# Schroedinger eq. for a charged particle in a magnetic field - translation operators

1. Feb 24, 2009

### FranzDiCoccio

Hi all,

there is something I cannot completely understand about the Schroedinger equation for a charged particle in a constant magnetic field.
I think I get most of it, hopefully, but there is something I still am unable to fit into place.

So we know that when a magnetic field is present the Hamiltonian becomes

$$H = \frac{{\mathbf P}^2 }{2 m}, \qquad {\mathbf P} = {\mathbf p} - \frac{q}{c}{\mathbf A}$$

where q is the charge of the particle and A is the magnetic vector potential, such that $$\nabla \times {\mathbf A} = {\mathbf B}$$.

Of course when the magnetic field is constant (say, along the z direction), the system is translationally invariant, but this can be made tricky by the gauge. Let me choose the "symmetric gauge" $$A_x = -B y /2$$, $$A_y = B x /2$$, so that the components of the kinetic momentum become

$$P_x = p_x +\alpha y,\qquad P_y = p_y -\alpha x,\qquad \alpha = \frac{q B}{ 2c}$$

My question is about the generators of the translations. As far as I can understand,
these operator are deceptively similar to the above components of the momentum. They
are

$$T_x = (p_x -\alpha y) = \alpha Y ,\qquad T_y = (p_y +\alpha x) = \alpha X$$

where $$(X,Y)$$ is the quantum analog of the center of the classical cyclotron motion.

Now, I see that these commute with the components of P, and therefore with the Hamiltonian. Hence they must be related to some kind of symmetry.
I also see that this symmetry is translational invariance, in a way. I mean, I can sort of derive them through the Noether's theorem from the (classical?) Lagrangian.
On the other hand, it seems to me that their action on a generic wavefunction is not a "simple" translation. There is also a phase factor dependent on the magnitude of the translation.
What's the point here? One should not be so fussy and call such a transformation a translation anyway, because the phase factor does not show up in a measurement?
Or is it a more tricky point, related to gauge fixing?
I'd say the second, but I'm not able to fit things into place...

I came up with this "picture", but I'm not sure about it...

As I mention, acting with $$T_x$$ on the wavefunction brings about a phase factor

$$\exp\left(i \frac{x_0}{\hbar} T_x \right)\psi(x,y) = \exp\left(-i \frac{x_0}{\hbar} \alpha y \right) \exp\left(i \frac{x_0}{\hbar} p_x \right)\psi(x,y) = \exp\left(-i \frac{q B}{2 \hbar c} x_0 y \right) \psi(x+x_0,y) = \exp\left(-i \frac{q \varphi}{\hbar c} \right) \psi(x+x_0,y)$$

Such phase factor is the same corresponding to a gauge transformation of the vector potential:

$${\mathbf A}(x,y) \mapsto {\mathbf A}(x,y) -\vec \nabla \varphi = {\mathbf A}(x,y)-\frac{B}{2} x_0 \hat e_y = {\mathbf A}(x-x_0,y)$$

i.e. the translation opposite to that we're considering, applied to the vector potential...

This seems to figure... However I'm not very sure about it...

Thanks for any insight you can provide

Franz

PS I've corrected a wrong (I think) sign in the last equation.

Last edited: Feb 24, 2009
2. Feb 24, 2009

### xepma

Re: Schroedinger eq. for a charged particle in a magnetic field - translation operato

The first thing that comes to mind is that the vector potential causes the presence of a curvature, which is the magnetic field. When a particles moves through a curved background it picks up a phase term, known as Berry's phase. My guess is that the phase you pick up through these translational operators is precisely this Berry phase.

Note that this is not just a minor phase factor which you can ignore. The terms causes the translational operator in the x and y-direction to no longer commute. In fact, Tx and Ty satisfy what is known as the magnetic translation algebra which has some interesting features.

3. Feb 24, 2009

### FranzDiCoccio

Re: Schroedinger eq. for a charged particle in a magnetic field - translation operato

Hi xepma,

and thanks for replying. Yes, I understand that all of the things you mention are related to my question. In fact, I stumbled on those translation operators in an algebraic proof that the magnetic flux must be quantized. But now I'm under the impression that I have misunderstood the sketch of the proof that a friend gave me.

I'll look for more information on "magnetic translations". Anyway the adjective "magnetic" somewhat satisfies me, because at least it warns you that there's more to them than the usual translations.

Thanks again

F

Thanks again

F

4. Feb 24, 2009

### per.sundqvist

Re: Schroedinger eq. for a charged particle in a magnetic field - translation operato

It becomes even more complicated when you consider more electrons in a magnetic field! You would need to fix (i.e., minimize) some constant terms in each A for each electron in order to obtain the global minimum of the quantum many-electron system.

5. Feb 24, 2009

### xepma

Re: Schroedinger eq. for a charged particle in a magnetic field - translation operato

Hi F,

I did some more browsing and thinking and I found just a little more on the subject.

Note that the Hamiltonian is given by

$H = \frac{1}{2m}\left(p-e A)\right)^2$

On the classical level the system is translational invariant, but on the quantum level we work with the vector potential. Hence, the system is no longer translational invariant when we introduce a gauge which is not translational invariant (like the symmetric gauge). To be more precise: the normal translational operators, x and y, do not commute with this Hamiltonian.

On the other hand, like you showed, the magnetic translational operators are more of use here. Specifically, these operators do commute with the Hamiltonian. But unlike normal translational operators, the action on the wavefunction also induces a phase factor. But apart from what I said earlier, I'm still not completely sure what else this implies...

6. Feb 24, 2009

### weejee

Re: Schroedinger eq. for a charged particle in a magnetic field - translation operato

Actually, the magnetic translation operator translates the guiding center(=center of the cyclotron motion).

7. Feb 25, 2009

### FranzDiCoccio

Re: Schroedinger eq. for a charged particle in a magnetic field - translation operato

Yes, I agree with that. They are translation operators, after all.

My comment was just meant to emphasize that if you take the classical center of the cyclotron motion (X,Y) and substitute classical position and velocity (or better, kinetic momentum) with their quantum versions you get the above operators (up to a multiplicative constant).

8. Feb 25, 2009

### FranzDiCoccio

Re: Schroedinger eq. for a charged particle in a magnetic field - translation operato

Hi xepma,

and thanks for the feedback. I'm not sure I agree with you that the classical hamiltonian is manifestly translationally invariant, though. It seems to me that you get the "gauge problem" also at the classical level. At a first glance I'd say that classically it is not a big deal because because the gauge transformation that you need for reabsorbing the extra term does not call for a "weird" phase factor.

Anyway, I have just found a book on the quantum Hall effect featuring an appendix on the magnetic translation operators. I hope it will clarify my doubts which, at the moment, I'm not even able to state clearly any more.

Thanks again

F

9. Feb 25, 2009

### xepma

Re: Schroedinger eq. for a charged particle in a magnetic field - translation operato

Hi F,

Yea, my statement referring to the classical <-> quantum level might not be the best one.

Anyways, if you find something new and you wanna discuss it, feel free to state it here. I'm also quite interested in the matter. Formulating the proper questions is always one of the bigger difficulties with these sort of things ;)

10. Feb 25, 2009

### weejee

Re: Schroedinger eq. for a charged particle in a magnetic field - translation operato

You are right. It isn't hard to verify it. (Note that X and Y are canonically conjugate variables. Try to evaluate the commutator [X,Y])