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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 travelling backwards in time, which is interpreted 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..?

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..?

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 ($-p_z$)?

§3. Feynman–Stueckelberg Interpretation

The spinor $\psi_s = v_s(p)\, \text{e}^{\text{i}p\cdot x}$ is a negative-energy particle travelling backwards in time, which is interpreted 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.

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

• 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..?

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 travelling backwards in time, which is interpreted 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..?

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 ani-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 ($-p_z$)?

§3. Feynman–Stueckelberg Interpretation

The spinor $\psi_s = v_s(p)\, \text{e}^{\text{i}p\cdot x}$ is a negative-energy particle travelling backwards in time, which is interpreted 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.

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

• 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..?

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 ($-p_z$)?

§3. Feynman–Stueckelberg Interpretation

The spinor $\psi_s = v_s(p)\, \text{e}^{\text{i}p\cdot x}$ is a negative-energy particle travelling backwards in time, which is interpreted 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.

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

• 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..?