### Create a StudySoup account

#### Be part of our community, it's free to join!

Already have a StudySoup account? Login here

# Introduction to Quantum Mechanics I PHY 567

Syracuse

GPA 3.93

### View Full Document

## 35

## 0

## Popular in Course

## Popular in Physics 2

This 27 page Class Notes was uploaded by Ms. Bryce Wisoky on Wednesday October 21, 2015. The Class Notes belongs to PHY 567 at Syracuse University taught by Staff in Fall. Since its upload, it has received 35 views. For similar materials see /class/225636/phy-567-syracuse-university in Physics 2 at Syracuse University.

## Popular in Physics 2

## Reviews for Introduction to Quantum Mechanics I

### What is Karma?

#### Karma is the currency of StudySoup.

#### You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

Date Created: 10/21/15

Lecture 3 Arguments for Schroedinger s equation Free particles We have seen that free particles are described by the relations E kw 1 P We 2 It is a reasonable guess that such particles should be associated with the simplest type of wave solution simple sine or cosine functions 11 sin kx at or cos kx at 3 Now k1 w the frequency These solutions describe waves moving in the direction of positive x at speed 1 Waves traveling in the opposite direction are obtained by simply switching the sign of 2 We will seek a linear wave equation that describes the time and space evolution of such waves This means any linear combination of sine and cosine functions will also be a solution Speci cally we can take the combination f kx wt cos kx wt isin kx It can be shown that this function fkx wt has a very special form fkx wt e 4 We will take this complex exponential function the free particle solution to the soughtfor wave equation 11 A expikx wt 5 Additionally notice the following result 2 8 7711 hklll l 6 i ax p Also 8 may hwll Elli 7 8t Thus very loosely the momentum of the particle can be found by differen tiating its wavefunction with respect to x and the energy by differentiating 1 with respect to t This means in turn that the classical formula for the en ergy of a free particle E 1722772 implies that the wavefunction of such a free particle will satisfy the wave equation h2 8211 811 if 1 if 2m 812 8t This is the free particle Schroedinger equation l 8 Interaction How can we generalize this to derive a wavefunction for a particle moving in some potential Just take the classical energy formula E 1722772 V and do the same replacements 2 2 187w V111 9 2m 812 at One must be careful We have not derived the Schroedinger equation in the previous pages rather we have written down the simplest linear differential equation that is consistent with conservation of energy and has simple sine or cosine like solutions in the absence of any potential energy It is an enlightened guess We must study the consequences of this equation and check them against experiment to be sure of the correctness of this equation So far this equation has proven fully consistent with all experiments notice that this equation treats time and space in distinct just did the classical expressions for energy so it is really only a nonrelativistic equation valid for speeds which are small compared to the speed of light Dirac the rst person to formulate the analogous relativistic equation which subsequently named after him Schroedinger s hopes With his wave equation Schroedinger hoped to show that quantum physics just classical phyiscs in disguise he hoped to show that particles were merely wavepackets and that the discrete energy levels of atoms were truly like the normal modes of classical waves when con ned to restricted geome tries His inspiration came not from Bohr and Heisenberg and others of the Copenhagen school but from Einstein and most principally de Broglie Ulti mately this position is not tenable it does appear to offer a bridge to the new physics from the old Certainly many physicists were much happier dealing with the Schroedinger equation which correct than with the new ma trix mechanics of Heisenberg even after they were shown to be equivalent Max Born the rst person to gure out the correct interpretation of I that its absolute square yields a probability density And the collapse of the wavepacket after measurement came later still Conservation of Probability We have argued that the wavefunction should be normalized to unity Physi cally this corresponds to the simple that the probability of nding the particle somewhere should be unity Whatever happens in the subsequent motion we expect that the total probability to nd the particle samewhere should still total to unity It is possible to prove this directly from Schroedinger s equa tion Consider i grde dt 87 Using Leibniz s rule a 8 1 8 1 6th l1 1 at at Using the Schroedinger equation for the time derivatives allows us to rewrite this II Army 7 7 8t 2m 812 812 We now notice that this expression can be rewritten 6PII a xaxp 6W 39 2 2x a m Waxy aw 8t 8x 2m 81 8x We can now integrate this over all x and see since I must go to zero at large 1 that we obtain the result 1 f 0 dt Thus if it is normalized at t 0 it will remain so probability is conserved This is an