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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?

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    $\begingroup$ Related post by OP: physics.stackexchange.com/q/576729/2451 $\endgroup$
    – Qmechanic
    Aug 31, 2020 at 17:32
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    $\begingroup$ 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 $|\Psi(t)>=U(t)|\Psi(t=0)>$. $\endgroup$ Aug 31, 2020 at 18:32

1 Answer 1

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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 $<\Psi(t)| = <\Psi(t=0)|U^\dagger$.

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