Quantum Mechanics II
Quantum Mechanics II PHY 662
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PHY662 Quantum Mechanics II HWK 8 Due Tues Mar 16 at the start of class 0 Reading 7 One more time read Ch 17 of Shankar skip the ne structure discus sion and or Ch 6 of Gri iths Introduction to Quantum Mcchanics 7 Read the rst few pages of Ch 18 of Shankar on time dependent 14 Elcctran an a circular track Let an electron of effec perturbation theory which we start on Mar 16 ive mass M be con ned to a one dimensional 1D ring in the zy plane with radius R the effective mass in a material is usually different than in a vacuum Con sider cvlindrical coordinates for the wavefunction 1 Assume that there is effectively no motion of the electron in the z direction or r direction so that the problem can be described using only the angular coordinate 94 a l A n 1 Let 1M6 be the wavefunction for the electron What condition or conditions result from the choice of cy indrical coordinates If there are no external electromagnetic elds or potentials what is the Hamiltonian for the electron What are the and eigenenergie for your b If a uniform magnetic eld of strength B is applied perpendicular to the plane of the ring in the 2 direction what is the Hamiltonian For convenience choose a gauge for the vector potential A so that A is uniform in magnitude along the ring See Ch 14 in Shankar for a review of the Hamiltonian in a magnetic eld if needed What are the exact and eigenenergie for the quot in c If you now add an clcctric eld of strength E in the plane of the ring perpendicular to B how is the Hamiltonian modi ed Compute the the energy eigenvalues for small E using the eigenfunc tions from e as the unperturbed wavefunctionsi Compute these cor rections to rst and second order in E for the case 0 lt B lt hceR2 Compute the rst order corrections to the energy eigenvalues for the case BR2 h BONUS If you get done early consider the reverse case where E is strong and B is weak what is the approximate energy of the unperturbed ground state for large E What is the second order correction to this ground state energy when B y 0 1101110 1121111011011210 0111 30 0311121 0111 0100 H 1010111121 1211 0111 3901011 09 1011110113 01111 0112 010111 11011121011 011111101 0111 01011 W 112111101011 1211112311111 0111 001113 112111101011 11111011101 0111 30 1101110 1121111011011110 11V 391111111 12 311131121 K11 11 01011101 110111 112111101011 11111011100 0111 01111 1101110 1121111011011110 1112 001113011111 01 01 110111 11101110111100 0110 39011111011100 112130111101111 013 131110110 01101101151 391 1 y A A w 1 Zazz 30 1111030111211 101111011 0111 01 11211111 01 112111 gag 0111121130 03112110 1301111 12 K11 130011120 01 112111 95 03112110 12 X11 1103 112111101011 11111011100 0111 30 1111030111211 101111011 0111 01 11211111 11 1y 001310 1111111 11011 0113011011 0111 111 0101111121140 0111 0101111211 0013 12 103 0011210110310 0111 0112 11211 12 01011 01111 111 110110010 1112 110 1113110 0111 112 1301011100 112111101011 11111011101 0111 112111101011 3111111111101112 30 100110 0111 1013101100 11111 1101 39w 51 I 10301111 103 grijr rIel 171grp 0110111131100 11112131111011 0113011011 001 3957 S 2 5 z gt 574 0311121 0111 111 01112 001121111311000 101001 1101110011 0111 111111100110 01101111111100 12 111103 01 1301013101100 011 11120 0011210110310 0111 112111 112111 113110110 031121 7 1111111 7 001310 30 X01 11211010110111113001111 12 111 95 03112110 30 0101111121140 1112 1013101100 0111 09729 g n 111 991721 110111911721 393 3900 lt 7 01 112111 0101111211 0013 X11111 12 30 1111111 0111 111 11013110 10111 01 quot011210 10 0011210 311110111 112111 0111100011 0111 0112 11211111 011210 311110111 12 01 130110110 011 11111 0101111211 0111 112111 01121 112101 0111 011211101120 01 01111 11013103 01111110 1 001 39110 130111111 011 gm 4 15030011 A 111103 0111 30 0012110 1112 10110 1301111013 112111101011 12 1011 391001 112 011 11112111111 0101111211 0111 1011 11 101011111211 01111 103 0011210110310 0111 0112 11211111 0110111131100 111112131111011 000111 1101111 39grjyfz gngz 1111111 911014110 4109 hunpunoq 9112011911 0011 01 10010120 01 11 0110111131100 1112131111011 101 39031121 0011100011 7 012 1111111 0111 011121 131112 7 11010110111113 30 11011 12 111 01011 11211 0013 0111 11111 01 11130011 110130 01 11 1101112211121111011 30 1101211 110011 01 12 01121111310004 0111 3110112 11010110111113 0110 111 10111211 01 0013 0101111211 12 1013101100 0111 g 39 p