What does it mean to act a Dirac field operator on the vacuum?

Question

The usual interpretation from my QFT courses is that when acting the scalar field operator onto the vacuum, we create particle:

$$ |x\rangle = \phi(x)|0\rangle. $$

If I have a multi-component field operator such as a Dirac spinor field, what does it mean to act that on the vacuum? The Dirac spinor field is expanded in terms of creation/annihilation operators with spinor valued coefficients,

$$ \psi(x) = \sum_{s=1}^2 \int \frac{d^3p}{(2\pi)^3}\bigg[b^s_pu^s(p)e^{-ip\cdot x}+c_p^{s \dagger}v^s(p)e^{-ip\cdot x} \bigg] $$

where $u^s(p)$ and $v^s(p)$ are 4-component spinors and are not operators on a Hilbert space. What is $ \psi(x) |0\rangle $ and what does it mean to multiply a ket vector by a 4-component spinor?

Answer 1

Indeed, you will get a collection of four states. But this is neither unusual or surprising.

Even in ordinary quantum mechanics, we have "vector operators" such as $\mathbf{p} = - i \nabla$, which have multiple components. When a vector operator acts on a quantum state, it produces a vector of three states, which formally lives in $\mathcal{H} \oplus \mathcal{H} \oplus \mathcal{H}$. In terms of group theory, if you start with a set of states with transform in a spin $s$ representation of the rotation group, and act with $\mathbf{p}$, you will get a set of states transforming in the representation $$\text{spin } s \, \otimes \text{spin } 1$$ by the Wigner-Eckart theorem. The fact that there are three components is necessary to make the group theory work out. For example, if we acted on a spin zero ground state, we should reach a representation of spin one, but we need three states for that; they are the three vector components.

Given that, you might wonder why we never talk about vectors of three states in undergrad quantum mechanics. The answer is that the Hamiltonian is a scalar, so the vector index in $\mathbf{p}$ has to be contracted with something. For example, in the free Hamiltonian it's $p^2$, while for an interaction with light it would be $\mathbf{p} \cdot \mathbf{A}$. Explicitly, we have $$\langle \phi | \mathbf{p} \cdot \mathbf{A} | \psi \rangle = \begin{pmatrix} \langle \phi| p_x, & \langle \phi| p_y, & \langle \phi | p_z \end{pmatrix} \begin{pmatrix} A_x |\psi \rangle \\ A_y |\psi \rangle \\ A_z |\psi \rangle \end{pmatrix} = \sum_i \langle \phi | p_i A_i | \psi \rangle.$$ This matrix element essentially gives the total transition amplitude between the initial and final states by emitting a photon in one of three channels.

The Dirac spinor field roughly does the same thing, but for spin $1/2$. Again, the Hamiltonian has to be a scalar, so we need to contract the spinor indices; that's why we focus so much on Dirac bilinears. The quantity $\bar{\psi} \psi$ is just analogous to $\mathbf{A} \cdot \mathbf{A}$. When we compute Feynman diagrams, the $\psi$'s either end up contracted with an external spinor, which means we are just picking out the state with the appropriate spin, or they end up contracted with each other in a fermion loop, in which case we need to sum over the four polarizations just like we summed over $i$ in $p_i A_i$.

Answer 2

I am not an authority in this area and I don't think what I have to say is particularly deep. Just a few thoughts I had upon reading your question.

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In all that follows I'll be referring to free fields.

The real scalar field has a unique spatial representation because there is only one polarization and one charge, which can be seen by realizing $|\vec{p}\rangle$ uniquely labels one-particle states (Fourier-transforming gives $|\vec{x}\rangle$ with a phase-factor).

The complex scalar field does not have a unique spatial representation because, although there is only one polarization, there are two charges (which we distinguish in canonical quantization by introducing separate creation and annihilation operators $\hat{a}/\hat{a}^{\dagger}$ and $\hat{b}/\hat{b}^{\dagger}$). This can be seen doubly by again realizing that the correct one-particle states need to also have a species label, say $|\vec{p},\pm\rangle$. The spatial representation must then also carry such a label, i.e. $|\vec{x},\pm\rangle$.

You can see how this logic will carry over to the spinor case. A Dirac field has two charges and, for each charge, two polarization states.

So when you ask:

"what does it mean to multiply a ket vector by a 4-component spinor?"

it means that you will get a four-component object out in the end, namely:

$$\begin{align} \psi(x)|0\rangle&=\sum_{s=1}^2 \int \frac{d^3p}{(2\pi)^3}\bigg[\require{cancel}\cancel{(b^s_p|0\rangle)}u^s(p)e^{-ip\cdot x}+(c_p^{s \dagger}|0\rangle)v^s(p)e^{-ip\cdot x} \bigg]\\ &=\sum_{s=1}^2 \int \frac{d^3p}{(2\pi)^3}\bigg[v^s(p)e^{-ip\cdot x} \bigg]|\vec{p},s,+\rangle \\ \end{align}$$

and now you can ask for its projection onto $|\vec{x},s,+\rangle$, or also less usefully but more straightforwardly $|\vec{x},i\rangle$ where $i$ is one of the components of the Dirac field, etc.. How these states and their projections along $|\vec{p},s,\pm\rangle$ are actually defined, I don't know, but I'm sure somebody here does. Nevertheless, you can see that we need to incorporate the different polarizations and charges $(s,\pm)$ in our definitions of state-projections, in contrast with the real scalar field where we simply had:

$$<\vec{x}|\vec{k}>=e^{-i\vec{k}\cdot\vec{x}}$$

If I had to guess a consistent way to define $\langle \vec{x},s',r'|\vec{p},s,r\rangle$, it would be simply:

$$\langle \vec{x},s',r'|\vec{p},s,r\rangle=\delta_{s,s'}\delta_{r,r'} e^{-i\vec{k}\cdot\vec{x}}$$