When doing canonical quantization for the Dirac field, one typically begins by expanding the field into the following form: $$\psi(x) = \int\frac{d^3\vec{p}}{\sqrt{(2\pi)^32E_\vec{p}}}\sum_{s} \left\{a(\vec{p},s)u(\vec{p},s)e^{-ipx}+ b^\dagger(\vec{p},s)v(\vec{p},s)e^{ipx} \right\}, $$ where $u(\vec{p},s)e^{-ipx}$ and $v(\vec{p},s)e^{ipx}$ are positive and negative-energy plane wave solutions to the Dirac equation that are also helicity eigen-spinors with eigenvalue $s$. By imposing equal-time anti-commutation relation for the field operator and the canonical momentum, one arrives at fermionic anti-commutation relations satisfied by $a(\vec{p},s)$, $b(\vec{p},s)$ and their Hermitian conjugates, where they acquire physical meaning as creation and annihilation operators.

After introducing some wise definition of what vacuum means, one can write down the expression for the Hamiltonian, momentum and charge of the field in terms of these creation and annihilation operators. For instance, the expression for momentum reads $$P^\mu = \int d^3\vec{p} \sum_{s}p^\mu[a^\dagger(\vec{p},s) a(\vec{p},s) + b^\dagger(\vec{p},s) b(\vec{p},s)].$$

However, when it comes to angular momentum, textbooks seem to be only willing to give it in terms of $\psi(x)$, i.e., $$M_{\mu\nu} = \int d^3x~\psi^\dagger(x)\left[i(x_\mu\partial_\nu-x_\nu\partial_\mu)+\frac{1}{2}\sigma_{\mu\nu}\right]\psi(x),$$ where $\sigma_{\mu\nu}$ is the set of matrices that control the behavior of the Dirac spinor under Lorentz transformation. I know that the above expression is a special case of a more general formula that come from Noether's theorem, and I am wondering if there is a neat expression for $M_{\mu\nu}$ in terms of $a(\vec{p},s)$, $b(\vec{p},s)$ and their Hermitian conjugates. I tried to substitute in the expansion of $\psi(x)$, but it did not lead me very far. In particular, since there is a derivative operator sandwiched between the spinors $u$ and $v$, I was not able to apply orthogonal relations, and cannot see how to proceed.

So my question is: Is there a neat expression for the total angular momentum $M_{\mu\nu}$ of the free Dirac field $\psi(x)$ in terms of creation and annihilation operators $a(\vec{p},s),~a^\dagger(\vec{p},s),~b(\vec{p},s),b^\dagger(\vec{p},s)$? If there is such an expression, then how can I arrive at it? Thanks for your help!

p.s. There is a related, but not identical question on this site, and it does not seem to have acquired a satisfactory answer.

UPDATE: For those who might be interested, here is what I have found following @Numrok 's answer: $$M_{\mu\nu} = \int d^3\vec{p}\sum_{s,s'}\left\{ \frac{a^\dagger(\vec{p},s')u^\dagger(\vec{p},s')}{\sqrt{2E_\vec{p}}} \left[i\left(\frac{\partial}{\partial p^\nu}p_\mu-\frac{\partial}{\partial p^\mu}p_\nu\right)+\frac{1}{2}\sigma_{\mu\nu}\right] \frac{a(\vec{p},s)u(\vec{p},s)}{\sqrt{2E_\vec{p}}} +\frac{b^\dagger(\vec{p},s')v^T(\vec{p},s')}{\sqrt{2E_\vec{p}}}\left[i\left(p_\mu\frac{\partial}{\partial p^\nu}-p_\nu\frac{\partial}{\partial p^\mu}\right)-\frac{1}{2}\sigma^T_{\mu\nu}\right]\frac{b(\vec{p},s)v^*(\vec{p},s)}{\sqrt{2E_\vec{p}}} \right\}.$$ Where ${\partial}/{\partial p^0}$ should be taken as zero, and all ${\partial}/{\partial p^i}$ with $i$ being a spatial index will act on any $p^0 = E_\vec{p}$ to the right of it. (I am using the $+,-,-,-$ signature.) Not sure if this is the simplest form because it still seems a bit complicated to me. I hope people won't hesitate to comment if they have any further idea or information.


1 Answer 1



  1. Plug in the first equation (expansion of $\psi(x)$) into the angular momentum integral. Be careful to label the spin on the ladder operators ($a$, $b$) and basis vector ($u$, $v$).
  2. Apply the spatial derivatives to the exponential terms.
  3. Use commutation relations of the ladder operators for the Dirac field.
  4. Perform one of the integrals to get rid of the resulting delta-functions.
  5. Use the completeness and orthonormality relations of the basis vectors, potentially the Dirac equation itself.

These hints should guide you through the conceptual steps, the rest is (hopefully) a simple calculation exercise.

  • $\begingroup$ Thanks! I did not realize that one can use anti-commutation relations of the ladder operators before performing any integral. Besides this, the vanishing of the terms involving both $a$ and $b$, or both $a^\dagger$ and $b^\dagger$, is rather non-trivial to me and I had to apply transformation properties of Dirac spinors to see this. $\endgroup$
    – Kaius
    Aug 4, 2016 at 0:12
  • 1
    $\begingroup$ My result is posted at the end of my original question. $\endgroup$
    – Kaius
    Aug 4, 2016 at 0:47

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