praharmitra
- 308
- 1
This doubt is about a text in Peskin Schroeder Pg 86. I reproduce it here.
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[tex]U(t,t')[/tex] satisfies the same differential equation (4.18),
[tex] i \frac{\partial}{\partial t} U(t,t') = H_I(t) U(t,t')[/tex]
but now with the initial condition [tex]U=1[/tex] for [tex]t=t'[/tex]. From this equation you can show that
[tex] U(t,t') = e^{iH_0(t-t_0)}e^{-iH(t-t')}e^{-iH_0(t'-t_0)}[/tex]
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Here [tex]H = H_0+H_{int} = H_{KG} + \int d^3x \frac{\lambda}{4!} \phi(\textbf{x})^4[/tex]
and
[tex]H_I = \int d^3x \frac{\lambda}{4!} \phi_I^4[/tex].
Can anyone explain how "one can show" the second statement that Peskin Schroeder makes?
--------------------------------
[tex]U(t,t')[/tex] satisfies the same differential equation (4.18),
[tex] i \frac{\partial}{\partial t} U(t,t') = H_I(t) U(t,t')[/tex]
but now with the initial condition [tex]U=1[/tex] for [tex]t=t'[/tex]. From this equation you can show that
[tex] U(t,t') = e^{iH_0(t-t_0)}e^{-iH(t-t')}e^{-iH_0(t'-t_0)}[/tex]
-----------------------------------
Here [tex]H = H_0+H_{int} = H_{KG} + \int d^3x \frac{\lambda}{4!} \phi(\textbf{x})^4[/tex]
and
[tex]H_I = \int d^3x \frac{\lambda}{4!} \phi_I^4[/tex].
Can anyone explain how "one can show" the second statement that Peskin Schroeder makes?