related to the famous theorem A quantum state cannot be cloned not covered in this class Phys 115 Lecture notes 2009 Andreas Albrecht 642009 13210 PM 61 combinations or coherent superpositions of the llligts will in general be changed More speci cally one could have written Eqn 1217 as Pz0SXA c111S c2 w2gts c3 M 0A 1220 Which would then evolve under a good measurement into Eqn 1218 with the same values for the CI s For this process the probability for finding the system in state lwigts E 2 is the same before and after the measurement 0139 With a measurement apparatus constructed in this way one can see that 3901 lzreally is the probability that the system will be measured to be in state llll gtS despite the fact that Griffiths tells you not to talk that way 1243 Quantum measurements as Hermitian operators From the System s point of view a measurement apparatus is represented by a set of pointer basis states in the system space which get correlated with the apparatus during the measurement process for an illustration of the properties of the Hamiltonian that are required to do this see my Following a collapsing wavefunction paper which is linked at the bottom of the course web page For a good measurement the correlations or entanglement must proceed without changing the probabilities of finding the system in the pointer basis states Without loss of generality on can think of the pointer basis states as eigenstates of a hermitian operator Furthermore one can assign each eigenstate the eigenvalue at given by the reading on the apparatus such as position charge volts etc corresponding to the state of the apparatus A that gets correlated with that state Thus once given the llligts s and the apparatus readings at one has all one needs to construct the hermitian operator Aging1 1221 Phys 115 Lecture notes 2009 Andreas Albrecht 642009 13210 PM 62 This is the operator the corresponds to the statement Observables are represented by hermitian operators in Griffiths section 3 2 But from the more modern point of viewl present here the relationship between measurements and Hermitian operators is a matter of convenience rather than a de ning principle One immediate benefit to defining A as in Eqn 1221 can be seen from ltWAWgtZlt leigtsaiSM IaFa Since lcl 392 E Klll is the probability of the apparatus achieving a 1222 resultai then l lA l 1 is the average or expectation value of the outcome of the experiment 1244 The measurement process Here is what happens when you measure a quantum system 1 The apparatus interacts with the system 2 The apparatus registers a single reading corresponding to at for one particular value of i The probability of a result at is given by 3901 l2 E I Igt2 where is the eigenstate corresponding to eigenvalue at and llgt is the state of the system before the measurement Contrary to the comments at the bottom of page 36 of Gri iths I do not think it is bad language to say 3901 lzis the probability of nding the system in state 3 From then on all subsequent calculations may assume that the system is in llllgts with the same value of igiven in 2 with unit probability this is the collapse of the wavefunction 1245 Interpretation Phys 115 Lecture notes 2009 Andreas Albrecht 642009 13210 PM 63 Physicists have different views of what additional words they like to associate with 3 These views are often passionately held but for the most part do not lead to practical differences in how they work with real data I ve outlined my impressions of these interpretational perspectives in the next paragraphs No doubt others will have their own versions I ve even had arguments with colleagues about which one of these to call the Copenhagen interpretation Many worlds 0r Everett The wavefunction only ever really changes in the sense of Eqn 1217 vs Eqn 1218 All possible outcomes are still represented This type of process can cascade through many subsystems such as the experimentalist noticing the outcome and recording it in her notebook But all these processes never change the fact that the full wavefunction continues to contain different branches or terms expressing different correlations such as the terms in 1218 which continue to re ect all possible outcomes The reassessment of the probabilities given in step 3 above only affects the realization of a given observer in a particular correlated state that she is on a particular branch Other states of the same observer are represented on other branches and they are busily reassessing their probabilities according to their experimental outcomes The collapse of the wavefunction only ever re ects an approximation that is convenient for a particular observer The lack of interference between branches which allows such an approximation to be a good one is a key aspect of classical behavior in a quantum system Such behavior is realized with abundance in our world and is strongly connected to the stability of the record mentioned in the section entitled Description of a quantum measurement A key aspect of this perspective is the notion that even though the state of the universe includes different copies of an observer or an apparatus which experience different outcomes of a measurement each of these copies can safely assume that their outcome is the only one and can count on their future interactions with the system to be consistent with that To understand how this works think about what practical actions an observer or apparatus might take to bolster or undermine hishersits confidence in the outcome of a measurement such as repeated remeasurement If one thinks about how these actions are re ected in the full quantum state one can see that re measurement just results in additional entanglement with new records that Phys 115 Lecture notes 2009 Andreas Albrecht 642009 13210 PM 64 are highly correlated on each branch It is this correlation of subsequent measurements that corresponds to the confidence of each observerapparatus that they may simply take a single outcome