important and necessary test of the correctness of Schroedinger s approach Averages We have argued that QM only gives us access to statistical aspects of a particles motion for example TINAx is the probability of nding the particle between 1 and x Ax How can we nd its average position Standard probability theory tells us this immediately multiply the position 1 by the probability of nding it near 1 ie WWIAx and integrate the result over all positions lt x gt dxIxtxIxt 10 Notice we have assumed that the wavefunction has been normalized This is all very well but suppose I want to know not the average position of the particle but its average momentum How should I calculate that This highlights an aspect of QM that we need to discuss I have stated that the wavefunction contains all the information that is available about the quantum particle but so far we only how to calculate information related to its position In general I should like to be able to calculate the probabilities of measuring speci c values for any physical observable the mean values of those observables etc etc The general question I want to postpone till later but for now I can give you the correct prescription for computing the average momentum Recall that differentiating the wavefunction with respect to position up to constant factors the same as multiplying by the momentum In general we that in QM the momentum of a particle is replaced by an operator which in this case is just the derivative operator Speci cally QM 7L 6 if 11 p z ax Operators are mathematical objects which when applied to functions yield other functions eg the operator 1 when applied to the function f x 1 2 yields another function f x f x 13 Similarly the operator when applied to f yields 1 21 The average value of the momentum in QM is now gotten by sandwiching its associated operator between Pxt and IxJ and integrating over all 7a 8 lt gt 13f iilI 137 12 p p lt gt Z ax lt gt lt gt This is the same prescription for the average position if we just replace the simple position operator 1 by the more complicated momentum 4 operator p We will return to this issue and its generalizations later At this point you may just consider eqn12 another postulate of QM What is meant by expectation value In the previous paragraph we introduced a formula for the expectation value of some observable or operator it is represented in QM What is this It is not the result of measuring that observable for a single particle many times On the contrary if the rst measurement of the particle s position yields 1 05 then every subsequent measurement of the particle s position will yield 1 05 we assume that no other measurements are made in between and the particle is subject to no new forces Rather lt x gt measures the average result for measurements of a ensemble of particles all in the same initial state Other observables For the simple situation we have discussed so far it is also to write down expressions for other mechanical observables such the kinetic energy or angular momentum eg if 62w 22 7 1 C Lecture 10 Angular Momentum Let us now turn to QM in more than one dimension In one dimension we know how to represent the position and momentum operators when referred to the basis of eigenvectors of the position operator In three dimensions we will immediately need to know how to represent the angular mementum of a particle This is a vector quantity in classical mechanics with components LI Ly Lz where Lx ypz zpy Ly 2px m La 5 m In QM we expect it to be represented by a set of three Hermitian operators wu r quot to these r By plugging in the 1D correspondance pfr etc we nd the operators 7 t3 13 L f 39 i 27 x z sz 8y 7 t3 13 L 7 2i J7 y z 33 T82 7 8 8 L3 The rst question we should ask is what are the eigenvalues and eigenfunc tions of these operators We know that the former are the possible values of a of the con r quot r of angular momentum L and the latter are important for extracting the probabilities of measuring any one such eigenvalue We can also ask the question what is the maximal amount of information we can have concerning the angular momentum of a particle in QM that is can we know all these components precisely and simultaneously This latter question can be answered by computing the commutator of these operators with each other We know if this is zero the operators are simultaneously diagonal in the basis of their common eigenvec tors Otherwise there will be a generalized uncertainty principle governing the minimum size of the uctuations in these two operators We can easily show that Lin Ly 2mg L3 L1 ihLy My L3 iszfr From these fundamental commutation relations the entire theory of angular momentum can be deduced Evidently the different components are incom patible observables h 010 2 lt Lz gt On the other hand the square of the total angular momentum L2 Li L LE does commute with any of these components eg lL2 Lzl 0 Thus we can choose eigenstates which are simultaneously eigenstates of both L2 and L3 L2 f M2 f and M uhf De ne the two operators Li L iLy The commutators with La are lea Lil le L113 i ile Ly ihLi and of course Li commutes with L2 Imagine some function f