This is likely one of the most fun problems I have encoutered so far in Griffiths'! We are essentially walked through a multi-step question on coherent states, where we have to
prove a bunch of stuff. Moving on to the problem, we start off with finding $\langle x \rangle$, $\langle x^2 \rangle$, $\langle p \rangle$ and $\langle p^2 \rangle$ for the coherent states
of the quantum harmonic oscillator, which are said to be linear combinations of the stationary states, and also be eigenfunctions of the lowering operator. This means they are given
by the equations:
$$a_{-}|\alpha\rangle \ = \ \beta|\alpha\rangle$$
I know they call the eigenvalue of the coherent states $\alpha$ in the book, but I'm going to call it $\beta$, so we can avoid confusion regarding the eigenfunctions and the eigenvalues.
Let us start off by finding $\langle x \rangle$ (Heads up, instead of writing $|\alpha\rangle$ while I'm solving the following integrals, I'll use $\alpha$).
Before we start, it is important to note that Griffiths makes a great reccomendation. He tells us to use an elegant method to solve these integrals by expression $x$ in terms of $a_{-}$ and $a_{+}$. We know that:
$$a_{\pm} \ = \ \frac{1}{\sqrt{2m\omega\hbar}}(\mp ip \ + \ m\omega x)$$
So we get:
$$(a_{-} \ + \ a_{+}) \ = \ \frac{2}{\sqrt{2m\omega\hbar}}(m\omega x) \ \Rightarrow \ (a_{-} \ + \ a_{+}) \ = \ \sqrt{\frac{2m\omega}{h}} \ x \ \Rightarrow \ x \ = \ \sqrt{\frac{h}{2m\omega}} \ (a_{-} \ + \ a_{+})$$
We can do the same thing for momentum!
$$(a_{+} \ - \ a_{-}) \ = \ -\frac{2ip}{\sqrt{2m\omega\hbar}} \ \Rightarrow \ p \ = \ i\sqrt{\frac{\hbar m \omega}{2}} (a_{+} \ - \ a_{-})$$
Now, let us do some calculations. We know that:
$$\langle x \rangle \ = \ \displaystyle\int_{-\infty}^{\infty} \ \alpha^{*} x \alpha \ \text{dx} \ = \ \sqrt{\frac{h}{2m\omega}} \ \displaystyle\int_{-\infty}^{\infty} \ \alpha^{*} (a_{-} \ + \ a_{+}) \alpha \ \text{dx}
\ = \ \sqrt{\frac{h}{2m\omega}} \ \displaystyle\int_{-\infty}^{\infty} \ \alpha^{*} a_{-} \alpha \ + \ \alpha^{*} a_{+} \alpha \ \text{dx}$$
It is important to now remember that $a_{-}|\alpha\rangle \ = \ \beta|\alpha\rangle$, so we get:
$$\sqrt{\frac{h}{2m\omega}} \ \displaystyle\int_{-\infty}^{\infty} \ \alpha^{*} a_{-} \alpha \ + \ \alpha^{*} a_{+} \alpha \ \text{dx} \ = \ \sqrt{\frac{h}{2m\omega}} \ \displaystyle\int_{-\infty}^{\infty} \ \alpha^{*} \beta \alpha \ + \ \alpha^{*} a_{+} \alpha \ \text{dx}$$
Ok, but what do we do about $a_{+}$? Well, luckily, we can take advantage of a very useful construction:
