I How does QM explain that we see electrons circulating in a magnetic field?

[Heidi]

Hi Pfs.
I think that QM can explain the classical things explained by classical physics. Using mean values and so on.
We know that in a constant magnetic field an electron will rotate on a circle (at the macroscopic scale approximation)
I have the answer for the Larmor precession but how to describe the whole thing with the quantum tools?

 

[Ssnow]

I don't have a precise answer but it is possible that the problem is connected with the quantization of the angular momentum $L$?
Ssnow

 

[Heidi]

As i see the problem we would consider \langle\psi^\dagger |q |\psi \rangle
where psi is the state of and electron at t = 0 and how it evolves in time when it is in a magnetic field. if can be done with its Hamiltonian but how to get the circle's radius ?

 

[HomogenousCow]

If you setup a localized wave packet in a constant magnetic field you will find that it behaves similar to the classical situation.

 

[Heidi]

I have no problem with that
My question is about QM where there is no notion of force.

 

[HomogenousCow]

There absolutely is the notion of forces, it's just not all that useful in the Schrodinger picture. If your question is, how are magnetic forces formulated in QM then the answer is through a momentum dependent potential in the Hamiltonian.
For a charged particle in a vector potential $\mathbf{A}$ the Hamiltonian is given by $$H = \frac{1}{2m}(\mathbf{p} - q \mathbf{A})^2.$$ If we choose the symmetric form of the vector potential for a constant magnetic field in the $\mathbf{z}$ direction $$\mathbf{A} = \frac{B_z}{2}(y \mathbf{i} - x \mathbf{j}),$$ then the above Hamiltonian can be written in a suggestive form:
$$H = \frac {\mathbf{p}^2}{2m} - \frac{qB_z }{2m} \mathbf{L}_z + \frac{q^2 {B_z}^2}{8 m} \rho^2$$ That this will give you rotational motion for a wavepacket should be evident.
To relate the average radial position of particle to its angular momentum you would look at the stationary states of definite angular momentum.

 

[andresB]

I have no problem with that
My question is about QM where there is no notion of force.

But the orbits of a classical particle can be encoded in the Lagrangian or the Hamiltonian without using the concept of force. If it can be done classically, what is the issue with QM?

 

[anuttarasammyak]

I would say let us have an initial state |i>, It will evolve with time $e^{iHt/\hbar}|i>$ with use of Hamiltonian. It would give us probability amplitude to find in a state <f| as
&lt;f|e^{iHt/\hbar}|i&gt;
Orbit does not appear here.

 

[Twigg]

@HomogenousCow nailed it in post #6. Just jumping into try and improve the signal-to-noise ratio of this thread. A wavepacket will behave, under expectation values, just the same as it would it a classical cyclotron (electron orbiting B-field).

This is actually a famous problem in quantum mechanics, at the intro graduate level. The solutions to the bound states of the Hamiltonian in post #6 are called the Landau levels. Just keep in mind that the coordinate system in that wikipedia article is different from what was used in post #6.

 

[Heidi]

We can use the Ehrenfest theorem to calculate the mean value <p> as a time derivative of <x> by using [x,H] and the time dependence of the operator x.
If we do the same thing with m we get something that we can call <F>
it is what we call the force in classical physics.
How can we show that il is orthognal to p, and has a constant norm?
of course with only quantum arguments.

 

[Twigg]

The math for Ehrenfest is a little hairy because $\vec{p}$ here means canonical momentum, NOT $m\vec{v}$, which is the kinetic momentum. However, if you do Ehrenfest on the Hamiltonian presented in post #6 (not the Hamiltonian in the asymmetric gauge used in the wikipedia article!), you get $$\frac{d \langle \vec{p} \rangle}{dt} = -\frac{\omega_c}{2} \hat{r} \times \langle\vec{p} \rangle + m \omega_c^2 \langle \vec{r} \rangle$$. If you start with a wavepacket with $$\langle \vec{p} \rangle \cdot \langle \vec{r} \rangle = 0$$, then you will stay in a circular orbit (i.e., $\langle \vec{F} \rangle \cdot \langle \vec{p} \rangle = 0$ for all time)

Maybe someone with a little more finesse could demonstrate a way to tease out the kinetic momentum and get back the Lorentz force equation, but I couldn't figure that out.

 

[vanhees71]

Indeed, one must be careful. Momentum at presence of a magnetic field is canonical momentum, which in this case is different from the usual kinetic momentum, which has to do with gauge invariance of the electromagnetic interaction. The most clear exposition I know is in the textbook by Cohen-Tannoudji and Laloe.

