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I'm trying to make an analogy with the phonon field. While preparing this answer I've learned that for a chain of atom-like entities, we have a probability density of the phonon field configuration: repeated measurements of the displacement of the atoms in a particular eigenstate of Hamiltonian would give something like this:

In coherent states we can see classical-like behavior of the chain of atoms as the evolution of expected value of the field of displacements:

But my attempts at stretching this analogy to fermions have come across some difficulties. Namely, unlike the easy to understand interpretation of the wavefunction of the phonon field, the wavefunctions of fermions like electrons are normally associated with probability density of detection of the fermion:

$$\rho(x)=|\psi(x)|^2.$$

If we try to stretch the analogy with phonon field, we'll get a probability density of having a field configuration, i.e. $\psi$, which, in turn, defines a probability density of detection of the particle. So, basically, probability of probability.

How should this be understood? Does this interpretation even make sense?

Ruslan
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  • In superconducting/superfluid states it makes sense to talk about the wave function amplitude (aka Landau-Ginzbourg equation). But canonical view is that probability amplitude is not observable - just a "mathematical trick". Phonons and photon fieldss, on the other hand, exist as classical fileds. – Roger V. Apr 22 '21 at 09:42
  • @ChiralAnomaly I'm not trying to mix up particle counting and field amplitudes. I'm trying to understand how to reconcile the Born rule for wavefunction of an electron with the field amplitude of corresponding quantum field (even if it's unobservable for fermions, it should be computable). – Ruslan Apr 22 '21 at 14:28
  • @Ruslan I assume you mean the Born rule for a position observable in a single-particle state. A single-particle state doesn't have a well-defined field amplitude, not even for a boson, but we can still calculate the expectation value of the field operator. Are you asking how to calculate the expectation value of a fermion field operator in a single-particle state? – Chiral Anomaly Apr 22 '21 at 19:02
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    @ChiralAnomaly yes, I mean exactly that. Now that I've thought a bit more, I've realized that there are no coherent states for fermionic field because creation operator squares to zero (at least no exact ones, although approximate ones with $\langle n\rangle<0.1$ could be OK), so there's no use in trying to reproduce the first-quantized wavefunction. I think my question then is how to reproduce the $|\psi(x)|^2$ probability density of the single-particle QM from the QFT single-particle state of the corresponding fermionic field. – Ruslan Apr 22 '21 at 22:20
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    @Ruslan To answer the question in the comment, use nonrelativistic QFT. Let $a^\dagger(x)$ and $a(x)$ denote the creation/annihilation operators for a fermion at $x$, which satisfy ${a(x),a^\dagger(y)}=\delta(x-y)$ and ${a(x),a(y)}=0$ with ${A,B}\equiv AB+BA$. A single-particle state is $|\psi\rangle\equiv \int dx\ \psi(x)a^\dagger(x)|0\rangle$ where $a(x)|0\rangle=0$. The observable $D(R)\equiv\int_R dx\ a^\dagger(x)a(x)$, integrated over a region $R$, represents a detector in $R$. The probability that it detects the particle is $\langle\psi|D(R)|\psi\rangle=\int_R dx\ |\psi(x)|^2$. – Chiral Anomaly Apr 23 '21 at 19:51
  • @ChiralAnomaly hmm, so uncertainty of the field value (that remains even in $|0\rangle$) doesn't appear to influence particle detection, although it is measurable in bosonic fields... Does this uncertainty have any physical significance in fermionic fields? – Ruslan Apr 23 '21 at 20:11
  • @Ruslan I'm not sure what you mean by "field value" in this context. Measurable values and their uncertainties are associated with observables, which in turn are represented by operators. What operator do you have in mind when you refer to uncertainty in the field value? Maybe something like $a^\dagger(x)+a(x)$? Fermion-odd operators like that aren't usually treated as observables, because that would contradict microcausality (it would allow action-at-a-distance). It would also cause other problems in models like quantum electrodynamics. – Chiral Anomaly Apr 23 '21 at 21:04
  • @ChiralAnomaly I suppose the proper term is "field amplitude" for what I mean by "field value". (I was trying to avoid the overloaded word "amplitude".) Yes, $a^\dagger(x)+a(x)$ seems to capture well what I meant. What bothers me is that, even if this is not an observable, we still can construct associated variance operator, and the expected value of this variance being nonzero doesn't appear to influence anything—at least particle detection. So I wonder whether it has any significance, or it's just an artifact of how we began with second-quantizing an already never-been-classical field. – Ruslan Apr 23 '21 at 22:15
  • @Ruslan Okay, I see what you mean now. (I didn't mean to pick on the words. Value and amplitude are both fine.) You can construct a variance-like operator $A^2-\langle A\rangle^2$ for any given operator $A$ and any given state. But if $A$ is not an observable, then I don't know of any special physical signifiance for it. In quantization, the classical model that we start with called "classical" because all of its observables commute with each other. It usually doesn't have it's own physical applications. If it involves fermions, which anticommute even in the classical version, then it doesn't. – Chiral Anomaly Apr 23 '21 at 23:03

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