Fall 2007 5 Qm

Consider a one-dimensional quantum particle with the Hamiltonian

(1)
\begin{align} H = T + V(x), \;\; T = - \frac{\hbar^2}{2 m}\frac{d^2}{dx^2} \end{align}

Suppose that m suddenly changes from m0 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 x2/2. Hint: For any a > 0,

(2)
\begin{align} \int_{- \infty}^{\infty} \exp{(- a x^2)} = \sqrt{\frac{\pi}{a}} \end{align}

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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)
\begin{align} \left< H \right> |_{t=-0}^{t=+0} = \left< T \right> |_{t=-0}^{t=+0} = (\lambda - 1) \left< T \right> |_{t=-0} \end{align}

(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)
\begin{align} \left< H \right> |_{t=-0}^{t=+0} = \frac{\lambda -1}{2}\frac{\pi^2 \hbar^2}{m_0 L^2} \end{align}

(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)
\begin{align} \left< \psi_1 | \psi_0 \right> = \frac{1}{\sqrt{\pi}} \frac{1}{(l_0l_1)^{1/2}} \int_{-\infty}^{\infty} dx \exp{\left[ - \frac{x^2}{2}\left(\frac{1}{l^2_0} + \frac{1}{l^2_1}\right)\right]} = \sqrt{\frac{2 l_0 l_1}{l_0^2+l_1^2}} \end{align}

The probability in question is

(6)
\begin{align} |\left< \psi_1 | \psi_0 \right>|^2 = \frac{2 l _0 l _1}{l_0^2+l_1^2} = \frac{2(l_1/l_0)}{1+(l_1^2/l_0^2)}= \frac{2\lambda^{1/4}}{1+\lambda^{1/2}} < 1 \end{align}

The change in energy is

(7)
\begin{align} \left< H \right> |_{t=-0}^{t=+0} = \frac{1}{4} \hbar \omega_0 (\lambda - 1) = \frac{\lambda - 1}{4} \hbar \sqrt{\frac{C}{m_0}} \end{align}
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