In the modern formalism you describe the physical time derivative of operators as
$$\mathring{\hat{O}}=\frac{1}{\mathrm{i} \hbar} [\hat{O},\hat{H}].$$
For the Hamiltonian of a particle in a magnetic field with a vector potential $\vec{A}$,
$$\hat{H}=\frac{1}{2m} [\hat{\vec{p}}-q \vec{A}(\hat{\vec{x}})],$$
you find the velocity operator, using the commutation relations,
$$\hat{\vec{v}}=\mathring{\hat{\vec{x}}}=\frac{1}{m} (\hat{\vec{p}}-q \vec{A}(\hat{\vec{x}}),$$
and finally
$$m \mathring{\hat{\vec{v}}}=\frac{q}{2} [\hat{\vec{v}} \times \vec{B}(\hat{\vec{x}}) - \vec{B}(\hat{\vec{x}})\times \hat{\vec{v}}],$$
where $\vec{B}=\vec{\nabla} \times \vec{A}$, which is just the correctly symmetrized form of the Lorentz-force law of a particle moving in a magnetic field.

 

[HomogenousCow]

An interesting point is that since the velocity operators $\frac{i }{\hbar}[H, Q_{\alpha}]$ don't in general commute with each other and in fact $$[v_\alpha, v_\beta] = i \frac{\hbar q}{m^2} \epsilon_{\alpha \beta \gamma} B_\gamma,$$
states of simultaneous kinetic momentum may or may not exist in the Hilbert space depending on the form of the magnetic field. For the aforementioned system in post #6 $v_x$ and $v_y$ are actually incompatible observables.

 

[Heidi]

Yes but when we consider Landau gauge
A = (0,Bx,0) one easily get one oscillator on one axis and the Ehrenfest theorem allows us to the mean value <F> = d <px> / dt. a classical "force" along the x axis.
why do we get another classical force along the orthogonal axis to get a circular classical movement?
is is easier wiht the symmetric gauge A = (-By,Bx,0) /2

 

[HomogenousCow]

Sorry but I don’t understand your question, are you wondering where the $y$ component of the magnetic force is in the Landau gauge?

 

[Heidi]

yes,
How does the circular classical movement emerge here?

 

[vanhees71]

Yes but when we consider Landau gauge
A = (0,Bx,0) one easily get one oscillator on one axis and the Ehrenfest theorem allows us to the mean value <F> = d <px> / dt. a classical "force" along the x axis.
why do we get another classical force along the orthogonal axis to get a circular classical movement?
is is easier wiht the symmetric gauge A = (-By,Bx,0) /2

The important point is that this is wrong, because the Ehrenfest theorem says
$$m \mathrm{d}_t \langle \vec{v} \rangle=q \langle \vec{v} \times \vec{B} \rangle.$$
Now the velocity operator is
$$\hat{\vec{v}}=\mathring{\hat{\vec{x}}}=\frac{1}{\mathrm{i} \hbar} [\vec{x},\hat{H}]=\frac{1}{m} (\hat{\vec{p}}-q \vec{A}(\hat{\vec{x}}).$$
That implies that the "kinetic momentum" is not $\hat{\vec{p}}$, which is rather the canonical momentum, and that's different in a magnetic field from the kinetic momentum.

Then you have
$$m \mathring{\vec{v}}=\frac{m}{\mathrm{i} \hbar} [\hat{\vec{v}},\hat{H}]=\frac{q}{2} [\hat{\vec{v}} \times \hat{\vec{B}}-\hat{\vec{p}} \times \hat{\vec{v}}],$$
which is what you expect. For $\hat{\vec{B}}=\vec{B} \hat{1}$ with $\vec{B}=\text{const}$ you get the circular motion for the expectation values as expected. It looks like the classical equations of motion, because the equations are linear in this case:
$$m \mathrm{d}_t^2 \langle \vec{x} \rangle=q \mathrm{d}_t \langle \vec{x} \rangle \times \vec{B}.$$

 

[Heidi]

thanks a lot.
my formula came fron a french wiki article

 

[Twigg]

@Heidi That version of the Ehrenfest theorem is correct 95% of the time. This situation, charged particles in an electric field, is the only commonplace example where you need to use a canonical momentum like in @vanhees71"s post. There are technically other cases, but they usually involve exotic Hamiltonians that depend on velocity.

 

[Heidi]

The Ehrenfest theorem says
$$m \mathrm{d}_t \langle \vec{v} \rangle=q \langle \vec{v} \times \vec{B} \rangle.$$

How to derive this from quantum rules?

