# Intuitive Description of Feynman Propagator?

1. Sep 26, 2014

### Xenosum

1. The problem statement, all variables and given/known data

The Feynman Propagator is given by

$$<0| T \phi(y)\phi(x) |0> ,$$

where T is the time-ordering operator. I understand that this turns out to be the solution to the inhomogeneous klein-gordon equation, etc., but is there any intuitive description of the propagator? Can this be interpreted, for example, as the propability amplitude for a particle to travel from point x to point y? If so, why? The only thing I have to work with is that $\phi(x)$ is to be interpreted as an operator which creates a particle at point x. But I don't see any notion of time evolution in this definition of the propagator, or anything which is to be interpreted as a 'propagating' particle.

Thanks.

2. Relevant equations

3. The attempt at a solution

2. Sep 27, 2014

### Orodruin

Staff Emeritus
The time evolution is already hidden inside the field operators themselves (we work in the Heisenberg picture). Note that the fields contain both creation and annihilation operators. For the right field (after time ordering) operator, only the creation operator survives when acting on the vacuum on the right while for the left one only the annihilation operator would survive acting on the vacuum to the left. Therefore, you can think of the propagator as the amplitude of creating a particle at x (assuming now that $x^0 < y^0$) and that particle being at $\vec y$ at $y^0$.

3. Sep 27, 2014

### Xenosum

I might be missing something extremely basic here, but what guarantees that $<0|\phi(y)$ corresponds to the same particle that was created by $\phi(x)|0>$? The way I understand it, $\phi(x)$ is an operator whose states live in the so-called fock space-- the multi particle space-- and the operator $\phi(x)$ can therefore create/annihilate any number of particles.

And even if they can be shown to correspond to the same particle, if the time-evolution is inherent in the operator itself, and not the state $\phi(x) |0>$, then why would the inner product $<0| \phi(y)\phi(x) |0>$ be nonvanishing? If the particle was created at $x$, we can expand it in terms of its basis vectors, and they should all 'point in the x-direction', so to speak, since the state has never actually been time evolved. The operator might have, and the bra vector $<0| \phi(y)$ might have some 'y components', but it seems like the inner product as a whole should vanish by orthogonality.

4. Sep 28, 2014

### Orodruin

Staff Emeritus
It is the same field, just time evolved. Therefore, $\phi(y)$ will try to annihilate a $\phi$ excitation at $y$. The field is a linear combination of annihilation and creation operators and therefore creates or annihilates a single particle in the Fock space. In order to create two you would need two insertions of the field, but that gives you a zero since that two particle state is orthogonal to the vacuum.

For the second question, compare this with the $\hat x$ operator of a harmonic oscillator in quantum mechanics. In general, the correlation $\langle 0| \hat x(t) \hat x(0) |0\rangle$ will be non zero. It all depends on whether the Hamiltonian evolves a particle from $x$ to time $y^0$ such that the wave functions have some overlap.

Last edited: Sep 28, 2014
5. Sep 28, 2014

### Xenosum

Ah, I think I understand. Thanks!