for the measurement and ignore the others Note that this confidence ie tight correlation of subsequent measurements is shared across all branches of the wavefunction even though each branch describes a different outcome If the observer wants to reexamine the situation in a more refined way it might be possible especially in specially designed situations to pick up quantum uctuations which represent interference from the other branches due to imperfections in the apparatus especially in quality of the record of the result In the many worlds picture the other branches are always there They can enable you to do a better calculation which takes into account imperfections in the apparatus vs assuming absolute wavefunction collapse However it typically the goal of designers of measurement apparatuses to avoid this effect basically errors due to quantum uctuations and indeed it is usually straightforward to reduce such errors to a completely negligible level Copenhagen There is a definite dividing line between the quantum and classical worlds Objects in the classical world generally macroscopic objects roll the dice upon the measurement of a quantum system and wind up with only one classical outcome If someone does a refined experiment of the sort described in the previous paragraph for which a many worlds advocate would need the other branches to explain the quantum uctuations the Copenhagen advocate would simply redraw the lines between quantum and classical to allow the subtle uctuations to be located on the quantum side In practice there are so many extremely classical systems that are very far from the boundary between quantum and classical behavior that the Copenhagen advocate need not worry about seeing the classical world completely consumed by these boundary shifts Assessment of the two views In practice there is no difference between the two pictures when it comes to interpreting most laboratory results There are many systems which are so solidly in the classical domain that we will never observe their quantum uctuations So while the many worlds person might say I am only Phys 115 Lecture notes 2009 Andreas Albrecht 642009 13210 PM 65 collapsing the wavefunction as an approximation that makes my calculation tractable while a Copenhagen person might say 1 am collapsing the wavefunction because this is the only correct way to describe the physical world There is in most cases no practical way of teasing out observable differences which could help us determine which is right Copenhagen advocates have to deal with the fact that the boundary between classical and quantum appears to be established by fiat and is not derivable from first principles Some hope that eventually someone will figure out how to make such a derivation while others insist that everything is fine just as it is and physicists should get used to using different rules in different domains Many worlds people need deal with the fact that when they describe the full fundamental calculation without imposing wavefunction collapse as a practical approximation they are describing a calculation which is preoccupied almost entirely with describing all the things that could have happened but did not These are all the branches that are not realized in our experience but which must be kept in the calculation at least in principle to fully describe the many worlds picture To many this seems very strange My personal view is that we should wish to understand this issue better My guess is that further insights into this issue will involve better understanding how small subsystems of the Universe which is what we are should view probabilities rather than any major change in the formulation of quantum mechanics A few physicists that 1 know express skepticism that we understand consciousness well enough to be sure that the picture of having one s brain and body split across different branches of the wavefunction and having different experiences on each branch is consistent with what we experience They not convinced that simple operations such as the repeated measurements described above that describe confidence building on the part of the apparatus or observer are the whole story in describing our conscious experience and are thus unconvinced that the many worlds are consistent with our consciousness that there is only one outcome Even though1 suspect there are plenty of exciting new discoveries probably involving novel collective phenomena of neurons on the path to better understanding consciousness my personal view is that these discoveries are highly unlikely to undermine the Everett interpretation Phys 115 Lecture notes 2009 Andreas Albrecht 642009 13210 PM 66 There are two areas where I feel the many worlds is at a distinct advantage One is the field of cosmology the study of the entire universe past present and future Many aspects of the cosmology appear to be well understood and have very successful theoretical calculations that agree well with data Many cosmology calculations involve novel boundaries between classical and quantum For example in the most successful model of the origins of cosmic structure all structure in the universe galaxies stars planets us originated as quantum uctuations of a particular field the so called in aton field The many worlds picture does fine in this context but I am much less impressed with how the Copenhagen picture manages I just don t think it is that easy to draw an absolute boundary between the classical and quantum domains in current cosmology theory However there is much we still do not understand in cosmology and it is certainly conceivable that changes to quantum physics will be needed before we get a more complete picture The other area of advantage is in quantum computation The basic point behind quantum computation is that the real world by being quantum mechanical is actually doing a much harder calculation to move forward in time as compared with than a classical one One can see this by