Physics 215 Quantum Mechanics I Assignment 8

Transcription

1 Physics 15 Quantum Mechanics I Assignment 8 Logan A. Morrison March, 016 Problem 1 Let J be an angular momentum operator. Part (a) Using the usual angular momentum commutation relations, prove that J = J z+ J + J J z (1) Recall that the angular momentum ladder operators are given by Thus, hence, J ± = J x ± ij y () J + J = (J x + ij y )(J x ij y ) () = J x + J y i(j xj y J y J x ) () = J J z i J x, J y (5) = J J z i(i J z ) (6) J = J z + J +J J z (8) (7) Part (b) Using (a), derive the relation J ψ j,m = c ψ j,m 1 (9)

2 Logan A. Morrison Assignment Problem 1 Page /8 First, let s show the state J j, m is an eigenstate of J z with eigenvalue (m 1) : J z (J j, m)= J z J x j, m ij z J y j, m (10) = (J x J z + J z, J x ) j.m i(jy J z + J z, J y ) j.m (11) = (J x J z + i J y ) j.m i(j y J z i J x ) j.m (1) = (m J x + i J y ) j.m i(m J y i J x ) j.m (1) = (mj x + ij y imj y J x ) j, m (1) = (m 1)J j, m (15) Which means J j, m j, m 1. Now let s determine the proportionality constant. To do this, let s calculate the normed squared of J j, m: Therefore, J j, m = j, m (J ) J j, m (16) = j, m J + J j, m (17) = j, m ( J J z+ J z ) j, m (18) = j, m ( j( j+1) m + m ) j, m (19) = j( j+1) m(m 1) (0) = c j, m (1) = j, m (c ) c j, m () = c () c = j( j+1) m(m 1) ()

3 Logan A. Morrison Assignment Problem 1 Page /8 Problem A particle in a spherically symmetrical potential is known to be in an eigenstate of L and L z with eigenvalues l(l+ 1) and m, respectively. Prove that the expectation values between l, m states satisfy Give a semiclassical interpretation of this result. L x = L y = 0 and L x = L y = l(l+ 1) m (5) First let s show that L x = L y = 0. First, note that L+ + L = L x and L + L = ij y. L x = lm L x lm (6) = 1 lm (L ++ L ) lm (7) = 1 (c + l, m l, m+1+c l, m l, m 1) (8) = 0 (9) where in the last line we used the fact that l, m l, m = δ mm and c + = l(l+1) m(m+1) and c = l(l+1) m(m 1). Similarly, L y = lm Ly lm (0) Now, for L x : = i lm (L + L ) lm (1) = 1 (c + l, m l, m+1 c l, m l, m 1) () = 0 () L x = lm L x lm () = 1 lm ( ) L ++ L + L + + L x L y lm (5) = 1 c + c + l, m l, m++c c l, m l, m (6) + l(l+1) m(m 1) l(l+ 1) m(m 1) l, m l, m (7) + l(l+1) m(m+1) l(l+ 1) m(m+1) l, m l, m (8) = l(l+ 1) m(m+1)+l(l+1) m(m 1) (9) = l(l+1) m = l(l+1) m (0) (1) ()

4 Logan A. Morrison Assignment Problem 1 Page /8 Now, for L y : Ly = lm L y lm () = lm ( L+ L ) lm () i = 1 lm ( ) L ++ L L + L L x L y lm (5) = 1 c + c + l, m l, m+ c c l, m l, m (6) + l(l+ 1) m(m 1) l(l+ 1) m(m 1) l, m l, m (7) + l(l+ 1) m(m+1) l(l+ 1) m(m+1) l, m l, m (8) = l(l+1) m(m+1)+l(l+1) m(m 1) (9) = l(l+1) m = l(l+1) m Above, c +, c +, c and c are normalization constants that come from acting with L + and L. We don t explictly write them since they aren t needed in the end. (50) (51) (5)

