Lorentz force in Dirac theory and its classical limit

Question

It is well known that in Dirac theory the time derivative of $P_i=p_i+A_i$ operator (where $p_i=∂/∂_i$, $A_i$ - EM field vector potential) is an analogue of the Lorentz force:

$\frac{dP_i}{dt} = e(E_i+[v×B]_i)$

On the other hand, in classical theory we have the same equation for $p_i$ instead of $P_i$. How comes that the effect of $A_i$ in Dirac theory vanishes in the classical limit?

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Comments

@Jon: I mean "Dirac equation in EM field". And I know what the "minimal coupling" is. The problem is that both classical and quantum mechanical Lagrangians are constructed using minimal coupling procedure. But the result is different, and the difference is not vanishing in classical limit.

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@MurodAbdukahimov: Indeed, any difference in the classical action must reduce to a full differential that is harmless in the action. In this way, the minimal coupling does work in any case. Just check this.

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@Jon: There is no need to check the ACTION. It is absolutely clear from the EQUATIONS that the difference is not vanishing.

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I don't think it's a research level question. This is standard textbook stuff on magnetism in QM

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@Squark, please flag questions you think need closing, by now it is too late.

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@MurodAbdukahimov: $P_i = p_i + A_i$ is used broadly to describe magnetic force - also in classical mechanics (its the way how you introduce magnetic force in Hamiltonians & Lagrangians). So it has nothing to do with a classical limit.

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@Jon: In Hamiltonian formalism the (full) time derivative of the operator is a sum of the partial time derivative of the operator and its commutator with Hamiltonian. Consider Dirac theory in Hamiltonian form. You will see that $\frac{dP_i}{dt}$ is given by the formula presented in my question. But it is not the same as the formula in classical mechanics, where $P=p+A$ need to be replaced by $p$ only. My question is why there is a difference. Normally the quantum equations can be reduced to classical ones by setting $h=0$. But this is not the case. Is my question clear now?

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Answer 1

I found the question completely clear and sort of important for this piece of learning about magnetism in quantum mechanics; the only problem is that it assumes a wrong answer to the question.

Nothing is vanishing; indeed, it would be very bad if a finite term such as $\vec v\times \vec B$ evaporated without a trace while taking the classical limit. However, one must be careful about the quantum mechanical commutators in order to see that everything is valid.

The full quantum mechanical (non-relativistic) Hamiltonian (which may be obtained as the non-relativistic limit of the Dirac equation) is $$ H = \frac{1}{2m}(\vec p+ \vec A)^2 + V(\vec x)$$ where $V(\vec x)$ primarily contains the electrostatic potential and generates the electric force $e\vec E = -i\nabla \Phi$ in a way you find uncontroversial. Your problem is localized to the terms containing $\vec A$ or its derivatives. I kept the conventions for the normalization and sign of $\vec A$ to be just like yours although it's a bizarre convention: one would usually add the factor of $q$ or $e$ in front of $\vec A$, too.

In these conventions of yours, $\vec p = -i\hbar \vec \nabla$, as you correctly wrote. However, this object isn't the usual $m\vec v$. Instead, the full $\vec P = \vec p + \vec A$ is equal to $m\vec v$, a multiple of the velocity. This is usually denoted as $\vec p$ in non-quantum physics but we obviously need to map it to $\vec P$: in the limit, once again, the usual non-quantum momentum is $\vec P$, not $\vec p$: $\vec P$ is gauge-covariant, $\vec p$ isn't, and the non-quantum momentum of a particle is clearly gauge-covariant so it can't be $\vec p$.

To quantum mechanically calculate the time-derivative of $P_i$ (the momentum as directly calculated from the velocity), we must compute $1/(-i\hbar)$ times the commutator of $P_i$ with the Hamiltonian (the Heisenberg equations of motion). The commutator of $P_i$ with $V(x)$, the electrostatic potential energy, gives us the usual electric force.

