Griffiths Quantum Mechanics 2nd Edition: Problem 2.22

This is a very gnarly integration question. While doing Part B, I spent probably around 30 minutes just tracking down a missing negative sign because there are so many terms to keep track of. However, this is still an important question, and helps us to further understand how to work with quantum systems whose stationary states are not normalizable, thus forcing us to take a continuous integral over the stationary states rather than a discrete sum. We can start with Part A, which in my opinion is an easy warm-up for Part B. We are given:

$$\Psi(x, \ 0) \ = \ Ae^{-ax^2}$$
We can normalize it easily:

$$\Psi(x, \ 0) \ = \ Ae^{-ax^2} \ \Rightarrow \ |A|^2 \displaystyle\int_{-\infty}^{\infty} \ e^{-2ax^2} \ \text{dx} \ = \ 1 \ \Rightarrow \ |A|^2 \ \sqrt{\frac{\pi}{2a}} \ = \ 1 \ \Rightarrow \ A \ = \ \Big( \frac{2a}{\pi} \Big)^{1/4}$$
Next comes the messy task of dealing with the time-dependent wavefunction. Luckily, Griffiths is very nice to us, and not only tells us what the final answer should be right in the problem set, he also gives us a hint as to how to deal with some of the integrals that will come up during our calculations. Specifically, he says that if we have an integral of the form:
$$\displaystyle\int_{-\infty}^{\infty} e^{-(ax^2 \ + \ bx)} \ \text{dx}$$

Then a strategy for attacking it is to recognize that we can "factor" the exponent of $e$ by setting $y \ = \ \sqrt{a}[x \ + \ (b/2a)]$, and then recognizing that $(ax^2 \ + \ bx) \ = \ y^2 \ - \ (b^2/4a)$. This piece of knowledge is very valuable, and I promise that we will make very good use of it. Let us start by writing out exactly the form of our wavefunction. As you may recall from the previous chapter on the free particle potential:
$$\Psi(x, \ t) \ = \ \frac{1}{\sqrt{2\pi}} \ \displaystyle\int_{-\infty}^{\infty} \ \phi(k) \ e^{i(kx \ - \ \frac{\hbar k^2}{2m}t)} \ \text{dk}$$

Ok, now by setting $t \ = \ 0$ and taking the inverse Fourier transform, we can find the "continuous expansion coefficients":
$$\phi(k) \ = \ \frac{1}{\sqrt{2\pi}} \displaystyle\int_{-\infty}^{\infty} \ \Psi(x, \ 0) e^{-ikx} \ \text{dx}$$

We are given $\Psi(x, \ 0)$, and we have already normalized it, so this should be fairly straightforward:
$$\phi(k) \ = \ \frac{1}{\sqrt{2\pi}} \ \Big( \frac{2a}{\pi} \Big)^{1/4} \displaystyle\int_{-\infty}^{\infty} \ e^{-(ax^2 \ + \ ikx)} \ \text{dx}$$

This is great! We now have an integral in the exact form of the one in the massive hint that Griffiths gives us! We calculate the following expression:
$$y \ = \ \sqrt{a}[x \ + \ (b/2a)] \ = \ \sqrt{a}[x \ + \ (ik/2a)] $$ $$\Rightarrow \ (ax^2 \ + \ bx) \ = \ y^2 \ - \ (b^2/4a) \ \Rightarrow \ (ax^2 \ + \ ikx) \ = \ a[x \ + \ (ik/2a)]^2 \ - \ ((ik)^2/4a)$$

Ok, now we can plug this back into the integral:
$$\frac{1}{\sqrt{2\pi}} \ \Big( \frac{2a}{\pi} \Big)^{1/4} \displaystyle\int_{-\infty}^{\infty} \ e^{-(ax^2 \ + \ ikx)} \ \text{dx} \ = \ \frac{1}{\sqrt{2\pi}} \ \Big( \frac{2a}{\pi} \Big)^{1/4} \displaystyle\int_{-\infty}^{\infty} \ e^{-a[x \ + \ ik/2a]^2} \ e^{-k^2/4a} \ \text{dx}$$

Now, $e^{-k^2/4a}$ is just a constant, since we are integrating with respect to $x$, so we can remove it:
$$\frac{1}{\sqrt{2\pi}} \ \Big( \frac{2a}{\pi} \Big)^{1/4} \ e^{-k^2/4a} \displaystyle\int_{-\infty}^{\infty} \ e^{-a[x \ + \ ik/2a]^2} \ \text{dx} \ \Rightarrow \ \frac{1}{\sqrt{2\pi}} \ \Big( \frac{2a}{\pi} \Big)^{1/4} \ e^{-k^2/4a} \sqrt{\frac{\pi}{a}}$$

We did it! Now, all we have to do is compute the original integral:
$$\frac{1}{2\pi} \ \Big( \frac{2a}{\pi} \Big)^{1/4} \ \sqrt{\frac{\pi}{a}} \displaystyle\int_{-\infty}^{\infty} \ e^{-k^2/4a} \ e^{i(kx \ - \ \frac{\hbar k^2}{2m}t)} \ \text{dk}$$