m 1091111117 91111 Homo2 guluod 391 0111001101011 11101011113 12 131112 11101101 103 11130011 011 11131111 0111111111 30 3901 131112 396 011011001 4 391110110113121110110010 110 1x81 39001 131201 01 111210 131112 112311112111 111 93981 39003 0101310 10113111 0111 3111131110111 1011 11111 01 1111 1312011 4 311113120H o SS1910 JO was 8 JdV Wm 91111 01 HMH 11 301111211an Imminent ZQQAHd ii To nd the Fourier transform of the pure Coulomb potential you want to nd the Fourier transform in the limit H A 0 limm o fdg39r Elk71W iii The result is independent of the direction of 1 so pick I k2 Of course work in spherical coordinates o you want to compute the integral 0 C dr OW sin6d6 27f dqbw Change vari ables y cos9 and do the integral over y rst iv After integrating over r take H A 0 to get the answer c Let an a particle of charge 2e be in motion with momentum 1539 his in a box o size L Let there be a very xed nucleus of charge Z e at the origin What is the rate for the particle to make a transition to another state with a given momentum 1539 hk Use Fermiquots golden rule with w 0 and 39 answer using a 6 function NOTE your answer for the transition rate will depend on the box size N I 3 a a a a gt4 1 N a r 4 O z N d Suppose the a particle of charge 2e is moving at the speed 100 km s along the Iaxis in a periodic box of size L 10m containing a nucleus of charge Z e at its center This corresponds to an e tron traveling in in nite space with protons present at a density of 1 01114 The initial speed de nes r39our initial state with wave tor hk 1539 m5 What is the rate measured in Hz at which the electron will make a transition to a state with a veloc y ne perpendicular to the original direction where nearly perpendicularquot means that the new direction is perpendicular to the original direc tion to within about 000 radians More precisel your nal state lf has kf M lt 00010 You may take kf 7 M to be constant over this range of nal states You will need to integrate over nal states using the 6 function and the density of states Ml ZNL 34 Gauge invariance 2 pts This is just Shankar s exercise 1814 a Given potentials fl and 15 write the Hamiltonian H b Given a gauge eld Mit write down the Hamiltonian H A for the transformed potentials AA qu c Show that if 11101 t is a solution to Schrodinger equation with the Hmniltonian H then 11m it e lqquot M is a solution to Schrodingerquots equation with the Hamiltonian H A PHY662 Quantum Mechanics II HWK 9 Due Tues Mar 30 at the start of class 0 Reading 7 Read up to but not including the higher orders Sec 183 in Shankar and start to read Sec 184 on electromagnetism4 7 Sections 91 and 101 of Gri iths might be useful for review and a different perspective 1 Practicc with crcation and annihilation opcrators 2 pts Consider a par ticle in a 1D harmonic well a This is just Ex 1821 in Shankar Let your particle be in the ground state 0 at large negative times t Expose the particle to the time dependent perturbation H1t iefX1 tr This corre sponds to an electric eld of time varying strength To rst order in the time dependent part compute the probability that the particle will be in the rst excited state of the well at large positive times t b Let your particle be in the third excited state 3 at large negative time t For the perturbation of part a what are the possible states the particle can be in at large positve t if you compute transition probabilities to rst order in the perturbation What are the prob abilities for the particle to be in these other states to rst order in 8 b3 4 Di crcnt typos of chanch in thc Hamiltonian 3 pts Again consider a particle in a 1D harmonic well with Hamiltonian H EX2 iP2 at large negative time it Let the particle initially be in the ground statc that is 1t 700 gt in each part belowl a If H is equal to H for all t lt 0 and H H X2 132 for t gt 0 what is the probability that the particle is in the ground state of the new Hamiltonian H at positive it b IfHt H for t lt 0 Ht 17 H H for 0 lt t lt T H t H for t gt T what is the probability that the particle is in the ground state of H at time T in the limit of large T If H t H uX26t what is the probability that the parti cle is in an excited state of H at large positive t Note that H t Compute your answer using time dependent perturbation the ory to rst order in u and the notes from class Please note that to rst order there is only one excited state that can be reached by this perturbation during your calculation