Is U(t,t_0) a Unitary Operator in Quantum Mechanics?

[jeebs]

the problem I am trying to solve is that, if we have say, $$|\Psi(t)\rangle = U(t,t_0)|\Psi(t_0)\rangle$$, then I need to prove that U is a unitary operator. I don't have a lot of confidence in what I've answered.

I know that a unitary operator is defined $$U^+ = U^-^1$$ (adjoint = inverse) so I need to somehow show this is happening.
I've said that U(t,t₀) takes your system from its state at time t₀ to time t, so that U(t,t) does nothing to your state and so it must be equal to the identity.
likewise, the adjoint U⁺(t,t) must do nothing and therefore be just the identity matrix too, as should U^-1, so that would mean the condition for a unitary operator is satisfied.

however, this seems very sketchy to me, I'm not convinced. have I done what's been asked here?

Actually I've just came up with another attempt, maybe this is better:

$$|\Psi(t)\rangle = U(t,t_0)|\Psi(t_0)\rangle$$ and $$\langle\Psi(t)| = \langle \Psi(t_0)|U^+$$

and $$\langle\Psi(t)|\Psi(t)\rangle = 1$$ and $$\langle\Psi(t_0)|\Psi(t_0)\rangle = 1$$

also $$\langle\Psi(t_0)|U^+U|\Psi(t)\rangle = \langle\Psi(t_0)|\Psi(t_0)\rangle$$

therefore and $$U^+U = I$$

is this valid?

 

[mark726]

Hello,

Thank you for sharing your thoughts and attempts in solving this problem. It seems like you have a good understanding of the concept of a unitary operator and have made some progress in your attempt to prove that U is a unitary operator.

Your first approach of showing that U(t,t) does nothing to your state and therefore must be equal to the identity is a good start. However, you are correct in feeling that it may be a bit sketchy. This is because you are assuming that U(t,t) is equal to the identity without any justification. Instead, you could try to use the definition of a unitary operator, which is U^+U=UU^+=I, and try to show that this is satisfied for U(t,t).

Your second approach is also valid. You have correctly used the definition of a unitary operator and have shown that U^+U=I. This is a more rigorous approach and shows that U is indeed a unitary operator.

In conclusion, your second approach is more valid and convincing. Keep up the good work!