 

[Twigg]

This was sketched out without the gory algebraic details in post #12. Props to @vanhees71 for explaining this the first time. Here's a reference that derives the same result in section 3.7 (page 6), but in the non-static case (time-varying E & B fields) and they use gaussian units. I did a little conversion to this specific case below:

Let's identify the kinetic momentum $$\vec{\Pi} = m \vec{v} = \vec{p} - q\vec{A}$$ The Hamiltonian is $$H = \frac{1}{2m} \Pi^2 $$. Ehrenfest theorem says $$\frac{d\langle \Pi_i \rangle }{dt} = m\frac{d \langle v_i \rangle}{dt} = \frac{1}{i\hbar} \frac{1}{2m} \langle [\Pi_i, \Pi_j \Pi_j] \rangle$$ where I'm using summation notation for the repeated indexes.

First, let's evaluate $[\Pi_i,\Pi_j]$. Just plugging in yields $$\begin{align*} [\Pi_i,\Pi_j] &= \require{cancel} \cancel{[p_i,p_j]} - [p_i,qA_j] - [qA_i,p_j] + \require{cancel} \cancel{[qA_i,qA_j]} \\ &= -\left( [p_i,qA_j] - [p_j,qA_i] \right)\end{align*} $$ If we solve for $[p_i,qA_j]$, we get $$[p_i,qA_j]\psi = \frac{\hbar}{i} q\left( \partial_i(A_j \psi) - A_j \partial_i \psi \right) = \frac{\hbar}{i} q (\partial_i A_j) \psi$$ Therefore, $$\begin{align*} [\Pi_i,\Pi_j] &= \frac{\hbar}{i} q \left( \partial_i A_j - \partial_j A_i \right) \\ &= \frac{\hbar}{i} q \epsilon_{ijk} B_k \end{align*}$$

Now we can evaluate the commutator from Ehrenfest:: $$\begin{align*} \frac{1}{i\hbar} \frac{1}{2m} [\Pi_i, \Pi_j \Pi_j] &= -\frac{1}{2m}\left([\Pi_i,\Pi_j] \Pi_j + \Pi_j [\Pi_i,\Pi_j] \right) \\ &= -\frac{1}{2m} \epsilon_{ijk} \left( B_k \Pi_j + \Pi_j B_k \right) \\ &= -\frac{1}{2} q \left( (\vec{v} \times \vec{B})_i - (\vec{B} \times \vec{v})_i \right) \end{align*}$$

Of course, I somehow picked up a rogue minus sign somewhere along the way. I ain't going to lose sleep over it, I'm just going to fudge it away . You saw nothin'!

Sticking that commutator result (with the extra minus sign fudged out) into Ehrenfest yields $$m\frac{d\langle \vec{v} \rangle}{dt} = \frac{q}{2} \langle \vec{v} \times \vec{B} - \vec{B} \times \vec{v} \rangle$$ This is what @vanhees71 had in post #12.

Why are there two terms and why did I leave the B-field inside the expectation value? Because as long as the B-field depends on the particle position (which is quantized!), it still behaves like a quantized operator. Now if the magnetic field is constant and doesn't depend on the particle position, then it can come outside the expectation value. In that case, $$\begin{align*} m\frac{d\langle \vec{v} \rangle}{dt} &= \frac{q}{2} (\langle \vec{v} \rangle \times \vec{B} - \vec{B} \times \langle \vec{v} \rangle) \\ &= q\langle \vec{v} \rangle \times \vec{B} \end{align*}$$

Voila!

 

[vanhees71]

The more general case of a time-dependent B-field is more subtle. Here one has to decompose the field in a static and a "wave" part to establish a gauge-invariant perturbation theory. In atomic physics, e.g., it makes a difference whether to use the dipole approximation with the electric field for the wave part or an apparently equivalent formulation with $\vec{A}$. However, both are only equivalent for the full solution, while in perturbation theory only the former leads to a correct gauge-invariant description of the approximated wave functions. For a thorough treatment see

https://doi.org/10.1119/1.13029

and the more detailed papers

https://doi.org/10.1016/0003-4916(76)90275-X
https://doi.org/10.1016/0003-4916(76)90276-1

Unfortunately I couldn't find freely available references of the same quality.

 

[Heidi]

I have another question about the |n,m> states. they are the degenacy states of the n Landau level.
How can we prepare particles in one of these states? i suppose that we will start with particles with the energy n (we choose B in the cyclotron and we inject in it electrons with the correct momentum. But after that , how to select those with the correct m (an integer that i choosed)? does "m" correspond to particular circles?

 

[Twigg]

The Landau levels do not look like classical orbits, just like harmonic oscillator eigenstates do not look like classical oscillations. The expectation values of $\vec{r}$ and $\vec{v}$ do not change with time in the Landau levels, since they are stationary states. Thus, a Landau level is not an orbit. It's just an energy eigenstate.

