# Propagators and Probabilities in the Heisenberg Picture

I'm trying to understand why

$$\Bigl|\langle0|\phi(x)\phi(y)|0\rangle\Bigr|^2$$

is the probability for a particle created at $y$ to propagate to $x$ where $\phi$ is the Klein-Gordon field. What's wrong with my reasoning below?

We can write

$$\phi(x) = \int \frac{\textrm{d}^3\mathbf{p}}{(2\pi)^3\sqrt{2E_{\mathbf{p}}}}(a_{\mathbf{p}}e^{-ip.x}+a_{\mathbf{p}}^{\dagger}e^{ip.x})$$

Then acting with the position operator (which presumably we can write in terms of the $a$ somehow) we find that $$\mathbf{\hat{x}}\phi(x)|0\rangle=\mathbf{x}\phi(x)|0\rangle$$ This tells us that $\phi(x)|0\rangle$ can be interpreted as a particle at $x$. Now define a state $|\psi(t)\rangle$ with $$|\psi(0)\rangle=\phi(y)|0\rangle$$ We have time evolution defined by $$|\psi(t)\rangle=e^{iHt}|\psi(0)\rangle=e^{iHt}\phi(y)|0\rangle$$

Pick some point $x=(T,X,Y,Z)$ (a four vector). The probability that the particle propagates from $y$ to $x$ is precisely the probability that we find $|\psi(T)\rangle$ in the state $\phi(x)|0\rangle$. But by the measurement postulate this is just the square of $$\langle0|\phi(x)e^{iHT}\phi(y)|0\rangle$$

Now I don't know how or why to get rid of the $e^{iHT}$. Since $H$ doesn't commute with $\phi$ we seem to be a bit stuck! Am I missing something simpler than this?

Many thanks in advance!

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This is a mixup between space-time propagation and space-propagation. Your formal argument is clarified by introducing explicit time on the fields:

$$|x\rangle = \phi(x,0)|0\rangle$$

This part is a definition--- you define the state on the left as the state on the right. It isn't quite right to interpret this as a localized state, as I will explain below, but first to clear up the formal confusion.

Then you want to ask, what is the probability for a particle at x at time 0 to propagate to y at time t. You apply the time evolution operator:

$$U(t) |x\rangle = e^{-iHt} \phi(x,0)|0\rangle$$

Then you ask what's the overlap of this with the state $|y\rangle$:

$$\langle y | U(t) |x\rangle = \langle 0 | \phi(y,0) e^{-iHt} \phi(x,0)|0\rangle$$

But, remember that the Heisenberg t-dependent field operator is:

$$\phi(y,t) = e^{iHt} \phi(y,0) e^{-iHt}$$

So that the above is equal to:

$$\langle y | U(t) |x \rangle = \langle 0 | e^{-iHt} \phi(y,t) \phi(x,0)|0\rangle$$

And now use the fact the the vacuum is an eigenstate with zero energy, and you conclude that the propagator is equal to the correlation function. This is the formal derivation.

The problem with this is only that the state I defined as the localized state

$$|x\rangle = \phi(x) |0\rangle = \int {d^dk\over (2\pi)^d (2\omega_k)} \sqrt{(2\pi)^d 2\omega_k} ||k\rangle$$

where $||k\rangle$ is the normalized k state (the reason I wrote it this way is that both parts, the measure and the k-state, are relativistically covariant this way--- this is relativistic normalization), cannot be interpreted as a 3 dimensional localized particle. This state doesn't have zero overlap with spacelike separated $\langle y|$.

$$\langle y | x \rangle \ne 0$$

for x different from y. The propagator doesn't vanish at spacelike separations. This means that the states $|x\rangle$ are not the eigenvalues of a position operator.

