Energy States of Electrons in a Plasma II

If $\vec{A}=\left(0,Bx,0\right)$, then

$$\bgroup\color{black}\vec{B}=\vec\nabla\times\vec{A}={\partial A_y\over{\partial x}}\hat{z}=B\hat{z}.\egroup$$

This $A$ gives us the same B field. We can then compute H for a constant B field in the z direction.

$$\bgroup\color{black}H={1\over {2 m_e}}\left(\vec{p}+{e\over c... ...^2_x+\left(p_y + {eB\over c} x\right)^2 + p^2_z\right) \egroup$$

$$\bgroup\color{black} = {1\over {2 m_e}}\left(p^2_x+p^2_y + {2... ... c}x p_y + \left(eB\over c\right)^2 x^2 + p^2_z\right) \egroup$$

With this version of the same problem, we have the

$$\bgroup\color{black}\left[H,p_y\right]=\left[H,p_z\right]=0.\egroup$$

We can treat $p_z$ and $p_y$ as constants of the motion and solve the problem in Cartesian coordinates.

$$\bgroup\color{black}\psi=v(x)e^{ik_yy}e^{ik_zz}\egroup$$

$$\bgroup\color{black}{1\over {2 m_e}}\left(-\hbar^2{d^2\over{d... ...ght)v(x) =\left(E-{\hbar^2k_z^2\over 2 m_e}\right)v(x) \egroup$$

This is the same as the 1D harmonic oscillator equation with $\omega={eB\over{ m_e c}}$ and $x_0={\hbar c k\over{eB}}$.

$$\bgroup\color{black}E=\left(n+{1\over 2}\right)\hbar\omega={\... ...c}}\left(n+{1\over 2}\right) +{\hbar^2k_z^2\over 2 m_e}\egroup$$

So we get the same energies with a much simpler calculation. The resulting states are somewhat strange and are not analogous to the classical solutions.