Let's start by rearranging the terms in a more suggestive way in the exponent within the integral:

$$\Rightarrow \ \frac{1}{2\pi} \ \Big( \frac{2a}{\pi} \Big)^{1/4} \ \sqrt{\frac{\pi}{a}} \displaystyle\int_{-\infty}^{\infty} \ e^{-\big(\big(\frac{i\hbar t}{2m} \ + \ \frac{1}{4a}\big)k^2 \ - \ ixk\big)} \ \text{dk}$$
Ok, we can use Griffiths' trick again! Let us define $\Omega$ as the following:

$$\Omega \ = \ \frac{i\hbar t}{2m} \ + \ \frac{1}{4a} \ = \ \frac{2ai\hbar t \ + \ m}{4ma}$$
We can now rephrase the exponent in the integral with $y$:

$$y \ = \ \sqrt{\Omega}[k \ + \ (-ix/2\Omega)] \ \Rightarrow \ \Omega k^2 \ - \ ixk \ = \ \Omega[k \ + \ (-ix/2\Omega)]^2 \ - \ (-ix)^2/4\Omega \ = \ \Omega[k \ + \ (-ix/2\Omega)]^2 \ + \ x^2/4\Omega$$ $$\Rightarrow \ \frac{1}{2\pi} \ \Big( \frac{2a}{\pi} \Big)^{1/4} \ \sqrt{\frac{\pi}{a}} \displaystyle\int_{-\infty}^{\infty} \ e^{-\Omega[k \ + \ (-ix/2\Omega)]^2} \ e^{-x^2/4\Omega} \ \text{dk} \ = \ \frac{1}{2\pi} \ \Big( \frac{2a}{\pi} \Big)^{1/4} \ \sqrt{\frac{\pi}{\Omega}} \ \sqrt{\frac{\pi}{a}} \ e^{-x^2/4\Omega}$$
Now, we just have to do a bit of rearranging to get the final answer:

$$\frac{1}{2\pi} \ \Big( \frac{2a}{\pi} \Big)^{1/4} \ \sqrt{\frac{\pi}{\Omega}} \ \sqrt{\frac{\pi}{a}} \ e^{-x^2/4\Omega} \ = \ \ \Big( \frac{2a}{\pi} \Big)^{1/4} \ \sqrt{\frac{4ma}{2ai\hbar t \ + \ m}} \ \sqrt{\frac{1}{a}} \ \frac{1}{2} \ e^{-\big( \frac{max^2}{2ai\hbar t \ + \ m} \big)} \ = \ \Big( \frac{2a}{\pi} \Big)^{1/4} \ \frac{ e^{-\big( \frac{ax^2}{2ai\hbar t/m \ + \ 1} \big)} }{\sqrt{2ai\hbar t/m \ + \ 1}}$$

So we have completed part $b$ of the problem. Now, we are tasked with finding the probability density corresponding to this wavefunction, $|\Psi(x, \ t)|^2$. Griffiths has asked us to express our answer in terms of the constant:

$$w \ = \ \sqrt{\frac{a}{1 \ + \ (2\hbar a t/m)^2}}$$
This problem is simply asking us to expand the following:
$$\Rightarrow \ \sqrt{ \frac{2a}{\pi} } \ \frac{ e^{-\big( \frac{ax^2}{2ai\hbar t/m \ + \ 1} \big)} }{\sqrt{2ai\hbar t/m \ + \ 1}} \ \frac{ e^{-\big( \frac{ax^2}{1 \ - \ 2ai\hbar t/m} \big)} }{\sqrt{1 \ - \ 2ai\hbar t/m}} \ = \ \sqrt{ \frac{2a}{\pi} } \ \frac{ e^{-\frac{ax^2}{1 \ - \ 2ai\hbar t/m} \ - \ \frac{ax^2}{1 \ + \ 2ai\hbar t/m} } }{\sqrt{(2ai\hbar t/m \ + \ 1)(1 \ - \ 2ai\hbar t/m)}}$$ $$\Rightarrow \ \sqrt{ \frac{2a}{\pi} } \ \frac{ e^{-\frac{ax^2}{1 \ - \ 2ai\hbar t/m} \ - \ \frac{ax^2}{1 \ + \ 2ai\hbar t/m} } }{\sqrt{(2ai\hbar t/m \ + \ 1)(1 \ - \ 2ai\hbar t/m)}} \ = \ w \ \sqrt{ \frac{2}{\pi} } \ e^{-\frac{2ax^2}{1 \ + \ (2\hbar a t/m)^2}} \ = \ w \ \sqrt{ \frac{2}{\pi} } \ e^{-2ax^2w^2}$$

Next, let us find all of the desired values, $\langle x \rangle$, $\langle p \rangle$, $\langle x^2 \rangle, \langle p^2 \rangle$:
$$\langle x \rangle \ = \ w \ \sqrt{ \frac{2}{\pi} } \ \displaystyle\int_{-\infty}^{\infty} \ xe^{-2aw^2 x^2} \ = \ \ w \ \sqrt{ \frac{2}{\pi} } \ \Big(-\frac{e^{-2aw^2 x^2}}{4aw^2}\Big) \biggr\rvert_{-\infty}^{\infty} \ = \ 0$$ $$\langle p \rangle \ = \ w \ \sqrt{ \frac{2}{\pi} } \ \displaystyle\int_{-\infty}^{\infty} \ xe^{-2aw^2 x^2}$$