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I'm having a hard time making sense of an expression like $$\left[\Psi(x), \bar{\Psi}(y)\right].$$ Up until now I imagined a spinor operator to be something like a column vector of operators, something along the lines of $$\Psi = \left( \matrix{\Psi_1 \\ \Psi_2 \\ \Psi_3 \\ \Psi_4}\right).$$ And in a similar way, the Dirac adjoint would be $$\bar{\Psi} = ( \matrix{\bar{\Psi}_1\, \bar{\Psi}_2\, \bar{\Psi}_3\, \bar{\Psi}_4}).$$ However, when I see terms containing operators, or commutators, all operators are just written down one after another, without any indication of what it means in terms of their matrix structures. For operators with just one "component", this works fine, since one can define something like multiplication without running into too much trouble. However, with a commutator: $$\left[\Psi, \bar{\Psi}\right]= \Psi \bar{\Psi} - \bar{\Psi} \Psi,$$ if I assume the the map between the two consecutive operators to be matrix multiplication, one yields a 4 $\times$ 4 matrix, while the other one yields just a single component.

How should I think of something written down like that? Or does this mean that any equation involving a spinor is just meant to be interpreted component-wise?

Buzz
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Quantumwhisp
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    You mean spinor anticommutator? – Cosmas Zachos Jul 20 '21 at 23:19
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    Yeah, there’s two implicit spinor indices on the right side. Just like how fields depend on position too, but you don’t usually write the two position arguments on the right side, it’s suppressed for brevity. – knzhou Jul 20 '21 at 23:21
  • Linked. – Cosmas Zachos Jul 20 '21 at 23:22
  • @CosmasZachos I just wrote the commutator to give an example, I could pose the same question about the anticommutator – Quantumwhisp Jul 20 '21 at 23:40
  • Why on earth didn't you use vector or isospined boson fields, then? – Cosmas Zachos Jul 21 '21 at 00:25
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    @CosmasZachos We didn't yet do isospined boson fields in the course, and vector fields don't have the same problem, because indices ($\mu$ $\nu$ and so on) are explicitly used all the time.

    I ran into the question when reading the section in peskin & schröder, where they try to use the commutator for quantisation (and afterwards see it doesn't work).

    TBH I don't see a problem here. The question doesn't get better / worse if I ask about the commutator instead of the anticommutator, although the anticommutator is used way more often.

     – Quantumwhisp Jul 21 '21 at 16:47

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When calculating commutators (or, more commonly, anticommutators) of fermion operators like this, you always calculate the (anti-)commutators of particular components of the fields. This makes the results unambiguous.

For example, you may with to calculate $\{\bar{\psi}'_{\alpha},\psi_{\beta}\}$. This is an object with residual $(\alpha,\beta)$ indices, and so it represents a matrix in spinor space. For example, if you calculate something like $\{\psi_{\alpha}^{\dagger}(\vec{x}),\psi_{\beta}(\vec{x}')\}$, you will find the components of a $4\times 4$ matrix that are diagonal in the spinor space, $\delta_{\alpha\beta}\delta^{3}(\vec{x}-\vec{x}')$. [Remember that $\psi_{\alpha}^{\dagger}$ can be further expanded as $\bar{\psi}_{\mu}(\gamma_{0})_{\mu\alpha}$.]

Buzz
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