you should determine what state the particle can be excited to A n 34 Lcnz s Law via quantum mechanics pts Clasi cally when you apply a time changing magnetic eld to a loop of conducting materia you create a current in the loop that opposes the change in the magne eld This is Lenzquots law Based on last wee s iomework 8 1 er the quantum dynamics of a spinless electron in a circular ring Spinless means we z t o d L a neglect the spin degree of freedom so that there is no 7 B part of the Hamiltonian You will compute how a changing magnetic eld might induc currents in a ring that con nes a single electron First for review of notation and concepts consider the general problem of a charged particle in the presence of electromagnetic elds The vector potential A is a eld such that the magnetic eld can be written as H t is po ihle to write H in this fashion using 0 and vector calculus The electric eld is then E 7Vq5 7 where 45 is the electrostatic potential Remember also that A can be transformed by adding a gradient of a scalar eld A 7 A Vx without affecting H or any other phy cal ly measurable quantity This insensitivity is called gauge invariance If A 0 the kinetic part of the Hamiltonian is and the current of a particle is given by 1 mm 7 V1J 2M quot is not electric current this is just the velocity times the y to get the trio current you need to multiply this particle current by the charge q For the Hamiltonian with A y 0 the kinetic part of the Hamiltonian is c 7 2 i 2M 5 When 15 y 0 the full Hamiltonian is g 7 2 W but for this problem we will take 15 0 This part of the Hamiltonian can be derived from a cor r espondence with classical mechanics or can be viewed as the simplest gauge invariant Hamiltonian for a particle interacting with the electromagnatic eld In the presence of a vector potential the locally conserved particle current is v 7 h t t 4 t a J v i v v fk It is easily shown from the Hamiltonian in a presence of a magnetic eld that a p 7 i 7V at J 7 where the particle density p This justi es the de nition of the current jr You can try to derive this yourself looking up dis c ions of this topic for hints We will also review how to do this in class A 53 A V 7 A d Recall the phy al setup and key from HWK electron con ned to a ring of radius R For arbitrary values of magnetic eld of strength B perpendicular to the ring compute the expectation value of the particle current ifor the exact eigenstates lmgtof the Hamiltonian for a armless electron The currents of course must he in the 3 dire you are to nd the strength and direction of these currents for a given value of m Note that any gradients of 1 or 1 are in the 3 direction O E 1 Suppose the electron is initially in the ground state for B 0 If you very quickly turn up the external magnetic eld to a nal value B What is the expectation value of the current around the ring after the eld is raised Suppose the electron is initially in the ground state for B 0 If you very 39lo wly turn up the external magnetic eld to a nal value B What is the expectation value of the current around the ring after you are done raising the eld In SI units the ux quantum which is hce in Gaussian units is he 14 X 10 15 T 1112 For reference note that 10 T is a strong eld in experiments 1 1 is not too hard to achieve and that the Earth s magnetic eld is of the order of 50 MT If you had a ring of radius 1 111 containing an electron in its ground state at B 0 and suddenly turned up the eld to 001 T 100 G What would the velocity of the electron be after raising the eld PHY662 Spring 2004 Feb 26 2004 2nd March 2004 1 Miscellaneous 1 HWK 8 due Tuesday Mar 9 2 Continue to read Ch 17 Shankar or Gri lths Ch 6 especially for 2ndorder perturbation theory Start Ch 18 for timedependent perturbation theory for Tuesday maybe Thursday this week 3 Today Review HWK 7 using maple to plot some results Then go over 2nd order perturbation theory 2 Perturbation theory Perturbation theory is an expansion about a solvable problem using a small parame ter Physicists often ask one another what is the small parameter Typical expansion parameters include the ratios of energies length scales or time scales Dimensionality is often used as an expansion parameter eg e 4 7 d as are dimensionless constants such as a 13704 Not all perturbation expansions converge nicely Most don t Here is a basic example of an aysmptotic expansion let s compute an expansion in A for CA draft The perturbative approach is to expand