Simple Harmonic Oscillator Ground State

A particle in one dimension is in the ground state of a simple harmonic oscillator (SHO). The Hamiltonian for the SHO is given by \(H = \frac{{{{\rm{P}}^{\rm{2}}}}}{{{\rm{2m}}}}{\rm{ + }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{m}}{{\rm{\omega }}^{\rm{2}}}{x^{\rm{2}}}\). The initial state of the particle is the ground state of this Hamiltonian.

Ground State Wavefunction

The normalized ground state wavefunction for a simple harmonic oscillator is:

\(\psi_0(x) = \left(\frac{m\omega}{\pi\hbar}\right)^{1/4} e^{-\frac{m\omega}{2\hbar}x^2}\)

Effect of Impulsive Force

At time \(t=0\), an impulsive force suddenly imparts a momentum \(P_0\) to the particle. In quantum mechanics, adding momentum \(P_0\) to a state described by a wavefunction \(\psi(x)\) transforms it into a new state proportional to \(e^{i P_0 x / \hbar} \psi(x)\).

So, immediately after the impulse, the state of the particle, let's call it \(\psi'(x)\), is proportional to:

\(\psi'(x) \propto e^{i P_0 x / \hbar} \psi_0(x)\)

We can use the unnormalized state for calculating probability overlap, as normalization factors will cancel out in the end probability calculation \(|\langle\psi_0|\psi'\rangle|^2\).

Probability of Remaining in Ground State

The probability that the particle remains in the original ground state immediately after the impulse is given by the square of the magnitude of the overlap integral between the initial ground state \(\psi_0(x)\) and the state after the impulse \(\psi'(x)\). The overlap is given by:

\(\langle \psi_0 | \psi' \rangle = \int_{-\infty}^{\infty} \psi_0^*(x) \left(e^{i P_0 x / \hbar} \psi_0(x)\right) dx\)

Substitute the expression for \(\psi_0(x)\):

\(\langle \psi_0 | \psi' \rangle = \int_{-\infty}^{\infty} \left(\frac{m\omega}{\pi\hbar}\right)^{1/4} e^{-\frac{m\omega}{2\hbar}x^2} e^{i P_0 x / \hbar} \left(\frac{m\omega}{\pi\hbar}\right)^{1/4} e^{-\frac{m\omega}{2\hbar}x^2} dx\)

\(\langle \psi_0 | \psi' \rangle = \left(\frac{m\omega}{\pi\hbar}\right)^{1/2} \int_{-\infty}^{\infty} e^{-\frac{m\omega}{\hbar}x^2 + \frac{i P_0}{\hbar}x} dx\)

Evaluating the Integral

This is a standard Gaussian integral of the form \(\int_{-\infty}^{\infty} e^{-ax^2 + bx} dx = \sqrt{\frac{\pi}{a}} e^{\frac{b^2}{4a}}\). Here, we have \(a = \frac{m\omega}{\hbar}\) and \(b = \frac{i P_0}{\hbar}\). The integral evaluates to:

\(\int_{-\infty}^{\infty} e^{-\frac{m\omega}{\hbar}x^2 + \frac{i P_0}{\hbar}x} dx = \sqrt{\frac{\pi}{m\omega/\hbar}} e^{\frac{(i P_0/\hbar)^2}{4(m\omega/\hbar)}} = \sqrt{\frac{\pi\hbar}{m\omega}} e^{\frac{-P_0^2/\hbar^2}{4m\omega/\hbar}} = \sqrt{\frac{\pi\hbar}{m\omega}} e^{\frac{-P_0^2}{4m\omega\hbar}}\)

Substitute this result back into the overlap integral:

\(\langle \psi_0 | \psi' \rangle = \left(\frac{m\omega}{\pi\hbar}\right)^{1/2} \sqrt{\frac{\pi\hbar}{m\omega}} e^{\frac{-P_0^2}{4m\omega\hbar}} = \sqrt{\frac{m\omega}{\pi\hbar}} \sqrt{\frac{\pi\hbar}{m\omega}} e^{\frac{-P_0^2}{4m\omega\hbar}} = e^{\frac{-P_0^2}{4m\omega\hbar}}\)

Using the Given Momentum \(P_0\)

The problem states that the impulsive force imparts a momentum \(P_0 = \sqrt{{\rm{hm\omega}}}\). Assuming 'h' here means \(\hbar\) (as is common in quantum mechanics contexts where \(\hbar\) appears), we have \(P_0^2 = \hbar m\omega\).

Substitute \(P_0^2\) into the exponent of the overlap integral:

\(\frac{-P_0^2}{4m\omega\hbar} = \frac{-(\hbar m\omega)}{4m\omega\hbar} = -\frac{1}{4}\)

So, the overlap integral is \(\langle \psi_0 | \psi' \rangle = e^{-1/4}\).

Calculating the Probability

The probability is the square of the magnitude of the overlap integral:

Probability \( = |\langle \psi_0 | \psi' \rangle|^2 = |e^{-1/4}|^2 = (e^{-1/4})^2 = e^{-1/2}\)

The probability that the particle remains in the original ground state is \(e^{-1/2}\).