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When we search for spinor solutions to the Dirac equation, we consider the 'positive' and 'negative' frequency ansatzes $$ u(p)\, e^{-ip\cdot x} \quad \text{and} \quad v(p)\, e^{ip\cdot x} \,,$$ where $p^0> 0$, and I assume the $(+,-,-,-)$ metric convention. If we take the 3-vector $\mathbf{p}$ to point along the positive $z$-direction, the first solution is supposed to represent a forwards moving particle, such as an electron. My question is simple to state:

If we take $\mathbf{p}$ to point along the positive $z$-direction, is the second solution supposed to represent a forwards or backwards moving positron?

I will give arguments in favour of both directions. I welcome an answer which not only addresses the question above, but also the flaws in some or all of these arguments.

Backwards:

  • Though we take $\mathbf{p} = |p|\mathbf{z}$ to point in the positive $z$-direction in both cases, a comparison of the spatial parts of the particle and antiparticle solutions shows that the former has the dependence $e^{i |p| z}$ whilst the latter has the dependence $e^{-i |p| z}$. These are orthogonal functions and one might imagine that they represent motion in opposite directions.
  • The total field momentum (see Peskin (3.105)) is given by $$ \mathbf{P} = \int_V \mathrm{d}^3 x\, \psi^\dagger(-i\boldsymbol{\nabla})\psi \,, $$ which yields a momentum $+|p| \mathbf{z}V u^\dagger u $ when evaluated on the particle solution, but $-|p|\mathbf{z}V v^\dagger v $ when evaluated on the antiparticle solution. This suggests that the given antiparticle solution corresponds in fact to a positron moving in the negative $z$-direction.

Forwards:

  • When we quantize the Dirac theory and write $\psi$ as a sum over creation and annihilation operators, the solution $v(p) \, e^{ip\cdot x}$ is paired up with the creation operator $\hat{b}_\mathbf{p}^\dagger$, the operator which creates a forwards moving positron. This suggests to me that the spinor $v(p)$ also represents a forwards moving positron.
  • In the quantum theory, we know that the 2-component spinors which correspond to 'up' and 'down' are interchanged for particle and antiparticle (see Peskin (3.112) and the preceding paragraph). One might imagine that the same is true for the spatial functions which correspond to 'forwards' and 'backwards', such that $e^{i|p|z}$ represents a forwards moving particle but a backwards moving antiparticle.

Bonus question:

It seems to me that a lot of the confusion surrounding these matters comes from the fact that we are trying to interpret negative energy solutions as, in some sense, the absence of positive energy particles, not actual negative energy states. David Tong, on page 101 of his QFT notes, states:

[Regarding positive and negative frequency solutions] It’s important to note however that both are solutions to the classical field equations and both have positive energy $$ E = \int \mathrm{d}^3 x \, T^{00} = \int \mathrm{d}^3 x \, i \bar{\psi}\gamma^0 \dot{\psi} \,.$$

However, it is clear that if one substitutes the negative energy (antiparticle) solution directly into this expression, one gets a negative number!

What's going on here?

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Dirac spinors are an infuriating subject, because there are about four subtly different ways to define phrases like "the direction a spinor is going" or "the charge conjugate of an spinor". Any two different sources are guaranteed to be completely inconsistent, and all but the best sources will be inconsistent with themselves. Here I'll try to resolve a tiny piece of this confusion. For more, see my answer on charge conjugation of spinors.

Classical field theory

Let's start with classical mechanics. We consider plane wave solutions of classical field equations, which generally have the form $$\alpha(k) e^{-i k \cdot x}$$ where $\alpha(k)$ is a polarization, e.g. a vector for the photon field, and a spinor for the Dirac field. The momentum of a classical field is its Noether charge under translations, so in general $$\text{a plane wave proportional to } e^{-ik \cdot x} \text{ has momentum proportional to } k$$ Now let's turn to the plane wave solutions for the Dirac field, $$\sum_{p, s} u^s(p) e^{-i p \cdot x} + v^s(p) e^{i p \cdot x}.$$ By comparison to what we just found, we conclude $$\text{classical plane wave spinors with polarization } \begin{cases} u^s(p) \\ v^s(p) \end{cases} \text{ have momentum } \begin{cases} p \\ -p. \end{cases}$$ That is, for Dirac spinors, the parameter $p$ doesn't correspond to the momentum of a classical plane wave solution. However, this doesn't tell us about how a wavepacket moves, because plane waves don't move at all. Instead, we need to look at the group velocity $$\mathbf{v}_g = \frac{d \omega}{d \mathbf{k}}$$ of a wavepacket. For the negative frequency solutions, both $\omega$ and $\mathbf{k}$ have flipped sign, so $$\text{wavepackets built around } u^s(p) \text{ and } v^s(p) \text{ both move along } \mathbf{p}.$$ I think this is the best way to define the direction of motion in the classical sense. (Some sources instead say that $v^s(p)$ moves along $- \mathbf{p}$ but backwards in time, but I think this is not helpful.)

