Why don't people discuss the eigenstate of the field operator? For example, the real scalar field the field operator is Hermitian, so its eigenstate is an observable quantity.
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$\begingroup$ These states are called coherent states, and you will find a discussion in any decent introductory text book on QFT. $\endgroup$– ThomasCommented Apr 21, 2014 at 14:03
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8$\begingroup$ @Thomas coherent state is the eigenstate of the annihilation operator. This is not the eigenstate of the field operator. $\endgroup$– 346699Commented Apr 21, 2014 at 14:16
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2$\begingroup$ Same thing. Formally $|\phi\rangle=\exp(\int dx \phi(x)\hat{\phi}(x))|0\rangle$. This is obviously analogous to $|z\rangle = \exp(za^\dagger)|0\rangle$. $\endgroup$– ThomasCommented Apr 21, 2014 at 16:36
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$\begingroup$ @Thomas Nogueira below asserts that coherent states are not eigenstates of the field operator. Is he correct? $\endgroup$– alexchandelCommented Jun 11, 2019 at 23:57
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$\begingroup$ see also physics.stackexchange.com/a/383473/84967 $\endgroup$– AccidentalFourierTransformCommented Jul 17, 2019 at 13:55
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3 Answers
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The answer is simple. They are not realized in nature too often. And more, they are not stationary states, so such a state evolves in time to a state that contains fluctuations of the field variable $\phi$ over all the space. The eigenstate of $\hat{\phi}$ evolves to a non-eigenstate of $\hat{\phi}$. These fluctuations increase and spread with time. To obtain this state, you need to measure $\phi$ over the space with considerable precision about Compton's field wavelength. We know that in this scale, fluctuations on the field start as the evolution through time begins.
The fields that we probe in classical mechanics are the coherent state: $$ |\phi_{cl}\rangle=\exp\left(\int \phi_{cl}(x)\hat{\phi}(x)\right)|0\rangle. $$ This state is not an eigenstate of $\hat{\phi}$, so the problems of the first paragraph can be avoided. This state has minimal fluctuations, and the uncertainty is constant in time. Therefore, the expectation value of the field operator is: $$ \langle\hat{\phi}(x)\rangle=\langle \phi_{cl}|\hat{\phi}(x)|\phi_{cl}\rangle=\phi_{cl}(x). $$ You can check this by the definition of $|\phi_{cl}\rangle$.
Note that the vacuum state $|0\rangle$ is also a coherent state associated with the trivial classical solution $\phi_{cl}(x)=0$.
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1$\begingroup$ Why the down vote? Can someone point out the reason. This will be of great help in improving my answer and my understanding on the subject? $\endgroup$– NogueiraCommented Feb 1, 2018 at 19:29
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$\begingroup$ It is misleading to say that the eigenstates are not realized in nature too often. The path integral requires the existence of complete eigenstates of the field operator for its derivation, so whether or not this means they are realized in nature, they are important. However it does seem unlikely (impossible) to find a physical particle in a perfect field eigenstate $|\phi\rangle$. The second issue you bring up is that the field is not stationary. This is not a problem. It is easy to relate the eigenvectors of the field at time 0 to the field at time t (they differ by e^itH). $\endgroup$ Commented Nov 28, 2018 at 9:50
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$\begingroup$ To add: the same holds in QM: It is unlikely if not impossible to find a perfect eigenstate $|x\rangle$ (barring measurement postulates, which most physicists do not take to be exact). And in the heisenberg picture, the eigenstates evolve in time exactly the same way that the field eigenstates do (e^itH) $\endgroup$ Commented Nov 28, 2018 at 9:52
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1$\begingroup$ @doublefelix This states are not realized in nature in the sense that after measurements and evolution this state is never achieved. But of corse, by the superposition principle, I can use this states as basis to represent others states. The path integral formulation for fields does exactly that. What I said is that the configuration of a classical fields is described at the quantum level by coherent states and not eigenstates of the fields, because of the fluctuations being not stationary. $\endgroup$– NogueiraCommented Nov 28, 2018 at 12:28
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$\begingroup$ Some states may not be realized in nature to often but nevertheles be useful as a set of basis to represent others state. This does not mean that these states are realized in nature, except if you like the many-world interpretation of quantum mechanics, that is garbage by the way. $\endgroup$– NogueiraCommented Nov 28, 2018 at 12:31
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As $\phi(f)$ and $\pi(f)$, which are self adjoint, satisfy the same commutation relations as $X$ and $P$, the closure of the space generated by polynomials of the former pair of operators applied to $\lvert 0\rangle$ is isomorphic to $L^2(\mathbb R)$. Therefore the spectrum of $\phi(f)$ and $\pi(f)$, is purely continuous and coincides to $\mathbb R$ and there are no proper eigenvectors, but they are just formal ones and isomorphic to $\lvert x\rangle$ and $\lvert p\rangle$.
answered Apr 21, 2014 at 14:05
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One thing missing in the other answers... people actually do discuss eigenstates of the field operator, or at least they are important in QFT. A complete set of field eigenstates are used to prove that n-point functions can be written in terms of a path integral, which is a critical result. But they are not used as "states after a field measurement", as $|x\rangle$ and $|p\rangle$ are used in quantum mechanics. At least not in mainstream applications / that I know of.
In case you're curious to know how exactly they are used, I'll sketch it out below, with the real scalar field as an example. In deriving the path integral, it is necessary to write the identity in a basis of field eigenstates
$$1 = \int \mathcal{D}\phi \, |\phi_t(\vec{x})\rangle \langle \phi_t(\vec{x}) |$$ with (operators have hats, numbers do not) $$\hat{\phi}(t, \vec{x})|\phi_t(\vec{x})\rangle = \phi_t(\vec{x})|\phi_t(\vec{x})\rangle \,\,\,\,\,\,\,\,\, \forall \vec{x}$$
So at a given time $t$, each eigenvector $|\phi_t(\vec{x})\rangle$ is a simultaneous eigenstate of all field operators of different $\vec{x}$ but equal $t$. The eigenvalue $\phi_t(\vec{x})$ depends on which $\vec{x}$ is chosen in the field argument, so we write it as a function of $\vec{x}$. This defines a classical field (a map from $\mathbb{R}^3\to \mathbb{R})$. Operators can be simultaneously diagonalized if they commute*, and luckily the usual QFT commutation relations give exactly this for equal times:
$$[ \phi(t,\vec{x}), \phi(t, \vec{y})] = 0$$
The diagonalization process can be repeated for any time because the commutation relation above holds for any $t$.
*I think that strictly there are more things to worry about for infinite-dimensional operators, so you might want to take the commutation relations as an indication that they can be simultaneously diagonalized, rather than a proof. I don't know enough to expand on this.
answered Jul 17, 2019 at 13:10