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As I understand , systems having large angular momenta relative to the planck constant (limit of large quantum numbers, e.g. $J/\hbar \to \infty$), can be treated as classical systems. Now in the case of cyclotron resonance type measurements, one often sees the classical equation of motion written down for the electron, e.g. in the presence of a magnetic field we have: $$m\mathbf{\dot{v}} = -e\mathbf{v} \times \mathbf{B} - \frac{m}{\tau}\mathbf{v} \tag{1}$$ With $\tau$ a relaxation time for the electron in its host material (e.g. a crystal), where we often have $\tau^{-1}\to 0$.

  1. Why is it physically allowed to assume the electron can be treated classically? What's the key idea behind this approximation in such contexts?

  2. Lastly, on a related note, if we add an external electric field to the above system, the first term on the rhs of (1) becomes $-e(\mathbf{E}+\mathbf{v}\times \mathbf{B}) $ and in order to solve this system, the planewave ansatz is usually used for the velocity, i.e. $v(t)=v_0 e^{-i\omega t}$, is this ansatz a sound choice here because we are treating the electron classically or there is another unrelated underlying reason?

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    $\begingroup$ sure you mean $\tau$ often close to zero? $\endgroup$ – jim Apr 4 '16 at 12:23
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    $\begingroup$ @jim its my fault: I wrote $\tau\approx0$ in an answer a couple of weeks ago. Actually, I meant $\tau^{-1}\approx 0$, so that the last term $m/\tau$ can be neglected. $\endgroup$ – AccidentalFourierTransform Apr 4 '16 at 14:08
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Before addressing your question, there is a point where I kind of disagree with Orca's answer that I'd like to discuss:

I will begin with part 2 of your question about plane waves. The use of this Ansatz is the first clue that you are actually treating the situation quantum-mechanically, but ending up with a result that exactly matches the classical result.

The ansatz $v=v_0 \mathrm e^{-i\omega t}$ has nothing to do with Quantum Mechanics. In fact, cyclotron resonance can be very well understood in the framework of the Drude Model, which is from 1900 (before QM was born). The use of this ansatz is in fact related to a mathematical theorem, with no physical insight whatsoever:

Theorem: the general solution of linear differential equations with constant coefficients are always exponentials. See Linear differential equation for the proof. This means: as your equation of motion is linear you know that the solution is an exponential. This is all there is: the ansatz $\mathrm e^{-i\omega t}$ has no physical explanation.

To illustrate my point, let's consider a different damping term, where the physics are the same but the mathematics different: $$ m\mathbf{\dot{v}} = -e\mathbf{v} \times \mathbf{B} - \alpha\ v^2\hat{\mathbf{v}} \quad \text{with} \quad \alpha\in\mathbb R $$ In this case, you also have a friction term, which is now quadratic. As this equation is non-linear, you cannot use the exponential ansatz. In this case, $v\neq \mathrm e^{-i\omega t}$, even though the physics are essentially the same.

A second point where I kind of disagree with Orca is in the statement

To begin with, I will take the simple case of a free electron of momentum...

By using a free particle, he/she is neglecting the most important point of your question: why can we treat the electrons as free, when in fact they are immersed in a lattice. Once we know that the electrons behave as free, we can study them Quantum mechanically or classically, depending on the temperature, carrier density, kinetic energy, etc. Orca is right that both QM and CM agree on their prediction for free particles, but we must argue why we are allowed to treat the electrons as free to begin with.

This is answered in any good book on Solid State Physics, so I won't explain the details here, but in a nutshell the conclusion is that (because of Bloch's Theorem, or $k\cdot p$ theory, etc) we know that the effective Hamiltonian, once we "integrate out" the interaction of the electrons with the lattice, is quadratic in $\boldsymbol k$. Therefore, the electrons behave as free particles, where we define the effective mass as the parameter that appears in this effective Hamiltonian: $$ H_\text{eff}\equiv\frac{\boldsymbol k^2}{2m^*}+\cdots $$ where this relation defines $m^*$. This means that the effect of the lattice on the electrons is to make them move as free particles with a different mass.

