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The equation for the photon propagator is straightforward $$ D_{ij} = \langle 0 |T \{ A_{i}(x')A_{j}(x) \}|0 \rangle $$ However, $A_{i}(x)$ is gauge-dependent and therefore unphysical (in the arguable sense). Then, since the propagator is dependent on the vector potential, the propagator is unphysical. Sadly, my whole understanding of what amplitudes mean may be skewed, but I would assume the probability amplitude for a photon to propagate between $x$ and $x'$ is something we would want to be gauge-independent.


I guess I wasn't clear enough. By computing the probability amplitude for a process, we obtain a complex number that when multiplied by it's complex conjugate we obtain a probability for such a process to occur (when normalized). Here, the physical process is propagation, and the probability is $|\langle 0 |T \{ A_{i}(x')A_{j}(x) \}|0 \rangle|^2$. However, this probability is gauge dependent, and hence, the usual physical interpretation of $|\langle 0 |T \{ A_{i}(x')A_{j}(x) \}|0 \rangle|^2$ is questionable to me. Where has my interpretation gone astray?

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So, what is your question? – Vladimir Kalitvianski Jan 4 '13 at 18:48
@VladimirKalitvianski I tried to clarify for you. – kηives Jan 4 '13 at 18:55
up vote 2 down vote accepted

The photon propagator $D_{\mu\nu}(x,y) = \langle 0 | A_\mu(x) A_\nu(y)|0\rangle$ is a building block for amplitudes, but it isn't necessarily an amplitude itself. The source for an electromagnetic field has to be a conserved current, which basically means that you create states from the vacuum using linear combinations of $A_\mu(x)$ operators whose coefficients are conserved currents. $$ |J\rangle = \int J^\mu(x) A_\mu(x) dx |0\rangle $$ where $\partial_\mu J^\mu = 0$.

You can show by direct computation that the amplitude $\langle J_1 | J_2 \rangle = \int\int J_1^\mu(x) D_{\mu\nu}(x,y)J_2^\nu(y)dxdy$ is gauge invariant if the currents $J_1$ and $J_2$ are conserved.

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I think this is the first time I have heard of this. Since the beginning of my physics education, anytime I have seen $\langle \; | \text{stuff}| \; \rangle$ I have thought of it as an amplitude. How will I know the difference? Gauge invariance? – kηives Jan 8 '13 at 5:10
@kηives: Yeah, it's an abuse of notation. You have to be suspicious when you're working with quantities which are not gauge invariant. But the end goal should always be to compute something which is gauge invariant. – user1504 Jan 8 '13 at 19:59
No doubt. Would you happen to know what $\langle 0 |A_\mu (x) A_\nu (y) |0\rangle$ is mathematically? Then I can at least learn more about this object. I thought it was just a complex number, like the normal inner product, but if the above is not the typical mapping in QM, I need to read more about it. – kηives Jan 8 '13 at 22:22
Mathematically, it is a complex number. It's the output of a sesquilinear form on the 'big' vector space of all not-necessarily-physical states (the Gupta-Bleuler space before the constraint is imposed, the complex on which the BRST operator acts, etc). If you want rigorous mathematical treatment, the place to start is probably Strocchi's Selected Topics on the General Properties of Quantum Field Theory. – user1504 Jan 11 '13 at 15:08

The probability amplitude is not a propagator, it's an S-matrix element. The latter is gauge-invariant.

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