$$\displaystyle\int \ f^{*}a_{\pm}g \ \text{dx} \ = \ \displaystyle\int \ (a_{\mp}f)^{*}g \ \text{dx}$$
See the textbook for the formal proof. Using this, we say that:
$$\sqrt{\frac{h}{2m\omega}} \ \displaystyle\int_{-\infty}^{\infty} \ \alpha^{*} \beta \alpha \ + \ \alpha^{*} a_{+} \alpha \ \text{dx} \ = \ \sqrt{\frac{h}{2m\omega}} \ \displaystyle\int_{-\infty}^{\infty} \ \alpha^{*} \beta \alpha \ + \ (a_{-}\alpha)^{*} \alpha \ \text{dx}
\ = \ \sqrt{\frac{h}{2m\omega}} \ \displaystyle\int_{-\infty}^{\infty} \ \alpha^{*} \beta \alpha \ + \ \beta^{*}\alpha^{*} \alpha \ \text{dx} \ = \ \sqrt{\frac{h}{2m\omega}} \ (\beta^{*} \ + \ \beta) \displaystyle\int_{-\infty}^{\infty} \ \alpha^{*} \alpha \ \text{dx}$$
We assume that $\alpha$ is normalized, so:
$$\langle x \rangle \ = \ \sqrt{\frac{h}{2m\omega}} \ (\beta^{*} \ + \ \beta)$$
We repeat this same procedure for the rest of the expectation values:
$$\langle x^2 \rangle \ = \ \frac{\hbar}{2m\omega} \ \displaystyle\int_{-\infty}^{\infty} \ \alpha^{*} (a_{+}^2 \ + \ a_{+}a_{-} \ + \ a_{-}a_{+} \ + \ a_{-}^2) \alpha \ \text{dx} \ = \
\frac{\hbar}{2m\omega} \ \displaystyle\int_{-\infty}^{\infty} \ \alpha^{*} a_{+}^2 \alpha \ + \ \alpha^{*} a_{+}a_{-} \alpha \ + \ \alpha^{*} a_{-}a_{+} \alpha \ + \ \alpha^{*} a_{-}^2 \alpha \ \text{dx}$$
$$\Rightarrow \ \frac{\hbar}{2m\omega} \ \displaystyle\int_{-\infty}^{\infty} \ \alpha^{*} a_{+}^2 \alpha \ + \ \alpha^{*} a_{+}a_{-} \alpha \ + \ \alpha^{*} a_{-}a_{+} \alpha \ + \ \alpha^{*} a_{-}^2 \alpha \ \text{dx} \ = \
\frac{\hbar}{2m\omega} \ \displaystyle\int_{-\infty}^{\infty} \ \alpha^{*} \beta^2 \alpha \ + \ \alpha^{*} \beta^{*}\beta \alpha \ + \ \alpha^{*} a_{-}a_{+} \alpha \ + \ \alpha^{*} (\beta^{*})^2 \alpha \ \text{dx}$$
Recall that:
$$[a_{-}, \ a_{+}] \ = \ a_{-}a_{+} \ - \ a_{+}a_{-} \ = \ 1 \ \Rightarrow \ a_{-}a_{+} \ = \ 1 \ + \ a_{+}a_{-}$$
So we can rewrite the integral to get:
$$\frac{\hbar}{2m\omega} \ \displaystyle\int_{-\infty}^{\infty} \ \alpha^{*} \beta^2 \alpha \ + \ \alpha^{*} \beta^{*}\beta \alpha \ + \ \alpha^{*} a_{-}a_{+} \alpha \ + \ \alpha^{*} (\beta^{*})^2 \alpha \ \text{dx} \ = \
\frac{\hbar}{2m\omega} \ \displaystyle\int_{-\infty}^{\infty} \ \alpha^{*} \beta^2 \alpha \ + \ \alpha^{*} \beta^{*}\beta \alpha \ + \ \alpha^{*} (1 \ + \ \beta^{*}\beta) \alpha \ + \ \alpha^{*} (\beta^{*})^2 \alpha \ \text{dx}$$
$$\Rightarrow \ \langle x^2 \rangle \ = \ \frac{\hbar}{2m\omega} (\beta^2 \ + \ 2\beta^{*}\beta \ + \ (\beta^{*})^2 \ + \ 1)$$
The momentum ones are basically the same thing:
$$\langle p \rangle \ = \ i\sqrt{\frac{\hbar m \omega}{2}} \ \displaystyle\int_{-\infty}^{\infty} \ \alpha^{*} (a_{+} \ - \ a_{-}) \alpha \ \text{dx} \ = \ i\sqrt{\frac{\hbar m \omega}{2}}(\beta^{*} \ - \ \beta)$$
$$\langle p^2 \rangle \ = \ -\frac{\hbar m \omega}{2} \ \displaystyle\int_{-\infty}^{\infty} \ \alpha^{*} (a_{+}^2 \ - \ a_{+}a_{-} \ - \ a_{+}a_{-} \ + \ a_{-}^2) \alpha \ \text{dx} \ = \
-\frac{\hbar m \omega}{2} \ (\beta^2 \ + \ (\beta^{*})^2 \ + \ 2\beta^{*}\beta \ - \ 1)$$
Now, we verify that the coherent states have minimized position-momentum uncertainty (you guys can do the expansion and substraction):
$$\sigma_x^2 \ = \ \langle x^2 \rangle \ - \ \langle x \rangle^2 \ = \ \frac{h}{2m\omega}$$
$$\sigma_p^2 \ = \ \langle p^2 \rangle \ - \ \langle p \rangle^2 \ = \ \frac{\hbar m \omega}{2}$$
$$\Rightarrow \ \sigma_x^2\sigma_p^2 \ = \ \Big( \frac{h}{2m\omega} \Big)\Big( \frac{\hbar m \omega}{2} \Big) \ = \ \frac{\hbar^2}{4} \ \Rightarrow \ \sigma_x\sigma_p \ = \ \frac{\hbar}{2}$$
And so we have verified that uncertainty is minimzed. Now, we are given that:
$$|\alpha\rangle \ = \ \displaystyle\sum_{n \ = \ 0}^{\infty} \ c_n|n\rangle$$
And we need to show that:
$$c_n \ = \ \frac{\beta^n}{\sqrt{n!}} c_0$$
Ok, now things are getting fun. We know by Fourier's trick, that:
$$c_n \ = \ \langle n | \alpha \rangle$$
And if you recall from the section on the quantum harmonic oscillator, we know that:
$$|n\rangle \ = \ \frac{1}{\sqrt{n!}} \ (a_{+})^n \ |0\rangle$$
So we get that:
$$c_n \ = \ \displaystyle\int_{-\infty}^{\infty} \ \Big( \frac{1}{\sqrt{n!}} \ (a_{+})^n \ |0\rangle \Big)^{*} \ \alpha \text{dx} \ = \ \displaystyle\int_{-\infty}^{\infty} \ \Big( \frac{1}{\sqrt{n!}} \ |0\rangle \Big)^{*} \ a_{-}^n \alpha \text{dx}
\ = \ \displaystyle\int_{-\infty}^{\infty} \ \Big( \frac{1}{\sqrt{n!}} \ |0\rangle \Big)^{*} \ \beta^n \alpha \text{dx} \ = \ \frac{\beta^n}{\sqrt{n!}} \displaystyle\int_{-\infty}^{\infty} \ |0\rangle^{*} \ \alpha \text{dx}$$
Notice how I used the same rule about moving around the raising/lowering operator under the integral sign as I did earlier in the question. We must now recall the original definition of $\alpha$, and remember that since we are taking the complex
conjugate of $|0\rangle$, the componenets of $\alpha$ will be orthonormal (yielding the Kronecker delta upon multiplication). Therefore, we get:
$$\frac{\beta^n}{\sqrt{n!}} \displaystyle\int_{-\infty}^{\infty} \ |0\rangle^{*} \ \alpha \text{dx} \ = \ \frac{\beta^n}{\sqrt{n!}} \displaystyle\int_{-\infty}^{\infty} \ |0\rangle^{*} \ \displaystyle\sum_{n \ = \ 0}^{\infty} \ c_n|n\rangle \text{dx} \ = \
