- #1

Romanopoulos Stelios

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$$ U(t,t')=e^{iH_0(t-t_0)} e^{-iH(t-t')} e^{-iH_0(t'-t_0)} $$

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- Thread starter Romanopoulos Stelios
- Start date

In summary: Yes, by using the Euler-Lagrange equation:$$ \frac{\partial}{\partial t}\left[e^{iH_0(t-t_0)}e^{-iH(t-t_0)}e^{-iH_0(t'-t_0)}+e^{iH_0(t-t_0)}(-iH)\right]= 0 $$

- #1

Romanopoulos Stelios

- 1

- 0

$$ U(t,t')=e^{iH_0(t-t_0)} e^{-iH(t-t')} e^{-iH_0(t'-t_0)} $$

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- #2

kontejnjer

- 72

- 21

$$i\frac{\partial}{\partial t}U(t,t')=i\left[e^{iH_0(t-t_0)}(iH_0)e^{-iH(t-t')}e^{-iH_0(t'-t_0)}+e^{iH_0(t-t_0)}(-iH)e^{-iH(t-t')}e^{-iH_0(t'-t_0)}\right]=\underbrace{e^{iH_0(t-t_0)}(H-H_0)e^{-iH_0(t-t_0)}}_{H_I(t)}\underbrace{e^{iH_0(t-t_0)}e^{-iH(t-t')}e^{-iH_0(t'-t_0)}}_{U(t,t')}$$

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.

- #3

jpman19t10

- 2

- 1

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.

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:

- #4

- 27,680

- 19,190

jpman19t10 said:You may not see this answer, but as an individual memo, I answer your question.

Note that this thread is five years old!

- #5

jpman19t10

- 2

- 1

Yeah, I know (ゝω・)PeroK said:

Note that this thread is five years old!

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

thx!

- #6

Vo Quang Chau

- 1

- 0

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]##kontejnjer said:

$$i\frac{\partial}{\partial t}U(t,t')=i\left[e^{iH_0(t-t_0)}(iH_0)e^{-iH(t-t')}e^{-iH_0(t'-t_0)}+e^{iH_0(t-t_0)}(-iH)e^{-iH(t-t')}e^{-iH_0(t'-t_0)}\right]=\underbrace{e^{iH_0(t-t_0)}(H-H_0)e^{-iH_0(t-t_0)}}_{H_I(t)}\underbrace{e^{iH_0(t-t_0)}e^{-iH(t-t')}e^{-iH_0(t'-t_0)}}_{U(t,t')}$$

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.

$$ i\frac{\partial}{\partial t}\exp\left[-iH(t-t_0)\right] = \frac{\partial H}{\partial t}(t-t_0)\exp\left[-iH(t-t_0)\right]+H\exp\left[-iH(t-t_0)\right] $$

And do we ensure that ##H## is independent on time?

A unitary operator is an operator in quantum mechanics that preserves the normalization of state vectors and the inner product between them. This means that the operator must be linear, invertible, and its inverse must be equal to its adjoint.

The differentiation of a unitary operator is important in quantum field theory because it allows us to calculate the time evolution of a quantum state. This is necessary for understanding the dynamics of quantum systems and making predictions about their behavior.

In Peskin and Schroeder, the unitary operator U(t,t') is defined as the time-ordered exponential of the time-dependent Hamiltonian. This means that U(t,t') is an operator that evolves a state from time t' to time t, following the time-dependent Hamiltonian.

The time-ordered exponential in the definition of U(t,t') ensures that the operator is unitary, meaning it preserves the normalization of state vectors and the inner product between them. It also takes into account the time dependence of the Hamiltonian, which is important for calculating the time evolution of a quantum state.

In Peskin and Schroeder, the differentiation of U(t,t') with respect to t is done using the time-ordered product of operators. This involves taking the commutator of U(t,t') with the time-dependent Hamiltonian and then integrating over all intermediate times.

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