However, we must also add the commutator of $P_i$ with the first term of the Hamiltonian which is $P_j P_j / 2m$. It is not zero because the different components of $P_j$ don't commute with one another. Instead, $$ \frac{1}{2m} [P_i,P_j P_j] = \frac{1}{m} [P_i,P_j] P_j + {\rm subleading\,\,in\,\,}\hbar $$ The commutator of $[P_i,P_j]$ is nonzero because $P_i$ depends both on $\vec p$ as well as $\vec x$: those two lowercase objects are the objects with the usual simple commutation relations. We have $$ [P_i,P_j] = [p_i,A_j] - [p_j,A_i] $$ in your conventions. You can see that the commutators on the right hand side are nothing else than $-i\hbar$ times the components of ${\rm curl} \vec A = \vec B$, more precisely $\epsilon_{ijk}B_k$. This is multiplied by $P_j/m = v_j$ above, so the total term in the commutator clearly gives $\epsilon_{ijk}v_j B_k = (v\times B)_i$, which is – after $-i\hbar$ is cancelled between the commutator and the factor in the Heisenberg equation (I ignored this factor) – exactly the magnetic force. Again, the usual convention would have the charge $q$ in front of it but I followed your conventions for the normalization of $\vec A$.

However, the correctly calculated classical limit obviously does generate and has to generate the full Lorentz force including the magnetic piece. The overall sign of $\vec A$ and the Lorentz force wasn't tracked very carefully above but believe me that it works (and has to work) as well when the calculation is done perfectly.

Because Murod repeated his or her doubts in the relativistic case, let me rerun the derivation above for the full relativistic Dirac equation. Its Hamiltonian is $$ H = \gamma_0 (P^i \gamma_i - m + A_0) $$ Note that if you multiply it by $\gamma_0$ and move everything to the same side, you get the simple and uniform operator $P^\mu \gamma_\mu -m$ which must annihilate the Dirac spinorial wave function. $A_0$ is the electric potential – normally one would write $q$ explicitly in front of this term as well.

The commutator $[H,P^i]$ which is what determines the change of the momentum of a relativistic classical (non-quantum) particle even though this energy-momentum vector is sometimes called $p_\mu$ in non-quantum relativistic physics of particles. However, this object comes from the limit (and should be identified with) $P_\mu$. Again, the commutator with $A_0$ produces the electric force. The commutator of $P^i$ with $P^j\gamma_j$ produces the Lorentz force except that $\gamma_j$ is here instead of $v_i$. But that's OK, $\gamma^\mu$ acts on the Dirac spinors just like the velocity 4-vector. Note that $\gamma^\mu$ is formally a vector that squares to one – a unit time-like vector – and it must be the velocity because it's the only gauge-invariant spacetime direction picked by the (quickly oscillating) plane wave. So in the relativistic case, the derivation of the non-quantum equation for the particle is almost identical, just with $v^\mu$ expressed as the matrix $\gamma^\mu$ instead of $\vec P / m$, a part of the reason why the Dirac equation manages to be a first-order equation (for the price of having many components and matrices).

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Comments

Thanks, Lubos. I know that there is no problem with non-relativistic hamiltonian. What about Dirac hamiltonian? $H=c[\alpha_1 P_1 + \alpha_2 P_2+\alpha_3 P_3]+m c^2 \alpha_4 - e\phi$, where $P_i=-ih\frac{∂}{∂_i}+\frac{e}{c} A_i$

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Dear @Murod, you obviously get the right limit from the relativistic Dirac equation, too. The Hamiltonian I wrote may be easily obtained as the non-relativistic limit of the Dirac equation. It's very clear that the momentum $\vec p$ always naturally combines with $\vec A$ to produce $\vec P$: the derivatives in the Dirac equation coupled to electromagnetism are the covariant derivatives and they stay this way in the non-relativistic limit as well. You can also avoid the non-relativistic limit and go from the Dirac equation to non-quantum but relativistic eqn of motion: $A$ is still there.

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At any rate, note that your original question had nothing to do with relativity vs. non-relativity. If the magnetic force dropped while taking the classical (non-quantum) limit, it would drop in the same way in the non-relativistic limit as well. It doesn't drop. Again, even in the full Dirac equation, $P_i$ including both $\partial$ as well as $A$ terms is what gets interpreted as the momentum (and/or energy) in the classical (non-quantum) limit, despite the fact that the capitalization isn't the most usual notation in classical physics. Also in relativity, be careful about nonzero $[P,P_j]$.