e MA in powers of A You then get oo 2 A2 CA dze 17Az4187 1600 7 k 00 Z A 6421 kl k0 00 x 8 Mg 1 x H o koo Arma k Fk1 A k0 The answer can also be written as C A koo7k 3 2k I k 8a a k0 Wigwam How do the coef cients of this series behave They diverge exponentially in k faster than Ark for any A So the terms in the series diverge The series is well behaved only in the limit A A 0 See the plot from maple for examples There is a clue to a problem here when A lt 0 CA diverges so the behavior cannot be smoothly expanded about A 0 in powers of A 3 Review of nondeg pert theory through 1st order Assume we have exact eigenstates for a Hamiltonian H 0 H w2 EM Very important note the assumption is that 11 is properly normalized f 1V 1 We wish to nd the eigenstates for the Hamiltonian H H 0 AH We can at least formally expand the new eigenstates and eigenvalues 1117 2A iA2 iw En E2AE A2E M Writing H 11 En n gives collecting equations to each order in A Horpg Egrpg 0th order in A Hort H rpg 13 Eirpg ist order inA m H m Wme The 0th order equation is not new Today we will just look at the lst order correction this is what will be needed for homework Take the rst order equation multiply by 2 and integrate over the space of the wavefunction to get lt lH l igt lt 3LlH l123gt E3lt 3Ll igt Eiwalwz Choosing m n gives Ei WWW 4 Secondorder perturbation theory 0 First order corrections to the energy are often straightforward to calculate 0 Higher order corrections usually involve in nite sums and are more dif cult in genera 0 Second order calculations are often the practical limit Here is a result that shows both the power and limits of perturbation expansion Re member the gyromagnetic ratio of the electron It can be computed using quantum electrodynamics QED in powers of a g m 13104 The answer for the QED part is 92 1 la 197 7r2 3 7an a 7 i 7 3 7 2 144l24lt 2 1 1 3 7 gm L1 g 1 lt1 43511122 7 4 72 18 2160 298 2 17101 2 28259 a 3 Tquot 1112 810 W T 5184ltgt T numerical estimate 34 m 1 05 7 0328478965 a 3 a 4 1181241456 g 7 1511 004 g 1001159652 Note that the third order term was solved analytically only in 1996 4 is the Riemann zeta function and Li4 is a logarithmic integral function The fourth order term is a numerical estimate that has required many years of work F or a review see Reviews of Modem Physics Vol 71 pp 5133139 Our goal will be a little more modest at rst computing 2nd order corrections in non relativistic quantum mechanics 5 2nd order The second order corrections to the energy require the rst order corrections to the wavefunction Note that the wavefuncl ians calculated using perturbation theory are generally less reliable than the energy estimates In any case the rst order corrections to the wavefunction are found by computing the inner product of the rst order terms in the Schrodinger equation for eigenstate n with the unperturbed eigenstate m lt 3LlH l igtlt 3llHW3gt lt 3LlE2l igtlt 3zlEil 3gt E3Llt 3Ll igt lt 3 lH l 3gt E3lt 31l igt Eilt 3nl 3gt lt 3AHWgt E9 7 E3 WW 0 H 0 WM lt1 E 3L Ei d A Since the set of all W2 form a complete basis 71 can be written as a sum over 3 Clearly Elm lt 3LlH l 3gtE3 E90 SO 0 H 0 W 2 Wm M n m 1 677171 Why did we leave W2 out of the sum Well the rst order approximation to W is W2 so that any part of that is proportional to W2 is redundant In fact this choice also keeps this rst order wave function normalized at least to rst order lt nl ngt lt 3llt ill 3gtl igt 1 0 0 terms second order in Given this we can now look at the second order correction to the energy by taking the inner product of the second order part of the Schrodinger equation with 9 imme diately canceling the rst terms on each side due to the same trick as for rst order ngHO ngEO lt 3lH l igt Eilt 3l igt E Z lt 3lH l gtlt lH l 3gt 7 0 E2 E3 mfn llt 3lH l 3ngtl n 7 Z 0 7 o 7 m n En Em an expression that again uses only matrix elements of H 5 1 Application Here is a sample taking problems 61 and 63 from Grif ths Let H 0 be the square well potential for a particle con ned to 0 lt z lt a so that the normalized unperturbed wavefunctions are 11 sinclflm with E2 f0 sinW dz n27r2h2 ma If the perturbing potential is a deltafunction at z a27 H 161 7 E Aaltw2ltzgtgtm6ltziggtw2ltzgtdz 2a 7 0 sin2na 6zig 7 270 nodd 7 07 neven The second order correction is a2 a mm mm a 2 E2 2 437M dz SlnTs1n a 617 25M m2 m n 7 2 4m2a2 71mn2 7 a 2amp2 n2 7 m2 7r m nmn bothodd 2 2 100 124721 a2 2 I 1 2h 21 1 7 2k 12 j1 k n2k1 2 a2 a 7 X constant W2h2n2