5 Logan A. Morrison Assignment Problem 1 Page 5/8 Problem The wave function of a particle subjected to a spherically symmetric potential V(r) is given by ψ( x)=(x+y+z) f (r) (5) Part (a) Isψan eigenfunction of L? If so, what is thel-value? If not, what are the possible values oflwe may obtain when L is measured? Note that Y 1±1 and Y 1,0 can be writen as Thus, Y 1±1 = 8π z Y 1,0 = π r x±iy r (5) (55) π ( ) x=r Y1, 1 Y 1,1 (56) π ( ) y=ir Y1, 1 + Y 1,1 (57) π z=r Y 1,0 (58) Hence, π (( ) ( ) ) ψ(r,θ,φ)=r Y1, 1 Y 1,1 + i Y1, 1 + Y 1,1 + Y1,0 f (r) (59) π = r f (r) (1+i)Y 1, 1 (1 i)y 1,1 + Y 1,0 (60) Since r,θ,φ l, m Y l,m, we can see that a wave function proportional to Y l,m will have a definite squared angular momentum of l(l+1). Since the wave function is in terms of spherical harmonics withl=1, this is indeed an eigenfunction of L with eigenvalue. Part (b) What are the possibilities for the particle to be found in various m l states? The state Y l,m is a state with definite angular momentum in the z direction with eigenvalue m. Since we have terms with m=±1, 0, the possible values of m l are±1, 0.

6 Logan A. Morrison Assignment Problem 1 Page 6/8 Part (c) Suppose it is known somehow that ψ( x) is an energy eigenfunction with eigenvalue E. Indicate how we may determine V(r). From class, we know that the Hamiltonian can be written in terms of the angular momentum operator: H= 1 m r r r r L + V(r) (61) r So, since our wave function is an eigenstate of L and since it is an energy eigenstate, we have that Thus, where g(θ,φ)= of derivatives: Hψ=Eψ (6) = 1 m r r r r ψ r ψ + V(r)ψ (6) V(r)ψ=Eψ+ 1 m r r r r ψ r ψ = Erg(θ,φ) f (r)+g(θ,φ) m 1 f (r)) r (r r r r r f (r) π (1+i)Y 1, 1 (1 i)y 1,1 + Y 1,0. Thus, the potential can be determine by taking a couple 1 d d(r f (r)) V(r)=E+ mr f (r) r r dr dr r f (r) 1 d( = E+ r f (r)+r f ) mr f (r) r dr r f (r) 1 = E+ mr f (r) r (r f (r)+r f+ r f (r)+r f (r)) r f (r) r f (r)+r f (r) = E+ m f (r) r (6) (65) (66) (67) (68) (69) (70) (71)

7 Logan A. Morrison Assignment Problem Page 7/8 Problem Consider an electron in a uniform magnetic field along the z direction, with a Hamiltonian H= µ B= ωσ z, ω=eb/mc (7) Let the result of a measurement be that the electron spin is along the positive y direction at t = 0. Find the Schrodinger state vector for the spin, and the average polarization (expectation value of S x ) along the x direction for t>0. If our system is initially in the state S y,+, then, ψ, t=0= 1 ( S z,++i S z, ) (7) Since the hamiltonian is time independent, we can time evolve the system simply by applying the following unitary operator: U(t, t 0 )=exp ih(t t 0) = exp iω(t t 0 )σ z = exp iω(t t 0)S z where we used the fact that S z = σ z. Acting with this operator on our state, we find that (7) ψ; t=u(t; 0) 1 ( S z,++i S z, ) (75) = 1 (exp iωts z S z,++i exp iωts ) z S z, (76) = 1 ) (exp iωt S z,++i exp iωt S z, (77) (78) Now, let s determine S x. To do this, we will expand the S z eigenstates in terms of S x eigenstates: S z,±= 1 S x ;+± S x, (79) Thus, 1 S x ψ; t=s x (exp 1 iωt ( exp iωt S x ;+ S x, 1 S x ;++ S x, + i exp iωt = 1 = ( ) exp iωt S z, +i exp iωt S z,+ + i exp iωt S x ;++ S x, ) S x ;+ S x, ) (80) (81) (8) (8) (8)

8 Logan A. Morrison Assignment Problem Page 8/8 Therefore, the expectation value is S x = ψ; t S x ψ, t = ( exp iωt S z,+ i exp = ) i exp iωt i exp iωt iωt )( ) S z, exp iωt S z, +i exp iωt S z,+ (85) (86) (87) and thus, the spin is precessing! = sin (ωt) (88)