solutions to problems quantum mechanics

Solutions to Problems in Quantum Mechanics P Saltsidis, additions by B Brinne 1995,1999 Most of the problems presented here are taken from the book Sakurai, J J., Modern Quantum Mechanics, Reading, MA: Addison-Wesley, 1985 Contents I Problems Fundamental Concepts Quantum Dynamics Theory of Angular Momentum Symmetry in Quantum Mechanics Approximation Methods II Solutions Fundamental Concepts Quantum Dynamics Theory of Angular Momentum Symmetry in Quantum Mechanics Approximation Methods 14 17 19 25 36 75 94 103 23 CONTENTS Part I Problems FUNDAMENTAL CONCEPTS Fundamental Concepts 1.1 Consider a ket space spanned by the eigenkets fja0ig of a Hermitian operator A There is no degeneracy (a) Prove that Y (A ; a0) a0 is a null operator (b) What is the signi cance of Y (A ; a00) 00 ? a00 6=a0 a ; a (c) Illustrate (a) and (b) using A set equal to Sz of a spin 21 system 1.2 A spin 12 system is known to be in an eigenstate of S~ n^ with eigenvalue h=2, where n^ 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 ; hSx i)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) 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 relation holds if the state in question satis es Aj i = B j 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 " (x0 ; hxi)2 # i h p i x ; = ; hx j i = (2 d ) exp h 4d2 satis es the uncertainty relation q q h( x)2i h( p)2i = h2 : 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) 1.4 (a) Let x and px be the coordinate and linear momentum in one dimension Evaluate the classical Poisson bracket x F (px)]classical : (b) Let x and px be the corresponding quantum-mechanical operators this time Evaluate the commutator x exp iphxa : (c) Using the result obtained in (b), prove that exp iphxa jx0i (xjx0i = x0jx0i) QUANTUM DYNAMICS is an eigenstate of the coordinate operator x What is the corresponding eigenvalue? 1.5 (a) Prove the following: (i) hp0jxj i = ih @p@ hp0j i Z @ (p0) (ii) h jxj i = dp0 (p0)ih @p where (p0) = hp0 j i and (p0) = hp0j i are momentum-space wave functions (b) What is the physical signi cance of exp ixh where x is the position operator and is some number with the dimension of momentum? Justify your answer Quantum Dynamics 2.1 Consider the spin-procession problem discussed in section 2.1 in Jackson It can also be solved in the Heisenberg picture Using the Hamiltonian eB S = !S H = ; mc z z write the Heisenberg equations of motion for the time-dependent operators Sx(t), Sy (t), and Sz (t) Solve them to obtain Sx y z as functions of time 2.2 Let x(t) be the coordinate operator for a free particle in one dimension in the Heisenberg picture Evaluate x(t) x(0)] : 2.3 Consider a particle in three dimensions whose Hamiltonian is given by H = 2p~m + V (~x): By calculating ~x p~ H ] obtain * + d h~x ~pi = p2 ; h~x r ~ V i: dt m 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 ;hipa j0i: You may use !23 x hx0j0i = ;1=4x;0 1=2 exp 4; 21 x 0 @x0 h m! !1=21 A: (b) Obtain a simple expression that the probability that the state is found in the ground state at t = 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 Evaluate the correlation function explicitly for the ground state of a one-dimensional simple harmonic oscillator 106 Together this gives V10 10 = V01 01 = y V10 01 = h! h j ax ay j i = y V01 10 = h! h j ax ay j i = The V -matrix becomes h! h! : (5.8) ! h! and so the eigenvalues (= 1) are h! : 1= To get the eigenvectors we solve ! ! ! x = x y y and get E+ = h!(2 + ) j i+ = p12 ( j i + j i) j i; = p12 ( j i ; j i) E; = h!