Or put it another way: for most materials, the band structure is approximately parabolic. If you search band structure on google you'll see that the energy looks like parabolas, that is, $$ H\approx a\boldsymbol k^2+\mathcal O(k^3) $$ where the approximation holds best if the excitation is not too high (that is, we are not far from the minima of these parabolas). This is not always true: in some materials, such as graphene, the band structure looks like $|\boldsymbol k|$ instead of $\boldsymbol k^2$ (google Dirac cones).

When $H= a \boldsymbol k^2+\cdots$ we can define $a=\frac{1}{2m^*}$ for a certain parameter $m^*$ with units of mass. With this definition, the Hamiltonian looks like the Hamiltonian of a free particle even though it really isn't. The electrons are not free - in fact, in general they are tightly bound - but if you approximate the energy with a parabola, the spectrum looks like that of a free particle, but with a different mass.

Note that the band structure can be both theoretically calculated and experimentally measured, and most of the times it indeed looks parabolic. This means that the "effective free particle approximation" is very well justified, both by theory and experiments. Once we know that electrons can be treated as free, we can ask ourselves whether we need to use Quantum Mechanics to study its dynamics, or we can use Classical Mechanics to get an approximate description. Orca's nice answer proves that actually both methods agree, so we can use whichever we like the most.

There is something weird about magnetic fields and Quantum Mechanics: we very often get the same prediction if we use Classical Mechanics or (first order perturbation theory of) Quantum Mechanics. For example, the Zeeman effect can be both studied with CM and QM.

Anyway, to answer your questions:

Why is it physically allowed to assume the electron can be treated classically? What's the key idea behind this approximation in such contexts?

In this case, the approximation is allowed because the temperature is such that $kT\gg \hbar\omega_c$, where $\omega_c$ is the cyclotron frequency. Therefore, the Landau levels $n_c$ are highly excited, $n_c\to\infty$, which means that the system is essentially classical. This is not always true: for example, to measure the electron magnetic moment in a Penning trap, they use a strong magnetic field and a very low temperature such that $kT\sim\hbar\omega_c$. In this case, the quantum effects are non-negligible, and we can't use CM. But to measure $m^*$ we use high temperatures so we can assume that the dynamics are classical.

Is the ansatz $v=v_0\; \mathrm e^{-i\omega t}$ a sound choice here because we are treating the electron classically or there is another unrelated underlying reason?

As I said above, the reason is mathematical: it has nothing to do with QM nor any other physical reason. It's simply because the differential equation is linear.

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  • $\begingroup$ Thank you for this! I misinterpreted what was meant by the $v(t) = v_0 \exp{(-iEt)}$ and should not have said that the Ansatz provides a clue that we are using Quantum Mechanics. I just wanted to show how the $v_0$ is interpreted in QM, from the decomposition of wavepackets into Fourier components. $\endgroup$ – Orca Apr 8 '16 at 11:38
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    $\begingroup$ @Orca That's what I thought, a simple misunderstanding! I mostly agree with your post, which is very nice and clear. I just wanted to point out that we can answer the 2) question without QM, because the answer is just mathematics, not physics. $\endgroup$ – AccidentalFourierTransform Apr 8 '16 at 11:46
  • $\begingroup$ in a minor sense you are not treating the electrons fully classically since you are making use of the effective mass $\endgroup$ – jim Apr 10 '16 at 18:54
  • $\begingroup$ @jim yes, sure, thank you for pointing that out :-) I didn't think it was necessary, but it is true that every treatment of electrons are, at most, semi-classical. We can never treat electrons fully classically in Solid State Physics. $\endgroup$ – AccidentalFourierTransform Apr 10 '16 at 19:03
  • $\begingroup$ @AccidentalFourierTransform should have written my comment "as a minor point" $\endgroup$ – jim Apr 10 '16 at 19:07
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We can treat this system classically because it is one of those nice situations in which the quantum-mechanical treatment produces the same results!

