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- Summary
- Under time-dependent unitary transformations, the Hamiltonian transforms in a different way than other operators, and I'm trying to understand why.

Hi!

I recently came across a quantum mechanics problem involving a change of basis to a rotating basis. As part of the solution, I wanted to transform the Hamiltonian operator into the rotating basis. Since the new basis is rotating, the basis change operator is time-dependent. This led to a different transformation law for the Hamiltonian than what I expected, and I'm trying to understand how one could expect this.

The transformation law for operators which I normally use is of course

$$A' = UAU^\dagger$$.

The proof that I remember for this is to introduce an eigenbasis ##|a\rangle## for ##A## so that ##A|a\rangle = a|a\rangle##, and then asking that, if ##|a'\rangle=U|a\rangle##, ##A'|a'\rangle = a|a'\rangle##. In that case:

$$U^\dagger A'U|a\rangle = U^\dagger A'|a'\rangle = aU^\dagger |a'\rangle = a|a\rangle = A|a\rangle \Rightarrow A' = UAU^\dagger$$.

However, if one assumes that ##U## is time-dependent and inserts ##|\psi'\rangle = U|\psi\rangle## into the Schrödinger equation for ##|\psi'\rangle##, one finds (##\hbar=1##):

$$i\frac{d}{dt}|\psi'\rangle = iU\frac{d}{dt}|\psi\rangle + i(\frac{d}{dt}U)|\psi\rangle = H'|\psi'\rangle=H'U|\psi\rangle$$

which after rearrangement is

$$i\frac{d}{dt}|\psi\rangle = U^\dagger(H'U - i(\frac{d}{dt}U))|\psi\rangle$$

which prompts the identification ##H = U^\dagger H'U - iU^\dagger\frac{d}{dt}U##, i.e. the transformation law

$$H' = UHU^\dagger + i(\frac{d}{dt}U)U^\dagger$$

which reduces to the "normal" law for time-independent transformations but is otherwise different.

I guess one has to accept this result, but my question is: Where does the proof of the "normal" law fail for the case of the Hamiltonian? I don't see the hole which lets the Hamiltonian be exempt from the first proof. Any clarification of this would be of much help!

Thanks!

I recently came across a quantum mechanics problem involving a change of basis to a rotating basis. As part of the solution, I wanted to transform the Hamiltonian operator into the rotating basis. Since the new basis is rotating, the basis change operator is time-dependent. This led to a different transformation law for the Hamiltonian than what I expected, and I'm trying to understand how one could expect this.

The transformation law for operators which I normally use is of course

$$A' = UAU^\dagger$$.

The proof that I remember for this is to introduce an eigenbasis ##|a\rangle## for ##A## so that ##A|a\rangle = a|a\rangle##, and then asking that, if ##|a'\rangle=U|a\rangle##, ##A'|a'\rangle = a|a'\rangle##. In that case:

$$U^\dagger A'U|a\rangle = U^\dagger A'|a'\rangle = aU^\dagger |a'\rangle = a|a\rangle = A|a\rangle \Rightarrow A' = UAU^\dagger$$.

However, if one assumes that ##U## is time-dependent and inserts ##|\psi'\rangle = U|\psi\rangle## into the Schrödinger equation for ##|\psi'\rangle##, one finds (##\hbar=1##):

$$i\frac{d}{dt}|\psi'\rangle = iU\frac{d}{dt}|\psi\rangle + i(\frac{d}{dt}U)|\psi\rangle = H'|\psi'\rangle=H'U|\psi\rangle$$

which after rearrangement is

$$i\frac{d}{dt}|\psi\rangle = U^\dagger(H'U - i(\frac{d}{dt}U))|\psi\rangle$$

which prompts the identification ##H = U^\dagger H'U - iU^\dagger\frac{d}{dt}U##, i.e. the transformation law

$$H' = UHU^\dagger + i(\frac{d}{dt}U)U^\dagger$$

which reduces to the "normal" law for time-independent transformations but is otherwise different.

I guess one has to accept this result, but my question is: Where does the proof of the "normal" law fail for the case of the Hamiltonian? I don't see the hole which lets the Hamiltonian be exempt from the first proof. Any clarification of this would be of much help!

Thanks!