which is an eigenfunction of L2 and La then we can show that Lif is also an eigenfunction of these two operators with eigenvalues A152 and u Uh L is termed a raising operator and L a lowering operator Thus for a given value of A the application of these two operators generates a ladder of states each one separated from its neighbors by a La eigenvalue different by h We can keep applying LL and move up the ladder Eventually however we must nd a state with the highest possible value of the Zcomponent of angular momentum For such a state 11 l and we have Lfz 0 Now we can show that LiL L L l i chLy Lny or L2 LiL L 2 15Lz Thus L2f 13212 h2l f Thus A W 1 In the same there is a bottom value of u if such that L ff 0 we nd similarly that 22 1 W 1 Thus 2 l Thus the eigenvalues of La are m7 where in runs from Z to l in 2i 1 integer steps Thus l must be integer mquot half integer Notice that the maximum component of the angular momentum l can never equal its total value W 1 since this would invalidate the generalized uncertainty principle since then we would precisely the values of Lfr and Ly zero The computation of the eigenfunctions takes a little more effort Eigenvectors We employ spherical coordinates a 7 sin 9 cos d y 7 sin 9 sin d z Tcos9 We nd a a L he cot 987 r3 3 L fie 1 0 cot 987 Notice that they are independent of the coordinate 7 thus so will be the eigenfunctions If we can solve the equation L11le 0 for the highest Lz eigenstate with L2 eigenvalue ll 1 we can always get the other 2i states by applying L Furthermore lets assume that the eigenstates can be written a product state fu F9G We nd that G ei w where m must be an integer in order for G to be single valued Furthermore since Lz ih we see in l The remaining equation for 9 reads dF 7 F d9 lcot9 VVritinglcotQ lnsinl dlnF d 71 1 d9 d9 ns1n 9 Thus 179 A sin 9 and the total eigenfunction looks like fucm sinl 9 By applying L we nd fat 1 N 8l1 sinl 19cos 9 These eigenfunctions are called spherical harmonics and play a crucial role in the states of 3D systems which are rotationally invariant Spin Usual orbital angular momentum corresponds to integral l But it is observed that particles may also possess intrinsic angular momentum analagous to rotation about an axis called spin This may be halfintegral For example the electron has spin 12 meaning that L2 12121h2 and its projection on the z axis is i h It is an example of a exact two state system similar to the ones introduced earlier to describe approximately chemical bonding In this case the representation we have derived for integral l is not valid Instead the spin operators can be represented by 2 X 2 matrices the 039 or Pauli matrices introduced in the homework A general spin state may be represented by a 2 component column vector or spinmquot 1gt X 2 2 4 The components of ix gt are the vector Z Also 01 SI10 0 1 Sym 10 Sz0 1 As we have seen these obey the fundamental commutation relations of angu lar momentum and yield the correct eigenvalues for S2 and 33 Suppose the system is known to be in the state 11 and we ask the question what is the probability of nding the particle spinup along the xaxis First we must confirm that Sfr has the same two eigenvalues i and then express the state in terms of the corresponding eigenvectors of Sm This yields 1gt 11 1 gtx a b 1gtfr X 2 2 Thus bi2 is the probability of nding the particle spin up along the xaxis Spin in a magnetic eld It is known that a particle with spin interacts with a magnetic eld B with Hamiltonian H yBS For an electron with magnetic eld in the Zdirection Schroedinger s equation takes the form d a 1 0 a mb lBh2o 1b The general state is then a linear combination X 0sa2givBt2 sin a2c i78t2 0 where we have automatically imposed a normalization condition on the state Thus lt 33 gt cos2 sin2 cos nth2 Similarly lt Sfr gt XleX with X above We nd lt Sfr gt gsin acos th and a similar result for the ycomponent We see that the spin vector classi cally precesses around the eld direction with angular frequency the Larmor frequency w 78 Addition of angular momentum Consider two spin 12 particles What is the total angular momentum for the system More generally consider two particles with angular momentum eigenvalues thl and l2m2 Think of them initially very far apart or uncoupled It is to see that there are 4 commuting angular momentum operators for this system L21 Lz1 L22 and L42 Any linear combi nations of them will also be commuting The obvious ones we are interested in are Lz Lz1 L42 and L2 L1 L2L1 L2 together with the squares of the individual angular momenta We can equally well choose these latter four the maximal commuting set The eigenstates of these will correspond to the total angular momentum of the combined system and its component along some axis The eigenstates of these operators will be built out of pmducts of eigenstates of the original angular momentum operators ilmlll2 gt Zcmlgmillml gt l2m2 gt If initially there are 2Z1 12l2 1 states for the two particles this will be preserved in this new representation Clearly the maximal value for m will be l1 Z2 This means the two particles can be found in a state with lmax 1 2 The