# 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

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.

Last edited:
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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!

berkeman
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?