(2 ; ): (5.9) The second excited state is also degenerate j i, j 1 i, j i, so we need the corresponding matrix elements However the only nonvanishing ones are: V11 20 = V20 11 = V11 02 = V02 11 = ph! (5.10) p (where the came from going from level to in either of the oscillators) and thus to get the eigenvalues we evaluate ; 0 = det B @ ; CA = ; ( ; 1) + = (2 ; 2) ; APPROXIMATION METHODS 107 which means that the eigenvalues are f0 as above we get the eigenvectors h!g By the same method p j i+ = 21 ( j i + j 1 i + j i) j i0 = p12 (; j i + j i) p j i; = 21 ( j i ; j 1 i + j i) E+ = h!(3 + ) E0 = 3h! E; = h!(3 ; ): (c) To solve the problem exactly we will make a variable change The potential is h i m!2 21 (x2 + y2) + xy = # " 2 2 = m! ((x + y) + (x ; y) ) + (x + y) ; (x ; y) ) : (5.11) Now it is natural to introduce x0 p1 (x + y) p0x p1 (p0x + p0y ) 2 1 y0 p (x ; y) p0y p (p0x ; p0y ): (5.12) 2 Note: x0 p0x] = y0 p0y ] = ih, so that (x0, p0x ) and (y0, p0y ) are canonically conjugate In these new variables the problem takes the form 02 02 H = 21m (p0x2 + p0y2) + m! (1 + )x + (1 ; )y ]: p p So we get one oscillator with !x0 = ! + and another with !y0 = ! ; The energy levels are: E0 = h! p E1 = h! + h!x0 = h!(1 + + ) = = h!(1 + + 12 + : : :) = h!(2 + 21 ) + O( 2) 108 E0 E2 E1 E0 = = = = h! + h!y0 = : : : = h!(2 ; 21 ) + O( 2) h! + 2h!x0 = : : : = h!(3 + ) + O( 2) h! + h!x0 + h!y0 = : : : = 3h! + O( 2) h! + 2h!y0 = : : : = h!(3 ; ) + O( 2): (5.13) So rst order perturbation theory worked! 5.2 A system that has three unperturbed states can be represented by the perturbed Hamiltonian matrix E a B@ E1 b CA a b E2 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 (a) First, nd the exact result by diagonalizing the Hamiltonian: E1 ; E a = E1 ; E b = a E2 ; E h b i = (E1 ; E ) (E1 ; E )(E2 ; E ) ; jbj2 + a ; a (E1 ; E )] = = (E1 ; E )2(E2 ; E ) ; (E1 ; E )(jbj2 + jaj2): (5.14) So, E = E1 or (E1 ; E )(E2 ; E ) ; (jbj2 + jaj2) = i.e E ; (E1 + E2)Es+ E1E2 ; (jaj2 + jbj2) = ) E1 + E2 ; E E + jaj2 + jbj2 = E = E1 +2 E2 2 APPROXIMATION METHODS s E1 ; E2 + jaj2 + jbj2: E + E = 109 (5.15) 2 Since jaj2 + jbj2 is small we can expand the square root and write the three energy levels as: E = E1 E = E1 +2 E2 + E1 ;2 E2 + 21 (jaj2 + jbj2)( E ;2 E )2 + : : : = 2 + jbj = E1 + jaEj ; E2 + jbj2 E = E1 +2 E2 ; E1 ;2 E2 (: : :) = E2 ; jaEj ; : E2 (5.16) (b) Non degenerate perturbation theory to 2'nd order The basis we use is 1 1 0 j i = B@ CA j i = B@ CA j i = B@ CA : 0 1 0 a The matrix elements of the perturbation V = B @ 0 b CA are a b h j V j i = a h j V j i = b h j V j i = h k j V j k i = 0: Since (1) k = h k j V j k i = 1'st order gives nothing But the 2'nd order shifts are (2) (2) (2) X jVk1 j2 jh j V j ij2 = jaj2 = 0 E1 ; E2 E1 ; E2 k6=1 E1 ; Ek 2 X jVk2 j jh j V j ij j b j = 0 = E1 ; E2 = E1 ; E2 k6=2 E2 ; Ek X jVk3 j2 jaj2 + jbj2 = ; jaj2 + jbj2 : = = 0 E2 ; E1 E2 ; E1 E1 ; E2 k6=3 E3 ; Ek = (5.17) 110 The unperturbed problem has two (degenerate) states j i and j i with energy E1, and one (non-degenerate) state j i with energy E2 Using nondegenerate perturbation theory we expect only the correction to E2 (i.e (2) ) to give the correct result, and indeed this turns out to be the case (c) To nd the correct energy shifts for the two degenerate states we have to use degenerate perturbation theory The V -matrix for the degenerate ! 