Particle oscillating between two wells

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SUMMARY

The discussion centers on a particle oscillating between two symmetric wells of finite potential, described by the initial state |\psi(0)\rangle = \frac{1}{\sqrt{2}}(|\psi_{s}\rangle + |\psi_{a}\rangle). For t > 0, the state evolves to |\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}). The expectation value of energy remains constant at (E_{s}+E_{a})/2, demonstrating that energy is conserved despite the oscillation. The initial state does not affect this conservation, as any superposition of energy eigenstates will yield a constant expectation value over time.

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maverick280857
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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?

Thanks in advance.
Cheers
Vivek
 
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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.
 
Avodyne said:
1. Energy is conserved!

Yes, of course ... I forgot :rolleyes:

Thanks Avodyne.
 

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