Historically, this confused people to no end, until Feynman and Schwinger explained the thing. The particle picture is not in space alone, it is in space-time, and then you can consider the state $|x\rangle$ as a space-time localized particle that propagates in spacetime, backwards and forwards. Only in the nonrelativistic limit is all the propagation forward in time, and in this case, you have a normal x operator and all the usual quantum stuff goes through. In the relativistic case, the free correlation function is the amplitude to go from x to y in a zig-zag in time manner, which is explained by Scwinger's representation.

$$\langle y | x\rangle = \int {1\over (2\pi\tau)^{d\over 2}} e^{-(x-y)^2\over 2\tau}\tau$$

Which holds in imaginary time. The Schwinger representation is the particle path formalism for relativistic quantum fields, and it is equivalent to Feynman's propagator picture for particle interactions, and it is equivalent to quantum field theory in the Hamiltonian or Lagrangian framework.

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Aha - I see my mistake now, I hadn't converted the operator $\phi(y)$ to the Schrodinger picture! Many thanks. Also it's really useful to see a proper argument for the interpretation of $\phi(x)\mid 0\rangle$ as a localised particle. I thought it didn't look like it was a position eigenstate! Glad that it confused early quantum physicists, and not just me. In short, my reasoning above was wrong in two places, both of which you've corrected. For this, thanks very much! –  Edward Hughes Oct 20 '12 at 20:11
@EdwardHughes: It's a miracly were able to get anything out of it considering the number of typos and tex errors I just fixed, but you got it anyway, despite my best attempts to be incomprehensible. –  Ron Maimon Oct 21 '12 at 0:13

The state $\phi(x)|0\rangle$, where $x=x^\mu$, is already evolving in time (or time dependent). Applying the evolution operator $e^{\pm i H t }$ just displaces the state in time. It's therefore unnecessary to carry out this step.

All of your other steps appear to be correct.

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So the argument essentially reduces to just the measurement axiom of QM then? In other words is the following correct? Let $\mid\phi\rangle$ be an initial particle state in the Heisenberg picture and $\mid\psi\rangle$ a state of definite $O$ for some observable $O$ at future time $t$. Then $\langle \phi\mid\psi\rangle$ is by definition the probability of the particle propagating from its initial state to a state with the particular observed value for $O$? –  Edward Hughes Oct 19 '12 at 22:12
That's pretty much right. Add "amplitude" after "probability" for the more correct statement. –  MarkWayne Oct 21 '12 at 3:38

The Lagrangian density for the Klein-Gordon field is:

$$\mathcal{L} = \frac{1}{2}(\partial_\mu \phi)^2 - \frac{1}{2}m^2\phi^2 = \frac{1}{2}\dot{\phi}^2-\frac{1}{2}(\nabla\phi)^2-\frac{1}{2}m^2\phi^2$$

and the Hamiltonian is given by:

$$H = \int d^3x [\pi(\textbf{x})\dot{\phi}(\textbf{x})-\mathcal{L}]$$

where

$$\pi(\textbf{x}) = \frac{\partial \mathcal{L}}{\partial \dot{\phi}(\textbf{x})}$$

there is an implicit time dependence in your Hamiltonian. Therefore, time-evolution is not as simple as applying $e^{iHT}$. You will run into complicated things such as Dyson series.

As to why $\langle0|\phi(x)\phi(y)|0\rangle$ is the propagator, the argument is very simple: think of the field operators is creation and annihilation operators (since they are nothing but linear combinations of them). The idea behind calling this quantity as the propagator is that you were destroying a state at $y$ and creating one at $x$. Hence the particle propagated from $y$ to $x$. But I have a hunch that you already knew that, and you simply wanted to reconcile that with quantum mechanics; I mentioned it just in case you didn't know.

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The statement, "Therefore, time-evolution is not as simple as applying e^{iHT}," isn't correct. This is the time evolution operator and, assuming that H is the full (or exact) Hamiltonian, it evolves states forward (or backward) in time as its name implies. The Dyson series is a consequence of working in the interaction (or Dirac) representation. Also the vacuum expectation value (VEV) of the product of two field operators at distinct points x and y is not the propagator. We require the VEV of the (Wick) time-ordered product for the propagator. –  MarkWayne Oct 19 '12 at 18:56
Yes, thanks for the correction. I shouldn't have said propagator. The "amplitude for a particle to propagate from $y$ to $x$" would have been more appropriate. –  NanoPhys Oct 19 '12 at 19:01