About the familiar formula $Q=I_3 + Y/2$. As an example, the left-handed electron has charge $Q=-1$, weak isospin $I_3 = -1/2$, and weak hypercharge $Y=-1$.

The electron is represented in the Dirac equation by bispinors. For example, Peskin and Schroeder's classic textbook "An Introduction to Quantum Field Theory" uses the Weyl or chiral representation where (see P&S equation 3.25): \begin{equation} \gamma^0 = \left(\begin{array}{cccc}0&0&1&0\\0&0&0&1\\1&0&0&0\\0&1&0&0\end{array}\right). \end{equation} The charge operator $Q$ is proportional to $\gamma^0$, so that, for example, particles are eigenstates of this operator with eigenvalue $+1$ while the antiparticles take eigenvalue $-1$. See P&S equation 3.47 for the particle (positive frequency) plane wave solution to the Dirac equation.

On the other hand, the operator for handedness in this representation is (see P&S equation 3.72): \begin{equation} \gamma^5 = \left(\begin{array}{cccc}-1&0&0&0\\0&-1&0&0\\0&0&1&0\\0&0&0&1\end{array}\right). \end{equation} and this anticommutes with $\gamma^0$.

In choosing a "complete of commuting observables" for the Dirac spinor we have the choice of "spin and charge" or "spin and handedness". We cannot use "charge and handedness" as these correspond to operators that do not commute.

P&S use the example of a spin-up and a spin-down state and give them a large boost to obtain (approximately) a pure right handed and pure left handed state in equations 3.52 and 3.53. You can see that before boosting they are eigenstates of charge and afterwards are eigenstates of $\gamma^5$, i.e. P&S equation 3.53: \begin{equation} u(p) = \left(\begin{array}{c} \sqrt{E-p^3\left(\begin{array}{c}0\\1\end{array}\right)}\\ \sqrt{E+p^3\left(\begin{array}{c}0\\1\end{array}\right)} \end{array}\right) \rightarrow \sqrt{2E}\left(\begin{array}{c}0\\1\\0\\0\end{array}\right). \end{equation} The above shows a spin-down electron being boosted to a left-handed electron.

Now boosting itself does not change the charge. But when the particle is no longer stationary it is no longer an eigenstate of charge. If you work it out you find out that while the (small) boosted states still have charge -1 in the sense that the average of charge measurement is $<Q> = -1$, a single measurement of charge will give numbers that go to $\pm \infty$. And of course using the pure left handed state on the right hand side of the above will give $<Q>=0$.

Another way of explaining the difficulty is to replace the spin-down electron with a spin-down positron. The result of a "large" boost is the same as the positron: \begin{equation} \bar{u}(p) = \left(\begin{array}{c} \sqrt{E-p^3\left(\begin{array}{c}0\\1\end{array}\right)}\\ -\sqrt{E+p^3\left(\begin{array}{c}0\\1\end{array}\right)} \end{array}\right) \rightarrow \sqrt{2E}\left(\begin{array}{c}0\\1\\0\\0\end{array}\right). \end{equation} In short, there are no states that have good quantum numbers for both charge and handedness so there are no states that have good quantum numbers for all three of charge, weak hypercharge and weak isospin. Then why can we find tables for these three quantum numbers?

Now this is essentially a quantum mechanical analysis of the relationship between charge, weak hypercharge and weak isospin. Is it that this problem goes away in a field theory analysis? And what exactly is the reason for the difference?

  • $\begingroup$ related to this, I am amazed that one seems to define Q over a chiral, Weyl, spinor, but that Nature only offers us Dirac, massive spinors. $\endgroup$ – arivero May 18 at 12:02
  • 2
    $\begingroup$ @arivero the joke is on you: all Standard Model spinors are Weyl. $\endgroup$ – Prof. Legolasov May 18 at 15:10
  • 1
    $\begingroup$ The operators $Q$, $I_3$ and $Y$ are defined on the 2-nd quantized Hilbert space of the QFT, not on the finite dimensional Dirac spinor representation. Have you tried writing them and computing their commutator? It should vanish for $I_3$ and $Y$ (and hence also for $Q$, which is a linear combination). $\endgroup$ – Prof. Legolasov May 18 at 15:15
  • $\begingroup$ Since $(1\pm \gamma^5)/2$ is the projection operator for left and right handedness, you can make any operator $K$ that depends on handedness by $K=k_L(1-\gamma^5)/2 + k_R(1+\gamma^5)/2$ where $k_L, k_R$ are the left and right values. Similarly for charge but with $\gamma^0$. So of course $I_3$ and $Y$ commute with $\gamma^5$ and similarly $Q$ commutes with $\gamma^0$. That you can do this is because $1$ also commutes with $\gamma^5$ and similarly $1$ commutes with $\gamma^0$. $\endgroup$ – Carl Brannen May 18 at 18:55
  • $\begingroup$ Related question: can a Weyl spinor emit, or absorb, a photon? $\endgroup$ – arivero May 20 at 6:04

For a quantum field, and comparably for a Dirac spinor, representing the electron, $$ \psi= \frac{1+\gamma^5}{2}\psi + \frac{1-\gamma^5}{2}\psi . $$

So, then, $$ I_3=-\frac{1}{2} \frac{1-\gamma^5}{2}~~, $$ (while $\frac{1}{2} \frac{1-\gamma^5}{2}$ for the neutrino), $$ Y= -1 \frac{1-\gamma^5}{2} - 2\frac{1+\gamma^5}{2} ~~, $$ whence $$ Q= I_3+Y/2= -(1/2+1/2) \frac{1-\gamma^5}{2} - 1\frac{1+\gamma^5}{2}= - 1\!\! 1. $$ So you may check the respective eigenvalues on $\psi$.

All three mutually commute and are Lorentz invariant.

| cite | improve this answer | |
  • $\begingroup$ I like this answer and am planning on "accepting" it in a few weeks. The problem of course, is that this definition of $Q$ is not "electric charge" in the sense of the conserved thingy that makes galvanometers move around. That is, this answer's $Q$ is an operator that assigns $Q=-1$ to the positron as well as the electron. And in pair production, this $Q$ increases by 2, one for the electron and one for the positron so this is an example of how it is not conserved. $\endgroup$ – Carl Brannen May 18 at 20:57
  • $\begingroup$ It assigns charge -1 to an electron and +1 to a hole. If ψ is a quantum field, it assigns charge -1 to a created electron and +1 to a destroyed electron, so a created positron. $\endgroup$ – Cosmas Zachos May 18 at 21:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.