I will begin with part 2 of your question about plane waves. The use of this Ansatz is the first clue that you are actually treating the situation quantum-mechanically, but ending up with a result that exactly matches the classical result.

Basically what we are doing when we represent a quantum-mechanical wavefunction $\psi({r})$ as a plane wave $\psi({r}) = e^{-i\mathbf{p} \cdot \mathbf{r}/\hbar}$ is treating a single component of its Fourier transform,

\begin{equation} \psi{(\mathbf{p})}=\int{d^3\mathbf{p} \: e^{-i\mathbf{p} \cdot \mathbf{r}/\hbar} \: \psi{(\mathbf{r})}}. \end{equation}

This is mathematically far easier to do than treating the entire wavepacket, and since Quantum Mechanical operators are linear, the behaviour of the plane wave under its operators tells us the behaviour of the entire wavefunction when we sum up all the plane waves under the integral.

Furthermore, we can always do this, since square-integrability, that is,

\begin{equation} \int^{\infty}_{-\infty}{d^3\mathbf{r} \: |\psi(\mathbf{r})|^2}< \infty \end{equation}

is a sufficient requirement for the Fourier transform of a function to exist, and obviously this must be true in Quantum Mechanics, since the LHS of the inequality represents the probability of finding our 'particle' anywhere in space, and so must equal $1$.

This is the underlying reason why the use of plane-waves is ubiquitous in Quantum Mechanics. It's easy, and it works.

In your case, you have written $v(t) = v_0 e^{-iEt+i\mathbf{p} \cdot \mathbf{x}}$ where $v_0$ is the classical velocity (the velocity of the midpoint of the quantum-mechanical wavepacket), obtained by taking the FT of $p_x \psi(\mathbf{r}) = \partial_x \psi(\mathbf{r})$, integrating by parts under the Fourier integral, and neglecting boundary terms, to get

\begin{equation} \int{d^3\mathbf{p} \: p_x e^{-i\mathbf{p} \cdot \mathbf{r}/\hbar} \: \psi{(\mathbf{r})}}. \end{equation}

Which we can then interpret classically as $mv_0$.

Now for the cyclotron resonance in Quantum Mechanics.

To begin with, I will take the simple case of a free electron of momentum $\mathbf{p}=(p_x,p_y,p_z)$ in a magnetic field along the z-axis, $\mathbf{B}=(0,0,B)$. We can treat this quantum-mechanically by setting the gauge of the electromagnetic vector potential to the Landau gauge

\begin{equation} \mathbf{A} = (-By,0,0) \end{equation}

Where $\nabla \times \mathbf{A}=\mathbf{B}$. The Schroedinger equation in this gauge can then be written down by taking the canonical momentum $\mathbf{p} \rightarrow \mathbf{p}-q\mathbf{A}$ in this gauge so that

\begin{equation} H \psi (\mathbf{r}) = \frac{\hbar^2}{2m}[(p_x+qBy)^2+p_y^2+p_z^2] \psi(\mathbf{r}) = E\psi(\mathbf{r}) \end{equation}

This Hamiltonian clearly commutes with both $p_x$ and $p_z$ which are therefore conserved quantities and so share a set of eigenstates. Taking our plane-wave Ansatz, the $x$ and $z$ parts can be commuted through $H$, so remain as $e^{i(p_x x + p_z z)/\hbar}$, while the $y$ part, which we may call $\chi(y)$, now obeys the new equation

\begin{equation} \Big[\frac{p_y^2}{2m} + \frac{p_z^2}{2m} + \frac{1}{2} m\omega_c^2 \Big(y+\frac{p_x}{qB}\Big)^2 \Big] \chi(y) = E\chi(y), \end{equation}

where $\omega_c = |q|B/m$, the classical cyclotron frequency.

