I am trying to understand the difference between the Dirac sea interpretation and the Feynman–Stueckelberg interpretation of the negative energy solutions of the Dirac equation. To do so, I would like to calculate the helicity of an antiparticle in both interpretations.

In particular, I would like to show that the left-chiral component has right helicity and vice versa.

§1. Preparation

First off, using the Weyl (or chiral) representation of the gamma matrices, we know that we can write a Dirac spinor using left-chiral and right-chiral components:

$$ v_s(p) \stackrel{\text{Weyl rep.}}{=} \begin{pmatrix}v_{L,s}\\v_{R,s} \end{pmatrix} = \begin{pmatrix} \sqrt{p\cdot \sigma}\,\eta_s\\-\sqrt{p\cdot \bar\sigma}\,\eta_s \end{pmatrix} \tag{Peskin & Schroeder: 3.62} $$

For an ultra-relativistic particle going in positive $z$ direction,

$$ v_s(p) \stackrel{\text{Weyl rep.}}{=} \begin{cases} \begin{pmatrix} 0\\-\sqrt{2p^0}\,\eta_1 \end{pmatrix}, & s=1 \text{ (right-chiral)} \\ \begin{pmatrix} \sqrt{2p^0}\,\eta_2 \\0\end{pmatrix}, & s=2 \text{ (left-chiral)} \end{cases} $$

Also, if we let the particle go in $z$ direction, it is convenient to choose eigenstates of $\sigma^3$ for the Weyl spinors $\eta_s$:

$$ \eta_1 = \begin{pmatrix}1\\0 \end{pmatrix},\qquad \eta_2 = \begin{pmatrix}0\\1 \end{pmatrix} $$

The helicity operator is

$$ \hat h = \frac{\textbf p \cdot \textbf S}{|\textbf p||\textbf S|} \stackrel{\text{moving in $z$ direction}}{=} \begin{pmatrix}\sigma^3 & 0 \\ 0 & \sigma^3\end{pmatrix} $$

If its eigenvalue is positive, $h=+1$, the particle has right helicity, if it's negative, $h=-1$, it has left helicity. For antiparticles, it's the other way round.

§2. Dirac Sea Interpretation

The spinor $\psi_s = v_s(p)\, \text{e}^{\text{i}p\cdot x}$ (with $p^0>0$) is a state of energy $-p^0<0$ and momentum $-\textbf p$. Here, $\eta_1$ stands for spin-up in the positive $z$ direction, as is the case with $u_s(p)$.

Assuming, the anti-particle moves in positive $z$ direction, its momentum is $-p_z$, therefore the helicity operator picks up a minus sign,

$$ \hat h \psi_1 = \begin{pmatrix}-\sigma^3 & 0 \\ 0 & -\sigma^3\end{pmatrix} \begin{pmatrix} 0\\-\sqrt{2p^0}\,\eta_1 \end{pmatrix} \, \text{e}^{\text{i}p\cdot x} = -\psi_1 $$

  • The anti-particle state $\psi_1$ is right-chiral (only lower components are non-zero) and has helicity eigenvalue $h=-1$. For an anti-particle, this means right-handed.
  • $\psi_1$ travels in positive $z$ direction and has spin in positive $z$ direction, which means they're parallel $\to$ it should be right-handed? But do I consider the direction it's travelling in (pos. $z$) or the momentum ($-p_z$)?

(Possible solution: We're treating this as a particle, instead of an anti-particle. Therefore $h=-1$ corresponds to left helicity. And if we're defining helicity to be dependent on the momentum and not of the actual direction of propagation, then spin and momentum are anti-parallel.)

§3. Feynman–Stueckelberg Interpretation

The spinor $\psi_s = v_s(p)\, \text{e}^{\text{i}p\cdot x}$ is a negative-energy particle. We claim that it is travelling backwards in time, which is mathematically the same as a positive-energy "antiparticle" travelling forwards in time. Antiparticle means that all charges are opposite: $\eta_1$ stands for spin-down in the positive $z$ direction.

$$ \hat h \psi_1 = \begin{pmatrix}\sigma^3 & 0 \\ 0 & \sigma^3\end{pmatrix} \begin{pmatrix} 0\\-\sqrt{2p^0}\,\eta_1 \end{pmatrix} \, \text{e}^{\text{i}p\cdot x} = +\psi_1 $$

  • The anti-particle state $\psi_1$ is right-chiral (only lower components are non-zero) and has helicity eigenvalue $h=+1$. For an anti-particle, this means left-handed, so this seems OK.
  • $\psi_1$ travels in positive $z$ direction, its spin points in negative $z$ direction $\to$ left helicity.

