Quantum motion of a charged particle in a magnetic field

[EightBells]

Homework Statement

Find eigenenergies for a charged particle in magnetic field B and in a parabolic central potential, $U(r) = \frac {kr^2} 2$. The particle has mass m and charge q.

Relevant Equations

Hamiltonian for charged particle in magnetic field $H = \frac 1 {2m} \left( p- \frac {qA} {c^2} \right)^2 +q \phi$

Once I know the Hamiltonian, I know to take the determinant $\left| \vec H-\lambda \vec I \right| = 0$ and solve for $\lambda$ which are the eigenvalues/eigenenergies.

My problem is, I'm unsure how to formulate the Hamiltonian. Is my potential $U(r)$ my scalar field $\phi$? I've seen equations relating B and E fields to the scalar and vector fields, $\vec B = \nabla {} \times \vec A$ and $\vec E = -\nabla \phi - \frac 1 c \frac {d \vec A} {dt}$, but I'm not sure how to incorporate them or if I even need to.

Do I need an equation for the vector potential $\vec A$? What is the momentum $p$ in this case?

Thanks!

 

[andresB]

First, $U(r)$ replaces the $q\phi$.

Now, my issue with your problem is that you are not giving us the actual magnetic field (or the vector potential). Without that the problem is ill stated and nothing can be done about it. It is an static, uniform Magnetic field?

 

[EightBells]

First, $U(r)$ replaces the $q\phi$.

Now, my issue with your problem is that you are not giving us the actual magnetic field (or the vector potential). Without that the problem is ill stated and nothing can be done about it. It is an static, uniform Magnetic field?

That's all the information provided, however the professor has since posted additional notes:

$\vec A = \left( \frac {By} 2 ; \frac {-Bx} 2 ; 0 \right)$
$H = H_{2d} + H_z$, the writing isn't entirely clear so I'm unsure if the "2d" subscript is correct, however $H_z = \frac {p_z^2} {2m} + \frac {kz^2} {2}$ and $H_{2d} = \frac {\hbar^2} {2m} \left[ \left( -i \frac {\partial} {\partial x} - \frac q {c\hbar} \frac {By} 2 \right)^2 + \left( -i \frac {\partial} {\partial y}+ \frac q {c\hbar} \frac {Bx} 2 \right)^2 \right] + \frac {k(x^2+y^2)} 2$.

From there I find $H = \frac {\hbar^2} {2m} \left[ \left( -i \frac {\partial} {\partial x} - \frac q {c\hbar} \frac {By} 2 \right)^2 + \left( -i \frac {\partial} {\partial y}+ \frac q {c\hbar} \frac {Bx} 2 \right)^2 + \left( -i \frac {\partial} {\partial z} \right)^2 \right] + \frac {k(x^2+y^2+z^2)} 2$

The professor however finds some expression $- \frac {\hbar^2} {2m} \frac {\partial^2} {\partial r^2} + \left[ \frac {\hbar^2} 2 \left( \frac {B^2} {4mc^2} + 1 \right) \left(x^2+y^2 \right) \right] + \frac {\hbar^2} {2m} \frac {qB} {c \hbar} \vec l_z$ where $\vec l_z = x \frac {\partial} {\partial y} - y \frac {\partial} {\partial x}$ is the angular momentum along the z-axis. We then look for wave functions $\Psi = e^{i {\varphi} {l_z}} \varphi(r)$ and $l_z = 0, \pm 1, \pm 2$ and $\phi(r)$ is the radial component.

I'm used to finding eigenenergies in matrix form using $|\vec H - E \vec I| = 0$, however given a wave function could I still use $H |\Psi \rangle = E |\Psi \rangle$ and work through the partial derivatives of $H \Psi$ to solve for the eigenenergies E? How do I find the appropriate wave function to use then?

 

[Dr Transport]

First, $U(r)$ replaces the $q\phi$.

Now, my issue with your problem is that you are not giving us the actual magnetic field (or the vector potential). Without that the problem is ill stated and nothing can be done about it. It is an static, uniform Magnetic field?

\vec{A} = \approx\frac{\vec{B}\times\vec{r}}{2}
I can never remember what the sign is...