Electron energy in magnetic field

For a problem early in a book I'm reading (Quantum Mechanics by Albert Messiah), I'm asked the following:

Consider an electron following a circular trajectory in a constant magnetic field $H$ (I'm guessing that I should also ignore synchrotron radiation). Apply to this rotational motion the de Broglie resonance condition. Show that the kinetic energy is quantized, and that the energy levels are equidistant, and that this distance is $\frac{e\hbar}{mc}H$. [this result differs from that of the rigorous quantum theory only by an overall displacement of all the eenrgy levels by an amount $\frac{e\hbar}{2mc}H$]

When I attempt this however, I get something different.

Let $r$ be the radius of orbit, $\mathbf{A}$ be the magnetic vector potential, and $m$ be the mass of an electron. All boldface symbols are vectors with their magnitude being represented by the corresponding not-boldface symbol.

The kinetic energy of the electron is $\frac{1}{2}mv^2$, but the restriction on the allowed velocities is that the radius of its orbit, a function of the velocity, must correspond to an integral multiple of the de Broglie wavelength, also a function of the velocity.

$$r=\frac{mv}{eH}$$

By the de Broglie resonance condition however,

$$2\pi r = n\lambda = \frac{nh}{p} \longrightarrow r=\frac{n\hbar}{p}$$

Equating the two, we get,

$$\frac{mv}{eH}=\frac{n\hbar}{p}=\frac{n\hbar}{\left|m\mathbf{v}-e\mathbf{A}\right|}$$

Squaring both sides and moving stuff around,

$$mv\left((mv)^2+(eA)^2-2me\mathbf{v}\cdot\mathbf{A}\right)=n\hbar e H$$

EDIT: From here, I know I could orient the z-axis in the direction of the field and choose the circular vector potential,

$$\mathbf{A}=\frac{-A}{2}y\hat{i}+\frac{A}{2}x\hat{j}$$

This happens to be in the $\hat{\phi}$ direction, just like $\mathbf{v}$. This makes the dot product above equal to

$$\mathbf{v}\cdot\mathbf{A}=\frac{Avr}{2}$$

This gives me another $r$, and so I'm now led to believe that my approach will go on in an infinite loop. If I were to somehow terminate this loop, I'd be left with solving a cubic polynomial or higher, which is quite tedious.

Have I made an incorrect assumption? Have I gone down a more difficult route? Please help me.

All of this is under the non-relativistic approximation.

• As a first step, I'd suggest that cylindrical coordinates would be the most useful choice here, due to the symmetries of the problem. As a second step, I'd note that $v$ is purely in the $\hat{\phi}$ direction, so it would be advisable to choose a gauge such that $A$ is also purely in the $\hat{\phi}$ direction. – ragnar Jun 28 '15 at 20:45
• Also, it looks like you're using inconsistent units. As a check, if $\frac{e}{c}A$ has the same units as $mv$, and $H=\nabla \times A$, then what units should $H$ have, and does this agree with your first equation for $r$? – ragnar Jun 28 '15 at 21:27
• Finally, in your definition of the canonical momentum $p$, keep in mind that you're talking about an electron, so the charge is negative. – ragnar Jun 28 '15 at 21:31
• Ok. Looks like you picked the wrong unit convention, since there's now no way to get a factor of c in there. But more importantly, you can write $A$ in terms of $H$. Something like $A=1/2 H r \hat{\phi}$. Plug that into the formula for the curl in spherical coordinates and see what you get. – ragnar Jun 29 '15 at 1:58

So this is problem is actually supposed to be done in a rather simple manner. The key phrase is "de Broglie resonance condition." If you refer back to earlier in the chapter, specifically the beginning paragraph on matter waves (page 49 in 2nd edition), Messiah discusses how we can associate $E=\hbar\omega$ with any given resonant wave. Reading on, de Broglie was able to show that starting with his matter-wave hypothesis, you can get the Bohr-Sommerfeld quantization condition.

This allows you to see that a given matter wave obeying Bohr-Sommerfeld can equivalently be modeled as a standing resonant wave of energy $n\hbar\omega$.

So now, working in cylindrical coordinates, B-S gives you $$\int pdq=nh$$$$p2\pi =nh$$ (since p is only in $\phi$ direction)$$p=n\hbar$$ and since $E=n\hbar\omega$, $E=p\omega$.

To find $\omega$, simply take advantage of the fact that the motion is a centripetal force due to the lorentz force:$$\frac{mv^2}{r}=\frac{evH}{c}$$and since $\omega=v/r$ $$w=\frac{eH}{mc}$$ and thus $$E=n \frac{e\hbar}{mc}H$$

The things that make this easier are working in cylindrical coordinates, and setting your vector potential to be purely along the $\phi$ direction, which also allows you to set it equal to the value of H, since it is the sole component of the field.

Some good references related to this problem and similar ones, take a look at these links: https://en.wikipedia.org/wiki/Landau_quantization https://en.wikipedia.org/wiki/Old_quantum_theory#De_Broglie_waves

I feel like a lot of the difficulty with these early quantum problems is that they require you to make a lot of sneaky assumptions based on approximations. Hope this was helpful!