§4. Question

Irrespective of which interpretation is "better", what is the mathematically correct way to calculate the helicity in both cases? I seem to get a wrong result for the Dirac sea interpretation, but the correct one for the Feynman–Stueckelberg interpretation..?

  • $\begingroup$ I don't use Dirac sea, but I think you answered your own question in the italics $\endgroup$ Commented Apr 1, 2020 at 7:56
  • $\begingroup$ It seems strange that the particle moves in $+z$ direction, but its momentum is $-|p_z|$..? $\endgroup$
    – ersbygre1
    Commented Apr 1, 2020 at 9:31

1 Answer 1


I think I figured it out, so I'm sharing my thoughts here in hope that it may help someone in the future.

The helicity operator is $\hat h$ and its eigenvalue is $h$. Just to be sure, right-handed helicity means spin and momentum are parallel, whereas left-handed helicity means that spin and momentum are antiparallel. As for the sign of $h$, this is ambiguous, as I will discuss.

  1. Solving the Dirac equation with a plane wave ansatz yields two sets of solutions, $$ \psi_\text{pos. energy}(x) = u(p)\text{e}^{-\text{i}p^0t+\text{i}\textbf{px}},\qquad \psi_\text{neg. energy}(x) = v(p)\text{e}^{+\text{i}p^0t-\text{i}\textbf{px}} $$ where the first one is called "positive energy solution" and the second one is called "negative energy solution". In both cases, $p^0 = +\sqrt{\textbf p^2 + m^2}>0$.

  2. Since a particle with negative energy is strange, we need some way to interpret this mathematical fact. So from here, we will focus on this state: $$ \psi_\text{neg. energy}(x) = v(p)\text{e}^{+\text{i}Et-\text{i}\textbf{px}} $$

  3. The Dirac sea interpretation (or hole theory) states that $\psi_\text{neg. energy}(x)$ has energy $E=-p^0<0$ and momentum $-\textbf p$, but is unobservable.

  4. This is because we assume that all negative energy states in our universe are already filled. So in order to "observe" one of these negative energy states (which are independent solutions of the Dirac equation, so they are different from $\psi_\text{pos. energy}(x)$!), we must annihilate them from the vacuum. So if we annihilate a state of energy, momentum and spin $(-p^0, -\textbf p, \textbf s)$, we get a state with $(+p^0, +\textbf p, -\textbf s)$. This is like saying, if we remove a charge of one Coulomb, it is basically the same as adding a charge of minus one Coulomb.

  5. Now we can talk about helicity. By choosing a basis for $\eta_{1,2}$, we decided a direction for its spin. In particular, we chose $(1,0)$ and $(0,1)$ and said this is a basis for $\sigma^3$. Therefore $\eta_1=(1,0)$ stands for spin-up (since the corresponding eigenvalue of $\sigma^3$ is $+1$) and $\eta_2=(0,1)$ stands for spin-down (since the corresponding eigenvalue of $\sigma^3$ is $-1$). This settles that our spin-operator is $$ \Sigma = \begin{pmatrix} \sigma^3 & 0\\0&\sigma^3 \end{pmatrix} $$

  6. As for the helicity operator, we have to project $\Sigma$ along the direction of the momentum. Since we chose $p^\mu = (p^0, 0,0,p_z)^\mu$, the momentum points in negative $z$ direction. (See item 3). Therefore, the helicity operator is $$ \hat h = \begin{pmatrix} -\sigma^3 & 0\\0&-\sigma^3 \end{pmatrix} $$

  7. In the ultrarelativistic limit (which we just take for easy mathematical calculation, this holds for any velocity), $v_1(p)$ is right-chiral (only lower components) and $v_2(p)$ is left-chiral (only upper components): $$ v_1(p) = \begin{pmatrix}0\\-\sqrt{2p^0}\eta_1\end{pmatrix},\qquad v_2(p) = \begin{pmatrix}\sqrt{2p^0}\eta_2\\0\end{pmatrix} $$ We can now determine the helicity in two ways: mathematically by evaluating the helicity operator $\hat h$, or physically by asking whether momentum and spin are parallel or antiparallel.

