# Particle oscillating between two wells

1. Nov 24, 2008

### maverick280857

Hi

Suppose we a pair of symmetric wells of finite potential and the particle is given to be in the initial state

$$|\psi(0)\rangle = \frac{1}{\sqrt{2}}(|\psi_{s}\rangle + |\psi_{a}\rangle)$$

(a = antisymmetric state, s = symmetric state)

For t > 0, we have

$$|\psi(t)\rangle = \frac{1}{\sqrt{2}}e^{-iE_{S}t/\hbar}(|\psi_{s}\rangle + e^{-it/\tau}|\psi_{a}\rangle)$$

where $\tau = \hbar\pi/(E_{a}-E_{s})$

We see that the particle oscillates between the two wells, but the expectation value of the energy

$$\langle\psi(t)|H|\psi(t)\rangle$$

is constant and equals $(E_{s}+E_{a})/2$.

I have two questions:

1. What is the physical significance of this?

2. Is this due to the specific initial state given?

Cheers
Vivek

2. Nov 24, 2008

### Avodyne

1. Energy is conserved!

2. No. Take any time-independent hamiltonian, and express the initial state as a superposition of energy eigenstates |n>, with coefficients c_n. The probability that the system has energy E_n is then |c_n|^2. Time evolution changes the phase, but not the magnitude, of each c_n. So the probability |c_n|^2 is constant in time. The expectation value of the energy is just the sum of |c_n|^2 E_n, so this is constant as well.

3. Nov 24, 2008

### maverick280857

Yes, of course ... I forgot

Thanks Avodyne.