Scattering Amplitude in Canonical Quantization

[CGH]

Hi there,

I have little question: reading zee 2nd edition, I.8 (pag 64) i came up with this:

start with
$$<k_1 k_2| e^{-iHT}| k_3 k_4>$$

and

$$H=H_0 +u$$

$$u=\lambda \int \phi^4$$

where $H_0$ is the usual hamiltonian for the free scalar field.

Then, zee says that "expanding in $\lambda$, we obtain $-i\lambda \int <k_1k_2|\phi^4|k_3k_4>$"

my question is: how did he get that?
You cannot do an expansion around $H$, because $\lambda$ is the small term, so, i do the following,

$$\exp(-iH_0T -iu T)=\exp(-iH_0 T)\exp(-iu)\exp(-T^2[H_0,u]/2)$$

then, by working a little, i find that $[H_0,u]=0$, so, i can perform a safe expansion around [itex]\lambda[\itex],

$$<k_1 k_2| e^{-iHT}| k_3 k_4>=<k_1 k_2| e^{-iH_0T}(1-iT u+O(\lambda^2))| k_3 k_4>$$

now, i still get something different: instead of $-i\lambda \int <k_1k_2|\phi^4|k_3k_4>$, i get $-i\lambda \int <k_1k_2|e^{-iH_0 T}\phi^4|k_3k_4>$ (note: $T=\int dx^0$, i omit the $dx$ in the integration),

my question: what happened to the $e^{-iH_0T}$ term?, i think that it is just a phase and you get $-i\lambda \int <k_1k_2|\phi^4|k_3k_4>e^{-iET}$, with $E$ a real number, and that why it doesn't matter.

another option is, by looking (15) you realize that the term [itex]\lambda^0[/tex] is normalize as 1, so, i guess, the S-matrix is normalize, and then, you don't get the factor outside.

which one is the answer?

Saludos!

 

[LightHero]

The answer is that the e^{-iH_0T} term does not matter because it is just a phase factor. When you expand in powers of \lambda, all terms with \lambda^n are normalized so that the \lambda^0 term is 1. This means that when you expand the S-matrix in powers of \lambda, the constant term is 1 and the other terms have coefficients that depend on the values of k_1, k_2, k_3, and k_4. Therefore, the result is -i\lambda \int <k_1k_2|\phi^4|k_3k_4>.