4 modern quantum mechanics (3) (20 points) Consider the spin-less three-dimensio

Question

4 modern quantum mechanics (3) (20 points) Consider the spin-less three-dimensional harmonic oscillator problem but now we add a magnetic field. (a) Which one of the following is the Hamiltonian in this case? p ep A 2m mc H-(p-eA / c)21 2m 2 (b) Using equation (2.2.23b), show that the Hamiltonian can be written as ieh 2mc 2m 2 mc 2mc (c) Using the symmetric gauge, A(By/2)i+ (Bx/2) (equation 2.7.40), reduce the above to 2m (4) (25 points) For the Hamiltonian in 3(c), answer the following questions: (a) Which quantity, L,, Ly or L,, can have a simultaneous eigenket with the Hamiltonian? Justify your answer. (b) If you were to find the energy eigenvalues and eigenkets, how would you proceed? You do not have to solve for anything, but some part(s) of the solution are commonly known from elementary quantum mechanics.

Explanation / Answer

Lets assume the central potential so we can compare to our later solution. We could have three different spring constants and the solution would be as simple. The Hamiltonian is

[egin{eqnarray*} H&=&{p^2over 2m}+{1over 2}momega^2 r^2 \ H&=&{p_x^2over ... ...+{p_z^2over 2m}+{1over 2}momega^2 z^2 \ H&=&H_x+H_y+H_z \ end{eqnarray*}]

The problem separates nicely, giving us three independent harmonic oscillators.

[groupcolor{black}$displaystyle E=left(n_x+n_y+n_z+{3over2} ight)hbaromega $egroup]
[groupcolor{black}$displaystyle psi_{nx,ny,nz}(x,y,z)=u_{nx}(x)u_{ny}(y)u_{nz}(z) $egroup]

This was really easy.

This problem has a different Fermi surface in [groupcolor{black}$n$egroup] -space than did the particle in a box. The boundary between filled and unfilled energy levels is a plane defined by

[egin{displaymath}groupcolor{black} n_x+n_y+n_z={E_Foverhbaromega}-{3over 2} egroupend{displaymath}]

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