Interpretation of Dirac equation states for moving electron

Question

I try to understand a physical interpretation of the four components of the Dirac 4-spinor for a moving electron (in the simplest case, a plane wave). There is a very good question and answer about the interpretations already at SE. Basically it is shown that going to the rest-frame of the electron (i.e. $p^\mu=(E,0,0,0)$), one finds the four different solutions:

$$\psi_1=N_1\left(\begin{array}{c}1\\0\\0\\0\end{array}\right)e^{-iEt}, \psi_2=N_2\left(\begin{array}{c}0\\1\\0\\0\end{array}\right)e^{-iEt}, \psi_3=N_3\left(\begin{array}{c}0\\0\\1\\0\end{array}\right)e^{iEt}\text{ and } \psi_4=N_4\left(\begin{array}{c}0\\0\\0\\1\end{array}\right)e^{iEt},$$

Where the $\psi_1$ and $\psi_3$ have positive helicity (projection of the spin into the direction of the momentum) while $\psi_2$ and $\psi_4$ have negative helicity.

In addition, the phase factor $e^{\pm i E t}$ shows whether the state has positive or negative energy, thus whether it is a particle or antiparticle.

For moving electrons (in the Dirac representation), the solutions get additional contributions. For instance $$ \psi_{move}(x)=N_1\left(\begin{array}{c}1\\0\\\frac{p_z}{E+m}\\\frac{p_x+ip_y}{E+m}\end{array}\right)\exp(-ip_\mu x^\mu) $$ It has a nonvanishing 3rd and 4th component. Here Dominique writes:

When the momentum is NOT equal to zero these different states mix up and you can't make such a simple identification. Usually one says that the electron becomes a mixture of an electron with positrons when it starts moving.

However, the time-dependent phase-factor $e^{-i p_{\mu} x^{\mu}}$ still corresponds to positive energy for all of the four components, thus it can not be interpreted as

$$ \psi_{move}(x)\neq N \left(N_1 \psi_1 + \frac{p_z}{E+m} \psi_3 + \frac{p_x+ip_y}{E+m} \psi_4 \right) $$

(A similar argument is given in these lecture-notes: The fact that the last two components are non-zero does not mean it contains ``negative energy'' solutions.)

Thus my question is:

For a moving electron with helicity $+\frac{1}{2}$ (i.e. $\psi_{move}(x)$), what is the interpretation of non-vanishing 3rd and 4th component (in the Dirac representation)?

This question becomes more physical relevant when one considers no plane wave. Then the 3rd and 4th component might have different intensity distribution than the 1st component.

I am interested in both explanations and literature which covers this question.

Answer 1

In order to check the spin projection on the $i$-th axis for the third and fourth component, you just need to compute the quantity $$ \Sigma_{i}\psi, \quad \Sigma_{i} = \text{diag}(\sigma_{i}, \sigma_{i}) $$ with (for simplicity) $\psi = (0,0,1,0)$ and $\psi = (0,0,0,1)$ correspondingly.

In order to check the helicity of these components, you just need to compute the quantity $$ \frac{(\Sigma \cdot \mathbf p)}{|\mathbf p|}\psi $$ again for $\psi = (0,0,1,0)$ and $\psi = (0,0,0,1)$.

...Therefore there are two independent SU(2) solutions $\phi_{R}$ and $\phi_{L}$ (which are connected to the Weyl spinors). $\phi_{R}$ has positive helicity and $\phi_{L}$ has negative helicity...

This sentence contains two incorrect statements, related to each other.

First, actually the "elementary" irreducible representations $\phi_{L/R}$ are defined as the eigenstates of the chirality matrix $\gamma_{5}$. The latter defines the way under which the spinors are lorentz-transformed, and has nothing to do with the helicity as long as the mass $m$ isn't zero. In the zero mass limit helicity and chirality formally coincide.

Second, $\phi_{L/R}$ are not Weyl spinors. The Weyl spinor is the one satisfying one of the equations $$ \sigma_{\mu}\partial^{\mu}\psi = 0, \quad \tilde{\sigma}_{\mu}\partial^{\mu}\psi = 0, \ \ \text{where }\ \ \sigma_{\mu} = (1,\sigma), \ \tilde{\sigma}_{\mu} = (1,-\sigma) $$ It is defined to describe massless particles with definite helicity, and has notning to do with the spinors $\psi_{L/R}$ inside the Dirac spinor as long as $m\neq 0$.

Answer 2

Consider the Dirac representation of the gamma matrices, there we have $$\gamma^0 = \begin{pmatrix} I_2 & 0 \\ 0 & -I_2 \end{pmatrix}$$ and the solutions corresponding to static particles and antiparticles $p_\mu = (\pm m,0,0,0)$ are $$\begin{pmatrix} \alpha \\ \beta \\ 0 \\ 0 \end{pmatrix} e^{-imt}\,,\; \begin{pmatrix} 0 \\ 0 \\ \epsilon \\ \delta \end{pmatrix} e^{imt}$$ where $\alpha, \beta, \epsilon, \delta$ are some constants. Of course, these eigenstates still correspond to two possible states for each particle/antiparticle and physically this corresponds to the possibility of the particle having two signs of intrinsic spin.

