Why is the expectation value of the quantum field in the vacuum state $$\phi_c(x)=\langle0|\hat{\phi}(x)|0\rangle_J=\frac{\delta W}{\delta J}$$ referred to as the classical field?
Why not the expectation $\langle\psi|\hat{\phi}(x)|\psi\rangle$, calculated in some other quantum state $|\psi\rangle$ (such as a $N$-particle state or in an arbitrary superposition of Fock states) be interpreted as the classical field?
Asked
Active
Viewed 629 times
6
SRS
- 26,333
-
2Got a source for this claim? I ask because it it my understanding that for any theory symmetric in $\hat{\phi} \rightarrow -\hat{\phi}$ the expectation in the first part vanishes identically. – Sean E. Lake Jun 06 '17 at 06:25
-
2@SeanE.Lake Yes it does vanish (or becomes a constant if the symmetry is spontaneously broken). But only when you set $J=0$. Before you set $J=0$, it is in general, a function of spacetime. You can can look at Peskin and Schroeder, or Ryder's book on QFT for example. – SRS Jun 06 '17 at 06:27
-
1possible duplicate: In what sense, the effective action $\Gamma[\phi_c]$ is quantum-corrected classical action $S[\phi]$?. – AccidentalFourierTransform Aug 07 '17 at 15:24
-
Related: https://physics.stackexchange.com/q/582964/226902 if we have the quantum Klein Gordon field, how can we define a classical KG field that satisfies the classical KG equation? See also this: https://physics.stackexchange.com/q/276599/226902 – Quillo Feb 20 '21 at 18:53
1 Answers
2
Unless stated otherwise, in quantum field theory we almost always assume that the system is in thermal equilibrium at zero temperature. This is usually an excellent approximation to the real world, because the characteristic temperature scale for elementary particle physics is the Hagedorn temperature of $\sim 10^{12} \text{ K}$, and almost all of the universe is effective at zero temperature relative to that scale. The zero-temperature thermal density matrix is just $\rho = | 0 \rangle \langle 0 |$ where $| 0 \rangle$ is the ground state, so thermal expectation values $\langle O(x) \rangle := \text{Tr } (\rho\, O(x)) = \langle 0 | O(x) | 0 \rangle$ of any field $O(x)$ are just given by the ground-state expectation values.
Occasionally people assume finite temperature instead (e.g. to describe the early universe moments after the Big Bang, or the very hot quark-gluon plasmas produced at heavy-ion colliders). People rarely consider expectation values with respect to highly excited pure states though, because there is no physically realistic way to get a real system into such a state. (One caveat: the "eigenstate thermalization hypothesis" suggests that for many realistic systems, thermal states at finite temperature are locally indistinguishable from energy eigenstates with finite energy density, so in that context people sometimes consider expectation values with respect to highly excited energy eigenstates.)
Also, note that in the formalism of second quantization, all states are created by applying creation operators to the ground state, so an expectation value with respect to any pure state can be equivalently expresses as a ground-state expectation value. E.g. if $|\psi\rangle = a^\dagger(x)^N | 0 \rangle$ is an $N$-particle Fock state, then $\langle \psi | O(x) | \psi\rangle = \langle 0 | a(x)^N\, O(x)\, a^\dagger(x)^N | 0 \rangle$ is equivalent to the vacuum expectation value (VEV) of the field $a(x)^N\, O(x)\, a^\dagger(x)^N$. So at least for pure states, there's no loss of generality in only considering VEVs.
tparker
- 47,418