Griffiths Quantum Mechanics 2nd Edition: Problem 2.14

We have a system where we know the potential is given as the harmonic osccilator potential. We are starting off with some particle in the ground state of the harmonic oscillator, and then suddenly, the spring constant quadruples (without affecting the initial state, only affecting the state's evolution after this happens). We are tasked with finding the probability that this system has total energy measuring $\hbar\omega/2$. Backing up a little bit, we know that in order to calculate the probability of a particle being in a certain state (therefore having a certain energy), we must simply calculate the value of the expansion coefficient of the time-independent portion of that state. More formally, we can recall the method that Griffiths has called Fourier's trick:

$$\displaystyle\int_{-\infty}^{\infty} \ \psi^{*}_m \ \Psi(x, \ 0) \ \text{dx} \ = \ \displaystyle\int_{-\infty}^{\infty} \ \psi^{*}_m \ \displaystyle\sum_{n \ = \ 0}^{\infty} c_n {dx} \ = \ \displaystyle\sum_{n \ = \ 0}^{\infty} c_n \ \displaystyle\int_{-\infty}^{\infty} \ \psi^{*}_m \ \psi_{n} \ \text{dx}$$
We know that for the quantum harmonic oscillator, orthonormality of solutions holds, so:

$$\displaystyle\sum_{n \ = \ 0}^{\infty} c_n \ \displaystyle\int_{-\infty}^{\infty} \ \psi^{*}_m \ \psi_{n} \ \text{dx} \ = \ \displaystyle\sum_{n \ = \ 0}^{\infty} c_n \ \delta_{mn} \ = \ c_m$$
Ok, so now that we know how to calculate the $n$-th expansion coefficient, we simply have to enter in our values. We know that the ground state of the quantum harmonic oscillator is given as:

$$\psi_{0}(x) \ = \ \Big( \frac{m\omega}{\pi\hbar} \Big)^{1/4} \ e^{-\frac{m\omega}{2\hbar} x^2}$$
Since $t \ = \ 0$ and we know that the particle starts off in this state, we get that $\Psi(x, \ 0) \ = \ \psi_0(x)$. We are attemtping to find the probability of the particle having energy equal to $\hbar\omega/2$. This is the energy of the ground state itself, so we know that we are trying to find $c_0$. Now, we know that as this state evolves, it will do so with the new spring constant. Given that $\omega \ = \ \sqrt{k/m}$, if we quadruple $k$, we double $\omega$. So our new wavefunction representing the $\psi^{*}_m$ in the expansion coefficient equation is:
$$\psi^{*}_m \ = \ \psi_{0}(x)^{'} \ = \ \Big( \frac{2m\omega}{\pi\hbar} \Big)^{1/4} \ e^{-\frac{m\omega}{\hbar} x^2}$$

Let us now set up our equation:
$$c_0 \ = \ \displaystyle\int_{-\infty}^{\infty} \ \Big( \frac{2m\omega}{\pi\hbar} \Big)^{1/4} \ e^{-\frac{m\omega}{\hbar} x^2} \ \Big( \frac{m\omega}{\pi\hbar} \Big)^{1/4} \ e^{-\frac{m\omega}{2\hbar} x^2} \ \text{dx} \ = \ 2^{1/4} \ \sqrt{\frac{m\omega}{\pi\hbar}} \ \displaystyle\int_{-\infty}^{\infty} \ e^{-\frac{3m\omega}{2\hbar} x^2} \ \text{dx}$$

This looks like it is going to work out nicely! The integral is evidently a Gaussian integral, meaning that it is an integral of the form:
$$\displaystyle\int_{-\infty}^{\infty} \ e^{-ax^{2}} \ \text{dx} \ = \ \sqrt{\frac{\pi}{a}}$$

This means that we get:
$$2^{1/4} \ \sqrt{\frac{m\omega}{\pi\hbar}} \ \displaystyle\int_{-\infty}^{\infty} \ e^{-\frac{3m\omega}{2\hbar} x^2} \ \text{dx} \ = \ 2^{1/4} \ \sqrt{\frac{m\omega}{\pi\hbar}} \ \sqrt{\frac{2\hbar\pi}{3m\omega}} \ = \ (2^{1/4}) \ \sqrt{\frac{2}{3}} \ \approx \ 0.97098$$
Finally, in order to calculate the probability of the particle being in the ground state (and having energy $\hbar\omega/2$), we have to find $|c_0|^2$, which is simply $(0.97098)^2 \ \approx \ 0.943 \ = \ 94.3\%$. Therefore, we have found that the probability of a particle in this system having energy $\hbar\omega/2$ is roughly $94.3\%$.

The textbook also asks us to comment on the probability of the particle having energy $\hbar\omega$. Seeing as the possibly energy states of the quantum harmonic oscillator are given by:

$$E \ = \ \Big( n \ + \ \frac{1}{2} \Big)\hbar\omega$$
Where $n$ is $0$ or a positive integer. From this, obviously we cannot get a physically realizable state in the quantum harmonic oscciallator that has energy $\hbar\omega$, so the probability is $0\%$.