0 subspace is 0 , so 1'st order pert.thy will again give nothing We have to go to 2'nd order The problem we want to solve is (H0 + V ) j l i = E j l i using the expansion j l i = j l0 i + j l1 i + : : : E = E + (1) + (2) + : : : (5.18) where H0 j l0 i = E j l0 i Note that the superscript index in a bra or ket denotes which order it has in the perturbation expansion Di erent solutions to the full problem are denoted by di erent l's Since the (sub-) problem we are now solving is 2-dimensional we expect to nd two solutions corresponding to l = Inserting the expansions in (5.18) leaves us with h i (E ; H0 ) j l0 i + j l1 i + : : : = h i (V ; (1) ; (2) : : :) j l0 i + j l1 i + : : : : (5.19) At rst order in the perturbation this says: (E ; H0) j l1 i = (V ; (1)) j l0 i where of course (1) = as noted above Multiply this from the left with a bra h k0 j from outside the deg subspace h k0 j E ; H0 j l1 i = h k0 j V j l0 i X 0 (5.20) ) j l1 i = j k Eih0k;jEV j l i : k6=D k This expression for j l1 i we will use in the 2'nd order equation from (5.19) (E ; H0) j l2 i = V j l1 i ; (2) j l0 i: To get rid of the left hand side, multiply with a degenerate bra h m0 j (H0 j m0 i = E j m0 i) h m0 j E ; H0 j l2 i = = h m0 j V j l1 i ; (2)h m0 j l0 i: APPROXIMATION METHODS 111 Inserting the expression (5.20) for j l1 i we get X h m0 j V j k0 ih k0 j V j l0 i = E ; Ek k6=D (2)h m0 j l0 i: To make this look like an eigenvalue equation we have to insert a 1: X X h m0 j V j k0 ih k0 j V j n0 i 0 h n j l i = (2)h m0 j l0 i: ; Ek E n2D k6=D Maybe it looks more familiar in matrix form X Mmn xn = (2)xm n2D where X h m0 j V j k0 ih k0 j V j n0 i E ; Ek k6=D = h m j l0 i Mmn = xm are expressed in the basis de ned by j l0 i Evaluate M in the degenerate subspace basis D = f j i j ig M12 = EV13;V32E = E ab; E M11 = EV13;V31E = E ja;j E 1 1 3 2 M21 = EV23;V31E = E a;bE M22 = EjV;23jE = E jb;j E : 3 With this explicit expression for M , solve the eigenvalue equation (de ne = (2)(E1 ; E2), and take out a common factor E1;1 E2 ) ! 2; j a j ab = det a b jbj2 ; = = (jaj2 ; )(jbj2 ; ) ; jaj2jbj2 = = ; (jaj2 + jbj2) ) = jaj2 + jbj2 + jbj2 j a j (2) (2) ) =0 = E ;E : (5.21) 112 From before we knew the non-degenerate energy shift, and now we see that degenerate perturbation theory leads to the correct shifts for the other two levels Everything is as we would have expected 5.3 A one-dimensional harmonic oscillator is in its ground state for t < For t 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 > Show that the t ! ( nite) limit of your expression is independent of time Is this reasonable or surprising? (b) Can we nd higher q excited p states?0 You may use hn jxjni = h=2m! ( n + n n+1 + pn n0 n;1 ):] (a) The problem is de ned by 2 H0 = 2pm + m!22x V (t) = ;F0xe;t= (F = ; @V @x ) At t = the system is in its ground state j i = j i We want to calculate X j t i = cn (t)e;Ent=h j n i En0 n = h!(n + 12 ) where we get cn (t) from its di eqn (S 5.5.15): @ c (t) = X V ei!nmtc (t) ih @t n nm m m Vnm = h n j V j m i !nm = En ;h Em = !(n ; m) We need the matrix elements Vnm Vnm = h n j ; F0sxe;t= j m i = ;F0e;t= h n j x j m i = p h (pm = ;F0e;t= 2m! n m;1 + m + n m+1 ): (5.22) APPROXIMATION METHODS 113 Put it back into (5.22) s @ h pn + 1e;i!tc (t) + pnei!tc (t) : ; t= ih @t cn(t) = ;F0e n+1 n;1 2m! (1) Perturbation theory means expanding cn(t) = c(0) n + cn + : : :, and to zeroth order this is @ c(0)(t) = ) c(0) = n0 n @t n To rst order we get Z t dt0 X V (t0)ei!nmt0 c(0) = c(1) ( t ) = n m ih m nm s h Z t dt0e;t= pn + 1e;i!t0 c(0) (t) + pnei!t0 c(0) (t) = ; Fih0 2m! n+1 n;1 We get one non-vanishing term for n = 1, i.e at rst order in perturbation theory with the H.O in the ground state at t = there is just one non-zero expansion coe cient s h Z t dt0ei!t0;t0 = p1 F c(1) ( t ) = ; 1;1 = ih s 2m! #t " F )t0 h ( i! ; = ; ih 2m! i! ; e s h 1 ; e(i!; )t = Fih0 2m! i! ; and X ;iE1 t ;iEn t (1) j t i = c(1) n (t)e h j n i = c1 (t)e h j i: n The probability