Particle oscillating between two wells

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
 
Physics news on Phys.org
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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If we release an electron around a positively charged sphere, the initial state of electron is a linear combination of Hydrogen-like states. According to quantum mechanics, evolution of time would not change this initial state because the potential is time independent. However, classically we expect the electron to collide with the sphere. So, it seems that the quantum and classics predict different behaviours!

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