# Problem based hermitian operator

1. Nov 3, 2007

### quantum_prince

problem based on hermitian operator

1. The problem statement, all variables and given/known data

A is an hermitian operator and as we know the eigenstates a of A with eigenvalues a satisfy A psi a = a psi a.

How do we show that lambda psi a (lambda is a non zero complex number) is an eigen state belonging to the same eigen value a. We also need to say whether the eigen value is non degenerate or not.

2. Relevant equations

We know the property of hermitian operator

(-infinity to + infinity) integral phi* A psi dx = (-infinity to + infinity) psi (A phi)* dx

3. The attempt at a solution

(-infinity to + infinity) integral phi* A (lambda psi a) dx =

(-infinity to + infinity) lambda psi a (A phi)* dx

Dont know how to proceed from here.

Regards.

QP.

Last edited: Nov 3, 2007
2. Nov 3, 2007

### Gokul43201

Staff Emeritus
Your notation is confusing. Please write down the original question, EXACTLY as it was given to you, and it would help if you used either clearer notation, or the latex capability built in here. From what I can make of it, there seems to be a mistake in the question.

Examples for latex:

$$A \psi_a = a \psi_a$$

$$A |a \rangle = a | a \rangle$$

3. Nov 3, 2007

### quantum_prince

Here it comes.Could you please tell me where the mistake is.

The eigenstates $$\psi_a$$ of A with eigenvalues a satisfy
the equation $$A \psi_a = a \psi_a$$. We need to show that then also $$\Lambda\psi_a$$ with an arbitrary complex number $$\Lambda$$ not equal to 0, is an eigenstate belonging to the same eigenvalue a.Need to check if the eigenvalue is nondegenerate or not

Last edited: Nov 3, 2007
4. Nov 3, 2007

### noospace

I don't understand the question, obviously if $\psi_a$ is an eigenstate of $\hat{A}$ with eigenvalue a then $\Lambda \psi_a$ is also an eigenstate with the same eigenvalue. This just follows from the fact that A is linear.

5. Nov 3, 2007

### quantum_prince

is there no proof for this. I mean cant we take a complex number of the form z=x+iy and prove the same.Since A is linear how can we conclude that,is there no mathematical proof for this.

6. Nov 3, 2007

### noospace

Sure. Let $\hat{A}$ be any linear operator with eigenvalue a and $\Lambda$ any complex number.

Then

$\hat{A} (\Lambda \psi) = \Lambda \hat{A} \psi = \Lambda a \psi = a (\Lambda \psi)$.

7. Nov 3, 2007

### quantum_prince

So as I understand Hermitian operator is also a linear operator, which is why we could perform this operation.

If B is a second commutation operator which commutes with A such that

[A,B] = 0.

How do we show that $$A \psi = a \psi$$ and $$B \psi = b \psi$$ ? i.e to show that these operators have simultaneous eigenstates.

All eigen values in this problem are degenerate.

Is this proof correct:
Since

if [A,B] = 0
AB = BA
AB-BA = 0

$$AB \psi = Ab \psi = bA\psi = ba\psi = ab\psi = aB\psi = Ba\psi = BA \psi$$

therefore

$$A \psi = a \psi$$ and $$B \psi = b \psi$$

8. Nov 3, 2007

### noospace

If A,B commute then AB = BA. Suppose $\psi$ is an eigenstate of A with corresponding eigenvalue a. Then

$A\psi = a \psi$
$BA\psi = a B\psi$
$A(B\psi) = a (B\psi)$

Thus $B\psi$ is an eigenstate of A with the same eigenvalue. Thus if the eigenvalue is non-degenerate $B\psi$ is proportional to $\psi$.

9. Nov 3, 2007

### quantum_prince

We know that the hermitian operator S interchanges coordinate x with −x. On consider
the Hamiltonian $$H = p^2/2m + V (x)$$ for a particle in a one dimensional
symmetric potential V (x) = V (−x). What can be said
about the wavefunctions in view of what we proved above i.e $$A \psi = a \psi and B \psi = b \psi$$.

How to proceed in this case?.