# How is the $\langle\psi| x(t')\, x(t)|\psi\rangle$ interpreted in the Heisenberg picture?

Expectation values of the form $$\langle \psi| x(t')\, x(t) |\psi\rangle$$ are very common in quantum field theory (Heisenberg picture operator $x(t)$). Focusing on $1$-dimensional QFT (i.e. ordinary quantum mechanics with just a time dimensions) for simplicity, what is the interpretation of that expectation value?

I suppose what's throwing me is that I'm used to thinking in the Schrödinger picture, and I was always told that translating between the two was equivalent to moving the time dependence between the operators and wave functions. In this case, though, there doesn't seem to be a clean way to move the time dependence of the $x(t')\, x(t)$ operator to the wave functions. The first version of this quantity in the Schrödinger picture is (assuming $H$ commutes with itself at all times): \begin{align} \langle \psi | x(t')\, x(t)|\psi\rangle_{\mathrm{Heis}} & = \langle\psi(t')| x_{\mathrm{Sch}} \operatorname{e}^{-iH(t'-t)}x_{\mathrm{Sch}}|\psi(t)\rangle. \end{align} which doesn't isolate the time evolution into the state vectors cleanly. At best, I read this as:

1. evolve $|\psi\rangle$ to time $t$,
2. measure its position,
3. evolve the result for time $t'-t$ (possibly backwards),
4. measure position again,
5. evolve state backwards by $t'$,
6. calculate overlap with $|\psi\rangle$, and
7. scale the the product by the results of the two position measurements.

That can't be right, though, because real "measurements" collapse superpositions, and that's not done here.

• It is a correlation function. By working it out in the Schroedinger representation it is an autocorrelation function for the Schr wave functions with a kernel specified by the hamiltonian. It is straightforward to work it out for free propagation, and the oscillator, no? – Cosmas Zachos Nov 22 '17 at 20:57
• So it's this: measure position at $t$, then measure it at $t'$, multiply those measurements and take the mean of repeating this process, and this is the average of that? – Sean E. Lake Nov 22 '17 at 21:00

Strictly in quantum mechanics (no connection to QFT), $$\langle \psi| \hat{x}(t')\, \hat{x}(t) |\psi\rangle$$ is a correlation function, in state ψ , that is the average of measurements of this operator in this state. Where, to be sure, the evolution of these operators is specified by the Hamiltonian.

For instance, for the SHO,
$$\hat{x}(t)= e^{it(\hat{p}^2 + \hat{x}^2)/2}~ \hat{x} ~e^{-it(\hat{p}^2 + \hat{x}^2)/2},$$ so that x̂(t)=x̂ cos t+ sin t, a rigid rotation of the ("Schroedinger") t= 0 operators (Have set $\hbar=1, ~ \omega=1$ for simplicity).

To transition to Schrödinger operators, $$\langle \psi| \hat{x}(t')\, \hat{x}(t) |\psi\rangle= \int\! dx dy ~ \psi^*(x) ~\psi(x) ~\langle x| \hat{x}(t')\, \hat{x}(t) |y\rangle ~,$$ so this can be also thought of as the Schrödinger wavefunction complex autocorrelation function, with the matrix element representing the convolution kernel for it.

For instance, for the above oscillator example, for $\psi(x)$ a wavefunction at t =0, it amounts to $$\int\! dx dy ~ \psi^*(x) ~\psi(y) ~\langle x| (\hat{x}\cos t'+ \hat{p} \sin t') (\hat{x}\cos t+ \hat{p} \sin t) |y\rangle \\ = \int\! dx dy ~ \psi^*(x) ~\psi(y) ~\langle x| (x\cos t'+ \hat{p} \sin t') (y\cos t+ \hat{p} \sin t) |y\rangle \\ =\! \int\!\! dx dy ~ \psi^*(x) \psi(y) \left (xy \cos t' \cos t -\sin t' \sin t ~\partial_x^2 -i(x\cos t' \sin t +y \cos t \sin t')\partial_x\right ) \delta(x-y)$$ where I've used the standard rule for the reduction of momentum operators acting on the bra. Integrating by parts, $$=\! \int\! dx ~ \psi(x) \left (x^2 \cos t \cos t'+i \cos t' \sin t-\sin t \sin t' \partial_x^2 + i x \sin (t+t') ~\partial_x\right ) \psi^*(x) ~,$$ etc, whose action on selected states, such as the ground state, is easy to monitor--this is an x-projection of the phase-space rotation. For real, normalized ψ, the last term may be integrated out by parts to merge with the "constant" to $i\sin(t-t')~/2$.