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Sorry, I missed an important point in your answer. You wrote that the full $P=p+A$ is equal to $mv$. This is what makes me confused. Why in classical theory it is not the "full" $P$, but just $p$?

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I've added special paragraphs rerunning the same derivation for the relativistic case.

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Dear @Murod, you are confusing yourself and you don't seem to be willing to stop. You wrote: "Why in classical theory it is not the "full" $P$, but just $p$?" - I just told you that $m\vec v$ is equal to $\vec P$, so why are you repeating the wrong statement that $m\vec v$ is equal to $\vec p$? The product $m \vec v$, the only one that may be identified with the momentum in classical physics (where no phases of wave functions exist), is always identified with $\vec P$ in quantum physics and never with $\vec p$. You just keep on repeating a wrong statement but the repetition won't make it true.

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Someone presenting classical physics may use the symbol $\vec p$ (lowercase) for the momentum. He may use a completely different letter, too. There is no article in the constitution that forces everyone to capitalize symbols for the momentum derived from the velocity. However, if you are comparing two situations and deriving limits, you must be careful about what symbols are mapped to one another and I've been telling you for an hour that the momentum in classical physics always corresponds to $\vec P$, whether the classical physicists call it $\vec p$, $\vec P$, or $\vec \mu_{HOHO}$.

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Let me paint a thought explaining why $m\vec v$ is $\vec P$ and not $\vec p$. $\vec p$ is linked to the partial derivatives e.g. to the inverse wavelength of the plane wave, OK? However, by a gauge transformation with the phase $\exp(ik\cdot x)$, you may change the wavelength and $\vec p$ by $\vec k$, clear? However, it's just a gauge transformation that doesn't change the physical, measurable quantities. And indeed, if you calculate $P=p+A$ and realize that $A$ changed under the gauge transformation, the action of $P$ on the plane wave (eigenvalue) didn't change by the gauge transformation.

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Thank you, Lubos. Your post was very helpful. I think the reason is that in classical theory the charge (or its "coordinate" function) is not affected by the gauge transformation, while in quantum theory the particle's wave function is affected. Since the $P$ operator acts on wave function, it has to include the $A$ that compensates the gauge transformation of the wave function. In classical theory there is no need to add $A$ to momentum.

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Dear @Murod, actually even in the classical (non-quantum) theory, both versions of the momenta are being used. Note that they differ by a finite, nonzero amount: they can't be the same thing so it would always be a mistake, even in classical theory, to confuse them. $\vec p$ is the "canonical momentum" while $\vec P$ is the kinetic momentum, see http://en.wikipedia.org/wiki/Kinetic_momentum - Both objects are also used in classical physics. After all, the prescription for the Hamiltonian is legit even in classical physics.

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Answer 2

The question(v1) seems to be caused by a misunderstanding. Let $\vec{p}^{kin}=\gamma m_0 \vec{v}$ denote the kinetic momentum (also known as the mechanical momentum), and let $$\vec{p}^{can}~=~\vec{p}^{kin}+q\vec{A}$$ denote the canonical momentum. Quantum mechanically, $$\vec{p}^{can}~=~\frac{\hbar}{i}\vec{\partial}.$$

The Lorentz force law

$$ \frac{d\vec{p}^{kin}}{dt}~=~ q(\vec{E} + \vec{v} \times \vec{B}) $$

applies also for the classical case, see e.g., Landau and Lifshitz, Vol.2, The Classical Theory of Fields, Chapter 3.

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Comments

Yes, It's indeed just a mix up of the canonical momentum and the kinetic momentum.

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Answer 3

As Qmechanic already pointed out: In order to obtain the kinetic momentum you have to take the derivatives (which give you the canonical momentum) and then subtract the interaction with $A^\mu$.

So everything is ok and the Dirac equation exactly reproduces the classical result. You can gain a deeper understanding of this if you write the Lorentz force in a more advanced way by using the electromagnetic field tensor.