Recall the symmetric-gauge Hamiltonian from Post #6, $$H = \frac{\mathbf{p}^2}{2m} - \frac{1}{2} \omega_c \mathbf{L}_z + \frac{1}{8} m \omega_c^2 \rho^2$$ where $\omega_c = qB_z/m$ is the cyclotron frequency. This Hamiltonian conserves angular momentum. That's a good starting point. A 1D gaussian wavepacket like you might find in Griffiths will not give you cyclotron orbits, because it doesn't have a well defined "radius"; such a wavepacket is only confined in the longitudinal axis. You need wavepacket that is confined in both the longitudinal and transverse directions to see cyclotron motion, and even then you need to assume that the Larmor radius is much larger than the transverse confinement or you will see very weird interference phenomena. And you probably have ignore dispersion. It's a mess. Just have a little faith in ol' Ehrenfest.

There is a mathematically clean way to derive states that look like the cyclotron orbits. They're called coherent states. I'm going to use the notation from the Landau levels wikipedia's section on the symmetric gauge. A coherent state $|\alpha,\beta \rangle$ is an eigenstate of both ladder operators: $$ \begin{align} \hat{a} |\alpha \beta \rangle &= \alpha |\alpha \beta \rangle \\ \hat{b} |\alpha \beta \rangle &= \beta |\alpha \beta \rangle \end{align}$$ My gut feeling (this stuff is tricky) is that a coherent state with large $\alpha$ and $\beta$ will look like a gaussian wavepacket with transverse confinement in a cyclotron orbit with large Larmor radius.

I couldn't derive the coherent states to prove it to you. That stuff's hard! Just getting the coherent states for the harmonic oscillator is usually an exercise for a graduate course. I looked for open-access papers on the topic, but the only articles I found had math that was way above my pay grade. If you're feeling ambitious, knock yourself out: here's the link. Maybe someone here knows a better article or can summarize the results?

 

[Heidi]

I am looking for something simpler. Is there a device for measuring the Lz of the electron in the magnetic field?

 

[Twigg]

Is there a device for measuring the Lz of the electron in the magnetic field?

AFAIK, not really, no. Especially not for electrons.

If there was some way, it'd be with electron microscopy. For example, these folks were able to image the radial wavefunctions of the Landau levels with an STM. To see the angular momentum quantum number, you'd need to be able to image the phase of the Landau level wavefunction, and my understanding is that phase-contrast electron imaging is still a developing technology. Don't quote me, I'm not an expert on electron microscopy!

For electrons, the degeneracy of the Landau levels is anyways https://www.researchgate.net/publication/1880630_Landau_level_mixing_by_full_spin-orbit_interactions. In theory, a spectroscopic measurement should be able to resolve them. However, the most sensitive spectroscopic measurement that I know (see here) is only barely able to resolve the non-degenerate (radial wavefunction) energy levels. There could be other more sensitive measurements that I'm not familiar with.

Now if you're just looking for some idealized, theoretical device to measure $L_z$, then yes we can come up with something. You just need something that can image the phase and amplitude of the electron wavefunction as a function of position. $L_z$ will be given by $\frac{\hbar}{i} \frac{1}{r} \frac{\partial \psi}{\partial \phi}$. This theoretical device could be an electron microscope with phase and amplitude contrast mechanisms.

I am looking for something simpler.

I wish I could help you, but the theory underlying the Landau levels is hard. If you want to keep it simple, you're better off doing numerical simulations of whatever you want to study.

 

[Twigg]

Also, if you do attempt numerical simulations, keep in mind that we haven't even mentioned cyclotron radiation in this thread. That math is definitely above my pay grade. The only thing I can tell you is that, just like how applying Ehrenfest's theorem to the Schrodinger-Pauli Hamiltonian gives you the Lorentz force law, applying Ehrenfest to the QED description of cyclotron radiation gives you the Larmor formula (in the non-relativistic limit, anyways).

 

[Heidi]

The Landau levels do not look like classical orbits, just like harmonic oscillator eigenstates do not look like classical oscillations. The expectation values of $\vec{r}$ and $\vec{v}$ do not change with time in the Landau levels, since they are stationary states. Thus, a Landau level is not an orbit. It's just an energy eigenstate.