Special limits. t =0, $$\! \int\! dx ~ \psi(x) \left (x^2 \cos t'+ i \sin t' ~x\partial_x\right ) \psi^*(x) ~,$$ with the last term reducing to $-i\sin t' /2$ for real ψ; and t=t', $$=\! \int\! dx ~ \psi(x) \left (x^2 \cos^2 t+\frac{i}{2} \sin (2t)-\sin^2 t ~\partial_x^2 +i\sin (2t)~x\partial_x\right ) \psi^*(x) ~,$$ with the 2nd and 4th terms cancelling for real ψ.

The issue of how to actually measure the expectation of this operator so computed far outranges the scope of the question: it has to fit into some observable quantity for a state to be chosen and prepared.

I've never seen the expression $x(t)$ used in quantum field theory. Position is not an operator in QFT, just a spacetime coordinate. Usually we see unequal-time correlation functions of quantum fields, e.g. $$C(x', t'; x, t) := \langle 0 | \varphi_H(x', t') \varphi_H(x, t) | 0 \rangle.$$ Any Heisenberg operator $A_H(t)$ is simply defined to be the conjugation $$A_H(t) := U^\dagger(t, 0)\, A_S\, U(t, 0)$$ of the Schrodinger operator $A_S$, where $U(t, 0)$ is the time-evolution operator. The vast majority of the time we consider Hamiltonians that don't depend explicitly on time, so $U(t,0)$ is simply $e^{-i H t}$ and $$A_H(t) := e^{i H t}\, A_S\, e^{-i H t}.$$ So the unequal-time correlation function $$C(x', t'; x, t) = \langle 0 | e^{i H t'} \varphi_S(x') e^{i H (t - t')} \varphi_S(x) e^{-i H t} | 0 \rangle.$$

To get a physical interpretation, it is useful to note that time-translational invariant implies that this correlation function can only depend on the time difference $\Delta t := t' - t$, so WLOG we can let $t \to 0$ and $t' \to \Delta t$ to get $$C(x', \Delta t; x) = \langle 0 | e^{i H \Delta t} \varphi_S(x') e^{-i H \Delta t} \varphi_S(x) | 0 \rangle.$$ Now $\langle 0 | e^{i H t \Delta t}$ is just the time-evolved ground state $\langle 0'|$ at time $\Delta t$. (If the ground state is unique, then $\langle 0' |$ is just $e^{i E_0 \Delta t} \langle 0 |$, the original ground state multiplied by an irrelevant overall phase factor that depends on your choice of energy zero.) So we have $$C(x', \Delta t; x) = \langle 0' | \varphi_S(x') e^{-i H \Delta t} \varphi_S(x) | 0 \rangle.$$

Recall that a (real) scalar field $\varphi_S(x) = \varphi_S^\dagger(x)$ creates a particle at position $x$. So $$\varphi_S(x) | 0 \rangle$$ is the state at time $t = 0$ with one particle at position $x$, $$e^{-i H \Delta t} \varphi_S(x) | 0 \rangle$$ is the state at time $t = \Delta t$ for which there was one particle at position $x$ back at time $t = 0$, and $$\langle 0'| \varphi_S(x')$$ is the bra corresponding to the state at time $t = \Delta t$ with one particle at position $x'$. So the physical meaning of $C(x', \Delta t; x)$ is simply the probability amplitude that a particle will be at position $x'$ at time $\Delta t$, given that it was at position $x$ at time $t = 0$ (and there were no other particles anywhere).

• In QFT the expressions are in terms of $\phi(x,t)$, with $\phi$ an operator. That's why ordinary QM is often called "zero (spatial) dimensional QFT". – Sean E. Lake Nov 23 '17 at 1:27
• Also, the interpretation of $\phi(x, t)$ creating a particle at $x,t$ is questionable. I believe that it applies to Dirac-type fermions, and that it doesn't apply to real scalar Klein-Gordon type fields. – Sean E. Lake Nov 23 '17 at 1:41
• @SeanE.Lake (a) It's true that ordinary QM is a zero-dimensional QFT in a formal mathematical sense, but I wouldn't use that duality for physical intuition for the higher-D QFT case, because they're conceptually rather different. In QFT, position is an argument of the quantum operator fields, while in QM position is itself a quantum operator. So IMO, considering QM as a zero-dimensional QFT is mathematically simple but conceptually subtle. And the conceptual subtlety outweighs the mathematical simplicity, so the zero-dimensional case is actually more confusing than the higher-dimensional case. – tparker Nov 23 '17 at 3:23
• @SeanE.Lake (b) No, $\varphi(x, t)$ definitely creates a particle from the vacuum at $x, t$ even for real scalar fields; we have $$\varphi(x) = \int \tilde{dk} \left[ a({\bf k}) e^{ik \cdot x} + a^\dagger({\bf k}) e^{-ik \cdot x} \right],$$ see Srednicki eqn. (3.19). It also annihilates a negative-frequency antiparticle, but there aren't any in the vacuum so the $a$ term doesn't matter when $\varphi(x)$ acts on the vacuum. – tparker Nov 23 '17 at 3:36