  8. Mathematically: $\hat h v_1(p) = -v_1(p)$, so $h=-1$. This means that right-chiral has left-handed helicity. Next, $\hat h v_2(p) = v_1(p)$, so $h=+1$. This means that left-chiral has right-handed helicity.

  9. Physically: the right-chiral state $v_1$ has spin pointing in the $+z$ direction. But its momentum is $\textbf p = -p_z \hat z$. Therefore they are antiparallel. Again, right-chiral has left-handed helicity. Next, the left-chiral state $v_2(p)$ has spin pointing in the $-z$ direction. And its momentum is $\textbf p = -p_z \hat z$. Therefore they are parallel. Therefore left-chiral has right-handed helicity.

  10. Fortunately, the mathematical and the physical way of obtaining the result agree.

  11. Finally, if we do annihilate the negative energy state from the vaacuum (as discussed in item 4), we would get a sign change in the momentum, but also the spin (which means that $\eta_1$ would stand for spin-down, and $\eta_2$ would stand for spin-up). The sign change in $p$ would give us an additional minus in the helicity operator, and the different interpretation of spin-up/down would give the opposite eigenvalues for the $\sigma^3$ inside the helicity operator. This means, the discussed results (both the mathematical one and the physical one) do not change: for "negative energy solutions", right-chiral means left-handed helicity, and left-chiral means right-handed helicity.

  12. The Feynman–Stueckelberg interpretation on the other hand introduces the concept of antiparticles.

  13. We claim that the negative energy solutions actually travel backwards in time. The thing is, this assumption is mathematically indistinguishable from saying it has positive energy and travels forwards in time: $$ \exp(\text{i}\underbrace{p^0}_{<0} \underbrace{t}_{<0}) = \exp(\text{i}\underbrace{p^0}_{>0} \underbrace{t}_{>0}) $$ however, it's not the same state, instead, we call it antiparticle. This is how we can keep the mathematical form $\exp(+\text{i}p^0t)$ (which is opposite to the "usual" phase factor $\exp(-\text{i}p^0t)$), and interpret it as positive energy, moving forwards in time, as if nothing happened.

  14. Antiparticle means that all "quantum charges" are reversed. So an electric charge of $e$ will be $-e$, spin-up will be spin-down. Similar things happen with hypercharge, lepton number etc.

  15. Let us consider helicity again. The thing that is represented by $v(p)$ (=the antiparticle) travels in $\textbf p$ direction, so in our example case, it travels in positive $z$ direction. But, being an antiparticle, $\eta_1=(1,0)$ now stands for spin-down and $\eta_2=(0,1)$ now stands for spin-up. This is a consequence of the before-mentioned flip of all charges. Now we can again calculate the helicity mathematically and physically.

  16. Mathematically: Since we chose $\eta_{1,2}$ as basis states of $\sigma^3$, the spin operator is $$ \Sigma = \begin{pmatrix} \sigma^3 & 0\\0&\sigma^3 \end{pmatrix} $$ and since we travel in positive $z$ direction, the helicity operator is $$ \hat h = \begin{pmatrix} \sigma^3 & 0\\0&\sigma^3 \end{pmatrix} $$ Therefore, $\hat h v_1(p) = v_1(p)$, so $h=+1$. It looks like right-handed helicity, but for antiparticles, we have to assign $h=\pm 1$ differently. In particular, $h=+1$ now means left-handed helicity. Similarly, $\hat h v_2(p) = -v_2(p)$, so $h=-1, which now means right-handed helicity.

  17. Physically: the right-chiral state $v_1$ has spin pointing in the $-z$ direction. Its momentum is $\textbf p = +p_z \hat z$. Therefore they are antiparallel. So, right-chiral has left-handed helicity. Next, the left-chiral state $v_2(p)$ has spin pointing in the $+z$ direction. And its momentum is $\textbf p = +p_z \hat z$. Therefore they are parallel. Therefore left-chiral has right-handed helicity.

  18. Both the Dirac sea interpretation as well as the Feynman–Stueckelberg interpretation lead to the same result concerning helicity. For completeness it should be mentioned that nowadays, the preferred way of thinking is the Feynman–Stueckelberg interpretation.

These answers were really helpful: one, two, three; as well as the QFT textbooks by Peskin & Schroeder, Lancaster & Blundell, Ohlsson and Thomson.


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