Solutions to Problems in Modern Quantum Mechanics 2nd Edition Sakurai, Exercises of Quantum Mechanics

Download Solutions to Problems in Modern Quantum Mechanics 2nd Edition Sakurai and more Exercises Quantum Mechanics in PDF only on Docsity! Solutions to Problems in Quantum Mechanics P. Saltsidis, additions by B. Brinne 1995,1999 0 Most of the problems presented here are taken from the book Sakurai, J. J., Modern Quantum Mechanics, Reading, MA: Addison-Wesley, 1985. Part I Problems 3 1. FUNDAMENTAL CONCEPTS 5 1 Fundamental Concepts 1.1 Consider a ket space spanned by the eigenkets fja0ig of a Her- mitian operator A. There is no degeneracy. (a) Prove that Y a0 (A a0) is a null operator. (b) What is the signi cance of Y a00 6=a0 (A a00) a0 a00 ? (c) Illustrate (a) and (b) using A set equal to Sz of a spin 1 2 system. 1.2 A spin 1 2 system is known to be in an eigenstate of ~S
n̂ with eigenvalue h=2, where n̂ is a unit vector lying in the xz-plane that makes an angle with the positive z-axis. (a) Suppose Sx is measured. What is the probability of getting +h=2? (b) Evaluate the dispersion in Sx, that is, h(Sx hSxi)2i: (For your own peace of mind check your answers for the special cases = 0, =2, and .) 1.3 (a) The simplest way to derive the Schwarz inequality goes as follows. First observe (h j + h j)
(j i + j i)  0 for any complex number ; then choose  in such a way that the preceding inequality reduces to the Schwarz inequility. 8 2.3 Consider a particle in three dimensions whose Hamiltonian is given by H = ~p2 2m + V (~x): By calculating [~x
~p;H] obtain d dt h~x
~pi = * p2 m + h~x
~rV i: To identify the preceding relation with the quantum-mechanical analogue of the virial theorem it is essential that the left-hand side vanish. Under what condition would this happen? 2.4 (a) Write down the wave function (in coordinate space) for the state exp ipa h  j0i: You may use hx0j0i = 1=4x1=20 exp 2 41 2 x0 x0 !235 ; 0 @x0  h m! !1=21A : (b) Obtain a simple expression that the probability that the state is found in the ground state at t = 0. Does this probability change for t > 0? 2.5 Consider a function, known as the correlation function, de ned by C(t) = hx(t)x(0)i; where x(t) is the position operator in the Heisenberg picture. Eval- uate the correlation function explicitly for the ground state of a one-dimensional simple harmonic oscillator. 2. QUANTUM DYNAMICS 9 2.6 Consider again a one-dimensional simple harmonic oscillator. Do the following algebraically, that is, without using wave func- tions. (a) Construct a linear combination of j0i and j1i such that hxi is as large as possible. (b) Suppose the oscillator is in the state constructed in (a) at t = 0. What is the state vector for t > 0 in the Schrodinger picture? Evaluate the expectation value hxi as a function of time for t > 0 using (i) the Schrodinger picture and (ii) the Heisenberg picture. (c) Evaluate h(
x)2i as a function of time using either picture. 2.7 A coherent state of a one-dimensional simple harmonic oscil- lator is de ned to be an eigenstate of the (non-Hermitian) annihi- lation operator a: aji = ji; where  is, in general, a complex number. (a) Prove that ji = ejj2=2eayj0i is a normalized coherent state. (b) Prove the minimum uncertainty relation for such a state. (c) Write ji as ji = 1X n=0 f(n)jni: Show that the distribution of jf(n)j2 with respect to n is of the Poisson form. Find the most probable value of n, hence of E. (d) Show that a coherent state can also be obtained by applying the translation ( nite-displacement) operator eipl=h (where p is the momentum operator, and l is the displacement distance) to the ground state. 10 (e) Show that the coherent state ji remains coherent under time- evolution and calculate the time-evolved state j(t)i. (Hint: di- rectly apply the time-evolution operator.) 2.8 The quntum mechanical propagator, for a particle with mass m, moving in a potential is given by: K(x; y;E) = Z 1 0 dteiEt=hK(x; y; t; 0) = A X n sin(nrx) sin(nry) E h2r2 2m n2 where A is a constant. (a) What is the potential? (b) Determine the constant A in terms of the parameters describing the system (such as m, r etc. ). 2.9 Prove the relation d(x) dx = (x) where (x) is the (unit) step function, and (x) the Dirac delta function. (Hint: study the e ect on testfunctions.) 2.10 Derive the following expression Scl = m! 2 sin(!T ) h (x20 + 2x 2 T ) cos(!T ) x0xT i for the classical action for a harmonic oscillator moving from the point x0 at t = 0 to the point xT at t = T . 2.11 The Lagrangian of the single harmonic oscillator is L = 1 2 m _x2 1 2 m!2x2 2. QUANTUM DYNAMICS 13 with ~j given by ~j = h m ! =( ~r0 )  e mc  ~Aj j2: 2.14 An electron moves in the presence of a uniform magnetic eld in the z-direction ( ~B = Bẑ). (a) Evaluate [x;y]; where x  px eAx c ; y  py eAy c : (b) By comparing the Hamiltonian and the commutation relation obtained in (a) with those of the one-dimensional oscillator problem show how we can immediately write the energy eigenvalues as Ek;n = h2k2 2m + jeBjh mc ! n+ 1 2  ; where hk is the continuous eigenvalue of the pz operator and n is a nonnegative integer including zero. 2.15 Consider a particle of mass m and charge q in an impenetrable cylinder with radius R and height a. Along the axis of the cylin- der runs a thin, impenetrable solenoid carrying a magnetic ux . Calculate the ground state energy and wavefunction. 2.16 A particle in one dimension (1 < x < 1) is subjected to a constant force derivable from V = x; ( > 0): 14 (a) Is the energy spectrum continuous or discrete? Write down an approximate expression for the energy eigenfunction speci ed by E. (b) Discuss brie y what changes are needed if V is replaced be V = jxj: 3 Theory of Angular Momentum 3.1 Consider a sequence of Euler rotations represented by D(1=2)( ; ; ) = exp i3 2  exp i2 2 ! exp i3 2  = ei( + )=2 cos 2 ei( )=2 sin 2 ei( )=2 sin 2 ei( + )=2 cos 2 ! : Because of the group properties of rotations, we expect that this sequence of operations is equivalent to a single rotation about some axis by an angle . Find . 3.2 An angular-momentum eigenstate jj;m = mmax = ji is rotated by an in nitesimal angle " about the y-axis. Without using the explicit form of the d (j) m0m function, obtain an expression for the probability for the new rotated state to be found in the original state up to terms of order "2. 