How to Show Potential Operator Acting on Eigenstates in Quantum Mechanics?

[gotjrgkr]

Hi! I was studying Shankar's Principles of Quantum Mechanics, but I was stuck to understand a relation concerned with operators.

Homework Statement

I've learned that in order to get the equation of motion, I should apply the Schrödinger's equation to the given Hamiltonian operator. In the book, it is said that a state function \lvert\psi(t)\rangle is represented by a product of the propagator and the initial state function \lvert\psi(0)\rangle for a time-independent Hamiltonian operator.

In p. 149 of the book, to get the propagator expressed as eigenstates of the Hamiltonian operator, a way is introduced to get those eigenfunctions. More specifically, the Hamiltonian operator given in p.149 is H=\frac{P^2}{2m} + \frac{1}{\cosh^2 X} where X is the position operator. Then the eigenstates \lvert E\rangle should satisfy H\lvert E\rangle = E \lvert E\rangle. If I put bra $\langle x \rvert$ to both sides of the equation, then $\langle x \rvert H \lvert E \rangle = \langle x \rvert E \lvert E\rangle$. This implies according to the book,
$$\left(-\frac{\hbar^2}{2m} \frac{d^2}{dx^2}+\frac{1}{\cosh^2 x}\right)\psi_E(x) = E\psi_E(x)$$ where $\psi_E(x) = \langle x \vert E \rangle$.

What I want to know is as follows. I think the above equation makes sense only if I show $\langle x \rvert \frac{1}{\cosh^2 X} \lvert E\rangle = \frac{1}{\cosh^2 x} \langle x \vert E\rangle$ (I've omitted other terms). However, I don't have an idea how to prove it. Even though I try to use completeness property, it's not easy. Could you give me a hint?

I think this kind of relation should hold generally. What I mean is for any potential operator V(X), if I put $\langle x \rvert$ and $\lvert E \rangle$ to both sides of it, I should get $V(x)\langle x \vert E\rangle$. Am I right? I want to check this as well. (Be careful in distinguishing the letters X, x.)

I hope somebody else answer my question. Thank you for reading my long question.

Homework Equations

The Attempt at a Solution

 

[Oxvillian]

Perhaps this trick can be of use to you:

\langle x \lvert V(X) \lvert \psi\rangle = \langle \psi\lvert V(X)\lvert x\rangle^* = V(x) \langle \psi\lvert x \rangle^* = V(x) \langle x\lvert \psi \rangle

 

[gotjrgkr]

For me, the second equality is still mystery... ;;
Could you give me more hint??

 

[Oxvillian]

All it is is the definition of the adjoint of V (which is V itself because V is self-adjoint).

Shankar p.26

 

[Oxvillian]

btw I find the "math notation" to be clearer than the Dirac notation for this kind of thing:

\langle x \lvert V \lvert \psi \rangle = (x, V\psi) = (V^{\dagger} x, \psi) = (Vx, \psi)