of nding the H.O in j i is As t ! jh j t ij2 = jc(1) (t)j : s F h c(1) ! ih 2m! i! ; = const: This is of course reasonable since applying a static force means that the system asymptotically nds a new equilibrium 114 (b) As remarked earlier there are no other non-vanishing cn's at rst order, so no higher excited states can be found However, going to higher order in perturbation theory such states will be excited 5.4 Consider a composite system made up of two spin 21 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 = S~1 S~2: h Suppose the system is in j + ;i for t 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 (b) By solving the problem assuming the validity of rst-order time-dependent perturbation theory with H as a perturbation switched on at t = Under what condition does (b) give the correct results? (a) The basis we are using is of course j S1z S2z i Expand the interaction potential in this basis: S~1 S~2 = S1x"S2x + S1y S2y + S1z S2z = fin this basisg = h4 ( j + ih ; j + j ; ih + j )1 ( j + ih ; j + j ; ih + j )2 + + i2(; j + ih ; j + j ; ih + j )1 (; j + ih ; j + j ; ih + # j )2 + + ( j + ih + j ; j ; ih ; j )1 ( j + ih + j ; j ; ih ; j )2 = 2" h = j + + ih ; ; j + j + ; ih ; + j + + j ; + ih + ; j + j ; ; ih + + j + + i ( j + + ih ; ; j ; j + ; ih ; + j + ; j ; + ih + ; j + j ; ; ih + + j ) + APPROXIMATION METHODS 115 + j + + ih + + j ; j + ; ih + ; j + # ; j ; + ih ; + j + j ; ; ih ; ; j = In matrix form this is (using j i = j + + i j i = j + ; i j i = j ; + i j i = j ; ; i) 01 0 01 B ;1 CC H= B B@ ;1 CA : (5.23) 0 This basis is nice to use, since even though the problem is 4-dimensional we get a 2-dimensional matrix to diagonalize Lucky us! (Of course this luck is due to the rotational invariance of the problem.) Now diagonalize the 2 matrix to nd the eigenvalues and eigenkets ! ; ; 2 = det ;1 ; = (;1 ; ) ; = + ; ) = ;3 =1: ! ! x = x y y ) ;x + 2y = x ) x = y = p12 ;1 2 ;1 = ;3 : ! ! ! x = ;3 x y y ) ;x + 2y = ;3x ) x = ;y = p12 So, the complete spectrum is: > < j + + i j ; ; i p2 ( j + ; i + j ; + i with energy > : p1 ( j + ; i ; j ; + i with energy ; ;1 2 ;1 ! 116 This was a cumbersome but straightforward way to calculate the spectrum A smarter way would have been to use S~ = S~1 + S~2 to nd S~ = S = S~12 + S~22 + 2S~1 S~2 ) S~1 S~2 = 21 S~ ; S~12 ; S~22 We know that S~12 = S~22 = h2 21 +1 S~1 S~2 = = 3h42 so S2 ; 3h2 ! Also, we know that two spin 21 systems add up to one triplet (spin 1) and one singlet (spin 0), i.e S = (3 states) ) S~1 S~2 = 21 (h21(1 + 1) ; 3h22 ) = 14 h2 : (5.24) 2 h ~ ~ S = (1 state) ) S1 S2 = (; ) = ; h Since H = 4h2 S~1 S~2 we get E(spin=1) = 14h = h E(spin=0) = ;34h = ;3 : h n (5.25) From Clebsch-Gordan odecomposition we know that j + + i j ; ; i p1 ( j + ; i + j ; + i) are spin 1, and p1 ( j + ; i ; j ; + i) is spin 0! 2 Let's get back on track and nd the dynamics In the new basis H is diagonal and time-independent, so we can use the simple form of tthe time-evolution operator: U (t t0) = exp ; hi H (t ; t0) : The initial state was j + ; i In the new basis n j i = j + + i j i = j ; ; i j i = p12 ( j + ; i + j ; + i) o j i = p12 ( j + ; i ; j ; + i) APPROXIMATION METHODS the initial state is 117 j + ; i = p12 ( j i + j i): Acting with U (t 0) on that we get j t i = p1 exp ; hi Ht ( j i + j i) = = p exp ; i t j i + exp 3i t j i = h h "2 = exp ;ih t p1 ( j + ; i + j ; + i)+ # i t +exp h p ( j + ; i ; j ; + i) = h ;i!t 3i!t i = (e + e ) j + ; i + (e;i!t + e3i!t) j ; + i where (5.26) h: The probability to nd the system in the state j i is as usual jh j t ij2 > h+ + j ti = h; ; j ti = > > < jh + ; j t ij2 = 14 (2 + e4i!t + e;4i!t) = 21 (1 + cos4!t) ' ; 4(!t)2 : : : > > > : jh ; + j t ij2 = (2 ; e4i!t ; e;4i!t) = (1 ; cos4!t) ' 4(!t)2 : : : ! (b) First order perturbation theory (use S 5.6.17): c(0) n = ni Z ;i t dt0ei!nit0 V (t0): c(1) ( t ) = ni n h t0 Here we have (using the original basis) H0 = 0, V given by (5.23) jii = j + ;i (5.27) 118 jf i = j ; +i Ei = fE = 0g = !ni = En ; n h Vfi = Vni = n 6= f: Inserting this into (5.27) yields (0) c(0) i = c j +; i = Zt i (1) (1) cf = c j ;+ i = ; h dt2 = ;2i!t: (5.28) as the only non-vanishing coe cients up to rst order The probability of nding the system in j ; ; i or j + + i is thus obviously zero, whereas for the other two states P ( j + ; i) = (2) 2 P ( j ; + i) = jc(1) f (t) + cf (t) + : : : j = j2i!tj = 4(!t) to rst order, in correspondence with the exact result The approximation breaks down when !t is no longer valid, so for a given t: h: !t ) t 5.5 The ground state of a hydrogen atom (n = 1,l = 0) is subjected to a time-dependent potential as follows: V (~x t) = V0 cos(kz ; !t): Using time-dependent perturbation theory, obtain an expression for the transition rate at which the electron is emitted with momentum ~p Show, in particular, how you may compute the angular distribution of the ejected electron (in terms of and de ned with respect to the z-axis) Discuss brie y the similarities and the di erences between this problem and the (more realistic) photoelectric e ect (note: For the initial wave function use Z ;Zr=a0 : n=1 l=0 (~x) = p a0 e APPROXIMATION METHODS 119 If you have a normalization problem, the nal wave function may be taken to be f (~x) = L 32 ei~p ~x=h with L very large, but you should be able to show that the observable e ects are independent of L.) To begin with the atom is in the n = l = state At t = the perturbation V = V0 cos(kz ; !t) is turned on We want to nd the transition rate at which the electron is emitted with momentum ~pf The initial wave-function is 1 3=2e;r=a0 i (~x) = p a0 and the nal wave-function is ei~p ~x=h: f (~x) = L3=2 The perturbation is h i V = V0 ei(kz;!t) + e;i(kz;!t) = V ei!t + V ye;i!t: (5.29) Time-dependent perturbation theory (S.5.6.44) gives us the transition rate wi!n = 2h Vniy (En ; (Ei + h!)) because the atom absorbs a photon h! The matrix element is Vniy = V40 eikz ni and Z ikz ikz ~ e ni = h kf j e j n = l = i = d3xh ~kf j eikz j x ih x j n = l = i = Z e;i~kf ~x 3=2 = d3x L3=2 eikx3 p1 a1 e;r=a0 = Z = 3=2p1 3=2 d3xe;i(~kf ~x;kx3 );r=a0 : (5.30) L a0 120 So eikz ni is the 3D Fourier transform of the initial wave-function (and some constant) with ~q = ~kf ; k~ez That can be extracted from (Sakurai problem 5.39) eikz ni = L643a5 h i4 a2 + (~kf ; k~ez )2 The transition rate is understood to be integrated over the density of states We need to get that as a function of ~pf = h~kf As in (S.5.7.31), the volume element is dn dp : n2dnd = n2d dp f f Using )2 p2 kf2 = f2 = n (2 L2 h we get dn = 2L2pf = h L2pf = L dp 2n (2 h)2 Lp (2 h)2 h which leaves f f L3kf2 L3p2f = (2 )3h d dpf = (2 h)3 d dpf and this is the sought density Finally, 2 L3p2f d dpf : wi!~pf = 2h V40 L643a5 h i a2 + (~kf ; k~ez )2 (2 h)3 n2dnd Note that the L's cancel The angular dependence is in the denominator: ~kf ; k~ez = (jkf jcos ; k) ~ez + jkf jsin (cos'~ex + sin'~ey )]2 = = jkf j2cos2 + k2 ; 2kjkf jcos + jkf j2sin2 = = kf2 + k2 ; 2kjkf jcos : (5.31) In a comparison between this problem and the photoelectric e ect as discussed in (S 5.7) we note that since there is no polarization vector involved, w has no dependence on the azimuthal angle On the other hand we did not make any dipole approximation but performed the x-integral exactly [...]