imagining programming a Schrodinger equation on a computer one wavefunction infinity real degrees of freedom vs a classical particle just two real degrees of freedom in 1D x and p If one designs a computer in terms of classical degrees of freedom one is missing out on the opportunity to more fully harness nature which is doing the harder quantum calculation The idea behind quantum computation is to hamess the full quantum nature of the physical world to make more powerful computers The theory has already been done to show that revolutionary increases in computing power are possible To do this one must design a computer that can preserve quantum coherence of bits Classically a bit can have two possible states 1 and 0 A bit in a quantum computer must be able to be in any state of the form lbit c0 0 c11gt 1223 and the quantum coherence must be preserved as the computer evolves That demand is technically challenging but the potential rewards are great and this is a very interesting area of ongoing research Such Qbits can get Phys 115 Lecture notes 2009 Andreas Albrecht 642009 13210 PM 67 in interesting entangled states with each other and each Everett world is effectively doing one piece of a parallel computation The trick is to limit interactions with the environment so that the different Everett branches do not get entangled with nonmeasurable microscopic environment degrees of freedom such as microscopic motions in the air or in states of emitted radiation If such entanglement is strictly controlled the many branches of the computation can be measured coherently in the end Dramatic computational benefits can in principle be derived from this type of quantum parallelism Technically one can think about quantum computers from the Copenhagen point of view by simply moving the boundary between the classical and quantum worlds outside of the quantum computer but there is no doubt that the freedom from worries about these issues enabled by the many worlds picture has been an essential part of some recent advances 1 know for a fact that the person who started the current revolution in quantum computation David Deutsch of Oxford is a serious adherent to the many worlds view and I suspect that way of thinking made it much more natural for him to eXplore that territory Probably the most general advantage to the many worlds picture is that it allows the practitioner to not be intimidated by novel boundaries between quantum and classical behavior The many worlds picture offers a complete set of tools for forging ahead in any situation Adherents to the Copenhagen picture are more likely to avoid situations where the boundaries between classical and quantum are not already established from experience such as in cosmology or quantum computation Start June 2 Lecture 0 Collect HW7 3 Discuss final exam info sheet bring your own hardcopy if you want one a Draw attention to additional of ce hours posted on web Course Evaluations Otherwise Thursday Draw attention to section 111 9 ID 1246 Can we see the many worlds Phys 115 Lecture notes 2009 Andreas Albrecht 642009 13210 PM 68 We build up our picture of the physical world by making a huge number of measurements and attempting to assemble a consistent story from the results To illustrate this process consider an observer who repeatedly observes a twostate system S and records the result of the measurement in other record subsystems of the physical world R1 R2 R3 These record subsystems might be ink in a lab notebook bits on a computer disk or human memories in our brains All these are very complicated systems but we will consider them in a highly idealized and simpli ed form below Let 1gtS 2gtS represent the basis in which S will be measured and let us describe the RN subspace by RN 1gtRN 2gtRN where the 0gtRN state represents the record subsystem before any record has been made To further simplify things consider a universe U in which only three measurements are made so U SgtltRlgtltR2gtltR3 Suppose time to the system is unmeasured and is in the state l1gtS 010 1gtS a 2S 1224 giving a state for U given by Wt0gtU a10 US a 2gts0gtR10gtR20gtR3 1225 after a measurement is made in R1 is made at time t1 the state for U will in general be given by Wt1gtU 05111 1gts 1gtR1 051121gts2gt121 all 2gts 1gtR1 all 2gts 2gtR10gtR2 0gtR3 1226 look carefully at the positions of the brackets in Eqn 1226 and also in Eqn 1228 below they are important but have showed up a bit small in this document If the measurement apparatus is a good one the coefficients in Eqn 1226 will obey 1 0 0L 20 1 1 1227 0 12 0 21 0 giving Wt1gtu a101gts1gt111 a 2gtS2gtR10gtR20gtR3 1228 In general after all three measurements are made the state of U will be Phys 115 Lecture notes 2009 Andreas Albrecht 642009 13210 PM 69 Wt3gtU a131111gts1gtR11gtR21gtR3 a21112gts 1gtR11 R2 1R3 a21122gts1gtR11gtR22gtR3 a 1212gts1gtR12gtR21gtR3 3 a21222gts1gtR12gtR22gtR3 a22112gts2gtR11gtR21gtR3 a232122gts2gtR11gtR22gtR3 1229 az32212gtS2gtR12gtR21gtR3 0632 2gts 2gtR1 2gtR2 2gtR3 What one expects for the different values of 05 depends on the details Suppose S has no inclination to evolve on its own so H S 1 If all three records represent new measurements of S then one expects 010 ijkll 05in a ijkl2 1230 0 otherwise Alternatively R3 might record a different sort of measurement A check whether RI and R2 recorded the same values If 1gtR3 means yes and 2gtR3 means no then assuming the records are good ones one again eXpects Eqn 1230 to hold leading again to iv t3 gtU a101gts1gt121 1gtR21gtR3 a 2gtS2gtR12gtR21gtR3 1231 a great simplification over the general fOI H l given by Eqn 1229 One could say Eqn 1231 represents two Everett worlds each of which contains observers who are completely confident in their own state for S Phys 115 Lecture notes 2009 Andreas Albrecht 642009 13210 PM 70 but each of which have different ideas of what that state is In a state like that given by Eqn 1231 every possible crosscheck bolsters the confidence in