Recall the symmetric-gauge Hamiltonian from Post #6, $$H = \frac{\mathbf{p}^2}{2m} - \frac{1}{2} \omega_c \mathbf{L}_z + \frac{1}{8} m \omega_c^2 \rho^2$$ where $\omega_c = qB_z/m$ is the cyclotron frequency. This Hamiltonian conserves angular momentum. That's a good starting point. A 1D gaussian wavepacket like you might find in Griffiths will not give you cyclotron orbits, because it doesn't have a well defined "radius"; such a wavepacket is only confined in the longitudinal axis. You need wavepacket that is confined in both the longitudinal and transverse directions to see cyclotron motion, and even then you need to assume that the Larmor radius is much larger than the transverse confinement or you will see very weird interference phenomena. And you probably have ignore dispersion. It's a mess. Just have a little faith in ol' Ehrenfest.

There is a mathematically clean way to derive states that look like the cyclotron orbits. They're called coherent states. I'm going to use the notation from the Landau levels wikipedia's section on the symmetric gauge. A coherent state $|\alpha,\beta \rangle$ is an eigenstate of both ladder operators: $$ \begin{align} \hat{a} |\alpha \beta \rangle &= \alpha |\alpha \beta \rangle \\ \hat{b} |\alpha \beta \rangle &= \beta |\alpha \beta \rangle \end{align}$$ My gut feeling (this stuff is tricky) is that a coherent state with large $\alpha$ and $\beta$ will look like a gaussian wavepacket with transverse confinement in a cyclotron orbit with large Larmor radius.

I couldn't derive the coherent states to prove it to you. That stuff's hard! Just getting the coherent states for the harmonic oscillator is usually an exercise for a graduate course. I looked for open-access papers on the topic, but the only articles I found had math that was way above my pay grade. If you're feeling ambitious, knock yourself out: here's the link. Maybe someone here knows a better article or can summarize the results?

I read this http://www.sa.infn.it/Massimo.Blasone/documents/cantrans.pdf.
Canonical transformations in quanatum field theory.
is it what you have in mind?
when there is a bosonic translation that moves the gaussian centered on 0,0 far away we have a greater number of occupation and it is more easy to measure the phase.

 

[Twigg]

I read this http://www.sa.infn.it/Massimo.Blasone/documents/cantrans.pdf.
Canonical transformations in quanatum field theory.
is it what you have in mind?
when there is a bosonic translation that moves the gaussian centered on 0,0 far away we have a greater number of occupation and it is more easy to measure the phase.

I have never seen this math before. No, it was not what I had in mind. If you want to see the math of coherent states, I recommend getting a copy of Shankar's Quantum Mechanics or any other graduate quantum textbook. Here's a nice introduction if you don't have access to those books.

 

[Heidi]

I found several links that puzzle me:
In the Landau quantization and in Landau's book.

In the wiki article there are two operators a ans b in the symmetric jauge. they write that the Landau levels are degenerate with the same energy H = \hbar \omega (a^\dagger a + 1/2) )
and i do not how to use a and b to find the energy ot the |n,m> state obtained from |0,0> with the creations operators .

there are also the formulas in the Landau book. n and m are definite and i read
E = hbar omega (n + m + 1/2)

are they talking about the same energy?

I do not see how to visualize latex formulas in this forum.

 

[George Jones]

I do not see how to visualize latex formulas in this forum.

See "Delimiting your LaTeX code" in Physics Forums help for LaTeX,

https://www.physicsforums.com/help/latexhelp/

 

[Twigg]

If you're not sure how to evaluate the energy from the Hamiltonian $H = \hbar \omega (a^\dagger a + \frac{1}{2})$, then you may want to review the harmonic oscillator in the ladder operator picture. It's discussed in pretty much any serious quantum textbook, undergraduate or graduate level. Griffiths (2nd Ed.) section 2.3.1 is a good place to start. (You can later derive the coherent states step-by-step with hints in Problem 3.35.)

there are also the formulas in the Landau book. n and m are definite and i read
E = hbar omega (n + m + 1/2)

are they talking about the same energy?

Yes, just in terms of quantum numbers that describe different basis states (sounds like eigenmodes in cartesian coordinates?). The energies are always the same, in any basis and any gauge. If you're quoting a result from Landau, please share which volume and section you are reading from so we can see what you're looking at.

 

[Heidi]

It is in vol 3 quantum mechanics (non relativistic)
chapter xv motion in a magetic field . problem 1 page 460
we have E depending on n and m and it is written that it is equivalent to 112.7 where there is no m.
we have here |m| + m which is null if m négative but is it so here?

 

[Heidi]

j bought the first part of the paper written by Kuo-Ho Yang . Thank you VanHees71. I ignored that an hamiltonian could be cut in 2 parts. A "basis" part with a basis of orthonormal eigenvectors and a residual part. Here the résidual part is omega times Lz i suppose.