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Proof: Given an eigenvalue-eigenvector equation, suppose that the state vector 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 $\hat{A}$ as our hermitian operator and $|\psi (t) \rangle$ for its time dependent eigenvector, and $\lambda$ for its eigenvalue, I suppose the simplest eigenvalue-eigenvector equation one can write would be

$$\hat{A}|\psi (t) \rangle = \lambda |\psi (t) \rangle$$

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

$$\hat{A} \sim \frac{\partial^4}{\partial t^4}, \ \ \ \lambda = \lambda^*$$

Now, based on the assumption I've been correct so far, the eigenfunctions are given by

$$\frac{d^4}{dt^4} \psi(t) = \lambda \psi(t)$$

And $\psi(t) = e^{\alpha t}$ as a possible solution immediately comes to mind, so

$$\frac{d^4}{dt^4}\left( e^{\alpha t} \right) - \lambda \left( e^{\alpha t} \right) = 0 \ \Rightarrow \alpha_{1,2} = \pm \sqrt[4]{\lambda}$$

And the general solution for the eigenfunctions is

$$\psi (t) = c_1 e^{\sqrt[4]{\lambda} t} + c_2 e^{-\sqrt[4]{\lambda}t}$$

Once again assuming I've been correct thus far, I can't find out from the provided information which boundary conditions would be required for this.

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  • $\begingroup$ I've been thinking, since the operator is hermitian and its eigenfunctions must orthogonal, the boundary conditions are given by an analysis of $\int dt \ \psi_1^*(t) \psi_2(t) =0$? $\endgroup$ – phyundergrad Jun 11 '15 at 7:06

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