3.3 The wave function of a patricle subjected to a spherically symmetrical potential V (r) is given by (~x) = (x+ y + 3z)f(r): 3. THEORY OF ANGULAR MOMENTUM 15 (a) Is an eigenfunction of ~L? If so, what is the l-value? If not, what are the possible values of l we may obtain when ~L2 is measured? (b)What are the probabilities for the particle to be found in various ml states? (c) Suppose it is known somehow that (~x) is an energy eigenfunc- tion with eigenvalue E. Indicate how we may nd V (r). 3.4 Consider a particle with an intrinsic angular momentum (or spin) of one unit of h. (One example of such a particle is the %- meson). Quantum-mechanically, such a particle is described by a ketvector j%i or in ~x representation a wave function %i(~x) = h~x; ij%i where j~x; ii correspond to a particle at ~x with spin in the i:th di- rection. (a) Show explicitly that in nitesimal rotations of %i(~x) are obtained by acting with the operator u~" = 1 i ~" h
(~L+ ~S) (3.1) where ~L = h i r̂  ~r. Determine ~S ! (b) Show that ~L and ~S commute. (c) Show that ~S is a vector operator. (d) Show that ~r ~%(~x) = 1 h2 ( ~S
~p)~% where ~p is the momentum oper- ator. 3.5 We are to add angular momenta j1 = 1 and j2 = 1 to form j = 2; 1; and 0 states. Using the ladder operator method express all 18 where a is the lattice constant, n labels the band, and the lattice momentum k is restricted to the Brillouin zone [=a; =a]. Prove that any Bloch function can be written as, n;k(x) = X Ri n(xRi)eikRi where the sum is over all lattice vectors Ri. (In this simble one di- mensional problem Ri = ia, but the construction generalizes easily to three dimensions.). The functions n are called Wannier functions, and are impor- tant in the tight-binding description of solids. Show that the Wan- nier functions are corresponding to di erent sites and/or di erent bands are orthogonal, i:e: proveZ dx?m(xRi)n(xRj)  ijmn Hint: Expand the ns in Bloch functions and use their orthonor- mality properties. 4.4 Suppose a spinless particle is bound to a xed center by a potential V (~x) so assymetrical that no energy level is degenerate. Using the time-reversal invariance prove h~Li = 0 for any energy eigenstate. (This is known as quenching of orbital angular momemtum.) If the wave function of such a nondegenerate eigenstate is expanded asX l X m Flm(r)Y m l (; ); what kind of phase restrictions do we obtain on Flm(r)? 4.5 The Hamiltonian for a spin 1 system is given by H = AS2z +B(S 2 x S2y): 5. APPROXIMATION METHODS 19 Solve this problem exactly to nd the normalized energy eigen- states and eigenvalues. (A spin-dependent Hamiltonian of this kind actually appears in crystal physics.) Is this Hamiltonian invariant under time reversal? How do the normalized eigenstates you ob- tained transform under time reversal? 5 Approximation Methods 5.1 Consider an isotropic harmonic oscillator in two dimensions. The Hamiltonian is given by H0 = p2x 2m + p2y 2m + m!2 2 (x2 + y2) (a) What are the energies of the three lowest-lying states? Is there any degeneracy? (b) We now apply a perturbation V = m!2xy where  is a dimensionless real number much smaller than unity. Find the zeroth-order energy eigenket and the corresponding en- ergy to rst order [that is the unperturbed energy obtained in (a) plus the rst-order energy shift] for each of the three lowest-lying states. (c) Solve the H0+V problem exactly. Compare with the perturba- tion results obtained in (b). [You may use hn0jxjni = q h=2m!( p n + 1n0;n+1 + p nn0;n1):] 5.2 A system that has three unperturbed states can be represented by the perturbed Hamiltonian matrix0 B@ E1 0 a0 E1 b a b E2 1 CA 20 where E2 > E1. The quantities a and b are to be regarded as per- turbations that are of the same order and are small compared with E2 E1. Use the second-order nondegenerate perturbation theory to calculate the perturbed eigenvalues. (Is this procedure correct?) Then diagonalize the matrix to nd the exact eigenvalues. Finally, use the second-order degenerate perturbation theory. Compare the three results obtained. 5.3 A one-dimensional harmonic oscillator is in its ground state for t < 0. For t  0 it is subjected to a time-dependent but spatially uniform force (not potential!) in the x-direction, F (t) = F0e t= (a) Using time-dependent perturbation theory to rst order, obtain the probability of nding the oscillator in its rst excited state for t > 0). Show that the t ! 1 ( nite) limit of your expression is independent of time. Is this reasonable or surprising? (b) Can we nd higher excited states? [You may use hn0jxjni = q h=2m!( p n+ 1n0;n+1 + p nn0;n1):] 5.4 Consider a composite system made up of two spin 12 objects. for t < 0, the Hamiltonian does not depend on spin and can be taken to be zero by suitably adjusting the energy scale. For t > 0, the Hamiltonian is given by H =  4
h2  ~S1
~S2: Suppose the system is in j + i for t  0. Find, as a function of time, the probability for being found in each of the following states j++i, j+i, j +i, j i: (a) By solving the problem exactly. Part II Solutions 23 1. FUNDAMENTAL CONCEPTS 25 1 Fundamental Concepts 1.1 Consider a ket space spanned by the eigenkets fja0ig of a Her- mitian operator A. There is no degeneracy. (a) Prove that Y a0 (A a0) is a null operator. (b) What is the signi cance of Y a00 6=a0 (A a00) a0 a00 ? (c) Illustrate (a) and (b) using A set equal to Sz of a spin 1 2 system. (a) Assume that j i is an arbitrary state ket. Then Y a0 (A a0)j i = Y a0 (A a0)X a00 ja00i ha00j i| {z } ca00 = X a00 ca00 Y a0 (A a0)ja00i = X a00 ca00 Y a0 (a00 a0)ja00i a 002fa0g = 0: (1.1) (b) Again for an arbitrary state j i we will have 2 4 Y a00 6=a0 (A a00) a0 a00 3 5 j i = 2 4 Y a00 6=a0 (A a00) a0 a00 3 5 1z }| {X a000 ja000iha000 j i = X a000 ha000j i Y a00 6=a0 (a000 a00) a0 a00 ja 000i = = X a000 ha000j ia000a0ja000i = ha0j ija0i )2 4 Y a00 6=a0 (A a00) a0 a00 3 5 = ja0iha0j  a0: (1.2) So it projects to the eigenket ja0i. 28 But we want also the eigenstate j~S
n̂; +i to be normalized, that is a2 + b2 = 1 ) a2 + a2 tan2 2 = 1) a2 cos2 2 a2 sin2 2 = cos2 2 ) a2 = cos2 2 ) a =  r cos2 2 = cos 2 ; (1.12) where the real positive convention has been used in the last step. This means that the state in which the system is in, is given in terms of the eigenstates of the Sz operator by j~S
