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

• I
• Ryder Rude
In summary: In that case, you can write$$|\phi\rangle=\int d^3x\, c_{\mu}({\bf x}) \psi({\bf x}) |0\rangle$$which is the same thing.The 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\
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

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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.

protonsarecool and vanhees71
Ryder Rude said:
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##.

HomogenousCow said:
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?

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Demystifier said:
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.

Ryder Rude said:
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.

## 1. What is the significance of interpreting ##A^{\mu}(x)|0\rangle## and ##\psi (x) |0\rangle## in quantum field theory?

The operators ##A^{\mu}(x)## and ##\psi (x)## represent the quantum fields for particles and antiparticles, respectively, in quantum field theory. Interpreting them in the context of the vacuum state, represented by the ket vector ##|0\rangle##, allows us to understand the behavior of these fields and their interactions with each other.

## 2. How does the interpretation of ##A^{\mu}(x)|0\rangle## and ##\psi (x) |0\rangle## relate to the concept of virtual particles?

In quantum field theory, virtual particles are represented by fluctuations in the quantum fields. The operators ##A^{\mu}(x)## and ##\psi (x)## act on the vacuum state to create and annihilate these virtual particles, allowing us to understand their behavior and effects on physical particles.

## 3. Can the interpretation of ##A^{\mu}(x)|0\rangle## and ##\psi (x) |0\rangle## be extended to interacting quantum field theories?

Yes, the interpretation of these operators can be extended to interacting quantum field theories by using perturbation theory. In this approach, the operators are expanded as a series of interactions and the vacuum state is used as a reference point for calculating the effects of these interactions.

## 4. How does the interpretation of ##A^{\mu}(x)|0\rangle## and ##\psi (x) |0\rangle## differ from that of single-particle quantum mechanics?

In single-particle quantum mechanics, the state of a particle is described by a wave function, while in quantum field theory, the state is described by a quantum field. This allows for the creation and annihilation of particles, as well as the concept of virtual particles, which are not present in single-particle quantum mechanics.

## 5. What are the implications of the interpretation of ##A^{\mu}(x)|0\rangle## and ##\psi (x) |0\rangle## for understanding the behavior of particles at high energies?

The interpretation of these operators is crucial for understanding the behavior of particles at high energies, as it allows us to describe and predict the interactions between particles and their antiparticles. This is especially important in the study of particle collisions and the creation of new particles in high-energy experiments, such as those conducted at the Large Hadron Collider.

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