Physics H7C Homework problem sets and solutions, Exercises of Physics

Download Physics H7C Homework problem sets and solutions and more Exercises Physics in PDF only on Docsity! Homework 9 Solutions May 6, 2016 1a) Multiplying the given ~z1, ~z2 vectors by the Sz matrix gives Sz~z1 = ~ 2 ( 1 0 0 −1 )( 1 0 ) = ~ 2 ( 1 0 ) = ~ 2 ~z1 (1) and Sz~z2 = ~ 2 ( 1 0 0 −1 )( 0 1 ) = ~ 2 ( 0 −1 ) = −~ 2 ~z2 (2) so we see that ~z1, ~z2 are indeed eigenvectors with respective eigenvalues sz,1 = ~/2 and sz,1 = −~/2 1b) Multiplying the given ~x1, ~x2 vectors by the Sx matrix gives Sx~x1 = ~ 2 ( 0 1 1 0 ) 1√ 2 ( 1 1 ) = ~ 2 1√ 2 ( 1 1 ) = ~ 2 ~x1 (3) and Sx~x2 = ~ 2 ( 0 1 1 0 ) 1√ 2 ( 1 −1 ) = ~ 2 1√ 2 ( −1 1 ) = −~ 2 ~z2 (4) so we see that ~z1, ~z2 are indeed eigenvectors with respective eigenvalues sz,1 = ~/2 and sx,1 = −~/2 so we see that ~x1, ~x2 are indeed eigenvectors with respective eigenvalues sx,1 = ~/2 and sx,1 = −~/2 1c) We see easily ~x1 = 1√ 2 ( 1 1 ) = 1√ 2 ( 1 0 ) + 1√ 2 ( 0 1 ) = 1√ 2 ~z1 + 1√ 2 ~z2 (5) and ~x2 = 1√ 2 ( 1 −1 ) = 1√ 2 ( 1 0 ) − 1√ 2 ( 0 1 ) = 1√ 2 ~z1 − 1√ 2 ~z2 (6) 1d) As described in the problem, if a state is written as a superposition of eigen- states like ~z = a~z1 + b~z2, then the probabilities of measuring each Sz eigenstate 1 is given by the square of the coefficients (or weight) of each eignenstate, so |a|2 and |b|2. So for the case of ~x1, we see from the above that a = b = 1/ √ 2 so the probability of measuring spin up or spin down in z are both = 1/2. 1e) If the particle is in the ~z1 state then it is in a superposition of Sx eigenstates, as we can write ~z1 = ( 1 0 ) = 1 2 ( 1 1 ) + 1 2 ( 1 −1 ) = 1√ 2 ~x1 + 1√ 2 ~x2 (7) so we see that the coefficients are a = b = 1/ √ 2 so the probability of measuring spin up or spin down in x are both = 1/2. 2a) Plugging our ansatz into the three dimensional Schrödinger’s equation, we find EX(x)Y (y)Z(z) = −~2 2m [ X ′′(x)Y (y)Z(z))+ X(x)Y ′′(y)Z(z) +X(x)Y (y)Z ′′(z) ] (8) inside the well, but we know that in the same region −~2 2m X ′′(x) = ExX(x) (9) and similarly for Y (y) and Z(z), hence EX(x)Y (y)Z(z) = (Ex + Ey + Ez)X(x)Y (y)Z(z) (10) thus the product of the solutions of one dimensional wells is a solution of the three dimensional Schrödinger’s equation with energy eigenvalue E = Ex + Ey + Ez. The wavefunction is then ψ(x, y, z) = 8√ L3 ∑ nx,ny,nz sin (nxπx L ) sin (nyπy L ) sin (nzπz L ) (11) simply the product of the individually normalized wavefunctions. 2b) Following the result in the previous part, we have E = (n2x + n 2 y + n 2 z)~2π2 2mL2 (12) 2c,d) Neglecting spin, there is 1 possible ground state (nx, ny, nz) = (1, 1, 1). There are three degenerate first excited states (2, 1, 1), (1, 2, 1), (1, 1, 2). 2