Quantum field theory

When we move to quantum field theory, we run into more sign flips. Recall that in quantum field theory, a plane wave solution $\alpha(k) e^{-i k \cdot x}$ is quantized into particles. To construct the Hilbert space, we start with a vacuum state and postulate a creation operator $a_{\alpha, k}^\dagger$ for every mode.

If we do this naively for the Dirac spinor, the raising operator for a negative frequency mode creates a particle with negative energy. This is bad, since the vacuum is supposed to be the state with lowest energy. But Pauli exclusion saves us: we can instead redefine the vacuum to have all negative frequency modes filled, and define the creation operator for such a mode to be what we had previously called the annihilation operator. This is the Dirac sea picture. Then $$\text{particles made by the creation operators for } u_s(p) e^{-ip \cdot x}, v_s(p) e^{ip \cdot x} \text{ have momentum } p.$$ Moreover, both of these particles move along the direction of their momentum $\mathbf{p}$. All other quantum numbers for the $v$ particles are flipped from what you would expect classically, such as spin and charge, but the direction of motion remains the same because the quantum velocity $\hat{\mathbf{v}}_g = d \hat{E} / d \hat{\mathbf{p}}$ stays the same.

Summary

To summarize, I'll quickly assess your arguments.

  1. Your first argument is wrong. Momentum doesn't correspond to propagation direction. You know this from second-year physics: a traffic jam is an example of a wave which moves backwards but has positive momentum.
  2. Your second calculation is correct, $v^s(p)$ indeed has momentum $-p$.
  3. The classical spinor does move the same direction as the quantum particle; it must, if we can take a classical limit. In both cases the classical/quantum spinor moves along $\mathbf{p}$.
  4. Indeed, spin up and down is interchanged by the implicit hole theory argument going on, along with everything else.
  5. Tong is generally a great source, but he messed up here. I've emailed Tong and he's agreed and fixed it in the latest version of the notes.

Other sources may differ from what's said here because of metric convention, gamma matrix convention, or whether they consider some subset of the objects to be "moving backwards in time".

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  • $\begingroup$ This answer has been extremely helpful, thank you. $\endgroup$ – gj255 Feb 8 '18 at 14:29
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Actually, I also struggled a lot the understanding of the "positron" solutions $v(p)e^{ipx}$ of the Dirac-equation, but I think, now I've understood it. I also agree that most literature sources don't clear up possible confusions of the topic, in their formulations they are often not precise enough which finally leads to a lot of questions marks. For instance, as I will show at the end, Tong is not completely wrong, he just forgot to add a counter term in the Hamiltonian of the Dirac-equation to get it right.

First of all, one has to stress that both solutions of the Dirac-equation can only be fully understood in the framework of second quantization. In this framework the field operator $\psi(x)$ and its counterpart $\psi^\dagger(x)$ are defined in the following way:

$\psi(x) = \Sigma_{p,s} ( a_{p,s} u(p)e^{-ipx} + b^\dagger_{p,s} v(p)e^{ipx})$

It is very important to recognize from this formula that so called "positron solution" get an creation operator attached in contrast to the normal solution which gets an annihilation operator attached.

This means that the "positron" solution is not just another additional solution, it has a property which really distinguishes it from the electron solution. Looking to $\psi^\dagger(x)$ we see it better:

$\psi^\dagger(x) = \Sigma_{p,s} ( b_{p,s} v^\dagger(p)e^{-ipx} + a^\dagger_{p,s}u^\dagger(p)e^{ipx})$

Actually, in scattering $u(p)e^{-ipx}$ describes an in-going particle whereas one would $v(p)e^{ipx}$ associate with the description of an out-going particle.