We can extend this argument to crossed electric and magnetic fields by introducing an electric potential $\phi = -Ey$ so that we have an electric field $\mathbf{E} = (0,E,0)$ in the $y$-direction, and the Hamiltonian becomes

\begin{equation} H = \frac{\hbar^2}{2m}[(p_x+qBy)^2+p_y^2+p_z^2] + qEy \end{equation}

Where we can still use the same Ansatz $e^{i(p_x x + p_z z)/\hbar}\chi(y)$ since H still commutes with $p_x$ and $p_z$. With this substitution we find that we can rearrange back to the cyclotron form we had before, if we switch to a frame in which $p_x \rightarrow p_x + mE/B$. That is:

\begin{equation} \Big[\frac{p_y^2}{2m} + \frac{p_z^2}{2m} + \frac{1}{2} m\omega_c^2 \Big(y+\frac{p_x-mE/B}{qB}\Big)^2 \Big] \chi(y) = E\chi(y), \end{equation}

Therefore, this predicts an additional 'drift velocity' in the $x$-direction of

\begin{equation} p_x = \frac{mE}{B}, \end{equation}

which we can interpret for the classical case of $\mathbf{p}=m\mathbf{v}$ as an actual 'drift' velocity (again the 'velocity' of the centre of the quantum-mechanical wavepacket),

\begin{equation} v_x = \frac{E}{B}. \end{equation}

This corresponds again, exactly to what you would get by using the classical Lorentz force equation, $\mathbf{F} = q(\mathbf{E}+\mathbf{v} \times \mathbf{B})$.

We can then introduce dielectrics with given relaxation times $\tau$ as in your question.

Note about the presence of $\hbar$: Notice here that when we arrive at the equation of motion from the Schroedinger equation, the $\hbar$ factors cancel, making our result independent of $\hbar$ so that we don't need to take the usual classical limit $\hbar \rightarrow 0$. This is not true in all situations, and is what makes this situation special.

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    $\begingroup$ I'm not completely convinced -- you say this is a special case where QM agrees with CM, but doesn't the kind of agreement here always work by Ehrenfest's Theorem? So I don't see where the QM is necessary. $\endgroup$ – knzhou Apr 10 '16 at 7:20
  • $\begingroup$ Sorry @knzhou I'm not sure I understand what you mean: I have not used any non-zero commutators here to arrive at the equations of motion for $\chi(y)$. By Ehrenfest, if I did take a non-zero commutator, then I could replace it by something like $i\hbar d\langle A \rangle /dt$ but then I would receive an $\hbar$ which would differentiate the equation from the classical case in all situations except for $\hbar \rightarrow 0$ which is the usual requirement for QM $\rightarrow$ CM since the version in CM has no $\hbar$. Is this the use of Ehrenfest's theorem that you meant? $\endgroup$ – Orca Apr 10 '16 at 8:01
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    $\begingroup$ I guess in summary, this is a special case where all the $\hbar$s cancel so we don't need $\hbar \rightarrow 0$. $\endgroup$ – Orca Apr 10 '16 at 8:07
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    $\begingroup$ @knzhou interesting! Would be great if you could elaborate a bit on how the Ehrenfest Thm can be used here. $\endgroup$ – user929304 Apr 14 '16 at 9:02
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The classic cyclotron with constant acceleration frequency is not well suited for relativistic velocities because the particle's cyclotron resonance frequency decreases while relativistic mass increases 1.

This implies that particles accelerated in a classic cyclotron could be treated classically.

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  • $\begingroup$ ...how does it imply that? $\endgroup$ – ACuriousMind Apr 4 '16 at 20:56
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    $\begingroup$ If the classical cyclotron does not accelerate particles to relativistic velocities, the particles in the cyclotron can be treated classically. $\endgroup$ – aventurin Apr 4 '16 at 21:02
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    $\begingroup$ The "classical" in the question is supposed to mean "not quantum" not "not relativistic". I don't see what how relativistic the velocity is has to do with the size of quantum effects. $\endgroup$ – ACuriousMind Apr 4 '16 at 21:05
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    $\begingroup$ OK. Seems that I couldn't think of quantum effects in the context of a cyclotron $\endgroup$ – aventurin Apr 4 '16 at 21:57

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