This question pertains to some notation in Zee's QFT book, Section II.2. The Dirac equation is

$$ (i\gamma^\mu\partial_\mu-m)\psi(x)=0, $$

which we can write in momentum space with the Fourier transform

$$ \psi(x)=\int\!\frac{d^4p}{(2\pi)^4}e^{-ipx}\psi(p)$$


\begin{align} (i\gamma^\mu\partial_\mu-m)\int\!\frac{d^4p}{(2\pi)^4}e^{-ipx}\psi(p)&=0\\ \int\!\frac{d^4p}{(2\pi)^4}(\gamma^\mu p_\mu-m)e^{-ipx}\psi(p)&=0\\[8pt] \implies(\gamma^\mu p_\mu-m)\psi(p)&=0, \end{align}

which is the Dirac equation in momentum space. In the rest frame, $p_i=0$ and $p_0=m$ so we have

$$ (\gamma^\mu p_\mu-m)\psi(p)=(\gamma^0p_0+\gamma^ip_i-m)\psi(p) =0\quad\implies\quad(\gamma^0-1)\psi(p)=0.$$ For two Dirac bispinors $u$ and $v$, the solutions are

$$\psi(p)=u(p,s)e^{-ipx}~~,\quad\text{and}\qquad \psi(p)=v(p,s)e^{ipx}.$$

We define $\bar u=u^\dagger\gamma^0$ and $\bar v=v^\dagger\gamma^0$, with dagger denoting the conjugate transpose, to form Lorentz invariants $\bar uu$ and $\bar vv$. Now I have gotten to my question. Zee gives a rest frame identity

$$ \sum_s u_\alpha(p,s)\bar u_\beta(p,s)=\left( \begin{matrix} 1&0&0&0\\0&1&0&0\\0&0&0&0\\0&0&0&0\end{matrix} \right)_{\!\alpha\beta}. $$

If it's not the spin $s$, which it is clearly not, then what do the $\alpha,\beta$ subscripts refer to? Is it two discrete positions $\alpha$ and $\beta$? What is the meaning of the subscript $\alpha\beta$ on the matrix? Does that turn it into a Kronecker matrix?


2 Answers 2


The expression in a general frame is $$ \frac 1{2m}(m+p_\mu \gamma^\mu)= \sum_{s=\pm} u(p,s)\bar u(p,s) $$ where $$ u(p,s)= \left[\matrix{u(p,s)_1\cr u(p,s)_2\cr u(p,s)_3\cr u(p,s)_4}\right], \\ \bar u(p,s)=[ \bar u(p,s)_1, \bar u(p,s)_2, \bar u(p,s)_3, \bar u(p,s)_4 ], $$ so $ u \bar u$ is a 4-by-4 matrix just like the $\gamma$'s.

  • $\begingroup$ What does the $\alpha\beta$ subscript on the matrix mean? This was the major thrust of my question, thanks. $\endgroup$ Nov 14, 2020 at 22:52
  • $\begingroup$ It means that the $\alpha, \beta$ entry in the 4-by4 matrix $u\bar u$ is the $\alpha,\beta$ entry in the 4-by-4matrix (m+\gamma^\mu p_\mu)/2m$ $\endgroup$
    – mike stone
    Nov 15, 2020 at 16:06
  • $\begingroup$ So that notation is completely irrelevant and redundant because we already know that's true for two matrices that are equal to each other? $\endgroup$ Nov 15, 2020 at 18:23

What do the $\alpha,\beta$ subscripts refer to?

On $u_\alpha$ and $\bar u_\beta$ they pick out one of the four components of these Dirac bispinors.

What is the meaning of the subscript $\alpha\beta$ on the matrix?

The $\alpha$ picks out one of the four rows and the $\beta$ picks out one of the four columns, so together they pick out an element of the matrix.

Does that turn it into a Kronecker matrix?

No. A Kronecker delta has four, not two, ones on the diagonal.

  • $\begingroup$ You wrote, "they pick out one of the four components of these Dirac bispinors." Since we are in the rest frame, I believe these only have two available components each. I thought $s$ tells us which one we use. Don't the two rest frame degree of freedom correspond to the two degrees of freedom: spin up and spin down? This is exactly the issue which was confusing me. Thanks! $\endgroup$ Nov 14, 2020 at 17:34
  • $\begingroup$ A Dirac bispinor has four components in every frame. Some may be zero. $\endgroup$
    – G. Smith
    Nov 14, 2020 at 17:47
  • $\begingroup$ $s$ doesn’t pick out a component of the bispinor. It specifies which of the two possible bispinors for a particular momentum you’re talking about. $\endgroup$
    – G. Smith
    Nov 14, 2020 at 17:51
  • $\begingroup$ Yes, G. Smith. Ones that are zero are not "available." Furthermore, $s$ certainly does pick out the components. If we have $u(p,s)=\left(\begin{matrix}u_1\\u_2\\0\\0\end{matrix}\right)$ then $u(p,s_1)=\left(\begin{matrix}u_1\\0\\0\\0\end{matrix}\right)$ and $u(p,s_2):=\left(\begin{matrix}0\\u_2\\0\\0\end{matrix}\right)$. I think these are called "eigenspinors." Would you please edit your answer? I am sure you are wrong. $\endgroup$ Nov 14, 2020 at 20:07
  • $\begingroup$ That is the wrong way to think about $s$ because it doesn’t work like that in an arbitrary frame. See Wikipedia. The notation is for an arbitrary Lorentz frame. $\endgroup$
    – G. Smith
    Nov 14, 2020 at 21:46

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