Differentiation of unitary operator U(t,t') in Peskin and Schroeder

[Romanopoulos Stelios]

How the authors came to the conclusion (eq. 4.25) that
$$ U(t,t')=e^{iH_0(t-t_0)} e^{-iH(t-t')} e^{-iH_0(t'-t_0)} $$

 

[kontejnjer]

You can plug the expression into equation 4.24 to verify it's correct. The left side is (taking into account the non-commutativity of $H_0$ and $H$):
$$i\frac{\partial}{\partial t}U(t,t')=i\left[e^{iH_0(t-t_0)}(iH_0)e^{-iH(t-t')}e^{-iH_0(t'-t_0)}+e^{iH_0(t-t_0)}(-iH)e^{-iH(t-t')}e^{-iH_0(t'-t_0)}\right]=\underbrace{e^{iH_0(t-t_0)}(H-H_0)e^{-iH_0(t-t_0)}}_{H_I(t)}\underbrace{e^{iH_0(t-t_0)}e^{-iH(t-t')}e^{-iH_0(t'-t_0)}}_{U(t,t')}$$
this is nothing but the right hand side of 4.25. In the second line I've used $1=e^{-iH_0(t-t_0)}e^{iH_0(t-t_0)}$, $H_{int}=H-H_0$ and the definition of the interaction picture Hamiltonian.

 

[jpman19t10]

You may not see this answer, but as an individual memo, I answer your question.

From eq. (4.18), you can get
$$ H_I(t) = \left( i \frac {\partial} {\partial t} U(t,t_0) \right) U^\dagger(t,t_0). $$
Then, substitute this eq. for the right-hand side of eq. (4.24).
If you do so,
\begin{align*}
& (\text{right-hand side of (4.24)}) \\
& = H_I(t)U(t,t') \\
& = \left( i \frac {\partial} {\partial t} U(t,t_0) \right) U^\dagger(t,t_0) U(t,t') \\
& = i \frac {\partial} {\partial t} \left( U(t,t_0)U^\dagger(t,t_0)U(t,t') \right)
- U(t,t_0) i \frac {\partial} {\partial t} \left( U^\dagger(t,t_0)U(t,t') \right).
\end{align*}
Since $U(t,t_0)U^\dagger(t,t_0)=1$, if you substitute this for (4.24) you get
$$ \frac {\partial} {\partial t} \left( U^\dagger(t,t_0)U(t,t') \right) = 0. $$
From this,
$$ U^\dagger(t,t_0)U(t,t') = C(t') ~~~~ (C(t')\text{ is an arbitrary function of }t'). $$
Now, since $U(t,t')=1$ for $t=t'$, $C(t') = U^\dagger(t',t_0)$.
Therefore,
$$ U(t,t') = U(t,t_0)C(t') = U(t,t_0)U^\dagger(t',t_0) = e^{iH_0(t-t_0)}e^{-iH(t-t')}e^{-iH_0(t'-t_0)}. $$

That's all.

 

[PeroK]

You may not see this answer, but as an individual memo, I answer your question.

Note that this thread is five years old!

 

[jpman19t10]

Note that this thread is five years old!

Yeah, I know (ゝω・)

This is for the people who couldn't find how to derive (4.25) from (4.24).

thx!

 

[Vo Quang Chau]

You can plug the expression into equation 4.24 to verify it's correct. The left side is (taking into account the non-commutativity of $H_0$ and $H$):
$$i\frac{\partial}{\partial t}U(t,t')=i\left[e^{iH_0(t-t_0)}(iH_0)e^{-iH(t-t')}e^{-iH_0(t'-t_0)}+e^{iH_0(t-t_0)}(-iH)e^{-iH(t-t')}e^{-iH_0(t'-t_0)}\right]=\underbrace{e^{iH_0(t-t_0)}(H-H_0)e^{-iH_0(t-t_0)}}_{H_I(t)}\underbrace{e^{iH_0(t-t_0)}e^{-iH(t-t')}e^{-iH_0(t'-t_0)}}_{U(t,t')}$$
this is nothing but the right hand side of 4.25. In the second line I've used $1=e^{-iH_0(t-t_0)}e^{iH_0(t-t_0)}$, $H_{int}=H-H_0$ and the definition of the interaction picture Hamiltonian.

The problem will occur if $H$ is dependent on time. We must perform derivative $H$ in the component $\exp\left[-iH(t-t_0)\right]$
$$ i\frac{\partial}{\partial t}\exp\left[-iH(t-t_0)\right] = \frac{\partial H}{\partial t}(t-t_0)\exp\left[-iH(t-t_0)\right]+H\exp\left[-iH(t-t_0)\right] $$
And do we ensure that $H$ is independent on time?