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Consider a one-dimensional quantum particle with the Hamiltonian

(1)Suppose that *m* suddenly changes from *m _{0}* to $m_1 = m_0 / \lambda$ at

*t*= 0. Assuming the particle was in the ground state at

*t*< 0, find: (i) probability to remain in the ground state at

*t*> 0 and (ii) change in the energy expectation value $\left< H \right >$. Consider two cases:

(a) Infinite square well, i.e., *V(x)* = 0 for 0 < *x* < *L* and infinite otherwise.

(b) Parabolic well, *V(x)* = *C x ^{2}/2*. Hint: For any

*a*> 0,

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Answer

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In a sudden perturbation the wavefunction does not change; hence, the probability in question is the squared overlap of the two groundstate wavefunctions, $\left< \psi_1 | \psi_0\right>^2$. The potential energy $\left < V \right >$ also does not change. Therefore,

(3)(a) In this case the ground-state wavefunction (0 = sin(&x/L) at t < 0 is also the ground state (1 at t > 0; hence the probability to remain in the ground state is 1. The change in energy is

(4)(b) The normalized ground-state wavefunction is $\psi = \pi^{-1/4}l^{-1/2}\exp{(-x^2/2l^2)}$, where $l = (\hbar^2/m C)^{1/4}$. The overlap is

(5)The probability in question is

(6)The change in energy is

(7)
1. Is the delta <H> for the HO potential case simply the difference in the new and old gnd. state energies? If so, then

(i) Where does the 1/4 factor come from? Shouldn't it be 1/2?

(ii) Shouldn't it involve sqrt(lambda) instead of lambda?

(iiI) Moreover, it shouldn't even be just the difference of gnd state energies, since the wavefunction is no longer an energy eigenstate, hence, there is a non-zero probability of finding the particle in the (new) excited states. Thus, shouldn't the <H> (t > 0) be a sum of energies of these excited states, weighed by the probabilities of finding it in the respective states? This is what we will get if we compute <H> as: <H> = <psi_existing|new H|psi_existing>

Would appreciate any clarification on this. Thanks.

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