Particle in A Box

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  • #1
Is the ground state wavefunction of a particle in a 3-D cubic box with V=0 an eigenstate of the z-direction orbital angular momentum operator, Lz?

I tried to determine this using cartesian coordinates, but I ended up with an imaginary answer for Lz(Psi), which is supposed to be a Hermitian operator.

Intuitively, I want to say that the ground state IS an eigenstate with an eigenvalue=0, but I can't get that to come out computationally.

Any help?
 

Answers and Replies

  • #2
Note: the ground state wavefunction on which I have Lz operating is (Psi) = Sqrt(8/V) sin(pi*x/a) sin(pi*y/a) sin(pi*z/a), where V=a^3

I have used the operator Lz = XPy-YPx = -ih(x*d/dy - y*d/dx)
 

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