- 200

- 12

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

[tex]H = \frac{{\mathbf P}^2 }{2 m}, \qquad {\mathbf P} = {\mathbf p} - \frac{q}{c}{\mathbf A}[/tex]

where

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" [tex]A_x = -B y /2[/tex], [tex]A_y = B x /2[/tex], so that the components of the kinetic momentum become

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

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

[tex]T_x = (p_x -\alpha y) = \alpha Y ,\qquad T_y = (p_y +\alpha x) = \alpha X[/tex]

where [tex](X,Y)[/tex] is the quantum analog of the center of the classical cyclotron motion.

Now, I see that these commute with the components of

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 [tex]T_x[/tex] on the wavefunction brings about a phase factor

[tex]\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)[/tex]

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

[tex]{\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)[/tex]

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.

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

[tex]H = \frac{{\mathbf P}^2 }{2 m}, \qquad {\mathbf P} = {\mathbf p} - \frac{q}{c}{\mathbf A}[/tex]

where

*q*is the charge of the particle and**A**is the magnetic vector potential, such that [tex]\nabla \times {\mathbf A} = {\mathbf B}[/tex].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" [tex]A_x = -B y /2[/tex], [tex]A_y = B x /2[/tex], so that the components of the kinetic momentum become

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

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

[tex]T_x = (p_x -\alpha y) = \alpha Y ,\qquad T_y = (p_y +\alpha x) = \alpha X[/tex]

where [tex](X,Y)[/tex] 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 [tex]T_x[/tex] on the wavefunction brings about a phase factor

[tex]\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)[/tex]

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

[tex]{\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)[/tex]

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: