A Interpretation of state created by the field in free QFT

[leo.]

Let us consider QFT in Minkowski spacetime. Let $\phi$ be a Klein-Gordon field with mass $m$. One way to construct the Hilbert space of this theory is to consider $L^2(\Omega_m^+,d^3\mathbf{p}/p^0)$ where $\Omega_m^+$ is the positive mass shell. This comes from the requirement that there be a unitary representation of the Poincare group as done in Weinberg's book.

To interpret a state here is simple: any $\Phi(\mathbf{p})$ is the probability amplitude for momentum, so that $|\Phi(\mathbf{p})|^2$ is the probability density of finding a single particle with momentum $\mathbf{p}$.

There is a second construction related to this one by Fourier transformation. Given $\Phi\in L^2(\Omega_m^+,d^3\mathbf{p}/p^0)$ We can consider $$\psi(t,\mathbf{x})=\int_{\Omega_m^+} e^{i p_\mu x^\mu} \Phi(\mathbf{p}) \dfrac{d^3\mathbf{p}}{p^0}$$
If $\mathfrak{F}$ denotes the Fourier transform, then we get an equivalent space of states $\mathcal{K}_m^+=\mathfrak{F}[L^2(\Omega_m^+,d^3\mathbf{p}/p^0)]$. Each such $\psi\in\mathcal{K}_m^+$ is a positive-frequency solution to the Klein-Gordon equation.

When we also expand the field into positive and negative frequency plane waves, it follows that $\psi(t,\mathbf{x})= \phi(t,\mathbf{x})|0\rangle$. In other words, we get such a state by acting on the vacuum with the field.

But what is the interpretation of $\psi(t,\mathbf{x})$? It doesn't seem a wavefunction because there's no position observable here. Then it also feels wrong to interpret $|\psi(t,\mathbf{x})|^2$ as probability amplitude at $t$ of locating a particle at $\mathbf{x}$.

So while $\Phi(\mathbf{p})$ has a clear interpretation as momentum amplitude, what is the interpretation of $\psi(t,\mathbf{x}) = \phi(t,\mathbf{x})|0\rangle$?

 

[Peter Morgan]

IMO, to combine both definitions and write $\hat\phi_f=\int\hat\phi(t,\mathbf{x})f(t,\mathbf{x})\mathrm{d}t\mathrm{d}\mathbf{x}$ (where I've put a $\hat\$ to indicate that your $\phi(t,\mathbf{x})$ is an operator-valued distribution; then $\hat\phi_f$ is an operator), which with different values of $f(x)$ acting on the vacuum vector will give you your Fourier transform field $\Phi(\mathbf{p})=\hat\phi_{f_p}|V\rangle$ or your $\psi(t,\mathbf{x})=\hat\phi_{f_x}|V\rangle$. We can call $f(x)$ a sampling function because it tells us where we're making a measurement and how each place where the measurement is being made should be weighted. [I edited the equations in Red. It's clear from the discussion below, however, that you prefer to engage with the question you put above narrowly.]
Then one finds that for the adjoint operation, $\hat\phi_f^\dagger=\hat\phi_{f^*}$, where $f^*(x)$ is the complex conjugate of the sampling function at $x$, and for the two-point Vacuum Expectation Value of the free field, $$\langle V|\hat\phi_f^\dagger\hat\phi_g|V\rangle=(f,g)=\hbar\int\tilde f^*(k)\tilde g(k)2\pi\delta(k{\cdot}k{-}m^2)\theta(k_0)\frac{\mathrm{d}^4k}{(2\pi)^4}.$$ We note that $(f,g)^*=(g,f)$ and $(f,f)\ge 0$, which makes $(f,g)$ a pre-inner product, the structure of which fixes the space-time properties of this free field theory. For the algebraic structure, for even powers, $\langle V|\hat\phi_f^{2n}|V\rangle=\frac{(2n)!}{2^n n!}(f^*,f)^n$, and for odd powers $\langle V|\hat\phi_f^{2n+1}|V\rangle=0$, which we can present neatly as a generating function as $$\langle V|\mathrm{e}^{\mathrm{i}\lambda\hat\phi_f}|V\rangle=\mathrm{e}^{-\lambda^2(f^*,f)/2}.$$ We find that the commutator can also be written neatly as $[\hat\phi_f,\hat\phi_g]=(f^*,g)-(g^*,f)$. When $f(x)$ is real, so that $f^*=f$ and $\hat\phi_f^\dagger=\hat\phi_{f^*}=\hat\phi_f$ is self-adjoint, then $\langle V|\mathrm{e}^{\mathrm{i}\lambda\hat\phi_f}|V\rangle=\mathrm{e}^{-\lambda^2(f,f)/2}$ is a Gaussian characteristic function, which we can inverse Fourier transform to obtain a Gaussian probability density $$\langle V|\delta(\hat\phi_f-v)|V\rangle=\frac{\mathrm{e}^{-v^2/2(f,f)}}{\sqrt{2\pi(f,f)}}.$$ This Gaussian noise, with variance $(f,f)$, proportional to Planck's constant, is different from thermal noise because it's Lorentz invariant.
To answer your question above, "what is the interpretation of $\psi(t,\mathbf{x})$", I'll suggest straight out that we can call $\hat\phi_g|V\rangle$, or $\hat\phi_g^n|V\rangle$, or $\mathrm{e}^{\mathrm{i}\lambda\hat\phi_g}|V\rangle$, or ..., a modulation of the vacuum vector $|V\rangle$, and then justify it. Note that you don't mention objects such as $\hat\phi_g^n|V\rangle$ in your OP, but they are vitally important. Suppose we measure $\hat\phi_f$ in the first of these modulated states: the characteristic function works out to be $$\frac{\langle V|\hat\phi_g^\dagger\mathrm{e}^{\mathrm{i}\lambda\hat\phi_f}\hat\phi_g|V\rangle}{\langle V|\hat\phi_g^\dagger\hat\phi_g|V\rangle}=\left(1-\frac{(g,f)(f,g)}{(g,g)}\lambda^2\right)\mathrm{e}^{-\lambda^2(f,f)/2},$$ which results in a probability density $$\frac{\langle V|\hat\phi_g^\dagger\delta(\hat\phi_f-v)\hat\phi_g|V\rangle}{\langle V|\hat\phi_g^\dagger\hat\phi_g|V\rangle}=\left(\!1-\frac{(g,\!f)(f,\!g)}{(g,\!g)(f,\!f)}+\frac{(g,\!f)(f,\!g)}{(g,\!g)(f,\!f)}\frac{v^2}{(f,\!f)}\right)\!\frac{\mathrm{e}^{-v^2/2(f,f)}}{\sqrt{2\pi(f,\!f)}}.$$ What we're modulating is not a signal, but probability densities. When the overlap $\frac{(g,f)(f,g)}{(g,g)(f,f)}$ is 0 (its minimum is 0 and its maximum is 1), we just have the Gaussian probability density, but when the overlap is 1, which happens when $f=g$, we have the probability density $\frac{v^2}{(f,f)}\frac{\mathrm{e}^{-v^2/2(f,f)}}{\sqrt{2\pi(f,\!f)}}$, which has peaks to either side of $v=0$ and is zero at $v=0$; the probability density changes smoothly as we move the sampling function $f$ nearer or further away from the modulation function $g$, so that the overlap moves smoothly between 0 and 1. For higher powers $\hat\phi_g^n$, giving us $\hat\phi_g^n|V\rangle$, we obtain probability densities that are more elaborately modulated, so that there are higher powers of $v$.
The Hilbert space vector $\mathrm{e}^{-\mathrm{i}\hat\phi_g}|V\rangle$ is especially important, because it results in a coherent state, for which we obtain the probability density $$\langle V|\mathrm{e}^{\mathrm{i}\hat\phi_g}\delta(\hat\phi_f{-}v)\mathrm{e}^{-\mathrm{i}\hat\phi_g}|V\rangle=\frac{\mathrm{e}^{-(v-2\omega(f,g))^2/2(f,f)}}{\sqrt{2\pi(f,f)}},$$ (where $\omega(f,g)=\mathrm{i}[(f^*,g)-(g^*,f)]/2$, and we've simplified the equation by setting $g^*=g$) which has the effect of moving the probability density around bodily, as one might say, introducing powers of $v$ only inside the exponential.
I hope the above is understandable, but the take-home story is that one way we can think of QFT is as a kind of signal analysis for probability densities. What I've called a sampling function above is similar in function to what is called a window function in signal analysis. That makes a simple-minded kind of sense insofar as all the data that comes out of an experimental apparatus comes into the computers that record all the data as electrical or optical signals. I can give you some more equations, for squeezed states, say, and for thermal states, and there can be other things to say that would be almost impossible to write down or decode if we tried to present them in the kind of book formalism you used for your OP, but for now I'll assume you're bored already.