n̂; +i = cos 2 j+i+ sin 2 ji: (1.13) (a) From (S-1.4.17) we know that jSx; +i = 1p 2 j+i+ 1p 2 ji: (1.14) So the propability of getting +h=2 when Sx is measured is given by hSx; +j~S
n̂; +i 2 = 1p 2 h+j+ 1p 2 hj ! cos 2 j+i+ sin 2 ji  2 = 1p2 cos 2 + 1p 2 sin 2 2 = 1 2 cos2 2 + 1 2 sin2 2 + cos 2 sin 2 = 12 + 1 2 sin = 1 2(1 + sin ): (1.15) For = 0 which means that the system is in the jSz; +i eigenstate we have jhSx; +jSz; +ij2 = 12(1) = 12 : (1.16) For = =2 which means that the system is in the jSx; +i eigenstate we have jhSx; +jSx; +ij2 = 1: (1.17) For =  which means that the system is in the jSz;i eigenstate we have jhSx; +jSz;ij2 = 12(1) = 12 : (1.18) 1. FUNDAMENTAL CONCEPTS 29 (b) We have that h(Sx hSxi)2i = hS2xi (hSxi)2: (1.19) As we know Sx = h 2 (j+ihj+ jih+j)) S2x = h2 4 (j+ihj + jih+j) (j+ihj+ jih+j)) S2x = h2 4 (j+ih+j + jihj)| {z } 1 = h2 4 : (1.20) So hSxi =  cos 2 h+j + sin 2 hj  h 2 (j+ihj+ jih+j)  cos 2 j+i+ sin 2 ji  = h 2 cos 2 sin 2 + h 2 sin 2 cos 2 = h 2 sin ) (hSxi)2 = h 2 4 sin2 and hS2xi =  cos 2 h+j + sin 2 hj  h2 4  cos 2 j+i+ sin 2 ji  = h2 4 [cos2 2 + sin2 2 ] = h2 4 : (1.21) So substituting in (1.19) we will have h(Sx hSxi)2i = h 2 4 (1 sin2 ) = h 2 4 cos2 : (1.22) and nally h(
Sx)2i =0;jSz;+i = h2 4 ; (1.23) h(
Sx)2i ==2;jSx ;+i = 0; (1.24) h(
Sx)2i =0;jSz;i = h2 4 : (1.25) 30 1.3 (a) The simplest way to derive the Schwarz inequality goes as follows. First observe (h j + h j)
(j i + j i)  0 for any complex number ; then choose  in such a way that the preceding inequality reduces to the Schwarz inequility. (b) Show that the equility sign in the generalized uncertainty re- lation holds if the state in question satis es
Aj i = 
Bj i with  purely imaginary. (c) Explicit calculations using the usual rules of wave mechanics show that the wave function for a Gaussian wave packet given by hx0j i = (2d2)1=4 exp " ihpix0 h (x 0 hxi)2 4d2 # satis es the uncertainty relation q h(
x)2i q h(
p)2i = h 2 : Prove that the requirement hx0j
xj i = (imaginary number)hx0j
pj i is indeed satis ed for such a Gaussian wave packet, in agreement with (b). (a) We know that for an arbitrary state jci the following relation holds hcjci  0: (1.26) This means that if we choose jci = j i+ j i where  is a complex number, we will have (h j + h j)
(j i + j i)  0) (1.27) h j i + h j i + h j i+ jj2h j i  0: (1.28) 1. FUNDAMENTAL CONCEPTS 33 is an eigenstate of the coordinate operator x. What is the corre- sponding eigenvalue? (a) We have [x; F (px)]classical  @x @x @F (px) @px @x @px @F (px) @x = @F (px) @px : (1.41) (b) When x and px are treated as quantum-mechanical operators we have x; exp  ipxa h  = " x; 1X n=0 (ia)n hn pnx n! # = 1X n=0 1 n! (ia)n hn [x; pnx] = 1X n=0 1 n! (ia)n hn n1X k=0 pkx [x; px] p nk1 x = 1X n=1 1 n! (ia)n hn (ih) n1X k=0 pkxp nk1 x = 1X n=1 n n! (ia)n1 hn1 pn1x (a) = a 1X n=1 1 (n 1)!  ia h px n1 = a exp  ipxa h  : (1.42) (c) We have now x  exp  ipxa h  jx0i (b)= exp  ipxa h  xjx0i a exp  ipxa h  jx0i = x0 exp  ipxa h  jx0i a exp  ipxa h  jx0i = (x0 a) exp  ipxa h  jx0i: (1.43) So exp  ipxa h  jx0i is an eigenstate of the operator x with eigenvalue x0 a. So we can write jx0 ai = C exp  ipxa h  jx0i; (1.44) where C is a constant which due to normalization can be taken to be 1. 34 1.5 (a) Prove the following: (i) hp0jxj i = ih @ @p0 hp0j i; (ii) h jxj i = Z dp0 (p 0)ih @ @p0  (p 0); where  (p0) = hp0j i and  (p0) = hp0j i are momentum-space wave functions. (b) What is the physical signi cance of exp  ix h  ; where x is the position operator and  is some number with the dimension of momentum? Justify your answer. (a) We have (i) hp0jxj i = hp0jx 1z }| {Z dx0jx0ihx0j i = Z dx0hp0jxjx0ihx0j i = Z dx0x0hp0jx0ihx0j i (S1:7:32)= Z dx0x0Ae ip0x0 h hx0j i = A Z dx0 @ @p0  e ip0x0 h  (ih)hx0j i = ih @ @p0 Z dx0Ae ip0x0 h hx0j i  = ih @ @p0 Z dx0hp0jx0ihx0j i  = ih @ @p0 hp0j i ) hp0jxj i = ih @ @p0 hp0j i: (1.45) (ii) h jxj i = Z dp0h jp0ihp0jxj i = Z dp0 (p 0)ih @ @p0  (p 0); (1.46) where we have used (1.45) and that h jp0i =  (p0) and hp0j i =  (p0). 1. FUNDAMENTAL CONCEPTS 35 (b) The operator exp  ix h  gives translation in momentum space. This can be justi ed by calculating the following operator  p; exp  ix h  = " p; 1X n=0 1 n!  ix h n# = 1X n=0 1 n!  i h n [p; xn] = 1X n=1 1 n!  i h n nX k=1 xnk[p; x]xk1 = 1X n=1 1 n!  i h n nX k=1 (ih)xn1 = 1X n=1 1 n!  i h n n(ih)xn1 = 1X n=1 1 (n 1)!  i h n1 xn1(ih)  i h  =  1X n=0 1 n!  ix h n = exp  ix h  : (1.47) So when this commutator acts on an eigenstate jp0i of the momentum oper- ator we will have p; exp  ix h  jp0i = p  exp  ix h  jp0i   exp  ix h  p0jp0i ) exp  ix h  = p  exp  ix h  jp0i  p0  exp  ix h  jp0i ) p  exp  ix h  jp0i  = (p0 + )  exp  ix h  jp0i  : (1.48) Thus we have that exp  ix h  jp0i  Ajp0 + i; (1.49) where A is a constant which due to normalization can be taken to be 1. 38 By calculating [~x
~p;H] obtain d dt h~x
~pi = * p2 m + h~x
~rV i: To identify the preceding relation with the quantum-mechanical analogue of the virial theorem it is essential that the left-hand side vanish. Under what condition would this happen? Let us rst calculate the commutator [~x
~p;H] [~x
~p;H] = " ~x
~p; ~p 2 2m + V (~x) # = 2 4 3X i=1 xipi; 3X j=1 p2j 2m + V (~x) 3 5 = X ij " xi; p2j 2m # pi + X i xi [pi; V (~x)] : (2.14) The rst commutator in (2.14) will give" xi; p2j 2m # = 1 2m [xi; p 2 j ] = 1 2m (pj [xi; pj ] + [xi; pj]pj) = 1 2m (pjihij + ihijpj) = 1 2m 2ihijpj = ih m ijpj : (2.15) The second commutator can be calculated if we Taylor expand the function V (~x) in terms of xi which means that we take V (~x) = P n anx n i with an independent of xi. So [pi; V (~x)] = " pi; 1X n=0 anx n i # = X n an [pi; x n i ] = X n an n1X k=0 xki [pi; xi]x nk1 i = X n an n1X k=0 (ih)xn1i = ih X n annx n1 i = ih @ @xi X n anx n i = ih @ @xi V (~x): (2.16) The right-hand side of (2.14) now becomes [~x
~p;H] = X ij ih m ijpjpi + X i (ih)xi @ @xi V (~x) = ih m ~p2 ih~x
~rV (~x): (2.17) 2. QUANTUM DYNAMICS 39 The Heisenberg equation of motion gives d dt ~x
~p = 1 ih [~x
~p;H] (2:17)= ~p 2 m ~x
~rV (~x)) d dt h~x
~pi = * p2 m + h~x
