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:

<br /> 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<br />
where
<br /> H_{\text{int}} = \int d^3x \frac{\lambda}{4!} \phi^4(\textbf{x})<br />
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:

<br /> 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<br />

 

[praharmitra]

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

<br /> 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<br />

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

<br /> \phi_I(t,\textbf{x}) = e^{iH_0(t-t_0)} \phi(t_0,\textbf{x}) e^{-iH_0(t-t_0)}<br />
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.