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

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!

• $\mathcal{U}$ 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. Jan 1 at 9:06
• @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. Jan 1 at 9:11
• $U$ may not be hermitian but it is unitary so the action on the left is well defined. Jan 1 at 14:41
• Essentially a duplicate of physics.stackexchange.com/q/502606/2451 Jan 2 at 11:32

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 $$\langle \hat{A}^{\dagger}\psi,\phi \rangle$$.

--

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

• Thanks! And $\langle \hat{A}^{\dagger}\psi,\phi \rangle=\langle \psi,\hat{A}\phi \rangle$ by definition! I'm sorry for forgetting that, haha. Jan 1 at 9:28
• $\uparrow$ Yes. Jan 1 at 9:29

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

• Right! I found that Sakurai has mentioned it in Chapter1! It's a basic property. Jan 2 at 2:58