Energy States of Electrons in a Plasma I

$$\bgroup\color{black}-{\hbar^2\over{2 m_e}}\nabla^2\psi + {eB\... ...e^2B^2\over{8 m_e c^2}}\left(x^2+y^2\right)\psi = E\psi \egroup$$

For uniform $\vec{B}$ field, cylindrical symmetry $\Rightarrow$ apply cylindrical coordinates $\rho$, $\phi$, $z$. Then

$$\bgroup\color{black}\nabla^2={\partial^2\over{\partial z^2}} ... ...}} + {1\over\rho^2} {\partial^2\over{\partial \phi^2}} \egroup$$

From the symmetry of the problem, we can guess (and verify) that $[H,p_z]=[H,L_z]=0$. These variables will be constants of the motion and we therefore choose

$$\begin{eqnarray*} \psi\left(\vec{r}\right) &=& u_{mk}\left(\rho\right)e^{im\phi}... ...{1\over\rho} {\partial u\over{\partial \rho}} e^{im\phi}e^{ikz} \end{eqnarray*}$$

$$\bgroup\color{black}{d^2u\over{d\rho^2}} + {1\over\rho}{du\ov... ..._e E\over{\hbar^2}}-{eBm\over{\hbar c}}-k^2\right)u = 0 \egroup$$

Let $x=\sqrt{eB\over{2\hbar c}}\rho$ (dummy variable, not the coordinate) and $\lambda={4 m_e c\over{eB\hbar}}\left(E-{\hbar^2k^2\over{2 m_e}}\right)-2m$. Then

$$\bgroup\color{black}{d^2u\over{dx^2}} + {1\over x}{du\over{dx}}-{m^2\over{x^2}}u-x^2u+\lambda = 0\egroup$$

In the limit $x\rightarrow\infty$,

$$\bgroup\color{black}{d^2u\over{dx^2}}-x^2u=0\qquad\Rightarrow\qquad u\sim e^{-x^2/2} \egroup$$

while in the other limit $x\rightarrow 0$,

$$\bgroup\color{black}{d^2u\over{dx^2}} + {1\over x}{du\over{dx}}-{m^2\over{x^2}}u = 0\egroup$$

Try a solution of the form $x^s$. Then

$$\bgroup\color{black}s(s-1)x^{s-2} + sx^{s-2} - m^2 x^{s-2} = 0 \qquad\Rightarrow\qquad s^2 = m^2 \egroup$$

A well behaved function $\Rightarrow$ $s\geq 0$ $\Rightarrow$ $s=\vert m\vert$

$$\bgroup\color{black}u(x)=x^{\vert m\vert} e^{-x^2/2} G(x)\egroup$$

Plugging this in, we have

$$\bgroup\color{black}{d^2G\over{dx^2}}+\left({2\vert m\vert+1\... ...over{dx}} + \left(\lambda-2 - 2\vert m\vert\right)G = 0\egroup$$

We can turn this into the hydrogen equation for

$$\bgroup\color{black}y=x^2\egroup$$

and hence

$$\bgroup\color{black}dy=2x \hspace{0.05 in} dx\egroup$$

$$\bgroup\color{black}{d\over{dy}}={1\over2x}\hspace{0.05 in} {d\over{dx}}.\egroup$$

Transforming the equation we get

$$\bgroup\color{black}{d^2G\over{dy^2}}+\left({\vert m\vert+1\o... ...r{dy}} + {\lambda - 2 - 2\vert m\vert\over{4 y}}G = 0. \egroup$$

Compare this to the equation we had for hydrogen

$$\bgroup\color{black}{d^2H\over{d\rho^2}}+\left({2\ell+2\over ... ...dH\over{d\rho}} + {\lambda - 1 - \ell\over{\rho}}H = 0 \egroup$$

with $\lambda=n_r+\ell+1$. The equations are the same if WE set our ${\lambda\over 4}=n_r + {1+\vert m\vert\over 2}\qquad\mbox{where }n_r = 0, 1, 2, \ldots$. Recall that our $\lambda={4 m_e c\over{eB\hbar}}\left(E-{\hbar^2k^2\over{2 m_e}}\right)-2m$. This gives us the energy eigenvalues

$$\bgroup\color{black}\Rightarrow\qquad E - {\hbar^2k^2\over{2 ... ...{ m_e c}} \left(n_r + {1+\vert m\vert+ m\over 2}\right).\egroup$$

As in Hydrogen, the eigenfunctions are

$$\bgroup\color{black}G(y)=L_{n_r}^{\vert m\vert}(y).\egroup$$

We can localize electrons in classical orbits for large E and $n_r\approx 0$. This is the classical limit.

$$\bgroup\color{black}n_r=0\qquad\Rightarrow\qquad L_0=\mbox{co... ...row\qquad\vert\psi\vert^2 \sim e^{-x^2}x^{2\vert m\vert}\egroup$$

Max when

$$\bgroup\color{black}{d\vert\psi\vert^2\over{dx}}= 0 =\left(-2... ...m\vert}+2\vert m\vert e^{-x^2}x^{2\vert m\vert-1}\right)\egroup$$

$$\bgroup\color{black}\vert m\vert=x^2\qquad\Rightarrow\qquad\rho=\left({2 c\over{eB}}\hbar m\right)^{1/2}\egroup$$

Now let's put in some numbers: Let $B\approx 20\mbox{ kGauss}=2\times 10^4\mbox{ Gauss}$. Then

$$\bgroup\color{black}\rho = \sqrt{{ 2\left(3\times 10^{10}{\mb... ...ght) }}m} \approx 2.5\times 10^{-6}\sqrt{m} \mbox{ cm}\egroup$$

This can be compared to the purely classical calculation for an electron with angular momentum $m\hbar$ which gives $\rho=\sqrt{m\hbar c\over Be}$. This simple calculation neglects to count the angular momentum stored in the field.