minimal value of m will be ill Z21 In fact we can show that this yields lmin the possible min imal value of the total angular momentum quantum number Since angular momentum is quantized we expect any angular momentum in between will also be seen One to see this is to see what value of lmin will yield the correct total number of states 121 1 21112z2 1 min To make things more concrete reconsider our original example of the spin of two electrons The possible conbined state with largest zcomponent of angular momentum m 1 is are Tgt1 Tgt2 Lets operate with L L1 L2 on this We get Lgt1 Tgt2ETgtIELgt2 This has m 0 Operate again with L and we nd igt1 igt2 which has m 1 expected for an l 1 state But wait this is only 3 states and I started with four base states Also my previous prescription implies there is also the possibility of a spin zero state l 0 which must have m 0 Of course there is indeed another m 0 state orthogonal to the one we have already written down It is ETgt1lgt2 igt1Tgt2 Thus the combined system can exist in a triplet l 1 or singlet l 0 state Lecture 1 Introduction and the essence of QM Purpose of these lectures is to give you an introduction to perhaps the most important theory of modern physics quantum mechanics This theory has revolutionized the we view the microscopic world and is perhaps the best tested scienti c theory ever devised certain quantities can be calculated and compared to experiment at the level of a part in a million or better That said it is a theory that has been contested since its inception Einstein always considered it a stop gap answer to a fundamental theory God does not play dice with the Universe We will hopefully have time to discuss why Einstein wrong in his view later in this course Philosophers still debate its true meaning and even practicing physicists nd it counter to intuition I think I can safely that nobody understands quantum me chanics Richard Feynman It forms the foundation with general relativity of all of twentieth cen tury physics and underpins most of modern chemistry As one of mankind s greatest intellectual achievements it really forms an important part of our culture although very few people have any understanding of it hopefully we can address that in these lectures OK then why did QM come to be Essentially it a response to a series of crises in physics at the turn of the century 1 Blackbody radiation 2 Photoelectric effect 3 Stability of atoms and discrete emission spectra It appeared that classical physics Newton and Maxwell incompatible with some of the new experimental results following from the discovery of the internal structure of atoms This the situation in the early years of this century it took till 1926 before a satisfactory new framework devel oped which could encompass and explain these problems That framework QM Unlike relativity QM owed its birth to a number of physicists Schroedinger Heisenberg Einstein de Broglie Bohr Born Ehrenfest Dirac and others Although QM may introduce some rather unfamiliar math its basic pos tulates are not too long or terribly complicated They can be written easily on the back of a Tshirt But they do signal a dramatic departure from New tonian physics Consider the motion of a particle subject to some force in one dimension The goal of classical physics is to calculate how the position coordinate varies with time The answer can be gotten by solving the differential equation 2 d at 1V m it d3 1 If we specify eg the initial position and velocity the resulting motion can be predicted from the solution of this equation In QM the analogous cal culation is phrased very differently Newton s second law 1 is replaced by Schroedinger s equation 2 2 81 Vi 2 2m 832 8t The wavefunction cc t plays the role of the coordinate in classical physics once we know it at some time we can use Schroedinger s eqn 2 to find it for all time Because xJ is a function of both a and t we must use partial derivatives in the equation Notice that a new fundamental constant has appeared Planck s constant 7 105 X 10 3 1Js The smallness of this constant is related to the observation that we do not need to invoke quantum methods until we study the realm of the very small If it were identically zero classical physics would work even at the smallest of scales although we wouldn t be here to observe it the very stability of the atoms in our bodies is the result of quantum effects It first measured by Planck in his work on the light emitted by radiant bodies the theoretical analysis of this problem led to the need for a quantum theory But what is Q and what does it tell us about the behavior of the original quantum particle The answer is simple and yet puzzling IJIJA3 gives the probability of finding the particle to be between a and a A3 Notice that because of the square root of minus one in eqn2 the wavefunction II is a complex number hence the need to multiply it by its complex conjugate to obtain a positive definite real probability In summary providing we are happy with partial derivatives complex numbers and probabilities QM gives us a clear prescription for calculat ing quantum phenomena But notice it is according to