I Interpreting ##A^{\mu}(x)|0\rangle## and ##\psi (x) |0\rangle##

[Ryder Rude]

I can understand how $\phi (x)|0\rangle$ represents the wavefunction of a single boson localised near $x$.I don't understand how the same logic appies to $A^{\mu}(x)|0\rangle$ and $\psi |0\rangle$. Both of these operators return a four component wavefunction when operated on the vaccuum, because of the vector/spinor indices in the expansion of these operators. However, the wavefunction of a photon or an electron is described by a one-component wavefunction as I show below:

A wavefunction of a single electron is written as $\sum_s \int C_s(p) |p, s\rangle dp$. The $s$ label represents the two spin states. A wavefunction of a single photon is written as $\sum_r \int C_r(p) |p,r\rangle dp$, $r$ labels polarisations.

The wavefunction returned by $\psi (x)|0\rangle$ is of the form $\sum_{\alpha} \sum_{s} \int C_{s,\alpha} (p) |p,s\rangle dp$. This has an extra index $\alpha$ which runs from 0 to 3.

$C$ is just a complex number in all of the above

 

[HomogenousCow]

You're right, the state $\psi(x)|0\rangle$ represents a specific superposition of localized fermion states which is exactly why the LSZ theorem for fermions includes factors of $\overline{u}_{s}(k)$ and $\overline{v}_{s}(k)$ to project out your desired states. It's probably better to think of objects like $\langle 0|\overline{\psi}(x)\psi(y)| 0 \rangle$ as the correlation function of the field rather than some transition function between one-particle states.

 

[Demystifier]

However, the wavefunction of a photon or an electron is described by a one-component wavefunction

No it isn't. You should distinguish the state in the Hilbert space from the wave function. The former is a vector so is a single object, the latter is one of components of the vector. Even for zero spin, the vector in the Hilbert space $|\psi\rangle$ has infinitely many components $\psi_x=\psi(x)=\langle x|\psi\rangle$.

 

[Ryder Rude]

You're right, the state $\psi(x)|0\rangle$ represents a specific superposition of localized fermion states which is exactly why the LSZ theorem for fermions includes factors of $\overline{u}_{s}(k)$ and $\overline{v}_{s}(k)$ to project out your desired states. It's probably better to think of objects like $\langle 0|\overline{\psi}(x)\psi(y)| 0 \rangle$ as the correlation function of the field rather than some transition function between one-particle states.

So I should think of $\psi (x) |0\rangle$ as just an absract object with spinor indices rather than a particle at a position $x$, right? And the same logic applies to $A^{\mu} (x)|0\rangle$?

Also, can you please explain your interpretation of this as a correlation function? Why is it called a correlation function? It's measuring correlation between what?

Should I think of $\langle 0| \phi(y) \phi(x) |0\rangle$ also as a correlation function, rather than an inner product between localised particle wavefunctions?

 

[Ryder Rude]

No it isn't. You should distinguish the state in the Hilbert space from the wave function. The former is a vector so is a single object, the latter is one of components of the vector. Even for zero spin, the vector in the Hilbert space $|\psi\rangle$ has infinitely many components $\psi_x=\psi(x)=\langle x|\psi\rangle$.

Okay. But how should I interpret $\psi (x)|0\rangle$ and $A^{\mu} (x) |0\rangle$? Both of these objects have have extra index $\mu$ or $\alpha$, which really shouldn't be there in the state-vector describing a single photon or an electron/positron.

 

[Demystifier]

Okay. But how should I interpret $\psi (x)|0\rangle$ and $A^{\mu} (x) |0\rangle$? Both of these objects have have extra index $\mu$ or $\alpha$, which really shouldn't be there in the state-vector describing a single photon or an electron/positron.

Strictly speaking, in a definition of state you should have both an integration over $x$ and a sum over spin indices. Something like
$$|\psi\rangle=\int d^3x\, c_{\mu}({\bf x}) A^{\mu}({\bf x}) |0\rangle$$
But you can still get a dependence on ${\bf x}$ and ${\mu}$ if $c_{\mu}({\bf x})$ is a Krorencker $\delta$ in the discrete label and Dirac $\delta$ in the continuous one.