~rV i; (2.18) where in the last step we used the fact that the state kets in the Heisenberg picture are independent of time. The left-hand side of the last equation vanishes for a stationary state. Indeed we have d dt hnj~x
~pjni = 1 ih hnj [~x
~p;H] jni = 1 ih (Enhnj~x
~pjni Enhnj~x
~pjni) = 0: So to have the quantum-mechanical analogue of the virial theorem we can take the expectation values with respect to a stationaru state. 2.4 (a) Write down the wave function (in coordinate space) for the state exp ipa h  j0i: You may use hx0j0i = 1=4x1=20 exp 2 41 2 x0 x0 !235 ; 0 @x0  h m! !1=21A : (b) Obtain a simple expression that the probability that the state is found in the ground state at t = 0. Does this probability change for t > 0? (a) We have j ; t = 0i = exp ipa h  j0i ) hx0j ; t = 0i = hx0 exp ipa h  j0i (Pr:1:4:c)= hx0 aj0i = 1=4x1=20 exp 2 412 x0 a x0 !235 : (2.19) 40 (b) This probability is given by the expression jh0j ; t = 0ij2 = jhexp ipa h  j0ij2: (2.20) It is hexp ipa h  j0i = Z dx0h0jx0ihx0j exp ipa h  j0i = Z dx01=4x1=20 exp 2 412 x0 x0 !2351=4x1=20  exp 2 412 x0 a x0 !235 = Z dx01=2x10 exp " 1 2x20  x02 + x02 + a2 2ax0 # = 1p x0 Z dx0 exp " 2 2x20 x02 2x0a 2 + a2 4 + a2 4 !# = exp a 2 4x20 ! 1p x0 p x0 = exp a 2 4x20 ! : (2.21) So jh0j ; t = 0ij2 = exp a 2 2x20 ! : (2.22) For t > 0 jh0j ; tij2 = jh0jU(t)j ; t = 0ij2 = jh0j exp  iHt h  j ; t = 0ij2 = eiE0t=hh0j ; t = 0i 2 = jh0j ; t = 0ij2: (2.23) 2.5 Consider a function, known as the correlation function, de ned by C(t) = hx(t)x(0)i; (2.24) where x(t) is the position operator in the Heisenberg picture. Eval- uate the correlation function explicitly for the ground state of a one-dimensional simple harmonic oscillator. 2. QUANTUM DYNAMICS 43 The average hxi in a one-dimensional simple harmonic oscillator is given by hxi = h jxj i = (c0h0j + c1h1j) x (c0j0i + c1j1i) = jc0j2h0jxj0i + c0c1h0jxj1i + c1c0h1jxj0i + jc1j2h1jxj1i = jc0j2 s h 2m! h0ja+ ayj0i+ c0c1 s h 2m! h0ja + ayj1i +c1c0 s h 2m! h1ja + ayj0i + jc1j2 s h 2m! h1ja + ayj1i = s h 2m! (c0c1 + c  1c0) = 2 s h 2m! <(c0c1) = 2 s h 2m! cos(1 0)jc0j q 1 jc0j2; (2.36) where we have used that x = q h 2m! (a+ ay). What we need is to nd the values of jc0j and 1 0 that make the average hxi as large as possible. @hxi @jc0j = 0 ) q 1 jc0j2 jc0j 2q 1 jc0j2 jc0j6=1) 1 jc0j2 jc0j2 = 0 ) jc0j = 1p 2 (2.37) @hxi @1 = 0 ) sin(1 0) = 0) 1 = 0 + n; n 2 Z: (2.38) But for hxi maximum we want also @2hxi @21 1=1max < 0) n = 2k; k 2 Z: (2.39) So we can write that j i = ei0 1p 2 j0i + ei(0+2k) 1p 2 j1i = ei0 1p 2 (j0i + j1i): (2.40) We can always take 0 = 0. Thus j i = 1p 2 (j0i + j1i): (2.41) 44 (b) We have j ; t0i = j i. So j ; t0; ti = U(t; t0 = 0)j ; t0i = eiHt=hj i = 1p 2 eiE0t=hj0i + 1p 2 eiE1t=hj1i = 1p 2  ei!t=2j0i+ ei!3t=2j1i  = 1p 2 ei!t=2  j0i+ ei!tj1i  :(2.42) (i) In the Schrodinger picture hxiS = h ; t0; tjxSj ; t0; tiS = " 1p 2  ei!t=2h0j+ ei!3t=2h1j # x " 1p 2  ei!t=2j0i + ei!3t=2j1i # = 12e i(!t=2!3t=2)h0jxj1i + 12ei(!3t=2!t=2)h1jxj0i = 12e i!t s h 2m! + 12e i!t s h 2m! = s h 2m! cos !t: (2.43) (ii) In the Heisenberg picture we have from (2.29) that xH(t) = x(0) cos!t+ p(0) m! sin!t: So hxiH = h jxH j i = " 1p 2 h0j + 1p 2 h1j # x(0) cos!t+ p(0) m! sin!t !" 1p 2 j0i+ 1p 2 j1i # = 1 2 cos !th0jxj1i+ 1 2 cos !th1jxj0i + 1 2 1 m! sin !th0jpj1i +12 1 m! sin!th1jpj0i = 12 s h 2m! cos!t+ 12 s h 2m! cos!t+ 1 2m! sin!t(i) s mh! 2 + 1 2m! sin!ti s mh! 2 = s h 2m! cos!t: (2.44) (c) It is known that h(
x)2i = hx2i hxi2 (2.45) 2. QUANTUM DYNAMICS 45 In the Schodinger picture we have x2 = 2 4 s h 2m! (a+ ay) 3 52 = h 2m! (a2 + ay 2 + aay+ aya); (2.46) which means that hxi2S = h ; t0; tjx2j ; t0; tiS = " 1p 2  ei!t=2h0j + ei!3t=2h1j # x2 " 1p 2  ei!t=2j0i+ ei!3t=2j1i # = h 1 2 ei(!t=2!t=2)h0jaayj0i + 1 2 ei(!3t=2!3t=2)h1jaayj1i + 1 2 h1jayaj1i i h 2m! = h 1 2 + 21 2 + 1 2 i h 2m! = h 2m! : (2.47) So h(
x)2iS (2:43)= h 2m! h 2m! cos2 !t = h 2m! sin2 !t: (2.48) In the Heisenberg picture x2H(t) = " x(0) cos!t+ p(0) m! sin!t #2 = x2(0) cos2 !t+ p2(0) m2!2 sin2 !t + x(0)p(0) m! cos!t sin!t+ p(0)x(0) m! cos !t sin!t = h 2m! (a2 + ay 2 + aay+ aya) cos2 !t mh! 2m2!2 (a2 + ay 2 aay aya) sin2 !t + i m! s hmh! 4m! (a+ ay)(ay a)sin 2!t 2 + i m! s hmh! 4m! (ay a)(a+ ay)sin 2!t 2 = h 2m! (a2 + ay 2 + aay+ aya) cos2 !t 48 = ejj 2 X n;m 1 n!m! ()nmh0janj(ay)mj0i [(ay)mj0i = p m!jmi] = ejj 2 X n;m p n! n! p m! m! ()nmhnjmi = ejj2X n 1 n! (jj2)n = ejj 2 ejj 2 = 1: (2.55) (b) According to problem (1.3) the state should satisfy the following relation
xji = c
pji; (2.56) where
x  x hjxji,
p  p hjpji and c is a purely imaginary number. Since ji is a coherent state we have aji = ji ) hjay = hj: (2.57) Using this relation we can write xji = s h 2m! (a+ ay)ji = s h 2m! (+ ay)ji (2.58) and hxi = hjxji = s h 2m! hj(a+ ay)ji = s h 2m! (hjaji + hjayji) = s h 2m! ( + ) (2.59) and so
xji = (x hxi)ji = s h 2m! (ay )ji: (2.60) Similarly for the momentum p = i q mh! 2 (a y a) we have pji = p i s mh! 2 (ay a)ji = i s mh! 2 (ay )ji (2.61) 2. QUANTUM DYNAMICS 49 and hpi = hjpji = i s mh! 2 hj(ay a)ji = i s mh! 2 (hjayji hjaji) = i s mh! 2 ( ) (2.62) and so
pji = (p hpi)ji = i s mh! 2 (ay )ji ) (ay )ji = i s 2 mh!
pji: (2.63) So using the last relation in (2.60)
xji = s h 2m! (i) s 2 mh!
pji = i m!| {z } purely imaginary
pji (2.64) and thus the minimum uncertainty condition is satis ed. (c) The coherent state can be expressed as a superposition of energy eigen- states ji = 1X n=0 jnihnji = 1X n=0 f(n)jni: (2.65) for the expansion coecients f(n) we have f(n) = hnji = hnjejj2=2eayj0i = ejj2=2hnjeayj0i = ejj 2=2hnj 1X m=0 1 m! (ay)mj0i = ejj2=2 1X m=0 1 m! mhnj(ay)mj0i = ejj 2=2 1X m=0 1 m! m p m!hnjmi = ejj2=2 1p n! n ) (2.66) jf(n)j2 = (jj 2)n n! exp(jj2) (2.67) whichmeans that the distribution of jf(n)j2 with respect to n is of the Poisson type about some mean value n = jj2. 50 The most probable value of n is given by the maximumof the distribution jf(n)j2 which can be found in the following way jf(n + 1)j2 jf(n)j2 = (jj2)n+1 (n+1)! exp(jj2) (jj2)n n! exp(jj2) = jj2 n+ 1  1 (2.68) which means that the most probable value of n is jj2. (d) We should check if the state exp (ipl=h) j0i is an eigenstate of the an- nihilation operator a. We have a exp (ipl=h) j0i = h a; e(ipl=h) i j0i (2.69) since aj0i = 0. For the commutator in the last relation we have h a; e(ipl=h) i = 1X n=0 1 n! il h !n [a; pn] = 1X n=1 1 n! il h !n nX k=1 pk1[a; p]pnk = 1X n=1 1 n! il h !n nX k=1 pn1i s mh! 2 = i s mh! 2 il h ! 