our intuitive ideas an apparently incomplete picture QM tells us only the probabilities of mea suring certain values for the position of the particle at a specific point in space and time And that is the most we can expect to be able to know I It is a radically different picture than that of classical physics and one that many people including Einstein were are unhappy with For suppose I ask the question where the particle just before I measured its position Einstein and others would have liked to believe that it had a welldefined position described by hidden variables and QM being incomplete cannot tell me about it In effect when one averages out the hidden variables a statistical theory results which can only tell me about probabilities QM In contrast the orthodox Copenhagen position on this is that the particle does not have a position before it is measured it is a meaningless concept there are no hidden variables this is the quantum nature of matter Recently a set of experiments following on theoretical work of John Bell have served to eliminate the possibility of hidden variables theories and strengthened the orthodox position It appears that many of of everyday intuitive concepts about the world fail to describe the behavior of the quantum world The only area of QM still in debate is the nature of a measurement on a quantum system and how that is affected by a socalled observer We will discuss this later Some math Lets introduce review a few things Partial derivatives Suppose we have a function of more than one variable eg f 3 t A partial derivative with respect to a is denoted and means differentiate with respect to x holding t constant An example if fa t 37253 then 8f 3 7 2m 83 T Similarly 6f 2 2 Complex numbers Generalize notion of a real number to accommodate square roots of negative numbers De ne V l i then v54 1x4wi2i Complex number is z a iy The real part is called a and the imaginary part is 3 Its complex conjugate is 2 a iy To add complex numbers just add real and imaginary components separately eg 2 aH39b cid acibd We can multiply complex numbers follows Zaibgtlt0idagtltci2bgtltdibgtltcagtltd Since 1392 1 we then have zagtltc bgtltdibgtltcagtltd Probability Suppose I were to look at snowfall for Syracuse in the month of January during this century I could imagine constructing a histogram bar chart having xaxis the number of inches and on the yaxis the number of times that number of inches fell during all 100 Januaries on record It is easy to convert this to a picture of the pmbability distribution for snowfall during a Syracuse January just divide the numbers on the yaxis by 100 The yaxis now runs between 0 and 1 and measures the probability of a certain number of inches of snow falling Notice now that the area under the histogram is now unity Of course snowfall does not really fall in exact inch amounts one January there might have been 74 inches Indeed when we constructed the original histogram we implicitly rounded snowfalls to their nearest integer We could improve on this by recording the snowfall in 12 inch increments Then 74 would be rounded into the bar corresponding to 75 rather than 70 inches Providing we have enough data we could imagine continuing this process counting the number of Januaries with snowfall in ever decreasing small intervals Suppose we do this to an increment of 1100 inch You 4 should see that tops of all the vertical bars start to approach a continuous curve the true probability distribution that describes the possibility of any possible snowfall amount Call this curve Notice that the total area under this curve will still be unity Suppose you now want to know the probability of having between 6 and 8 inches this be read off from the area under the curve from x26 and x8 More generally suppose you want to know the probability of having a snowfall between a 6 and a 6 A3 where A3 01 This will be approximately 01 X P6 In the limit where A3 gt 0 this is exact That is A3 is the probability that 3 lies in the range a gt a A3 A3 gt 0 Back to QM Thus in QM the probability distribution Dattllatt and tells us the probability of a measurement of the particle s position resulting in a value within the range a gt a Act Notice that we now have an additional constraint on the wavefunction am we must have Wm tliatdat 1 We that the wavefunction must be narmal zed It is to see that it always possible to modify any solution to the Schroedinger equation to make this true It is also possible to show that this feature is preserved in time that is if we normalize a solution at some initial time and then evolve it in time in accord with the Schroedinger equation eqn2 it will always be so normalized This is an important consistency check on the framework 0 Lecture 6 Solving the Schroedinger equation on the computer Types of solutions We have seen that nding solutions to the full Schroedinger equation reduces to solving the time independent Schroedinger equation H IIEII where H is the hamiltonian or energy operator The solutions to this equation are of two types 0 Bound state solutions 0 Scattering solutions The former are exempli ed by solutions of the infinite