Determining Hermitian Operators: Cases 1 and 2

[James R]

My memory is fading. Can somebody please remind me how I would go about determining in each of the following cases whether the operator A is Hermitian or not?

Case 1.

A\psi(x) = \psi(x+a)

Case 2.

A\psi(x) = \psi^*(x)

where the star indicates complex conjugation.

 

[vanesch]

My memory is fading. Can somebody please remind me how I would go about determining in each of the following cases whether the operator A is Hermitian or not?

Case 1.

A\psi(x) = \psi(x+a)

Case 2.

A\psi(x) = \psi^*(x)

where the star indicates complex conjugation.

take two vectors, f and g. In your case, these are functions of x.
Now calculate ff = A f and gg = A g.
If A is hermitean, then <gg,f> = <g,ff> for all f and g.

 

[dextercioby]

Assume for simplicity that \hat{A} is an linear operator continuous on the Hilbert space L^{2}\left(\mathbb{R}\right) in which the scalar product between 2 arbitrary vectors is

\langle \psi,\phi \rangle =\int_{-\infty}^{+\infty} dx \ \psi^{*}(x) \phi (x).

If an operator described above is symmetric, then

\langle \psi, \hat{A}\phi\rangle =\langle \hat{A}\psi, \phi\rangle , \forall \psi,\phi \in \mathcal{H}

So check both operators now.

Daniel.

 

[James R]

Thanks.....