1X n=1 1 (n 1)! ilp h !n1 = l r m! 2h e(ipl=h); (2.70) where we have used that [a; p] = i s mh! 2 [a; ay a] = i s mh! 2 : (2.71) So substituting (2.70) in (2.69) we get a [exp (ipl=h) j0i] = l r m! 2h [exp (ipl=h) j0i] (2.72) which means that the state exp (ipl=h) j0i is a coherent state with eigen- value l q m! 2h . (e) Using the hint we have j(t)i = U(t)ji = eiHt=hji (2:66)= eiHt=h 1X n=0 ejj 2=2 1p n! njni 2. QUANTUM DYNAMICS 53 = (x)f(x) +11 Z +1 0 df(x) dx dx = lim x!+1 f(x) f(x) +1 0 = f(0) = Z +1 1 (x)f(x)dx) d(x) dx = (x): (2.80) 2.10 Derive the following expression Scl = m! 2 sin(!T ) h (x20 + 2x 2 T ) cos(!T ) x0xT i for the classical action for a harmonic oscillator moving from the point x0 at t = 0 to the point xT at t = T . The Lagrangian for the one dimensional harmonic oscillator is given by L(x; _x) = 1 2 m _x2 1 2 m!2x2: (2.81) From the Lagrange equation we have @L @x d dt @L @ _x = 0 (2:81)) m!2x d dt (m _x) = 0) x+ !2x = 0: (2.82) which is the equation of motion for the system. This can be solved to give x(t) = A cos!t+B sin!t (2.83) with boundary conditions x(t = 0) = x0 = A (2.84) x(t = T ) = xT = x0 cos!T +B sin!T ) B sin!T = xT x0 cos!T ) B = xT x0 cos!T sin!T : (2.85) 54 So x(t) = x0 cos!t+ xT x0 cos!T sin!T sin!t = x0 cos!t sin!T + xT sin!t x0 cos!T sin!t sin!T = xT sin!t+ x0 sin!(T t) sin!T ) (2.86) _x(t) = xT! cos!t x0! cos!(T t) sin!T : (2.87) With these at hand we have S = Z T 0 dtL(x; _x) = Z T 0 dt  1 2m _x 2 12m!2x2  = Z T 0 dt " 1 2m d dt (x _x) 12mxx 12m!2x2 # = 1 2 m Z T 0 dtx[x+ !2x] + m 2 x _x T 0 (2:82) = m 2 [x(T ) _x(T ) x(0) _x(0)] = m 2  xT! sin!T (xT cos!T x0) x0! sin!T (xT x0 cos!T )  = m! 2 sin!T h x2T cos!T x0xT x0xT + x20 cos!T i = m! 2 sin!T h (x2T + x 2 0) cos!T 2x0xT i : (2.88) 2.11 The Lagrangian of the single harmonic oscillator is L = 1 2 m _x2 1 2 m!2x2 (a) Show that hxbtbjxatai = exp  iScl h  G(0; tb; 0; ta) where Scl is the action along the classical path xcl from (xa; ta) to (xb; tb) and G is G(0; tb; 0; ta) = 2. QUANTUM DYNAMICS 55 lim N!1 Z dy1 : : : dyN  m 2ih"  (N+1) 2 exp 8< : ih NX j=0  m 2" (yj+1 yj)2 1 2 "m!2y2j 9= ; where " = tbta (N+1) . [Hint: Let y(t) = x(t) xcl(t) be the new integration variable, xcl(t) being the solution of the Euler-Lagrange equation.] (b) Show that G can be written as G = lim N!1  m 2ih"  (N+1) 2 Z dy1 : : : dyNexp(nTn) where n = 2 664 y1 ... yN 3 775 and nT is its transpose. Write the symmetric matrix . (c) Show that Z dy1 : : : dyNexp(nTn)  Z dNnen T n = N=2p det [Hint: Diagonalize  by an orthogonal matrix.] (d) Let  2ih" m N det  det0N  pN . De ne j  j matrices 0j that con- sist of the rst j rows and j columns of 0N and whose determinants are pj . By expanding  0 j+1 in minors show the following recursion formula for the pj : pj+1 = (2 "2!2)pj pj1 j = 1; : : : ; N (2.89) (e) Let (t) = "pj for t = ta+ j" and show that (2.89) implies that in the limit "! 0; (t) satis es the equation d2 dt2 = !2(t) with initial conditions (t = ta) = 0; d(t=ta) dt = 1. 58 (b) For the argument of the exponential in the last relation we have i h NX j=0  m 2" (yj+1 yj)2 1 2 "m!2y2j  (y0=0) = i h NX j=0 m 2" (y2j+1 + y 2 j yj+1yj yjyj+1) i h NX i;j=1 1 2 "m!2yiijyj (yN+1=0) = m 2"ih NX i;j=1 (2yiijyj yii;j+1yj yii+1;jyj) i"m! 2 2h NX i;j=1 yiijyj :(2.96) where the last step is written in such a form so that the matrix  will be symmetric. Thus we have G = lim N!1  m 2ih"  (N+1) 2 Z dy1 : : : dyNexp(nTn) (2.97) with  = m 2"ih 2 6666666664 2 1 0 : : : 0 0 1 2 1 : : : 0 0 0 1 2 : : : 0 0 ... ... ... ... ... 0 0 0 : : : 2 1 0 0 0 : : : 1 2 3 7777777775 + i"m!2 2h 2 6666666664 1 0 0 : : : 0 0 0 1 0 : : : 0 0 0 0 1 : : : 0 0 ... ... ... ... ... 0 0 0 : : : 1 0 0 0 0 : : : 0 1 3 7777777775 :(2.98) (c) We can diagonalize  by a unitary matrix U . Since  is symmetric the following will hold  = UyDU ) T = UTD(Uy)T = UTDU = ) U = U: (2.99) So we can diagonalize  by an orthogonal matrix R. So  = RTDR and detR = 1 (2.100) which means thatZ dNnen T n = Z dNnen TRTRn Rn== Z dNe T  = Z d1e 21a1  Z d2e 22a2  : : : Z dN e 2 N aN  = s  a1 s  a2 : : : s  aN = N=2qQN i=1 ai = N=2p detD = N=2p det (2.101) 2. QUANTUM DYNAMICS 59 where ai are the diagonal elements of the matrix D. (d) From (2.98) we have 2ih" m !N det = det 8>>>>>< >>>>>: 2 6666666664 2 1 0 : : : 0 0 1 2 1 : : : 0 0 0 1 2 : : : 0 0 ... ... ... ... ... 0 0 0 : : : 2 1 0 0 0 : : : 1 2 3 7777777775 "2!2 2 6666666664 1 0 0 : : : 0 0 0 1 0 : : : 0 0 0 0 1 : : : 0 0 ... ... ... ... ... 0 0 0 : : : 1 0 0 0 0 : : : 0 1 3 7777777775 9>>>>>= >>>>>; = det0N  pN : (2.102) We de ne j  j matrices 0j that consist of the rst j rows and j columns of 0N . So det0j+1 = det 2 666666666664 2 "2!2 1 : : : 0 0 0 1 2 "2!2 : : : 0 0 0 0 1 : : : 0 0 0 ... ... ... ... ... 0 0 : : : 2 "2!2 1 0 0 0 : : : 1 2 "2!2 1 0 0 : : : 0 1 2 "2!2 3 777777777775 : From the above it is obvious that det0j+1 = (2 "2!2) det0j det0j1 ) pj+1 = (2 "2!2)pj pj1 for j = 2; 3; : : : ; N (2.103) with p0 = 1 and p1 = 2 "2!2. (e) We have (t)  (ta + j")  "pj ) (ta + (j + 1)") = "pj+1 = (2 "2!2)"pj "pj1 = 2(ta + j") "2!2(ta + j") (ta + (j 1)") ) (t+ ") = 2(t) "2!2(t) (t "): (2.104) 60 So (t+ ") (t) = (t) (t ") "2!2(t)) (t+")(t) " (t)(t") " " = !2(t)) lim "!0 0(t) 0(t ") " = !2(t)) d 2 dt2 = !2(t): (2.105) From (c) we have also that (ta) = "p0 ! 0 (2.106) and d dt (ta) = (ta + ") (ta) " = "(p1 p0) " = p1 p0 = 2 "2!2 1! 1: (2.107) The general solution to (2.105) is (t) = A sin(!t+ ) (2.108) and from the boundary conditions (2.106) and (2.107) we have (ta) = 0) A sin(!ta + ) = 0)  = !ta + n n 2 Z (2.109) which gives that (t) = A sin!(t ta), while d dt = A! cos(t ta)) 0(ta) = A! (2:107)) A! = 1) A = 1 ! (2.110) Thus (t) = sin!(t ta) ! : (2.111) (f) Gathering all the previous results together we get G = lim N!1 " m 2ih" (N+1) Np det #1=2 2. QUANTUM DYNAMICS 63 2.13 (a) Verify the relation [i;j] = ihe c ! "ijkBk where ~  m ~x dt = ~p e ~A c and the relation m d2~x dt2 = d~ dt = e " ~E + 1 2c d~x dt  ~B ~B  d~x dt !# : (b) Verify the continuity equation @ @t + ~r0
