electrons have Magnetic bound states ?

[Widdekind]

Please ponder the Wave Function (WF), of an electron, spiraling around a field line, in a uniform magnetic field [math]\vec{B} = B \hat{z}[/math]. This field corresponds to a vector potential [math]\vec{A} = \hat{\phi} B r / 2[/math]. The SWE, for a time-independent, 'magnetically bound' state, is then:

 

[math]E \Psi = \frac{\left( \vec{p} - q \vec{A} \right)^2}{2 m} \Psi[/math]

Now, the vector potential commutes with the momentum operator (since the former's azimuthal-angle component, is independent of that variable). Thus, defining [math]e \equiv -q[/math], w.h.t.:

 

[math]2 m E \Psi = \left( -\hbar^2 \nabla^2 + 2 e A_{\phi} p_{\phi} + (e A)^2 \right) \Psi[/math]

And now, if we assume solutions of the form [math]\Psi \equiv R® e^{i n \phi}[/math], then w.h.t.:

 

[math]2 m E = -\hbar^2 \left( \frac{1}{r R} \frac{\partial}{\partial r} \left( r \frac{\partial R}{\partial r} \right) - \frac{n^2}{r^2} \right) + n \hbar e B + \frac{ \left( e B \right)^2}{4}r^2 [/math]

 

[math]2 m E = -\hbar^2 \left( \frac{1}{r R} \frac{\partial R}{\partial r} + \frac{1}{R} \frac{\partial^2 R}{\partial r^2} - \frac{n^2}{r^2} \right) + n \hbar e B + \frac{ \left( e B \right)^2}{4}r^2 [/math]

Assuming that [math]E < 0[/math], w.h.t.:

 

[math]\frac{2 m | E |}{\hbar^2} = \left( \frac{1}{r R} \frac{\partial R}{\partial r} + \frac{1}{R} \frac{\partial^2 R}{\partial r^2} - \frac{n^2}{r^2} \right) - n \left( \frac{e B}{\hbar} \right) + \left(\frac{1}{2} \frac{e B}{\hbar} \right)^2 r^2 [/math]

 

[math]\frac{2 m | E |}{\hbar^2} + n \left( \frac{e B}{\hbar} \right) = \left( \frac{1}{R} \frac{\partial^2 R}{\partial r^2} + \frac{1}{r R} \frac{\partial R}{\partial r} - \frac{n^2}{r^2} \right) + \left(\frac{1}{2} \frac{e B}{\hbar} \right)^2 r^2 [/math]

 

[math]\left[ \frac{2 m | E |}{\hbar^2} + n \left( \frac{e B}{\hbar} \right) \right] r^2 R = \left( r^2 \frac{\partial^2 R}{\partial r^2} + r \frac{\partial R}{\partial r} - n^2 R \right) + \left(\frac{1}{2} \frac{e B}{\hbar} \right)^2 r^4 R [/math]

Now, defining the 'Larmor energy' [math]E_L \equiv \hbar \omega_L[/math]; and, the 'reduced mass' [math]\mu \equiv m / \hbar^2[/math]; then, w.h.t.:

 

[math]\mu \left( \, 2 \, |E_n| \, + \, n \, E_L \, \right) r^2 R = \left( r^2 \frac{\partial^2 R}{\partial r^2} + r \frac{\partial R}{\partial r} - n^2 R \right) + \left( \frac{\mu \, E_L}{2} \right)^2 r^4 R [/math]

or:

 

[math]\left( r^2 \frac{\partial^2 R_n}{\partial r^2} + r \frac{\partial R_n}{\partial r} - n^2 R_n \right) - \mu \left( \, 2 \, |E_n| \, + \, n \, E_L \, \right) r^2 R_n + \left( \frac{\mu \, E_L}{2} \right)^2 r^4 R_n = 0 [/math]

where we have explicitly added in subscripts, denoting the dependence on rotation-rate index "n".

 

Now, one could solve, for the resulting Bessel-function-resembling [math]R_n[/math] functions, by laboriously calculating their infinite-sum-of-terms Taylor expansion. For example,

 

[math]R_1® \approx r + \frac{\mu}{8} \left( 2 |E_1| + E_L \right) r^3 + \frac{\mu^2}{192} \left( \left( 2 |E_1| + E_L \right)^2 - 2 E_L^2 \right) r^5 + ...[/math]

Note, too, that, as expected, there is no non-rotating, n=0, solution ([math]a^0_k = 0 \, \forall k[/math]). And, [math]|E_1| < E_L/2[/math], if we demand that [math]a^1_5 < 0[/math], in analogy to the 'alternation of terms' characteristic of Bessel functions.

 

A priori, this formula seemingly suggests, that the electro-magnetic attraction, of electrons, to magnetic field lines, produces a spectrum of 'magnetic bound states', loosely analogous to the Hydrogen WFs, from the electro-static potential, produced by protons. These WFs 'clad' magnetic field lines, and phase-rotate around them, like cylindrical 'x-zylo' frisbee footballs. Synchrotron radiation could be viewed, as loosely analogous to Hydrogen de-excitation emissions, to ground state, as cycling electrons "gear shift down" the magnetic-bound-state energy spectrum.

Edited July 31, 2011 by Widdekind