0
$\begingroup$

The time-evolution operator $\hat U$ is defined so that $\Psi(x,t)=\hat U(t)\Psi(x,0)$. In terms of the Hamiltonian, it is expressed as $\hat{U}(t)=\exp \left(-\frac{i t}{\hbar} \hat{H}\right)$. I'm trying to calculate the adjoint conjugate $\hat U^\dagger(t)$.


My attempt at a solution

It must satisfy $\langle \hat{U}(t) \Psi(x,0) | \Phi(x,0) \rangle=\langle \Psi(x,0) | \hat{U}^\dagger(t)\Phi(x,0) \rangle$, so

$$\int _{-\infty}^{+\infty} \hat{U}^\star(t) \Psi^\star(x,0) \Phi(x,0) dx=\int _{-\infty}^{+\infty} \Psi^\star(x,0)\hat{U}^\dagger(t)\Phi(x,0)dx$$

I know that $\hat U$ is unitary, so $\hat U^\dagger(t)=\hat U^{-1}(t)=\hat U^{\star}(t)$, but, without using this information, could the expression of $\hat U^\dagger(t)$ be deduced from the expression above?

$\endgroup$
  • 1
    $\begingroup$ Related post by OP: physics.stackexchange.com/q/576729/2451 $\endgroup$ – Qmechanic Aug 31 at 17:32
  • 1
    $\begingroup$ Notice that for computing the adjoint, you don't need the vector state in a specific basis. An equivalent and in fact more general definition is $|\Psi(t)>=U(t)|\Psi(t=0)>$. $\endgroup$ – user2820579 Aug 31 at 18:32
2
$\begingroup$

Probably it is cleaner to do it by series.

\begin{equation} \begin{split} U^\dagger(t)&=\left(\sum_{n=0}^\infty \frac{1}{n!}\left(\frac{-it}{\hbar} \right)^n H^n\right)^\dagger\\ &=\sum_{n=0}^\infty \frac{1}{n!}\left(\left(\frac{-it}{\hbar} \right)^n \right)^\dagger (H^n)^\dagger\\ &=\sum_{n=0}^\infty \frac{1}{n!}\left(\frac{it}{\hbar} \right)^n H^n\\ &=\exp(it H/\hbar), \end{split} \end{equation}

since $H$ is Hermitian.

Another possibility is to start with Schrödinger equation, compute the adjoint and finally derive and solve an equation for $U^\dagger$ provided that $<\Psi(t)| = <\Psi(t=0)|U^\dagger$.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.