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In the book Modern Quantum Mechanics by Sakurai, it said that:

We see that the expansion coefficients of a state ket in terms of base kets are the same in both pictures: $$ \begin{aligned} & c_{a^{\prime}}(t)=\underbrace{\left\langle a^{\prime}\right|}_{\text {base bra }} \cdot \underbrace{\left(\mathscr{U}\left|\alpha, t_0=0\right\rangle\right)}_{\text {state ket }} \text { (the Schrödinger picture) } \\ & c_{a^{\prime}}(t)=\underbrace{\left(\left\langle a^{\prime}\right| \mathscr{U}\right)}_{\text {base bra }} \cdot \underbrace{\left|\alpha, t_0=0\right\rangle}_{\text {state ket }} \text { (the Heisenberg picture). } \end{aligned} $$

In this case, it seems that the operator $\mathscr{U}$ could left- or right- multiply on the bra or ket, that looks like saying that this operator is Hermitian.

But $$\mathscr{U}=\exp\left\{iHt/\hbar\right\}$$ is not seems to be Hermitian.

I can not figure out where is the mistake!

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  • $\begingroup$ $\mathcal{U}$ is an unitary operator, but here we have to see what is evolving in the respective picture. Since base kets evolve in Heisenberg picture, the unitary operator should act on the base ket but not necessarily like $e^{iHt/\hbar}|\psi\rangle$, but this refers the time evolution of the base kets. $\endgroup$ Commented Jan 1, 2023 at 9:06
  • $\begingroup$ @TanmoyPati You are right! But more precisely, I think$<a'|(\mathscr{U}|a(0)>)\neq(<a'|\mathscr{U})|a(0)>$ because of $\mathscr{U}$ is not hermitian. $\endgroup$
    – PhyDuck
    Commented Jan 1, 2023 at 9:11
  • $\begingroup$ $U$ may not be hermitian but it is unitary so the action on the left is well defined. $\endgroup$ Commented Jan 1, 2023 at 14:41
  • $\begingroup$ Essentially a duplicate of physics.stackexchange.com/q/502606/2451 $\endgroup$
    – Qmechanic
    Commented Jan 2, 2023 at 11:32

2 Answers 2

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Dirac (and Sakurai) are thinking of a bra $$\langle \psi|~=~|\psi \rangle^{\dagger}$$ as a Hermitian adjoint ket $|\psi \rangle$. So by that logic an operator $\hat{A}$ acting from the right on a bra $\langle \psi |$ means $$\langle \psi | \hat{A}~=~(\hat{A}^{\dagger} |\psi \rangle)^{\dagger}.$$

TL;DR: The Dirac bra-ket physics notation $( \langle \psi | \hat{A}) ~|\phi \rangle$ translates to the mathematical inner product$^1$ notation $\langle \hat{A}^{\dagger}\psi,\phi \rangle$.

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$^1$ Be aware that by convention the sesquilinear form $\langle \cdot,\cdot \rangle$ is conjugated $\mathbb{C}$-linear in the first (not the second) entry.

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    $\begingroup$ Thanks! And $\langle \hat{A}^{\dagger}\psi,\phi \rangle=\langle \psi,\hat{A}\phi \rangle$ by definition! I'm sorry for forgetting that, haha. $\endgroup$
    – PhyDuck
    Commented Jan 1, 2023 at 9:28
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    $\begingroup$ $\uparrow$ Yes. $\endgroup$
    – Qmechanic
    Commented Jan 1, 2023 at 9:29
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For any linear operator, hermitian or not, we have $$ (\langle \psi |A)|\chi\rangle = \langle \psi |(A|\chi\rangle). $$ See the discussion here

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  • $\begingroup$ Right! I found that Sakurai has mentioned it in Chapter1! It's a basic property. $\endgroup$
    – PhyDuck
    Commented Jan 2, 2023 at 2:58

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