# Time dependent eigenvalues?

1. Nov 15, 2007

### pellman

Given an operator $$\hat{Q}$$ (in the Schrodinger picture) in non-relativistic quantum mechanics and a state $$|\psi(t)\rangle$$ such that

$$\hat{Q} |\psi(t)\rangle=q(t)|\psi(t)\rangle$$

where q(t) is explicitly time-dependent, can we properly say that $$|\psi(t)\rangle$$ is an eigenstate of Q with a time-dependent eigenvalue. That is, $$|\psi(t)\rangle$$ remains a eigenstate of Q for all times but its eigenvalue is different depending on when you measure it?

For example, suppose we had a wave function of the form

$$\psi(x,t)\propto e^{ixg(t)+h(t))}$$

then applying the momentum operator we find

$$-i\frac{\partial}{\partial x}\psi(x,t)=g(t)\psi(x,t)$$.

Would you say that $$\psi(x,t)$$ is momentum eigenstate with momentum = g(t)?

Last edited: Nov 15, 2007
2. Nov 15, 2007

### Magister

This wave function is a eigenfunction of the Hamiltonian, and because the momentum operator commutes with the hamiltonian, it is also a eigenfunction of the momentum. You can write this wave function as

$$\psi(x,t)\propto e^{i(p x-E t))}$$

Where p corresponds to the eigenvalue of the momentum operator and E corresponds to the eigenvalue of the energy operator.