~j = 0 with ~j given by ~j = h m ! =( ~r0 )  e mc  ~Aj j2: (a) We have [i;j] =  pi eAi c ; pj eAj c  = e c [pi; Aj] e c [Ai; pj] = ihe c @Aj @xi ihe c @Ai @xj = ihe c @Aj @xi @Ai @xj ! = ihe c ! "ijkBk: (2.114) We have also that dxi dt = 1 ih [xi;H] = 1 ih 2 4xi; ~2 2m + e 3 5 = 1 ih 2 4xi; ~2 2m 3 5 = 1 ih2m f[xi;j] j +j [xi;j]g = 1 ih2m f[xi; pj ]j +j [xi; pj]g = 2ih 2ihm jij = i m ) 64 d2xi dt2 = 1 ih " dxi dt ;H # = 1 ihm 2 4i; ~2 2m + e 3 5 = 1 ih2m2 f[i;j] j +j [i;j]g+ e ihm [i; ] (2:114) = 1 2m2ih " ihe c "ijkBkj + ihe c "ijkjBk # + e ihm  pi eAi c ;   = e 2m2c ("ikjBkj + "ijkjBk) + e ihm [pi; ] = e 2m2c m  "ijk xj dt Bk "ikjBkxj dt  e m @ @xi ) m d2xi dt2 = eEi + e 2c " ~x dt  ~B ! i ~B  ~x dt ! i # ) m d2~x dt2 = e " ~E + 1 2c ~x dt  ~B ~B  ~x dt !# : (2.115) (b) The time-dependent Schrodinger equation is ih @ @t hx0j ; t0; ti = hx0jHj ; t0; ti = hx0j 1 2m 0 @~p e ~A c 1 A2 + ej ; t0; ti = 1 2m 2 4ih~r0 e ~A(~x0) c 3 5
2 4ih~r0 e ~A(~x0) c 3 5 hx0j ; t0; ti+ e(~x0)hx0j ; t0; ti = 1 2m " h2~r0
~r0 + e c ih~r0
~A(~x0) + ihe c ~A(~x0)
~r0 + e 2 c2 A2(~x0) # (~x0; t) +e(~x0) (~x0; t) = 1 2m  h2r02 (~x0; t)0 + e c ih  ~r0
~A  (~x0; t) + e c ih ~A(~x0)
~r0 (~x0; t) + ih e c ~A(~x0)
~r0 (~x0; t) + e 2 c2 A2(~x0) (~x0; t) # + e(~x0) (~x0; t) = 1 2m " h2r02 + e c ih  ~r0
~A  + 2ih e c ~A
~r0 + e 2 c2 A2 # + e : (2.116) Multiplying the last equation by  we get ih  @ @t = 2. QUANTUM DYNAMICS 65 1 2m " h2 r02 + e c ih  ~r0
~A  j j2 + 2ihe c ~A
~r0 + e 2 c2 A2j j2 # + ej j2: The complex conjugate of this eqution is ih @ @t  = 1 2m " h2 r02  e c ih  ~r0
~A  j j2 2ihe c ~A
~r0  + e 2 c2 A2j j2 # + ej j2: Thus subtracting the last two equations we get h 2 2m h r02 r02  i +  e mc  ih  ~r0
~A  j j2 +  e mc  ih ~A
( ~r0 + ~r0 ) = ih  @ @t + @ @t  ! ) h 2 2m ~r0
h ~r0 ~r0  i +  e mc  ih  ~r0
~A  j j2 +  e mc  ih ~A
(~r0j j2) = ih @ @t j j2 ) @ @t j j2 = h m ~r0
h =( ~r0 ) i +  e mc  ~r0
h ~Aj j2 i ) @ @t j j2 + ~r0
" h m =( ~r0 )  e mc  ~Aj j2 # = 0) @ @t + ~r0
~j = 0 (2.117) with ~j =  h m  =( ~r0 )  e mc  ~Aj j2: and  = j j2 2.14 An electron moves in the presence of a uniform magnetic eld in the z-direction ( ~B = Bẑ). (a) Evaluate [x;y]; where x  px eAx c ; y  py eAy c : 68 If we write (; ; z) = ()R()Z(z) and k2 = 2mE h2 we will have ()Z(z) d2R d2 + ()Z(z) 1  dR d + R()Z(z) 2 d2 d2 +R()() d2Z dz2 + k2R()()Z(z) = 0) 1 R() d2R d2 + 1 R() dR d + 1 2() d2 d2 + 1 Z(z) d2Z dz2 + k2 = 0(2.128) with initial conditions (a; ; z) = (R;; z) = (; ; 0) = (; ; a) = 0. So 1 Z(z) d2Z dz2 = l2 ) d 2Z dz2 + l2Z(z) = 0) Z(z) = A1eilz +B1eilz (2.129) with Z(0) = 0) A1 + B1 = 0) Z(z) = A1  eilz eilz  = C sin lz Z(a) = 0) C sin la = 0) la = n) l = ln = n a n = 1;2; : : : So Z(z) = C sin lnz (2.130) Now we will have 1 R() d2R d2 + 1 R() dR d + 1 2() d2 d2 + k2 l2 = 0 )  2 R() d2R d2 +  R() dR d + 1 () d2 d2 + 2(k2 l2) = 0 ) 1 () d2 d2 = m2 ) () = eim: (2.131) with (+ 2) = ()) m 2 Z: (2.132) So the Schrodinger equation is reduced to 2 R() d2R d2 +  R() dR d m2 + 2(k2 l2) = 0 2. QUANTUM DYNAMICS 69 ) d 2R d2 + 1  dR d + " (k2 l2) m 2 2 # R() = 0 ) d 2R d( p k2 l2)2 + 1p k2 l2 dR d( p k2 l2) + " 1 m 2 (k2 l2)2 # R() = 0 ) R() = A3Jm( p k2 l2) +B3Nm( p k2 l2) (2.133) In the case at hand in which a ! 0 we should take B3 = 0 since Nm ! 1 when ! 0. From the other boundary condition we get R(R) = 0) A3Jm(R p k2 l2) = 0) R p k2 l2 = m (2.134) where m is the -th zero of the m-th order Bessel function Jm. This means that the energy eigenstates are given by the equation m = R p k2 l2 ) k2 l2 =  2 m R2 ) 2mE h2  n  a 2 = 2m R2 ) E = h 2 2m " 2m R2 +  n  a 2# (2.135) while the corresponding eigenfunctions are given by nm(~x) = AcJm( m R )eim sin  n z a  (2.136) with n = 1;2; : : : and m 2 Z. Now suppose that ~B = Bẑ. We can then write ~A = B2a 2 ! ̂ =  2 ! ̂: (2.137) The Schrodinger equation in the presence of the magnetic eld ~B can be written as follows 1 2m 2 4ih~r e ~A(~x) c 3 5
2 4ih~r e ~A(~x) c 3 5 (~x) = E (~x) ) h 2 2m " ̂ @ @ + ẑ @ @z + ̂ 1  @ @ ie hc  2 !#
" ̂ @ @ + ẑ @ @z + ̂ 1  @ @ ie hc  2 !# (~x) = E (~x): (2.138) 70 Making now the transformation D  @@ iehc 2 we get h 2 2m " ̂ @ @ + ẑ @ @z + ̂ 1  D #
" ̂ @ @ + ẑ @ @z + ̂ 1  D # (~x) = E (~x) ) h 2 2m " @2 @2 + 1  @ @ + 1 2 D2 + @2 @z2 # (~x) = E (~x); (2.139) where D2 =  @ @ ie hc  2 2 . Leting A = e hc  2 we get D2 = @2 @2 2ie hc  2 @ @ A2 ! = @2 @2 2iA @ @ A2 ! : (2.140) Following the same procedure we used before (i.e. (; ; z) = R()()Z(z)) we will get the same equations with the exception of" @2 @2 2iA @ @ A2 #  = m2) d 2 d2 2iAd d + (m2 A2) = 0: The solution to this equation is of the form el. So l2el 2iAlel + (m2 A2)el = 0) l2 2iAl+ (m2 A2) ) l = 2iA q 4A2 4(m2 A2) 2 = 2iA 2im 2 = i(Am) which means that () = C2e i(Am): (2.141) But (+ 2) = ()) Am = m0 m0 2 Z ) m = (m0 A) m0 2 Z: (2.142) This means that the energy eigenfunctions will be nm(~x) = AcJm( m R )eim 0 sin  n z a  (2.143) but now m is not an integer. As a result the energy of the ground state will be E = h2 2m " 2m R2 +  n  a 2# (2.144) 2. QUANTUM DYNAMICS 73 These are the Hamiltonian eigenstates in momentum space. For the eigen- functions in coordinate space we have (x) = Z dphxjpihpjEi (2:150)= 1 2h p  Z dpe ipx h e i h  p3 6mEp  = 1 2h p  Z dp exp " i p3 h6m i h  E  x  p # : (2.151) Using now the substitution u = p (h2m)1=3 ) p 3 h6m = u3 3 (2.152) we have (x) = (h2m)1=3 2h p  Z +1 1 du exp " iu3 3 i h  E  x  u(h2m)1=3 # = 2 p  Z +1 1 du exp " iu3 3 iuq # ; (2.153) where =  2m h2 1=3 and q = h E  x i . So (x) = 2 p  Z +1 1 du cos u3 3 uq ! =  p  Z +1 0 cos u3 3 uq ! du since R+1 1 sin  u3 3 uq  du = 0. In terms of the Airy functions Ai(q) = 1p  Z +1 0 cos u3 3 uq ! du (2.154) we will have (x) = p  Ai(q): (2.155) For large jqj, leading terms in the asymptotic series are as follows Ai(q)  1 2 p q1=4 exp  2 3 q3=2  ; q > 0 (2.156) Ai(q)  1p (q)1=4 sin  2 3 (q)3=2+  4  ; q < 0 (2.157) 74 Using these approximations in (2.155) we get (q)   p  1 q1=4 sin  2 3 q3=2 +  4  ; for E > V (x) (q)  2 p  1 (q)1=4 exp  2 3 (q)3=2  ; for E < V (x) (2.158) as expected from the WKB approximation. (b) When V = jxj we have bound states and therefore the energy spec- trum is discrete. So in this case the energy eigenstates heve to satisfy the consistency relationZ x1 x2 dx q 2m[E jxj] =  n+ 12  h; n = 0; 1; 2; : : : (2.159) The turning points