each world that they know the state of S This idea can be extended to any number of records Of course no measurements and records are perfect and in principle some of the crossterms in Eqn 1229 will not be precisely zero in a realistic situation Then one will need the full form of Eqn 1229 to get correct answers and thus need the many worlds to eXplain one s results Whenever the role of the crossterms is measurable we do see them but those are typically situations that are very specially constructed in the lab In most situations we encounter in the macroscopic classical world the role of the crossterms is vastly below any measurable level and they can be neglected A believer in the many worlds picture will be pleased to note that one can quantitatively predict when the crossterms would be measurable in this picture and thus eXpress con dence that the quantumclassical boundary is well understood A skeptic of the many worlds picture will object to the fact that two worlds are represented when we only experience one and may not be happy with the above argument that we have no way of telling that the other worlds are there 13 More about measurements and uncertainties 131 Compatibility of measurements One can follow one measurement with another measurement possibly with a different apparatus Let us consider the case where the 2nd measurement occurs rapidly enough that the time evolution of the state between the two measurements can be neglected In general the new measurement could have a pointer basis that is different from that of the first measurement In that case the state of the system could be different possibly very different after the 2nd measurement takes place This situation is said to describe incompatible measurements and the consequence of this incompatibility is that it is meaningless to talk about the system having simultaneous values of the observables corresponding to both measurements Phys 115 Lecture notes 2009 Andreas Albrecht 642009 13210 PM 71 Mathematically the statement that two measurement devices have the same pointer bases is equivalent to that statement that the two hermitian operators that represent those measurements say A and B commute AB0 131 Eqn 131 is equivalent to saying that the measurements corresponding to A and B are compatible or equivalently that the observables represented by A and B are simultaneously measurable Following one measurement by another will not change the state of the system Example We have encountered this situation in a partial sense with the free particle Since Hp 0 the energy and momentum can be determined simultaneously If you measure E you can predict with certainty that a subsequent measurement of p would yield p 2 ix ZmE Aside from the sign of the momentum the state is already well determined by the energy measurement If we measured p first then the result of subsequent energy 2 P measurement would be completely determined to be E Similarly if AB 0 132 The corresponding measurements are incompatible Following one measurement by the other will change the state of the system See Homework 8 Example Position and moment are the classic example of incompatible observables Since Xp 2 I39ll measuring the position of a particle after measuring its momentum will dramatically change the state of the particle 132 Generalized uncertainty principle As we discussed earlier in this course and as proved in detail in section 351 of Griffiths the commutator is related to an inequality for the variances of any pair of observables Using the definition aj EltA2gt A2 133 one can derive Phys 115 Lecture notes 2009 Andreas Albrecht 642009 13210 PM 72 2 00 2 AB 134 This is a generalization consequence of observables being incompatible and gives some intuition which applies to the case where the wavefunction is not an eigenstate of either observable Start June 4 Lecture 0 Return HW7 Plan for the day Regular lecture short General questions Course evaluations of ce hours 9 i9 3 Q 9 a Draw attention to section 111 again re HW 7 133 The energytime uncertainty principle From section 353 of Griffiths eqn 372 eqn 372 eqn 373 One can differentiate the eXpectation value of hermitian operator Q to get ltQgtltWQwgtgmjelwwlth1lrgtltq lQ quotgtl 135 Using a H 1 all m 1 136 and its conjugate a H 1 1 all lt m 37 gives 0 l39 5Q H 138 dIltQgt ha Qgtltalgt lt Phys 115 Lecture notes 2009 Andreas Albrecht 642009 13210 PM 73 Note that in the case where there is no explicit time dependence of the 6Q operator Q that is 0 then Eqn 138 makes sense in a I straightforward way If 0 then the eigenstates of Q should also be eigenstates of H the stationary states so should be time dependent It seems intuitive that the degree of time dependence of should depend on the degree to which i 0 One can now set A H and B Q in Eqn 134 and combine the result with Eqn 138 to get 02p 2 T T139 or Ft 010 a a 2 1310 H Q 2 dz now if we define AE E 7H and a At Q 1311 dltQgtdrl Then Eqn 1310 can be written AEAz 2 1312 This is an uncertainty principle for E and I but note that due to the somewhat imprecise definitions especially the definition of AZ by 1311 which depends on the choice of Q this is not as concrete a result as Eqn 134 However it can be very helpful in building intuition For example an interesting perspective can be gained by rewriting the above as dltQdz SEAE 1313 a h Q Phys 115 Lecture notes 2009 Andreas Albrecht 642009 13210 PM 74 This gives a bound on the rate of change of the expectation value of any hermitian operator Q in units of its standard deviation in terms of the standard deviation of the energy distribution AE The more narrowly the energies are spread the more slowly the expectation values of all observables in fact all hermitian operators can evolve Of course in the limit of totally sharply peaked energy AE 0 nothing evolves at all The system is in a stationary state Phys 115 Lecture notes 2009 Andreas Albrecht 642009 13210 PM 75

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