square well potential we see a discrete spectrum of normalizablc allowed states A particle in such a state has zero chance of being found at in nity it is essentially confined to a certain region of the xaxis The reason for this is simply that its energy E is less than the potential at large distance E lt loo even quantum effects cannot allow it to tunnel over in nite distances The scattering solutions on the other hand allow the particle to escape to large distances and are characterized by E 2 V One example of such a state is the free particle We have seen that such states have a continuous range of possible energies E and can be represented by nonnormalizable wavefunctions like To achieve physical scattering wavefunctions we must then superpose many of these mathematical solutions to Schroedinger s equation leading to the creation of wavepackets Such wavepackets contain a set of momenta and energies centred around some central classical value and hence satisfy the uncertainty principle They are then normalizable For the moment we will concentrate on the bound states for potentials like the infinite square well these are the only possibility For most realistic potentials we will nd that is impossible to solve the time independent Schroedinger equation to find the allowed states and energies and so we must turn to numerical methods to find approximate solutions Discretization The simplest method to solve the onedimensional problem for bound states is called the shunting method First we replace derivatives with finite di cr cnccs d3 a n1 n 12 A2 Here 05 represents the value of at 2 nA2 and A2 is a small interval on the xaxis The idea is that for A2 small enough the approximation for the derivative will be accurate enough and the nal solution 05 will approximate closely the solution to the continuum equations To proceed we rst write the Schroedinger equation two rst order equations 115 d2 p dp 2m 7 7 V E 05 13 a2 I Next we apply our simple discretizatian recipe and rearrange the two equa tions into the simple form n1 nApn 2m 7Z2 This of writing the Schroedinger equation makes it clear what we have to do if we specify an initial value at 2 0 for 090 and M we can use these equations to predict their values at 2 A2 Having their values at 2 A2 we can use the equations again to get their values at 2 2A2 etc etc In this we can generate the wavefunction for all 2 What value should we choose at 2 0 If the potential is symmetric about 2 0 ie V2 V 2 it is to see that the nal wavefunction obeys one of two conditions q5 2 the socalled even parity solutions or q5 2 the add parity solutions The former are even functions and hence have gradient 17 0 at 2 0 The latter have 050 0 Furthermore we can adjust the scale of arbitrarily at this stage so that a suitably general set of initial conditions are pn1 pn39l39AJ ln E n o 050 1 p0 0 even parity 0 050 0 p0 1 odd parity Allowed energies shooting and bisection OK so now we have initial conditions and a recipe eqn 1 to generate the wavefunctions At this point you may be wondering what determines the allowed values of E We will see that only for rather special values of the energy E will iteration of these equations yield a normalizable wavefunction in general the numerical solution will yield wavefunctions which diverge very rapidly at large 2 Thus the shooting technique consists of guessing a value for E iterating the equations determining 05quot and seeing whether in some region far from the center of the potential the wavefunction is small if not we use some criterion for improving on our guess for E If we do this carefully gradually increasing the energy E we can nd all the allowed energies and associated wavefunctions One good to locate the energies precisely is called biscctian By experimenting with the local Java applet you will notice that if you have found two values of E E1 and E2 which straddle an allowed energy they will diverge with opposite signs at large a some xed point a 21 Thus if you overshoot the allowed energy the wavefunction gets large in a positive direction gt 0 while undershooting yields a wavefunction which diverges in a negative direction lt 0 Now consider the mean energy E1 E22 If its wavefunction at J 13L has the same sign the wavefunction computed at E1 you can use this midpoint energy a new guess for E If not it must have the same sign as that corresponding to E2 and you may use the midpoint energy the new E2 In either case you have now halved the region in which the true allowed energy is located By iterating this procedure many times you can locate the energy to arbitrary precision General Strategy 0 Decide a region of E in which you want to search for allowed energies Decide also the minimum possible energy 0 Guesstimate a typical energy level separation using dimensional analysis 0 Decide on a value for m the potential will contain a length scale use some multiple of this 0 Set a lower energy E to the minimum possible energy Set the upper energy E2 to the same Compute 0522L for this energy 0 Scan upward in the energy E2 using the average energy level separation as a guide until you nd changing sign an allowed energy