are x1 = E and x2 = E . So  n + 1 2  h = Z E= E= dx q 2m[E jxj] = 2 Z E= 0 q 2m[E x]dx = 2 p 2m Z E= 0  E  x 1=2 d(x) = 2 p 2m 2 3  E  x 3=2 E= 0 = 2 p 2m 2 3  E  3=2 )  E  3=2 = 3  n+ 1 2  h 4 p 2m )  E   = [3  n+ 1 2  h]2=3 42=3(2m)1=3 ) En = 2 43  n + 12  h 4 p 2m 3 5 2=3 : (2.160) 3. THEORY OF ANGULAR MOMENTUM 75 3 Theory of Angular Momentum 3.1 Consider a sequence of Euler rotations represented by D(1=2)( ; ; ) = exp i3 2  exp i2 2 ! exp i3 2  = ei( + )=2 cos 2 ei( )=2 sin 2 ei( )=2 sin 2 ei( + )=2 cos 2 ! : Because of the group properties of rotations, we expect that this sequence of operations is equivalent to a single rotation about some axis by an angle . Find . In the case of Euler angles we have D(1=2)( ; ; ) = ei( + )=2 cos 2 ei( )=2 sin 2 ei( )=2 sin 2 e i( + )=2 cos 2 ! (3.1) while the same rotation will be represented by D(1=2)(; n̂) (S3:2:45)= 0 @ cos   2  inz sin   2  (inx ny) sin   2  (inx + ny) sin   2  cos   2  + inz sin   2  1 A : (3.2) Since these two operators must have the same e ect, each matrix element should be the same. That is ei( + )=2 cos 2 = cos  2 ! inz sin  2 ! ) cos  2 ! = cos ( + ) 2 cos 2 ) cos  = 2 cos2 2 cos2 ( + ) 2 1 )  = arccos " 2 cos2 2 cos2 ( + ) 2 1 # : (3.3) 3.2 An angular-momentum eigenstate jj;m = mmax = ji is rotated by an in nitesimal angle " about the y-axis. Without using the 78 and 1 sin  @ @ sin  @ @ ! (~x) = rf(r) sin  @ @ h 3 sin2  + (cos+ sin) sin  cos  i = rf(r) sin  h 6 sin  cos  + (cos+ sin )(cos2  sin2 ) i :(3.15) Substitution of (3.14) and (3.15) in (3.11) gives h~xj~L2j i = h2rf(r)  1 sin  (cos + sin)(1 cos2  + sin2 ) 6 cos   = h2rf(r)  1 sin  2 sin2 (cos + sin) + 6 cos   = 2h2rf(r) [sin  cos + sin  sin+ 3 cos ] = 2h2 (~x)) L2 (~x) = 2h2 (~x) = 1(1 + 1)h2 (~x) = l(l+ l)h2 (~x) (3.16) which means that (~x) is en eigenfunction of ~L2 with eigenvalue l = 1. (b) Since we already know that l = 1 we can try to write (~x) in terms of the spherical harmonics Y m1 (; ). We know that Y 01 = s 3 4 cos  = s 3 4 z r ) z = r s 4 3 Y 01 Y +11 = q 3 8 (x+iy) r Y 11 = q 3 8 (xiy) r 9= ;) 8< : x = r q 2 3  Y 11 Y +11  y = ir q 2 3  Y 11 + Y +1 1  So we can write (~x) = r s 2 3 f(r) h 3 p 2Y 01 + Y 1 1 Y +11 + iY +11 + iY 11 i = s 2 3 rf(r) h 3 p 2Y 01 + (1 + i)Y 1 1 + (i 1)Y +11 i : (3.17) But this means that the part of the state that depends on the values of m can be written in the following way j im = N h 3 p 2jl = 1;m = 0i + (1 + i)jl = 1;m = 1i + (1 i)jl = 1;m = 1i i and if we want it normalized we will have jN j2(18 + 2 + 2) = 1) N = 1p 22 : (3.18) 3. THEORY OF ANGULAR MOMENTUM 79 So P (m = 0) = jhl = 1;m = 0j ij2 = 9
2 22 = 9 11 ; (3.19) P (m = +1) = jhl = 1;m = +1j ij2 = 2 22 = 1 11 ; (3.20) P (m = 1) = jhl = 1;m == 1j ij2 = 2 22 = 1 11 : (3.21) (c) If E(~x) is an energy eigenfunction then it solves the Schrodinger equation h2 2m " @2 @r2 E(~x) + 2 r @ @r E(~x) L 2 h2r2 E(~x) # + V (r) E(~x) = E E(~x) ) h 2 2m Y ml " d2 dr2 [rf(r)] + 2 r d dr [rf(r)] 2 r2 [rf(r)] # + V (r)rf(r)Y ml = Erf(r)Y ml ) V (r) = E + 1 rf(r) h2 2m " d dr [f(r) + rf 0(r)] + 2 r [f(r) + rf 0(r)] 2 r f(r) # ) V (r) = E + 1 rf(r) h2 2m [f 0(r) + f 0(r) + rf 00(r) + 2f 0(r)]]) V (r) = E + h2 2m rf 00(r) + 4f 0(r) rf(r) : (3.22) 3.4 Consider a particle with an intrinsic angular momentum (or spin) of one unit of h. (One example of such a particle is the %- meson). Quantum-mechanically, such a particle is described by a ketvector j%i or in ~x representation a wave function %i(~x) = h~x; ij%i where j~x; ii correspond to a particle at ~x with spin in the i:th di- rection. (a) Show explicitly that in nitesimal rotations of %i(~x) are obtained by acting with the operator u~" = 1 i ~" h
(~L+ ~S) (3.23) 80 where ~L = h i r̂  ~r. Determine ~S ! (b) Show that ~L and ~S commute. (c) Show that ~S is a vector operator. (d) Show that ~r ~%(~x) = 1 h2 ( ~S
~p)~% where ~p is the momentum oper- ator. (a) We have j%i = 3X i=1 Z j~x; iih~x; ij%i = 3X i=1 Z j~x; ii%i(~x)d3x: (3.24) Under a rotation R we will have j%0i = U(R)j%i = 3X i=1 Z U(R) [j~xi jii] %i(~x)d3x = 3X i=1 Z jR~xi jiiD(1)il (R)%l(~x)d3x detR=1= 3X i=1 Z j~x; iiD(1)il (R)%l(R1~x)d3x = 3X i=1 Z j~x; ii%i0~x)d3x) %i0(~x) = D(1)il (R)%l(R1~x)) ~%0(~x) = R~%(R1~x): (3.25) Under an in nitesimal rotation we will have R(; n̂)~r = ~r + ~r = ~r + (n̂ ~r) = ~r + ~" ~r: (3.26) So ~%0(~x) = R()~%(R1~x) = R()~%(~x ~" ~x) = ~%(~x ~" ~x) + ~" ~%(~x ~" ~x): (3.27) On the other hand ~%(~x ~" ~x) = ~%(~x) (~" ~x)
~r~%(~x) = ~%(~x) ~"
(~x ~r)~%(~x) = ~%(~x) i h ~"
~L~%(~x) (3.28) 3. THEORY OF ANGULAR MOMENTUM 83 In the same way we have Jjj = 2;m = 1i = 1p 2 (J1 + J2)j0+i + 1p 2 (J1 + J2)j+ 0i ) p 6jj = 2;m = 0i = 1p 2 hp 2j +i + p 2j00i i + 1p 2 hp 2j00i + p 2j+i i ) p 6jj = 2;m = 0i = 2j00i + j +i + j+i ) jj = 2;m = 0i = s 2 3 j00i + 1p 6 j+i+ 1p 6 j +i (3.38) Jjj = 2;m = 0i = s 2 3 (J1 + J2)j00i + 1p 6 (J1 + J2)j+i+ 1p 6 (J1 + J2)j +i ) p 6jj = 2;m = 1i = s 2 3 hp 2j 0i + p 2j0i i + 1p 6 p 2j0i+ 1p 6 p 2j 0i ) jj = 2;m = 1i = 2 6 p 2j0i + 2 6 p 2j 0i + 1 6 p 2j0i+ 1 6 p 2j 0i ) jj = 2;m = 1i = 1p 2 j0i+ 1p 2 j 0i (3.39) Jjj = 2;m = 1i = 1p 2 (J1 + J2)j0i+ 1p 2 (J1 + J2)j 0i ) p 4jj = 2;m = 2i = 1p 2 p 2j i + 1p 2 p 2j i ) jj = 2;m = 2i = j i: (3.40) Now let us return to equation (3.35). If j = 1, m = 1 we will have jj = 1;m = 1i = aj+ 0i + bj0+i (3.41) This state should be orthogonal to all jj;mi states and in particular to jj = 2;m = 1i. So hj = 2;m = 1jj = 1;m = 1i = 0) 1p 2 a+ 1p 2 b = 0) a+ b = 0) a = b : (3.42) 84 In addition the state jj = 1;m = 1i should be normalized so hj = 1;m = 1jj = 1;m = 1i = 1) jaj2 + jbj2 = 1 (3:42)) 2jaj2 = 1) jaj = 1p 2 : By convention we take a to be real and positive so a = 1p 2 and b = 1p 2 . That is jj = 1;m = 1i = 1p 2 j+ 0i 1p 2 j0+i: (3.43) Using the same procedure we used before Jjj = 1;m = 1i = 1p 2 (J1 + J2)j+ 0i 1p 2 (J1 + J2)j0+i ) p 2jj = 1;m = 0i = 1p 2 hp 2j00i + p 2j+i i 1p 2 hp 2j +i + p 2j00i i ) jj = 1;m = 0i = 1p 2 j+i 1p 2 j +i (3.44) Jjj = 1;m = 0i = 1p 2 (J1 + J2)j+i 1p 2 (J1 + J2)j +i ) p 2jj = 1;m = 1i = 1p 2 p 2j0i 1p 2 p 2j 0i ) jj = 1;m = 1i = 1p 2 j0i 1p 2 j 0i: (3.45) Returning back to (3.35) we see that the state jj = 0;m = 0i can be written as jj = 0;m = 0i = c1j00i + c2j+i+ c3j +i: (3.46) This state should be orthogonal to all