Understanding Eq (4.19) of Peskin-Schroeder

[praharmitra]

Can someone explain to me how the authors got the second equation of eq (4.19), Page 84, of Peskin Schroeder.

The equation is:

$$H_I(t) = e^{iH_0(t-t_0)}(H_{\text{int}}) e^{-iH_0(t-t_0)} = \int d^3x \frac{\lambda}{4!} \phi_I(t,\textbf{x})^4$$
where
$$H_{\text{int}} = \int d^3x \frac{\lambda}{4!} \phi^4(\textbf{x})$$
I do not understand how the second part of this eq is equal to the third. Please explain.

 

[matonski]

Since $e^{iH_0(t-t_0)}e^{-iH_0(t-t_0)} = 1$, you can insert it between each factor of $\phi$:

$$e^{iH_0(t-t_0)}\phi^4 e^{-iH_0(t-t_0)}<br /> = \left[ e^{iH_0(t-t_0)}\phi e^{-iH_0(t-t_0)} \right]<br /> \left[ e^{iH_0(t-t_0)}\phi e^{-iH_0(t-t_0)} \right]<br /> \left[ e^{iH_0(t-t_0)}\phi e^{-iH_0(t-t_0)} \right]<br /> \left[ e^{iH_0(t-t_0)}\phi e^{-iH_0(t-t_0)} \right]<br /> =\phi_I^4$$

 

[praharmitra]

Since $e^{iH_0(t-t_0)}e^{-iH_0(t-t_0)} = 1$, you can insert it between each factor of $\phi$:

$$e^{iH_0(t-t_0)}\phi^4 e^{-iH_0(t-t_0)} \\<br /> = \left[ e^{iH_0(t-t_0)}\phi e^{-iH_0(t-t_0)} \right]<br /> \left[ e^{iH_0(t-t_0)}\phi e^{-iH_0(t-t_0)} \right]<br /> \left[ e^{iH_0(t-t_0)}\phi e^{-iH_0(t-t_0)} \right]<br /> \left[ e^{iH_0(t-t_0)}\phi e^{-iH_0(t-t_0)} \right] \\<br /> =\phi_I^4$$

Of course that would be the natural thing to do. However, $$e^{iH_0(t-t_0)}\phi e^{-iH_0(t-t_0)}$$ is a vague statement since you have not examined the arguments of $$\phi$$.

In our current context of $$H_{int}$$, the argument is $$\phi(\textbf{x})$$. However, the definition of $$\phi_I$$ is

$$\phi_I(t,\textbf{x}) = e^{iH_0(t-t_0)} \phi(t_0,\textbf{x}) e^{-iH_0(t-t_0)}$$
where $$\phi(t_0,\textbf{x}) = e^{iHt_0}\phi(\textbf{x})e^{-iHt_0}$$. This surely does not reproduce the same result that has been written.

What am I doing wrong?

 

[muppet]

$$\phi(t_0,\mathbf{x}) = \phi(\mathbf{x})$$.

Schroedinger picture operators are defined at some reference time $$t_0$$.