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This might be a stupid question, but how does one derive the TDSE (time-dependent Schrödinger equation) for the dual state $\langle \Psi(t)|$?

Here is my try:

TDSE: $i\hbar\ \partial_t |\Psi(t)\rangle = \hat{H} |\Psi(t) \rangle$. Now we just take the adjoint on both sides:

  • $(\hat{H}|\Psi(t)\rangle)^\dagger\stackrel{?}{=} |\Psi(t)\rangle^\dagger\hat{H}^\dagger=\langle\Psi(t)| \hat{H}$

  • $(i\hbar\ \partial_t|\Psi(t)\rangle)^\dagger\stackrel{?}{=}|\Psi(t)\rangle^\dagger \partial_t\ ^\dagger(i\hbar)^\dagger= \langle\Psi(t)|\partial_t \ (-i\hbar)=\frac{\hbar}{i}\langle\Psi(t)|\partial_t$

So, I get: $$(i\hbar\ \partial_t |\Psi(t)\rangle)^\dagger = (\hat{H} |\Psi(t) \rangle)^\dagger \qquad\Rightarrow\qquad\frac{\hbar}{i}\langle\Psi(t)|\partial_t=\langle\Psi(t)| \hat{H}$$ which seems weird? Is this correct or wrong? Shouldn't $\langle \Psi(t) |$ evolve in time as well? I would expect something like $(\cdot\cdot\cdot)\partial_t \langle\Psi(t)| = (\cdot\cdot\cdot)$. Or do I have to further manipulate the equation to get the final result?

The step $\stackrel{?}{=}$ is where I'm uncertain. When can I swap the order and when can't I? I just assumed the operator rule: $(\hat{A}\hat{B})^\dagger=\hat{B}^\dagger \hat{A}^\dagger$ could be used like this $(\hat{A}|\Psi\rangle)^\dagger=\langle \Psi| \hat{A}^\dagger$, which seems right? $(\langle a|\hat{A}| a\rangle)^\dagger=|a\rangle^\dagger \hat{A}^\dagger \langle a|^\dagger=\langle a|\hat{A}^\dagger|a\rangle$

Can $\langle \Psi | \partial_t$ be interpereted as $\partial_t$ acting on $\langle \Psi |$? I think the thing is that when we have $\langle a|\hat{A}| a \rangle$ we can do either $\langle a, \hat{A}| a \rangle$ or $\langle a | \hat{A},a \rangle$, where we interperet $\langle a, \hat{A}|=\langle a|\hat{A}$ as one bra or $|\hat{A},a\rangle=\hat{A}|a\rangle$ as one ket.Is this the solution? That we interperet $\langle \Psi|\partial_t$ as the bra $\langle\Psi,\partial_t|$?

Any comments to clear up my misunderstanding would be helpful. I think this probably is very simple and I'm just missing something obvious here (?).

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    $\begingroup$ This for sure has been asked here before...in any case, my comments here should (perhaps some modifications are needed if e.g. the Hamiltonian is unbounded; but the idea is the same). The problem you have seems partly/mostly due to the use of Dirac notation, which IMHO is not good, especially if one doesn't know what exactly is happening... $\endgroup$
    – Tobias Fünke
    Commented Dec 7 at 17:10
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    $\begingroup$ ...to add: You should not think of the derivative as an operator on the Hilbert space (this has been asked and answered/discussed here on PSE also quite often; if you search for it you should find a few good posts); so I'd recommend to not write something like $\partial_t^\dagger$. $\endgroup$
    – Tobias Fünke
    Commented Dec 7 at 17:21

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