# What is the correct way of looking at the Dirac field?

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 $$\hat{\phi}$$?

• 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. Commented Apr 29, 2022 at 18:08
• @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? Commented Apr 29, 2022 at 18:13

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.

• 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? Commented Apr 29, 2022 at 18:06
• I'm not sure what kind of answer you are looking for other than what I provided.
– hft
Commented Apr 29, 2022 at 18:07
• 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. Commented Apr 29, 2022 at 18:22
• 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.
– hft
Commented Apr 29, 2022 at 18:29

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.