states jj;mi and in particulat to jj = 2;m = 0i and to j = 1;m = 0i. So hj = 2;m = 0jj = 0;m = 0i = 0) s 2 3 c1 + 1p 6 c2 + 1p 6 c3 ) 2c1 + c2 + c3 = 0 (3.47) hj = 1;m = 0jj = 0;m = 0i = 0) 1p 2 c2 1p 2 c3 ) c2 = c3: (3.48) 3. THEORY OF ANGULAR MOMENTUM 85 Using the last relation in (3.47), we get 2c1 + 2c2 = 0) c1 + c2 = 0) c1 = c2: (3.49) The state jj = 0;m = 0i should be normalized so hj = 0;m = 0jj = 0;m = 0i = 1) jc1j2 + jc2j2 + jc3j2 = 1) 3jc2j2 = 1 ) jc2j = 1p 3 : (3.50) By convention we take c2 to be real and positive so c2 = c3 = 1p 3 and c1 = 1p3 . Thus jj = 0;m = 0i = 1p 3 j+i + 1p 3 j +i 1p 3 j00i: (3.51) So gathering all the previous results together jj = 2;m = 2i = j++i jj = 2;m = 1i = 1p 2 j0+i+ 1p 2 j+ 0i jj = 2;m = 0i = q 2 3 j00i + 1p6 j+i+ 1p6 j +i jj = 2;m = 1i = 1p 2 j0i+ 1p 2 j 0i jj = 2;m = 2i = j i jj = 1;m = 1i = 1p 2 j+ 0i 1p 2 j0+i jj = 1;m = 0i = 1p 2 j+i 1p 2 j +i jj = 1;m = 1i = 1p 2 j0i 1p 2 j 0i jj = 0;m = 0i = 1p 3 j+i+ 1p 3 j +i 1p 3 j00i: (3.52) 3.6 (a) Construct a spherical tensor of rank 1 out of two di erent vectors ~U = (Ux; Uy; Uz) and ~V = (Vx; Vy; Vz). Explicitly write T (1) 1;0 in terms of Ux;y;z and Vx;y;z . (b) Construct a spherical tensor of rank 2 out of two di erent vectors ~U and ~V . Write down explicitly T (2) 2;1;0 in terms of Ux;y;z and Vx;y;z. 88 (3:52) = s 2 3 U0V0 + s 1 6 U1V+1 + s 1 6 U+1V1 = s 2 3 UzVz s 1 6 1 2 (Ux iUy)(Vx + iVy) s 1 6 1p 2 1 2 (Ux + iUy)(Vx iVy) = s 1 6  2UzVz 12UxVx i 2 UxVy + i 2 UyVx 12UyVy 12UxVx + i 2 UxVy i 2 UyVx 12UyVy  = s 1 6 (2UzVz UxVx UyVy) (3.65) T (2) 1 = h11; 0 1j11; 2 1iU0V1 + h11;10j11; 2 + 1iU1V0 (3:52) = 1p 2 U0V1 + 1p 2 U1V0 = 12(UzVx + UxVz iUzVy iUyVz) (3.66) T (2) 2 = h11;1 1j11; 2 2iU1V1 (3:52)= U1V1 = 12(Ux iUy)(Vx iVy) = 12(UxVx UyVy iUxVy iUyVx): (3.67) 3.7 (a) Evaluate jX m=j jd(j)mm0( )j2m for any j (integer or half-integer); then check your answer for j = 12 . (b) Prove, for any j, jX m=j m2jd(j)m0m( )j2 = 12j(j + 1) sin +m0 2 + 12(3 cos 2 1): [Hint: This can be proved in many ways. You may, for instance, examine the rotational properties of J2z using the spherical (irre- ducible) tensor language.] 3. THEORY OF ANGULAR MOMENTUM 89 (a) We have jX m=j jd(j)mm0( )j2m = jX m=j mjhjmjeiJy =hjjm0ij2 = jX m=j mhjmjeiJy =hjjm0i  hjmjeiJy =hjjm0i  = jX m=j mhjmjeiJy =hjjm0ihjm0jeiJy =hjjmi = jX m=j hjm0jeiJy =hmjjmihjmjeiJy =hjjm0i = 1 h hjm0jeiJy =hJz 2 4 jX m=j jjmihjmj 3 5 eiJy =hjjm0i = 1 h hjm0jeiJy =hJzeiJy =hjjm0i = 1 h hjm0jD( ; êy)JzD( ; êy)jjm0i: (3.68) But the momentum ~J is a vector operator so from (S-3.10.3) we will have that D( ; êy)JzD( ; êy) = X j Rzj( ; êy)Jj: (3.69) On the other hand we know (S-3.1.5b) that R( ; êy) = 0 B@ cos 0 sin 0 1 0 sin 0 cos 1 CA : (3.70) So jX m=j jd(j)mm0( )j2m = 1 h [ sin hjm0jJxjjm0i+ cos hjm0jJzjjm0i] = 1 h  sin hjm0jJ+ + J 2 jjm0i + hm0 cos  = m0 cos : (3.71) 90 For j = 1=2 we know from (S-3.2.44) that d (1=2) mm0 ( ) = cos 2 sin 2 sin 2 cos 2 ! : (3.72) So for m0 = 1=2 1=2X m=1=2 jd(j)m1=2( )j2m = 12 sin2 2 + 12 cos 2 2 = 12 cos = m 0 cos (3.73) while for m0 = 1=2 1=2X m=1=2 jd(j)m1=2( )j2m = 12 cos2 2 + 12 sin 2 2 = 1 2 cos = m0 cos : (3.74) (b) We have jX m=j m2jd(j)m0m( )j2 = jX m=j m2jhjm0jeiJy =hjjmij2 = jX m=j m2hjm0jeiJy =hjjmi  hjm0jeiJy =hjjmi  = jX m=j m2hjm0jeiJy =hjjmihjmjeiJy =hjjm0i = jX m=j hjm0jeiJy =hm2jjmihjmjeiJy =hjjm0i = 1 h2 hjm0jeiJy =hJ2z 2 4 jX m=j jjmihjmj 3 5 eiJy =hjjm0i = 1 h2 hjm0jeiJy =hJ2z eiJy =hjjm0i = 1 h hjm0jD( ; êy)J2zDy( ; êy)jjm0i: (3.75) 3. THEORY OF ANGULAR MOMENTUM 93 (3:82) = p 6eh ; j;m = jjT (2)0 j ; j;m = ji (W:E:)= hj2; j0jj2; jji h jkT (2)k jip 2j + 1 p 6e ) h jkT (2)k ji = Qp 6e p 2j + 1 hj2; j0jj2; jji : (3.84) So e h ; j;m0j(x2 y2)j ; j;m = ji (3:83) = eh ; j;m0jT (2)+2 j ; j;m = ji+ eh ; j;m0jT (2)2 j ; j;m = ji = e 0z }| { hj2; j2jj2; jm0i h jkT (2)k jip 2j + 1 + em0;j2hj2; j 2jj2; jj 2ih jkT (2)k jip 2j + 1 (3:84) = Qp 6 hj2; j;2jj2; j; j 2i hj2; j; 0jj2; j; ji m0;j2: (3.85) 94 4 Symmetry in Quantum Mechanics 4.1 (a) Assuming that the Hamiltonian is invariant under time reversal, prove that the wave function for a spinless nondegenerate system at any given instant of time can always be chosen to be real. (b) The wave function for a plane-wave state at t = 0 is given by a complex function ei~p
~x=h. Why does this not violate time-reversal invariance? (a) Suppose that jni in a nondegenerate energy eigenstate. Then Hjni = Hjni = Enjni ) jni = eijni ) jn; t0 = 0; ti = eitH=hjni = eitEn=hjni = eitEn=hjni = ei(Enth +)jni = ei(2Enth +)jn; t0 = 0; ti )  Z d3xj~xih~xj  jn; t0 = 0; ti = ei( 2Ent h +) Z d3xj~xih~xj  jn; t0 = 0; ti ) Z d3xh~xjn; t0 = 0; tij~xi = Z d3xei( 2Ent h +)h~xjn; t0 = 0; tij~xi ) n(~x; t) = ei( 2Ent h +)n(~x; t): (4.1) So if we choose at any instant of time  = 2Enth the wave function will be real. (b) In the case of a free particle the Schrodinger equation is p2 2m jni = Ejni ) h 2 2m ~rn(x) = En(x) ) n(x) = Aei~p
~x=h +Bei~p
~x=h (4.2) The wave functions n(x) = e i~p
~x=h and 0n(x) = e i~p
~x=h correspond to the same eigenvalue E = p 2 2m and so there is degeneracy since these correspond to di erent state kets j~pi and j ~pi. So we cannot apply the previous result. 4.2 Let (~p0) be the momentum-space wave function for state j i, that is, (~p0) = h~p0j i.Is the momentum-space wave function for the 4. SYMMETRY IN QUANTUM MECHANICS 95 time-reversed state j i given by (~p0), (~p0), (~p0), or (~p0)? Justify your answer. In the momentum space we have j i = Z d3p0h~p0j ij~p0i ) j i = Z d3p0(~p0)j~p0i ) j i = Z d3p0 [h~p0j ij~p0i] = Z d3p0h~p0j ij~p0i: (4.3) For the momentum it is natural to require h j~pj i = h~ j~pj~ i ) h~ j~p1j~ i ) ~p1 = ~p (4.4) So ~pj~p0i (4:4)= ~pj~p0i ) j~p0i = j ~p0i (4.5) up to a phase factor. So nally j i = Z d3p0h~p0j ij ~p0i = Z d3p0h~p0j ij~p0i ) h~p0jj i = ~(~p0) = h~p0j i = (~p0): (4.6) 4.3 Read section 4.3 in Sakurai to refresh your knowledge of the quantum mechanics of periodic potentials. You know that the en- ergybands in solids are described by the so called Bloch functions n;k full lling, n;k(x+ a) = e ika n;k(x) where a is the lattice constant, n labels the band, and the lattice momentum k is restricted to the Brillouin zone [=a; =a]. Prove that any Bloch function can be written as, n;k(x) = X Ri n(xRi)eikRi