Jump to content

electrons have Magnetic bound states ?

Widdekind
in Quantum Theory

Featured Replies

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 by Widdekind

Archived

This topic is now archived and is closed to further replies.

Go to topic listing

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.

Configure browser push notifications
Chrome (Android)
  1. Tap the lock icon next to the address bar.
  2. Tap Permissions → Notifications.
  3. Adjust your preference.
Chrome (Desktop)
  1. Click the padlock icon in the address bar.
  2. Select Site settings.
  3. Find Notifications and adjust your preference.