 

[A. Neumaier]

When we also expand the field into positive and negative frequency plane waves, it follows that $\psi(t,\mathbf{x})= \phi(t,\mathbf{x})|0\rangle$. In other words, we get such a state by acting on the vacuum with the field.

But what is the interpretation of $\psi(t,\mathbf{x})$? It doesn't seem a wavefunction because there's no position observable here. Then it also feels wrong to interpret $|\psi(t,\mathbf{x})|^2$ as probability amplitude at $t$ of locating a particle at $\mathbf{x}$.

So while $\Phi(\mathbf{p})$ has a clear interpretation as momentum amplitude, what is the interpretation of $\psi(t,\mathbf{x}) = \phi(t,\mathbf{x})|0\rangle$?

It is a single-particle state, the same that you would get from the Klein-Gordon equation in the single-particle (nonfield) picture. It is well-known to suffer from interpretational problems (wrong speeds, Zitterbewegung, etc.). To get a probability interpretation you need to consider an additional Foldy-Wouthuisen transformation which makes the Newton-Wigner position operator diagonal!

 

[leo.]

It is a single-particle state, the same that you would get from the Klein-Gordon equation in the single-particle (nonfield) picture. It is well-known to suffer from interpretational problems (wrong speeds, Zitterbewegung, etc.). To get a probability interpretation you need to consider an additional Foldy-Wouthuisen transformation which makes the Newton-Wigner position operator diagonal!

So in the end, $\psi(t,\mathbf{x})$ can be seen as the free evolution - with the Klein-Gordon equation - of a single particle in the initial state $\psi(0,\mathbf{x})$ as one would do with relativistic quantum mechanics without fields?

As you say this has a few interpretational problems. Also, I admit I have no experience with the Newton-Wigner position operator, but AFAIK it also has a fews problems, is it right?

In that case, it seems in the end it is better to just think of $\psi(t,\mathbf{x})$ as just encoding the momentum amplitude through a Fourier transform.

 

[A. Neumaier]

it seems in the end it is better to just think of ψ(t,x) as just encoding the momentum amplitude through a Fourier transform.

Yes. Indeed, this works also for photons, which can have well-defined momentum but can be proved not to have a good position operator.

 

[Demystifier]

But what is the interpretation of $\psi(t,\mathbf{x})$? It doesn't seem a wavefunction because there's no position observable here. Then it also feels wrong to interpret $|\psi(t,\mathbf{x})|^2$ as probability amplitude at $t$ of locating a particle at $\mathbf{x}$.

Anything that satisfies the axioms of probability, that is $0\le p_k \le 1$ and $\sum_k p_k =1$, can, in principle, be a probability under appropriate circumstances. In quantum mechanics, the "appropriate circumstances" means appropriate measurement procedure. For generalized (POVM) quantum measurements, a measurement does not need to be a measurement of an "observable". So your quantity above can indeed be a probability amplitude under appropriate measurement conditions, despite the fact that there is no "position observable".

For an introduction to generalized quantum measurements see e.g. Sec. 3.6 of https://arxiv.org/abs/1805.11965