... ii correspond to a particle at ~x with spin in the i:th direction (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~ = hi r^ r~ Determine S~ ! (b) Show that L~ and S~ commute (c) Show that S~ is a vector operator (d) Show that r~ %~(~x) = h1 (S~ ~p)%~ where p~ is the momentum operator 2 3.5 We are to add angular... T T !T ) 0 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 = 21 mx_ 2 ; 12 m!2x2 2 QUANTUM DYNAMICS 11 (a) Show that hxbtbjxatai = exp iShcl 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) = Z m lim dy : : : dy 1 N... applying the translation ( nite-displacement) operator e;ipl=h (where p is the momentum operator, and l is the displacement distance) to the ground state 10 (e) Show that the coherent state j i remains coherent under timeevolution and calculate the time-evolved state j (t)i (Hint: directly apply the time-evolution operator.) 2.8 The quntum mechanical propagator, for a particle with mass m, moving in a potential... spherical (irreducible) tensor of rank 2 4 SYMMETRY IN QUANTUM MECHANICS 17 (b) The expectation value Q eh j m = j j(3z2 ; r2)j j m = j i is known as the quadrupole moment Evaluate eh j m0j(x2 ; y2)j j m = j i (where m0 = j j ; 1 j ; 2 : : : )in terms of Q and appropriate ClebschGordan coe cients 4 Symmetry in Quantum Mechanics 4.1 (a) Assuming that the Hamiltonian is invariant under time reversal, prove that... this problem exactly to nd the normalized energy eigenstates 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 obtained transform under time reversal? 5 Approximation Methods 5.1 Consider an isotropic harmonic oscillator in two dimensions The Hamiltonian is given by... corresponding quantum- mechanical operators this time Evaluate the commutator x exp iphxa : (c) Using the result obtained in (b), prove that exp iphxa jx0i (xjx0i = x0jx0i) 1 FUNDAMENTAL CONCEPTS 33 is an eigenstate of the coordinate operator x What is the corresponding eigenvalue? (a) We have x F (px)]classical @x @F (px) ; @x @F (px) @x @px @px @x = @F (px) : @p x (1.41) (b) When x and px are treated as quantum- mechanical... simple harmonic oscillator is de ned to be an eigenstate of the (non-Hermitian) annihilation operator a: aj i = j i where is, in general, a complex number (a) Prove that j i = e;j j =2e ay j0i is a normalized coherent state (b) Prove the minimum uncertainty relation for such a state (c) Write j i as 1 X j i = f (n)jni: 2 n=0 Show that the distribution of jf (n)j2 with respect to n is of the Poisson... Justify your answer 4.3 Read section 4.3 in Sakurai to refresh your knowledge of the quantum mechanics of periodic potentials You know that the energybands in solids are described by the so called Bloch functions n k full lling, ika n k (x + a) = e n k (x) 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... equivalent to a single rotation about some axis by an angle Find 3.2 An angular-momentum eigenstate jj m = mmax = j i is rotated by an in nitesimal angle " about the y-axis Without using the (j ) explicit form of the dm0m 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... 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 ... unit vector n^ makes an angle with the positive z-axis and is lying in the xz-plane, it can be written in the following way n^ = e^z cos + e^x sin (1.6) So S~ n^ = S" z cos + Sx sin = # (S-1.3.36),(S-1.4.18)]... 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... Find the zeroth-order energy eigenket and the corresponding energy to rst order that is the unperturbed energy obtained in (a) plus the rst-order energy shift] for each of the three lowest-lying