$\frac{\partial j^\mu}{\partial \tau} ~~=~~ \frac{q}{mc}\,F^{\mu}_{~\nu}\,j^\nu~$

Which couples the E field with the boost generators K and the B field with the rotation generators J

$ F^{\mu}_{~\nu} ~~=~~ \Big(\,\mathsf{E}^i\,\hat{K}^i + \mathsf{B}^i\,\hat{J}^i\,\Big) \ =\ \left( \begin{array}{rrrr} ~\ 0\ \ & ~~\mathsf{E}_x & ~~\mathsf{E}_y & ~~\mathsf{E}_z \ \\ ~ \mathsf{E}_x & \ 0\ \ & ~~\mathsf{B}_z & - \mathsf{B}_y \ \\ ~ \mathsf{E}_y & - \mathsf{B}_z & \ 0\ \ & ~~\mathsf{B}_x \ \\ ~ \mathsf{E}_z & ~~\mathsf{B}_y & - \mathsf{B}_x & \ 0\ \ \ \end{array} \right)$

For spinors the equivalent interaction generator of time evolution is:

${\cal F}^\mu_{~\nu}\,\varphi ~=~ \left(\,\vec{E}\cdot\hat{\mathbb{K}} + \vec{B}\cdot\hat{\mathbb{J}}\,\right)\varphi$

$\mathbb{K}^i ~=~ -\tfrac12\,\gamma^i\gamma^o, ~~~~~~~~~~ \mathbb{J}^i ~=~ \tfrac{i}{2}\,\gamma^5\gamma^i\gamma^o$

Again the electric field boosts while the magnetic field rotates.

The classical time evolution due to the classical electromagnetic field tensor $F$ operating on the current is exactly the same as when the Spinor field tensor ${\cal F}$ operates on the spinor.

$\exp(F^{\mu}_{~\nu}\,t)\,\bar{\varphi}\,\gamma^\nu\varphi ~~=~~ \overline{\Big(\exp({\cal F}^\mu_{~\nu}\,t)\varphi\Big)}\,\gamma^\mu \, \Big(\exp({\cal F}^\mu_{~\nu}\,t)\varphi\Big)$

If you work out the series expansion of the exponential functions you can find all kind of beauties like.

$\begin{aligned} &\dot{\bar{\varphi}}\gamma^\mu\dot{\varphi} &=~~~ &\tfrac12\,T^\mu_{~\nu}~\bar{\varphi}\,\gamma^\nu\varphi \\ &\dot{\bar{\varphi}}\gamma^5\gamma^\mu\dot{\varphi} &=~~~ &\tfrac12\,T^\mu_{~\nu}~\bar{\varphi}\,\gamma^5\gamma^\nu\varphi \\ &\dot{\bar{\varphi}}~\mathbb{K}^\mu\,\dot{\varphi} &=~~~ &\tfrac12\,T^\mu_{~\nu}~\bar{\varphi}\,\mathbb{K}^\nu\,\varphi \\ &\dot{\bar{\varphi}}~\mathbb{J}^\mu\,\dot{\varphi} &=~~~ &\tfrac12\,T^\mu_{~\nu}~\bar{\varphi}\,\mathbb{J}^\nu\,\varphi \\ \end{aligned}$

Where T is the symmetric stress energy tensor of the electromagnetic field.

The term ${\cal F}^\mu_{~\nu}\,\varphi$ is just the extra term which occures if you square the Dirac equation with interaction. (although its role there is generally poorly interpreted) The squared Dirac equation contains the second order derivative in time so it should include a term which accounts for the spinor boosts and spinor rotates due to the electromagnetic field. The Klein Gordon equation does not need such a term because it describes a scalar field and scalars are per definition Lorentz invariant.

Regards, Hans

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Comments

Thank you Hans! That's exactly what I'm looking for. The action of electromagnetic field on charged spinors in QED need to be reduced to rotation and boost, and I just wanted to see how it works. Could you please advise any literature on this? I still need to check how this can be derived from Dirac equation: $\frac{\partial j^\mu}{\partial \tau} ~~=~~ \frac{q}{mc}\,F^{\mu}_{~\nu}\,j^\nu~$.

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I see that there is something on your web page.

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