# Quantum unitary transformation

In quantum mechanics, we know $$\dot{\psi}=-\frac{i}{\hbar}H\psi$$,

but why is $$U\dot{\psi}=-\frac{i}{\hbar} \left(UHU^\dagger \right) U\psi$$?

Does it mean $$UHU^\dagger = H$$ ? I think $$UU^\dagger H = H$$, but why can we change the order of the matrices here?

You're overthinking this, assuming $$U$$ is unitary:

$$U\dot\psi= -\frac{i}{\hbar} UH\psi=-\frac{i}{\hbar} UH\mathbb 1\psi= -\frac{i}{\hbar} UHU^\dagger U\psi.$$

$$U$$ need not be the time evolution operator and it need not commute with $$H$$ for this to work, it can be any unitary. This is just saying that if you write $$\psi$$ in another basis then it evolves with the Hamiltonian written in the new basis. (Or equivalently that a rotated vector evolves with the rotated Hamiltonian).

1. If the Hamiltonian $$\hat{H}$$ does not depend on time, and $$U$$ is supposed to be the time-evolution operator, then $$\hat{U}~=~\exp\left(-\frac{i}{\hbar}\hat{H}\Delta t\right),\tag{A}$$ which commutes$$^1$$ with $$\hat{H}$$, so that $$UHU^{\dagger} ~=~ H,\tag{B}$$ cf. OP's question.

2. If the Hamiltonian $$\hat{H}$$ do depend on time, then eqs. (A) & (B) need to be modified, cf. e.g. this Phys.SE post.

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$$^1$$ A function $$f(\hat{H})$$ of $$\hat{H}$$ commutes with $$\hat{H}$$, cf. e.g. this & this Phys.SE posts.

• But OP's calculation would be correct even if the Hamiltonian depended on time, right? Nov 30, 2020 at 14:31
• Moreover OP never mentioned $U$ is time evolution Nov 30, 2020 at 14:32
• Thanks! Could you please explain a little about why H and U commute if U is the time-evolution operator and H does not depend on time? Nov 30, 2020 at 17:18
• I updated the answer. Nov 30, 2020 at 17:35

user2723984 is correct. However, the 2nd part of your question is unresolved: if the Hamiltonian commutes with itself at different times, then the only operator in $$U$$ is $$H$$ and, as $$H$$ commutes with itself, the order of the operators may then be changed.

• OP never mentioned $U$ is time evolution I think Nov 30, 2020 at 14:32
• Yes, I assumed this since it is standard notation. Nov 30, 2020 at 14:34