Is the Displacement Operator Tψ(x)=ψ(x+a) Hermitian?

[Rude]

Consider the displacement operator Tψ(x)=ψ(x+a). Is T Hermitian?

 

[The_Duck]

To get you started, what's the definition of a Hermitian operator?

 

[Rude]

here is the definition: <f│Ag>=<Af│g> always if A is Hermition.
Don't know how to start.

 

[The_Duck]

here is the definition: <f│Ag>=<Af│g> always if A is Hermition.

Suppose the wave function of the state |g> is $\psi_g(x)$ and the wave function of state |f> is $\psi_f(x)$. Then what does the above definition of an operator A being Hermitian say if you translate the inner product into an integral of wave functions, using

$\langle a | b \rangle = \int dx \psi_a(x)^* \psi_b(x)$

and plug in A = T?

 

[Rude]

Is this what you are suggesting?

<f│Tg>=∫dxΨ(x)*ψ(x+a)

If so I don't know how to proceed.

It does not look like this would give <Tf│g> but don't know why.

 

[The_Duck]

Is this what you are suggesting?

<f│Tg>=∫dxΨ(x)*ψ(x+a)

Yup.

It does not look like this would give <Tf│g> but don't know why.

If you think they aren't equal, perhaps you can find an explicit counterexample?