Why all operators in QM have a Hermitian Matrices

1. Nov 24, 2012

mwalmasri

Why all operators in QM have a Hermitian Matrices ?

2. Nov 24, 2012

dextercioby

They don't. Only the self-adjoint do.

3. Nov 24, 2012

mwalmasri

yes so all operators have a self-adjoint Matrices, which operator can be Represented as anti-Hermitian operators?

4. Nov 24, 2012

cattlecattle

It's not true that all operators are Hermitian, The simplest example would be i times any Hermitian operator, it's anti-Hermitian.
You may be asking why any operator representing observable is Hermitian. There is no proof for this because this is simply one of the postulates of QM. But if it's not true, you will have an observable that has complex-value eigenvalue, which doesn't make any physical sense.

5. Nov 24, 2012

mwalmasri

yes,when I asked the question I mean a physical operator like a Hamiltonian.... maybe I must be clear enough in my question. Hermitian is used because its have a real eigenvalue that is right....
Thanks

6. Nov 25, 2012

andrien

The eigenvalues of a hermitian operator are real,like hamiltonian which should be hermitian operator because it's eigenvalues are simply energy,which should be a real quantity.So all observables are associated with hermitian operator.
Assume a hermitian operator,and the eigenvalue eqn
A|a>=a|a>,assuming normalization of eigenstates,multiplying by <a|
<a|A|a>=a
taking complex conjugate of both sides,
<a|A|a>*=a*,
By hermiticity condition,<a|A|a>=<a|A|a>*,so a=a* implying reality of eigenvalues.

Last edited: Nov 25, 2012
7. Nov 25, 2012

bhobba

Basically it's a postulate of QM.

The way I like to look at it however is as follows. Suppose we have some observational apparatus with n possible outcomes that have some real number yi assigned to each outcome. List them out as a vector and write it as sum yi |bi>. Now we come to a problem - its not basis independent - change to another basis and the yi change - but since the choice of basis is entirely arbitrary we expect nature to be independent of that choice. To get around that problem QM simply replaces the |bi> by |bi><bi| to give sum yi |bi><bi| which is the same regardless of basis. It is a Hermitian operator whose eigenvalues are the possible outcomes of the observation.

Thanks
Bill