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All quantum fields are operators in QFT. However, the Dirac field operator $\hat{\psi}$ has the following difference with the scalar field operator $\hat{\phi}$:

For the $\hat{\psi}$, it makes sense to define the operation $$M \cdot \hat{\psi}$$ where $M$ is any $4\times 4$ matrix and "$\cdot$" denotes matrix multiplication. Such an operation is not defined for a scalar field $\hat{\phi}$. It makes sense for the Dirac field $\hat{\psi}$ because, in the Fourier mode expansion of $\hat{\psi}$, each creation and annihilation operator is multiplied by $4\times 1$ column matrix called $u$-spinor and $v$-spinor (which are not operators).

This brings me to the following question. How should we think of the mathematical object $\hat{\psi}$ as opposed to $\hat{\phi}$? I agree that both are operators but $\hat{\psi}$ is more than an operator. It is also a matrix. What is the correct way to think about $\hat{\psi}$ so as to distinguish it from $\hat{\phi}$?

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    $\begingroup$ It's a set of 4 operators. It's not any weirder than the fact that in nonrelativistic quantum mechanics, $\hat{\mathbf{r}}$ is not technically an operator on a Hilbert space, but a vector of operators. $\endgroup$
    – knzhou
    Apr 29 at 18:08
  • $\begingroup$ @knzhou Do you mean that like $\hat{\vec r}$ stands for a set of $3$ operators $\hat{x},\hat{y},\hat{z}$ in QM, $\hat{\psi}$ stand for the set of 4 fields $\{\hat{\psi}_{\alpha}\}=\hat{\psi}_{1},\hat{\psi}_{2},\hat{\psi}_{3},\hat{\psi}_{4}$ where $\alpha$ is the spinor index? $\endgroup$ Apr 29 at 18:13

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I agree that both are operators but $\hat{\psi}$ is more than an operator. It is also a matrix. What is the correct way to think about $\hat{\psi}$ so as to distinguish it from $\hat{\phi}$?

For starters, you can think of $\hat \psi$ as a set of operator fields instead of just one operator field (like $\hat \phi$). This is analogous to how a vector is different from a scalar. The set of $\hat\psi$ operators rotate into one another under transformation (like rotations), whereas the $\hat\phi$ just stays as it is.

In addition, the $\hat \phi$ and $\hat \psi$ usually describe particles of different statistics (bosons vs fermions, respectively). This is reflected in the commutation vs anti-commutation properties of the partial creation/annihilation operators.

In terms of path integrals, the "classical" $\psi$ fields are anti-commuting number fields, whereas the classical $\phi$ fields are regular numbers.

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  • $\begingroup$ I know the fermionic, bosonic, commutator-anticommutator stuff. That's not a problem. My question is what sort of mathematical object it is i.e, operator or matrix. What do you mean by set of $\hat{\psi}$ operators? $\endgroup$ Apr 29 at 18:06
  • $\begingroup$ I'm not sure what kind of answer you are looking for other than what I provided. $\endgroup$
    – hft
    Apr 29 at 18:07
  • $\begingroup$ Dear @hft , I'm more interested in the 1st paragraph of your answer. By the "set of 4 operators" do you mean $\{\hat{\psi}_{\alpha}\}=\hat{\psi}_{1},\hat{\psi}_{2},\hat{\psi}_{3},\hat{\psi}_{4}$ where $\alpha$ is the spinor index, and by "transformation" do you mean Lorentz transformation? I am trying to verify if I really understand this. $\endgroup$ Apr 29 at 18:22
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    $\begingroup$ Basically, yes. You can see this because the Dirac matrices $\gamma^\mu$ are 4x4 matrices. E.g., $\gamma^0$ is a 4x4 matrix, $\gamma^1$ is a 4x4 matrix, etc. $\endgroup$
    – hft
    Apr 29 at 18:29
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To add to what hft said, $\hat \psi$ is valued in a tensor product of operator densities and the spinor bundle. If you want to get an operator, you need to pair it with a smooth section of the dual bundle and integrate. In flat space, that can be a constant section times a bump function, which corresponds to creating or destroying a particle with a polarized spin at the bump.

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