momentum - What spinor field corresponds to a forwards moving positron?

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?

Answer

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".