Srednicki Ch5 creation operator time dependence

[cedricyu803]

Hi folks, originally I read Peskin & Schroeder but then I realized it was too concise for me.
So I switched to Srednicki and am reading up to Chapter 5.

(referring to the textbook online edition on Srednicki's website)
Two questions:

1. In the free real scalar field theory, the creation operator given by eq. 5.2 is time-independent.
But in eq. 4.5, when $\phi^{+}$ evolves with time, the time evolution acts on $a^{\dagger}(k)$, which is time-independent, and it gets an extra phase.
I know it follows directly from the commutation relation with H0,
but it looks to me that $a^{\dagger}(k)$ becomes time-dependent.
Can anyone explain it?

2. Srednicki argues that $\left \langle 0|\phi (0)|0 \right \rangle = 0$, which makes sense to me if I consider eq. 5.15.

But for $\left \langle p|\phi (x)|0 \right \rangle=e^{-ipx}\left \langle p|\phi (0)|0 \right \rangle$
I have trouble with justifying that $\left \langle p|\phi (0)|0 \right \rangle=1$ ensures the 1-particle state normalisation,
if I consider eq. 5.15 again, (looks like) $-\partial ^2+m^2 (e^{-ipx}\left \langle p|\phi (0)|0 \right \rangle)$ gives me zero.

What's wrong with this logic and what should be the correct one?

Reading a QFT book alone is never easy... #sigh#
hope I can finish this book before the next June.

Thanks for your generous help

 

[Lapidus]

1. the k in a(k) is boldface, that means it is a three-vector, it can be transformed into integral over space (3.21), so it is time-independent

the integral and the exponential in 4.5 run over a four-vector instead

2. we want it to be one, it is not automatically one for the interacting case! We have to rescale/ renormalize the field in order to have this amplitude set to one

the reason why it should be one, is because it was required so for the derivation of the LSZ formula but for which we used free fields which create one-particle states out of the vacuum with amplitude one without any field rescaling

 

[cedricyu803]

1. the k in a(k) is boldface, that means it is a three-vector, it can be transformed into integral over space (3.21), so it is time-independent

the integral and the exponential in 4.5 run over a four-vector instead

2. we want it to be one, it is not automatically one for the interacting case! We have to rescale/ renormalize the field in order to have this amplitude set to one

the reason why it should be one, is because it was required so for the derivation of the LSZ formula but for which we used free fields which create one-particle states out of the vacuum with amplitude one without any field rescaling

1. OK now k in a(k) is the operator in the Schrodinger Picture

In $\varphi^+ (x)=e^{iH_{0}t}\varphi^+ (\mathbf{x},0)e^{-iH_{0}t}=\int \widetilde{dk}e^{ikx}a(\mathbf{k})$ follows from $[a(\mathbf{k}),H_{0}]=\omega a(\mathbf{k})$, so $e^{iH_{0}t}a(\mathbf{k})e^{-iH_{0}t}$ is the Heisenberg picture operator.
Now I got it. (almost forgot what I learned in QM...)

and the line above eq. 5.8 says, $a^\dagger(\mathbf{k})$ becomes time-dependent in the interacting theory, in contrast to the t-indp. in free field theory.
So it means that eq. 5.6 is already time-dependent, without the need of applying the Heisenberg Picture time evolution like in eq. 4.5, am I correct?

2. I understand the reason qualitatively, XXXXX but could you/ anyone show it explicitly in terms of the formulas we have in this chapter?XXXX

Ah I recalled LHS of eq. 5.18 is particle state in the momentum space. Now it makes perfect sense

Thanks very much