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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.

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    $\begingroup$ Are you sure the statements in the first half are with respect to the same gamma-matrix representation as in the second half? Anyways, if you choose the chiral, Majorana, Dirac, or any other representation, your statements in terms of components will look very differently. The right way to do this is not to refer to spinor components but to find relativistic-QM representations of operators corresponding to concrete physical measurements and sort the states according to their eigenstates. $\endgroup$ – Void Sep 19 '17 at 8:56
  • $\begingroup$ @Void Thanks for the comment. I would find it odd that the Spinor components have no physical interpretation. In that case, why would one choose that representation? However, if you are right and it depends on specific representations, than it might be difficult to interpret them as general physical properties. Can you point to literature that discusses that? If so, then your comment would actually be the perfect answer to my question. (i.e. "No interpretation because representation-dependent"). $\endgroup$ – Mario Krenn Sep 19 '17 at 16:18
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    $\begingroup$ +1, NiceDrean, I post a related but more general question, physics.stackexchange.com/questions/359904/…, I will come back to answer your question later. $\endgroup$ – wonderich Sep 29 '17 at 4:52
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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 $m\neq 0$.

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    $\begingroup$ thank you for the answer, it is quite useful. Just to be sure, could you please give me an explicit interpretation for the 3rd and 4th component? I think it follows from the first part of your answer, but I am not completely sure. Also do you have a reference for a paper or book which explains this in a bit more detail? Thanks a lot! $\endgroup$ – Mario Krenn Sep 17 '17 at 7:12
  • $\begingroup$ @NiceDean : my answer at lease clarifies how exactly to determine the values of the spin and helicity to which these components correspond, so if this is the thing in which you're interested in (as it seems for me), then the explicit interpretation follows. As for the reference, I'm not sure, but this may be discussed in Peskin's QFT (in the paragraph called like "The solution of the free Dirac equation"). $\endgroup$ – Name YYY Sep 17 '17 at 15:06
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    $\begingroup$ Thank you for the reference, that helps. My main question is "For a moving electron with helicity +1/2 (i.e. ψ1 is at rest), what is the interpretation of non-vanishing third and fourth component?" I see in your answer some useful properties of electrons explained, but not the interpretation of the non-vanishing third and fourth component. This was the reason for starting the bounty. I am sure that for you it is clear - for me it is unfortunatly still not clear. So could be please explicitly answer this question or point to the literature where this is explicitly discussed? thanks so much! $\endgroup$ – Mario Krenn Sep 17 '17 at 21:06
  • $\begingroup$ @NiceDean : the free massive particle is characterized by its 4-momentum, orbital momentum and spin (and, probably, by its internal group charges). For the plane wave spinor (which doesn't have definite orbital momentum) the only way to interpret the non-vanishing 3rd and 4th component is to calculate the corresponding spin projection and/or helicity. So I don't understand what else do you want to clarify. $\endgroup$ – Name YYY Sep 18 '17 at 19:14
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    $\begingroup$ Unfortuantly this does not answer my question. My question is about the interpretation of the non-zero 3rd and 4th component of the Dirac spinor $\psi_{move}(x)$ in the Dirac representation. I will clarify the question and restart the bounty. $\endgroup$ – Mario Krenn Sep 24 '17 at 21:54
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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.

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  • $\begingroup$ Thank you for your answere. It is important what you wrote about the different representations of the Dirac equation. I used in my question the Dirac representation. Then in the third part of your question you give an interpretation of the four possible spinors, but you don't answer my question about the interpretation of the third and fourth component of $\psi_1(x)$. I will clarify the question and restart the bounty. I hope you could then answer directly to my question about the interpretation of the non-zero 3rd and 4th component in of $\psi_1(x)$ in the Dirac representation for p>0. $\endgroup$ – Mario Krenn Sep 24 '17 at 21:52
  • $\begingroup$ It is not clear to me what is it exactly that you are looking for. You see that 1) single components have no representation-invariant meaning, and 2) in the Dirac representation and the pure, sharp-momentum state the "other" components such as the third and fourth component of $\psi_1(x)$ have (up to phase and normalization) the physical meaning of $p_z/(E+m)$ or $(p_x + i p_y)/(E+m)$. $p_x,p_y,p_z,E,$ and $m$ are all hard physical observables and give you a direct meaning of the component with no wiggle space for any other interpretation. $\endgroup$ – Void Sep 25 '17 at 12:32
  • $\begingroup$ I.e., at a given point in momentum space and in a given representation the meaning and interpretation of the components is absolutely and positively fixed. But there is no universal meaning of a single component across the momentum space and representations. $\endgroup$ – Void Sep 25 '17 at 12:33
  • $\begingroup$ As I mentioned, I'm interested what these components mean in the Dirac representation. Of course, they have some physical meaning there -- but I do not see it yet. (I think the interpretation should be clear from the derivation of the Dirac equation in that representation). $\endgroup$ – Mario Krenn Sep 25 '17 at 12:49
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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.

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