lies now between E and E2 Bisect to nd exact energy Reset lowest energy E to just above this allowed energy 0 Repeat last three steps until you have exhausted the initial energy region of interest The C code you will use employs this strategy To change the potential you are examining you just edit the function potentialo Harmonic Oscillator We will rst use this technique to nd the allowed energies and wavefunctions of the harmonic oscillator potential V mw2r2 First we simplify the dis crete equations by working in terms of rescaled energy 6 27E Also the po tential is now of the form 1222 Initially in the code we set a 12 First we set PARITY0 even parity and compile the code When we run it the energies are approx 050 250 450 650 850 Similarly when we set PARITY1 odd parity we nd the energies 150 350 550 750 950 Thus we see that the energies 6 of the harmonic oscillator are equally spaced and for oz 12 separated by unity The ground state has nonzero energy required by the uncertainty principle The plotted wavefunctions oscillate and then decay rapidly to zero at large 2 As the energy increases the number of oscillations increases just for the in nite square well Furthermore we can show that the product of any two such states integrated over the interval is zero as required by orthogonality If we vary oz we will nd that the energy level splitting varies 2a In fact we can demonstrate numerically that 1 En 7m 71 7 2 We will see later that we can derive this result analytically However by repeating this calculation for the anharmonic oscillator V 124 we can use 4 these numerical methods to solve a system which is not tractable by analytic methods Similarly we can use these techniques to study the bound states of the nite square well for a deep well the low lying states will look like those of the in nite well although the wavefunction will now extend outside the range of the well decaying exponentially out to large distance Also a nite well possesses only a nite number of bound states for large energies the particle can move off to in nite distance with nite probability it exhibits a continuum of socalled scattering states These are not accessible with the current code C Lecture 8 The Uncertainty Principle As we have described for any observable A we will associate an hermitian operator A The expectation value of the operator in the state 31 gt is then naturally given by the expression lt A gtlt gt Consider two such observables A and B The product of the squared un certainty in A will be 0 A lt A gtIIA lt A gtII lt if gt Similarly for the observable B with f replaced by g Therefore 030 lt fif gtlt yiy gt2 i lt yif gt i2 This is called the Schwarz inequality It is a theorem for all vector spaces equipped with a notion of dot product l39ow the RHS of this expression is always bigger than the square of the imaginary part of lt f g gt so we can also write this as 1 2 030 2 27k fig gt lt yif gt1 But lt fgg gtlt II A lt A gtB lt B gtII gt This is simply lt AB gt lt A gtlt B gt Similarly ltg fgtltBAgt ltBgtltAgt So we nd 2 1 230 Z lt A313 gtgt where the square brackets are de ned by AB AB BA 1 This is the Uncertainty Principle in its most general form It says that for any pair of observables whose quantum operators do not commute there will be an associated uncertainty relation for the product of the uctuations in their expectation values This will be true for any quantum state vector II gt If we choose A L39 and B a we nd wapl 15 and hence 7 0301 2 i This is the famous Heisenberg uncertainty principle which we have encoun tered several times already Operators which do not commute do not share a complete set of common eigenvectors thus if we make a measurement of one of them we will by the generalized statistical interpretation described above collapse the state vector to a particular eigenvector of that observablev which will not then be an eigenvector of the other observable Indeed if we choose to expand this collapsed state vector on the basis of eigenvectors of the second observable we will generate a range of values ie a measurement of that second observable would be uncertain The magnitude of that uncertainty would be given by the uncertainty relation Finite dim representations matrix mechanics Let us now derive a concrete realization of these ideas by thinking of a nite dimensional system endowed with some orthonormal basis set a gt i 1 n An operator T takes every vector into some other vector Thus acting on a particular basis vector el gt we would nd T 61gt T11 61gt T21 62 gt 39 39 39 Tube gt and similarly for all the other basis vectors Thus the effect of the transfor mation can be encoded in n2 coef cients T15 where we can see that Tij lt gt Furthermore if gt is an arbitrary vector gt a1 61gt a2 62 gt an 6n gt then T gt 2 Tier gt Z 271 iez gt i 139 i Evidently T takes a vector with components a on into a vector with con gnyn s is easy to see that a compound operator 0 ST is just represented by a set of n2 numbers Cg given by Co Z Sikaj k Thus if we represent the state vector by its components in a particular basis then operators may be represented by matrices Furthermore if we think