But first let us see what happens if we choose another representation. In the chiral representation, we have $$\gamma^0 = \begin{pmatrix} 0 & I_2 \\ I_2 & 0 \end{pmatrix}$$ When we compute the particle/antiparticle stationary states, we now get $$\begin{pmatrix} \alpha \\ \beta \\ -\alpha \\ -\beta \end{pmatrix} e^{-imt}\,,\; \begin{pmatrix} \epsilon \\ \delta \\ \epsilon \\ \delta \end{pmatrix} e^{imt}$$ where $\alpha,\beta,\epsilon,\delta$ are again some constants. So you see that even though "top" and "bottom" components in the Dirac representation correspond to particles vs. antiparticles, the picture is quite different in the chiral representation of gamma matrices.

There is an infinite amount of possible representations of the gamma matrices related by unitary transformations and anything we discuss about spinorial components will be dependent on the chosen representation.

---

To make the argument complete, let us sort the energy eigenstates in the Dirac representation according to their projection of spin into a certain axis. The intrinsic spin operator is given by $\Sigma^i = i \gamma^0 \gamma^i \gamma^5$ in any representation. In the Dirac representation (as well as in the other two major chiral and Majorana representations), the spin operator ends up as $$\Sigma_i = \begin{pmatrix} \sigma_i & 0 \\ 0 & \sigma_i \end{pmatrix}$$ We choose to sort our states according to their z-component, because most conveniently we have $$\sigma_z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$$ We then easily see that the upper component of every energy eigenstate in the Dirac representation has positive projection of spin into the z axis, whereas the lower component corresponds to a particle with a negative projection of spin into z. But I will repeat once again that in the chiral or any other representation we would interpret the components quite differently.

---

As for the answer you cite, there one obtains the solutions for the moving particles by picking these stationary $\pm 1/2$ spin states and boosting them by a Lorentz transform into moving solutions.

In other words, the interpretation of the solutions $$ \psi_1(x)=N_1\left(\begin{array}{c}1\\0\\\frac{p_z}{E+m}\\\frac{p_x+ip_y}{E+m}\end{array}\right)\exp(-ip_\mu x^\mu) \,,\; \psi_2(x)=N_2\left(\begin{array}{c}0\\1\\\frac{p_x-ip_y}{E+m}\\\frac{-p_z}{E+m}\end{array}\right)\exp(-ip_\mu x^\mu) $$ $$ \psi_3(x)=N_3\left(\begin{array}{c}\frac{p_z}{E-m}\\\frac{p_x+ip_y}{E-m}\\1\\0\end{array}\right)\exp(ip_\mu x^\mu)\,,\; \psi_4(x)=N_4\left(\begin{array}{c}\frac{p_x-ip_y}{E-m}\\\frac{-p_z}{E-m}\\0\\1\end{array}\right)\exp(ip_\mu x^\mu) $$ is that they correspond to particles/antiparticles which, in their rest frame, have a negative/positive projection of spin into the z axis.

This is a common construction for the basis of general solutions of the Dirac equation, in other cases people often choose instead to sort the solutions according to their chirality.

Answer 3

The Dirac equation is solved using Clifford algebra, in particular the gamma matrices. There are 16 4x4 gamma matrices that form the required basis. You can choose which basis to use (mainly depending on the physics problem being solved) and it's possible to transform between basis.

In the Weyl basis (chiral representation of the Dirac equation) the gamma matrix conventionally known as $\gamma^5$ is diagonal $\gamma^5 = \begin{pmatrix} I & 0 \\ 0 & -I \end{pmatrix}$. When applied to the Dirac 4 component spinor, in any basis it will decompose it into two parts the left hand and right hand. In the Weyl basis because it is diagonal it will not mix up the two parts of the spinor so we can say that the top two components of the Dirac spinor represents the left handed field and the bottom two the right i.e.

$\gamma^5\begin{pmatrix} \psi_R \\ \psi_L \end{pmatrix} = \begin{pmatrix} \psi_R \\ -\psi_L \end{pmatrix}$

In the Dirac basis, used in your question, $\gamma^0$ is diagonal (and $\gamma^5$ is non-diagonal).

$\gamma^0 = \begin{pmatrix} I & 0 \\ 0 & -I \end{pmatrix}$

$\gamma^0$ is the parity operator (in all basis). So the Dirac spinor is split into two parts in the Dirac basis, one with even parity and one odd parity.

So to answer the question, the 3rd and 4th components of $\psi_{move}$ represent the part of the spinor with odd parity in this solution.

In your question you hint that perhaps a moving electron contains a combination of positive and negative energy states. This is not the case. The questions and answers on zitterbewegung discuss this especially when constructing wavepackets from states.