# Tag Info

3

Peskin & Schroeder, An Intro to QFT, are using that$^1$ $$i\Delta(x-y)~:=~\langle 0 | [\phi(x), \phi(y)] |0\rangle \tag{K}$$ vanishes for space-like vectors, see below eq. (2.53) on p. 28. In particular for equal times $x^0=y^0$, we have $$i\Delta(0,{\bf x}-{\bf y})~=~0.\tag{L}$$ Therefore at the physics level of rigor ...

3

If at every time $t$, $\phi(\mathbf{x},t)$ is a nice enough function that it has a Fourier transform, then $$\phi(\mathbf{x},t)=\int_{-\infty}^{\infty}\frac{d^{3}k}{(2\pi)^{3}}\widetilde{\phi}(\mathbf{k},t)e^{i\mathbf{k}\cdot\mathbf{x}},$$ where $\widetilde{\phi}(\mathbf{k},t)$ is just the Fourier coefficient at that time $t.$ But now you ask that the whole ...

2

No, the Lagrangian density is different: $$\mathcal{L} = \pm \frac{1}{2} \partial^{\mu} \phi \partial_{\mu} \phi.$$ The Hamiltonian density is actually the same in both conventions. However, this has no physical meaning. The choice of the signature is purely conventional.

1

Conventions do not change physics. If they would, we would not call them conventions. When studying Lagrangian mechanics, you may have noticed that you can multiply a lagrangian by any constant, and receive the same dynamics. Thus, we often (Or always) choose the constant such that the term $(\partial_0\phi)^2$ appears with a positive sign. (And often with ...

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The question is not entirely precise but I can at least clarify that photons emitted by one atom can be absorbed by another atom, even if those atoms are moving at different speeds. The likelihood of excitation is simply maximized when the atoms have the same velocity. In general, the law which states that transitions are only possible between initial and ...

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You are actually not calculating the same thing. The momentum space Feynman of the propagator is used for internal momentum, which is off-shell. So your first calculation is correct. While in your second case, you are used the LSZ formula to force the momentum to be on-shell. One way to get the same result is as follow: Consider the following diagram: ...

1

The correct commutation relations for creation/annihilation operators are the following: $$[a(k), a^{\dagger} (k')] = (2\pi)^3 \delta^{(3)} (k - k')$$ $$[b(k), b^{\dagger} (k')] = (2\pi)^3 \delta^{(3)} (k - k')$$ Everything else commutes, including $[a, b^{\dagger}]$. From this, it is clear that $[H, a]$ can not depend on $b$.

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