# Proof on Operator Properties

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1. May 30, 2015

### Je m'appelle

Hello, could you please give me an insight on how to get through this proof involving operators?

Proof:
Given an eigenvalue-eigenvector equation, suppose that the vectorstate depends on an external parameter, e.g. time, and that over it acts an operator that is the fourth derivative w.r.t. time. If this operator is hermitian, find the most general operator possible that satisfies these conditions and what are the boundary conditions on the eigenfunctions that are needed.

Attempt:

Using $A$ as our hermitian operator and $|\psi (t) \rangle$ for its time dependent eigenvector, and $a(t)$ for its eigenvalue, I suppose the simplest eigenvalue-eigenvector equation one can write would be

$$A|\psi (t) \rangle = a(t) |\psi (t) \rangle$$

As for the operator, I'd say it follows directly that

$$A \sim \frac{\partial^4}{\partial t^4}$$

$$a(t) = a^*(t)$$

Now, based on the assumption I've been correct so far, comes the part where I'm stuck. I understand the general concept of eigenvalues and eigenfunctions, i.e. $a(t)$ would generate a set of eigenfunctions $\psi(t)$, however I'm not fully aware which boundary conditions would be necessary from the provided information aside from the fact that $a(t)$ is a real number. Any hints?

2. May 30, 2015

### nrqed

Bonjour "Je m'appelle".... En ce qui me concerne, moi je m'appelle Patrick :-)

Write the operator as $C \partial^4/\partial t^4$ where C is a constant. What is the basic condition for such an operator to be hermitian?

3. May 30, 2015

### Je m'appelle

Bonjour Patrick! Merci pour ta réponse. Ça va? As for your question, I'd say it's the operator being self-adjoint, i.e. $A = A^{\dagger}$, also I believe I know where you're going at, I should have written $A = i \frac{\partial^4}{\partial t^4}$ instead?