Taking the Curl of (zB, 0, 0) it returns B = (0, B, 0), which agrees with the B field specified. But also, we can take the curl of (0, 0, -xB) and this yields B = (0, B, 0). Is the zB one correct or -xB one correct or both are correct? I'm thinking it has something to do with the symmetry of the system.
However, if I do proceed with using A = (zB, 0, 0) and the scalar potential \phi = -zE. (Note: the time partial derivative term in the expression relating E and the scalar potential is canceled out since A isn't not dependent on time since B is constant, there E = -\nabla \phi)
Thus, the generalised time-dep. Schrodinger eq. is:
i\hbar\frac{\partial}{\partial t} \Psi\left( \vec{r}, t\right) = \left[ -\frac{\hbar^{2}}{2m}\nabla^{2} + i\hbar\frac{e}{2m} \left(\vec{A}.\vec{\nabla}+ \vec{\nabla}.\vec{A}\right) + \frac{e^{2}}{2m} \vec{A}^{2} + e\phi\right]\Psi\left( \vec{r}, t\right)
where e is the charge of the particle, m is the mass of the particle. Using the Coulomb gauge: \vec{\nabla} . \vec{A}\right) = 0. Therefore, the above equation can be reduced to:
i\hbar\frac{\partial}{\partial t} \Psi\left( \vec{r}, t\right) = \left[ -\frac{\hbar^{2}}{2m}\nabla^{2} + i\hbar\frac{e}{m} \left(\vec{A}.\vec{\nabla}\right) + \frac{e^{2}}{2m} \vec{A}^{2} + e\phi\right]\Psi\left( \vec{r}, t\right)
where \nabla. \left( \vec{A}\Psi \right) = \vec{A}. \left( \vec{\nabla}\Psi \right) + \left(\vec{\nabla} . \vec{A}\right)\Psi = A. \left(\vec{\nabla}} \Psi\right). (This eliminates the 2 in the denominator of e/2m of the middle term of the square brackets).
Thus, finally substituting the vector and scalar potential into the time-dep. Schrodinger equation with the Coulomb gauge applied, I obtain:
i\hbar\frac{\partial}{\partial t} \Psi\left( \vec{r}, t\right) = \left[ -\frac{\hbar^{2}}{2m}\nabla^{2} + i\hbar\frac{e}{m} \left(zB \frac{\partial}{\partial x} \right) + \frac{e^{2} z^{2} B^{2}}{2m} - eEz\right]\Psi\left( \vec{r}, t\right)
I was wondering if the logic was correct in obtaining the final Schrodinger equation?
Thanks.