of the components of a ket vector a gt as just a column matrixv then the dot product lt be gt is just the matrix product bla where the dagger operation transposes the column vector gb gt into the row vector lt b and takes its complex conjugate Now consider the scalar lt WW gt Z v y lt i A 6i gt 1 Li Z lAii i 2 Li If A is Hermitian we must have that the LHS equal lt H at gt This will only be true when Ari A This then is the de nition of a Hermitian matrix It is a result in matrix theory that the eigenvalues of a hermitian matrix are purely real and that the eigenvectors are orthonormal and span the original space just as for the abstract operators Unitary matrices are then simply those matrices whose matrix inverse is just equal to its matrix hermitian conjugate as you would expect Thus any observable can be represented by either an operator or a possi bly in nite dimensional hermitian matrix The possible real values which 3 can result from measurement of that observable are just the eigenvalues of that matrix and the probability of measuring any such value is just the mod square of the component of its state vector along the eigenvector correspond ing to that eigenvalue Thus most of day to day business of doing a QM calculationv written in this language corresponds to nding the eigenvalues and eigenvectors of some Hermitian matrix A For small systems this can be done by setting the determinant of A A equal to zero This generates a polynomial equation with n roots the eigenvalues Aiii 0 n Notice that a matrix M will take on a diagonal form in a basis of its own eigenvectors llid lt gt lt Hie gt j51 j The transformation to this basis from the original basis must be a unitary matrix transformation since it must preserve the length of the state vector Thus 1 l J V gci gt 31 c gt The form of the matrix eigenvalue equation may then be preserved if the matrix Mij undergoes a so called similarity transformation Adiagonal 331 The only remaining question is how is the unitary matrix S determined The unitary matrix can be built from the eigenvectors of the matrix by assembling them into successive columns of S Thus if we know the eigenvectors of an given matrix in a given basis we can construct the unitary matrix that effects the change of basis which renders the matrix diagonal It is also clear then that if two operators have a common set of eigen vectors they can be simultaneously diagonalized put in diagonal form by a common unitary transformation But diagonal matrices commute hence there will be no uncertainty relation holding between the two corresponding observables So such observables are often said to be compatible The oppo site reasoning is also true incompatible operators are those which do not commute and have a non trivial mutual uncertainty relation They cannot be simultaneously diagonalized and have different eigenvectors Numerical methods for solving matrix mechanics problems often turn this around they typically focus on finding iteratively a transformation S which is capable of rendering A diagonal The diagonal elements are then the eigenvalues and the matrix which effects the diagonlization yields the eigenvectors The Schroedinger equation just bCCOIIICS a matrix equation now This can be seen by taking the vector form and expanding 31 gt on a time independent set of basis states a gt IIgtZlt6139 IIgt 6139gt i Inserting this into the Schroedinger equation we nd gtlt 6431 gt i Take dot product of this with lt ej Orthogonality of the basis says ltej IIgt if H Zltej H eigtlt ei Igt i or equivalently lt 6131 gt ci etc dc ihi H H 12 2 where Hg is the Hamiltonian matrix If we look for solutions of this equation in which all the c vary with time in the simple way c N 6 1Eth the station ary solutions we nd that the allowed energies E are just the eigenvalues of the Hamiltonian matrix and the eigenveetors the allowed stationary states 0

### BOOM! Enjoy Your Free Notes!

We've added these Notes to your profile, click here to view them now.

### You're already Subscribed!

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

## Why people love StudySoup

#### "I was shooting for a perfect 4.0 GPA this semester. Having StudySoup as a study aid was critical to helping me achieve my goal...and I nailed it!"

#### "When you're taking detailed notes and trying to help everyone else out in the class, it really helps you learn and understand the material...plus I made $280 on my first study guide!"

#### "There's no way I would have passed my Organic Chemistry class this semester without the notes and study guides I got from StudySoup."

#### "It's a great way for students to improve their educational experience and it seemed like a product that everybody wants, so all the people participating are winning."

### Refund Policy

#### STUDYSOUP CANCELLATION POLICY

All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email support@studysoup.com

#### STUDYSOUP REFUND POLICY

StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here: support@studysoup.com

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to support@studysoup.com

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.