Linear Operators: Relationship Between Action on Kets & Bras

[aaaa202]

This is basically more of a math question than a physics-question, but I'm sure you can answer it. My question is about linear operators. If I write an operator H as (<al and lb> being vectors):
<alHlb>
What is then the relationship between H action the ket and H action on the bra. Is this for example true:
<alHlb> = H <alb>
Which rules determine how the operator acts?
Can it be thought of as a scalar such that instead:
<alHlb> = H* <alb>
My question came naturally as I was applying the position-state <xl to the left side of the schrödinger-equation (H being the hamiltonian and lψ(t)> the state vector:
<xlHlψ(t)>
does this equal to:
H<xlψ(t)>
I reckon it must because I've seen it work the other way around, but why is it precisely so. Doesn't it have to do with the fact that H is hermitian?

 

[strangerep]

This is basically more of a math question than a physics-question, but I'm sure you can answer it. My question is about linear operators. If I write an operator H as (<al and lb> being vectors):
<alHlb>
What is then the relationship between H action the ket and H action on the bra. Is this for example true:
<alHlb> = H <alb>
Which rules determine how the operator acts?
Can it be thought of as a scalar such that instead:
<alHlb> = H* <alb>
My question came naturally as I was applying the position-state <xl to the left side of the schrödinger-equation (H being the hamiltonian and lψ(t)> the state vector:
<xlHlψ(t)>
does this equal to:
H<xlψ(t)>
I reckon it must because I've seen it work the other way around, but why is it precisely so. Doesn't it have to do with the fact that H is hermitian?

You can't move the operator outside the inner product like that.
An operator is a mapping between vector spaces. Think of it like
a machine: you give it one vector and it gives you another vector back.
However, the inner product $$\langle a | b \rangle$$ is a scalar,
so in your equation

$$\langle a | H | b \rangle ~=~ H \langle a | b \rangle$$

you have a scalar on the LHS but an operator on the RHS,
which is mathematical nonsense, in general.

 

[aaaa202]

But in the question about the schr. equation. Does this not equal to?
<xlHlψ(t)>
<=>
H<xlψ(t)>
<xl is a position eigen-state and lψ(t)> the state vector.