As well as we would associate $v^\dagger(p)e^{-ipx}$ with an in-going particle and $u^\dagger(p)e^{ipx}$ with an out-going particle. So we can interprete that $v^\dagger(p)e^{-ipx}$ is actually an in-going positron to be annihilated which has to be compared to $u(p)e^{-ipx}$ since it is an in-going electron solution also to be annihilated. To make the analogy complete we check out energy and momentum of this solution $v^\dagger(p)e^{-ipx}$ by acting the 1-particle operators $P$ and $H$ on it and get positive values.

But what is $v(p)e^{ipx}$ then ? It is a description of an out-going "positron". However, in (typically non-relativistic) scattering matrix elements like $\psi^*V\psi$ such an expression is symbolized as $\psi^*$ on which normally operators $P$ and $H$ are not applied from the left side, if done inspite one might not wonder about surprising results like $(-{\bf p},-E)$ as eigenvalues. The last 2 paragraphs mainly serve as intuitive explanations, for more rigorous explanation see the QFT-formalism to which I come now.

But the formalism of 2. quantization (or just QFT-formalism) describes this elegantly by putting an creation operator as coefficient of $v(p)e^{ipx}$ (and not an annihilation operator) and one is not obliged to make act an operator $P$ or $H$ on $\psi^*$ which is actually very awkward.

The QFT-formalism can actually more: The momentum operator $P$ composed of field operators is:

$P = \int d^3x\, \psi^\dagger (-i {\bf\nabla})\psi =\int \frac{d^3p}{(2\pi)^3} \sum_s {\bf p} \left(a^\dagger_{p,s} a_{p,s} + b^\dagger_{p,s} b_{p,s}\right)$

If it is applied on a 1-antiparticle (positron) state $\overline{|{\bf p}>}$ (the bar over the state is supposed to mark it as antiparticle state) we get $+{\bf p}$ as eigenvalue. So the momentum of the positron state is positive. Of course we can apply it to 1-particle state $|{\bf p}>$ and also get $+{\bf p}$ as eigenvalue.

The same can be done with the Hamilton operator: $H=\int d^3x\, \psi^\dagger (i \frac{\partial}{\partial t})\psi + 4E_0 V =\int \frac{d^3p}{(2\pi)^3} \sum_s E_p \left(a^\dagger_{p,s} a_{p,s} + b^\dagger_{p,s} b_{p,s}\right)$.

And why did you get an negative result? Because the counter term was forgotten in Tong's expression (apart from other aspects explained in the following).

To reach the expression containing the creation and annihilation operators the anti-commutator rules for fermions have to be applied which lead to a negative zero-point energy which has to be compensated by the counter term $4E_0 V$ where $E_0= \frac{1}{2}\int\frac{d^3p}{(2\pi)^3}E_p$ and $V=\int d^3 x\,\,$.

(These manipulations and this formalism is very well documented in the books on QFT, so I don't go here in longer explanations). So Tong is not so wrong, but may be he did not stress in particular that the $\psi$'s and $\psi^\dagger$ are supposed to be field operators and not 1-particle solutions (and he forgot the counter term). With the corresponding (anti)-commutator rules applied the correct result comes out quasi automatically.

Let's recap: The general rule to keep in mind is that any result to be obtained in relativistic QM or QFT has to be done with the field operators (and NOT with the 1-particle solutions) and applying the corresponding (anti)-commutator rules.

The second important aspect to keep in mind is that $v(p)e^{ipx}$ actually does not describe a "standard", i.e. in-going positron, but an out-going positron which explains a couple of its "awkward" properties. However, $v^\dagger(p)e^{-ipx}$ describes an in-going, i.e. kind of "normal" positron.
If you don't like this, you can still interpret the out-going positron $v(p)e^{ipx}$ as in-going electron with $(-{\bf p},-E)$ running backwards in time.

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  • $\begingroup$ Thanks a lot for your answer, it has been very helpful. I gave the bounty to the other answer because it talked more about how to make sense of all this in the context of classical field theory, not QFT, but I am grateful for the insights you've given. $\endgroup$ – gj255 Feb 8 '18 at 14:30

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