# Adjoint of the time-evolution operator

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?

• Related post by OP: physics.stackexchange.com/q/576729/2451 – Qmechanic Aug 31 at 17:32
• 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 $|\Psi(t)>=U(t)|\Psi(t=0)>$. – user2820579 Aug 31 at 18:32

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