Why can the time evolution operator be left or right multiplied on bra or ket? [duplicate]

Question

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!

Answer 1

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.

Answer 2

For any linear operator, hermitian or not, we have $$ (\langle \psi |A)|\chi\rangle = \langle \psi |(A|\chi\rangle). $$ See the discussion here