2
$\begingroup$

Consider the Dyson series of the $S$-matrix of Quantum Electrodynamics. The second term in this series predicts non-zero probability amplitudes of observing photons that are off-shell.

For an incoming state of a positron and an electron, the second term of the series gives a non-zero amplitude that the final product is an off-shell photon.

So why don't we observe these photons? Is it simply because Einstein's energy relation forbids their existence? But then that means the probability of observing them is 0, which contradicts the non-zero probability predicted by theory. If you simply set it to zero in an ad-hoc manner, then the other probabilities wouldn't add to 1.

Edit I mean the operator $$A_{\mu}^-(x) \psi^ {'+} (x)\gamma ^{\mu} \psi^+ (x) $$

$\psi ^{'}$ means the spinor adjoint. This term is part of the Dyson series. It's been normal ordered using Wick's theorem. This creates off-shell photons when operated on $|e^+e^-\rangle$, and just leaves them there! The off-shell photons are the final products of this operator.

Improve this question
$\endgroup$
2
  • 3
    $\begingroup$ I don't know what you're talking about - the S-matrix is something you sandwich between an input and an output state and your space of output states should not contain "an offshell photon" to begin with. Please be more explicit what term you think represents this off-shell amplitude. $\endgroup$
    – ACuriousMind ♦
    Jul 6 at 8:07
  • $\begingroup$ @ACuriousMind But it's the operator that I wrote that's producing the off-sheell state when operated on $|e^+ e^-\rangle$. Since this operator is part of the S-matrix expansion, that means that the S matrix produces off-shell states. The two annihilation operators make it a vaccuum, and then the A at the left produces photons. Good so far. But conservation of four-momentum demands that the photon produced must be off-shell, because the total momentum of the incoming particles does not fulfill the energy-momentum relation of a real photon. $\endgroup$
    – Ryder Rude
    Jul 6 at 8:16

2 Answers 2

Reset to default
3
$\begingroup$

Virtual photons are not real photons - they exist only in theory / perturbation expansions. In terms of the usual perturbation theory, they are energetically inaccessible intermediate states that we sum over to calculate real quantities.

Improve this answer
$\endgroup$
8
  • 1
    $\begingroup$ I specifically asked about the second term of the Dyson series (the one after identity). There, they exist as a final product of experiments, rather than something in between. $\endgroup$
    – Ryder Rude
    Jul 6 at 7:40
  • $\begingroup$ @RyderRude please spell it explicitly - there are many S-matrices and expansions, some for systems with no photons at all - it is not clear what you are talking about. $\endgroup$
    – Roger Vadim
    Jul 6 at 7:44
  • $\begingroup$ @RyderRude also, though the matrix element is non-zero, they are still off-shell - just like the intermediate states in the usual PT. $\endgroup$
    – Roger Vadim
    Jul 6 at 7:45
  • $\begingroup$ Please read the edit $\endgroup$
    – Ryder Rude
    Jul 6 at 8:01
  • 2
    $\begingroup$ I got it. The delta function is 0 for any photon state that's part of the Fock space. pathetic mistake. Thanks. $\endgroup$
    – Ryder Rude
    Jul 6 at 8:35
3
$\begingroup$

Virtual photons by definition are not observes.

When you set up a scattering problem, you first decide on initial and final states. The photons in the initial\final states are called "real photons", and they are on-shell.

When you calculate the probability you can expand in a perturbation series which includes amplitudes with virtual particles. But- all of these diagrams have the same initial and final states, and therefore all of them agree that the initial and final particles are real (on-shell).

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.