How Gupta-Bleuler condition implies $(a_p^3-a_p^0)| \phi \rangle=0$?

Question

Gupta-Bleuler condition is $$\partial^\mu A_\mu^+ | \phi \rangle=0\tag{6.54}$$ where $$A_\mu^+= \int\frac{d^3\mathbf p}{(2\pi)^3 \sqrt{2|\mathbf p|}} \sum_{\lambda=0}^3 \epsilon^\lambda_\mu a_p^\lambda e^{-ip\cdot x}.$$

David Tong's QFT lecture note claims that this implies $$(a_p^3-a_p^0)| \phi \rangle=0.\tag{6.56}$$ I can't see how to obtain that. When I apply the Gupta-Bleuler condition and substitute the above equation, I obtain $$\int\frac{d^3\mathbf p}{(2\pi)^3 \sqrt{2|\mathbf p|}} \sum_{\lambda=0}^3 (-ip^\mu)\epsilon^\lambda_\mu a_p^\lambda e^{-ip\cdot x}| \phi \rangle=0$$ Since $\epsilon^1, \epsilon^2$ are transverse photon and they are orthogonal to $p$, only $\lambda=0$ and $\lambda=3$ term contributes. Then a sufficient condition for Gupta-Bleuler condition is that $$\{(p\cdot \epsilon^0)a_p^0-(p\cdot\epsilon^3)a_p^3\}|\phi\rangle=0$$ but this is different from the lecture note!

Answer 1

You are forgetting that $\epsilon^\lambda = \epsilon^\lambda(p)$.

That basis is constructed this way, referring to a fixed Minkowskian reference frame (I adopt the signature (-,+,+,+) and $\vec{a}$ indicates your ${\bf a}$)

(1) $\epsilon^3(p) = \frac{\vec{p}}{|\vec{p}|}$,

(2) $\epsilon^{1}(p), \epsilon^2(p)$ are of unit length, mutually orthogonal, and also orthogonal to $\vec{p}$.

(3) Finally, the triple of three-vectors $\epsilon^1(p),\epsilon^2(p), \epsilon^3(p)$ viewed as spacelike four-vectors, form a pseudo-ortonormal (future-oriented, positive) basis of Minkowski space when adding $\epsilon^0(p)$ to them.

Hence $\epsilon^0(p)$ must have components $(1,0,0,0)$. This way $$p\cdot \epsilon^0(p) = p^0\:,\quad p\cdot \epsilon^3(p) = \frac{\vec{p}\cdot\vec{p}}{|\vec{p}|}\:.$$ Since $p$ is light-like (we are dealing with photons) $$p^0= |\vec{p}|=\frac{\vec{p}\cdot\vec{p}}{|\vec{p}|}$$ This identity inserted in your final identity gives rise to David Tong's assertion.