\frac{\beta^n}{\sqrt{n!}} \displaystyle\int_{-\infty}^{\infty} \ |0\rangle^{*} \ (c_0|0\rangle \ + \ c_1|1\rangle \ + \ c_2|2\rangle \ + \ ...) \ \text{dx}$$
$$\Rightarrow \ \frac{\beta^n}{\sqrt{n!}} \displaystyle\int_{-\infty}^{\infty} \ |0\rangle^{*} \ (c_0|0\rangle \ + \ c_1|1\rangle \ + \ c_2|2\rangle \ + \ ...) \ \text{dx} \ = \
\frac{\beta^n}{\sqrt{n!}} \Big( \ c_0 \ \displaystyle\int_{-\infty}^{\infty} \ |0\rangle^{*}|0\rangle \text{dx} \ + \ c_1 \ \displaystyle\int_{-\infty}^{\infty} \ |0\rangle^{*}|1\rangle \ \text{dx} \ + \ c_2 \ \displaystyle\int_{-\infty}^{\infty} \ |0\rangle^{*}|2\rangle \ \text{dx} \ + \ ... \Big)$$
$$\Rightarrow \ \frac{\beta^n}{\sqrt{n!}} \ (c_0 \ (1) \ + \ c_1 \ (0) \ + \ c_2 \ (0)) \ + \ ... \ \Rightarrow \ c_n \ = \ \frac{\beta^n}{\sqrt{n!}} \ c_0$$
Moving along, we must normalize $|\alpha\rangle$, to show that $c_0 \ = \ \text{exp}(-|\beta|^2/2)$. Since we have a discrete sum, the best way to go about normalizing $|\alpha\rangle$ is by remembering that the modulus squared of all the expansion coefficients of a wavefunction add up to $1$:
$$\displaystyle\sum_{n \ = \ 0}^{\infty} \ |c_n|^2 \ = \ 1$$
This means that we have:
$$\displaystyle\sum_{n \ = \ 0}^{\infty} \ \frac{|\beta|^{2n}}{n!} \ c_0^2 \ = \ 1$$
The key insight here is remembering the Taylor series of the form:
$$e^x \ = \ \displaystyle\sum_{n \ = \ 0}^{\infty} \ \frac{x^n}{n!}$$
This leads us to rewrite our equation as:
$$\displaystyle\sum_{n \ = \ 0}^{\infty} \ \frac{|\beta|^{2n}}{n!} \ c_0^2 \ = \ 1 \ \Rightarrow \ e^{|\beta|^2} c_0^2 \ = \ 1 \ \Rightarrow \ c_0 \ = \ e^{-|\beta|^2/2}$$
Ok, now we have to prove that when we introduce time-dependence, the coherent state remains an eigenstate of the lowering operator, but the eigenvalue evolves in time. Let us consider the original
eigenvector-eigenvalue equation with time dependence:
$$\hat{a}_{-} |\alpha(t)\rangle \ = \ \alpha(t) |\alpha(t)\rangle \ \Rightarrow \ \hat{a}_{-} \ \displaystyle\sum_{n \ = \ 0}^{\infty} \ \frac{\alpha^n}{\sqrt{n!}} c_0 e^{-iEt/\hbar} |n\rangle \ = \ \alpha(t) \ \displaystyle\sum_{n \ = \ 0}^{\infty} \ \frac{\alpha^n}{\sqrt{n!}} c_0 e^{-iEt/\hbar} |n\rangle$$
By definition of the lowering operator, we have $\hat{a}_{-} |n\rangle \ = \ \sqrt{n}|n \ - \ 1\rangle$, thus we have:
$$\Rightarrow \ \ \displaystyle\sum_{n \ = \ 0}^{\infty} \ \frac{\alpha^n}{\sqrt{(n \ - \ 1)!}} c_0 e^{-iEt/\hbar} |n \ - \ 1\rangle \ = \ \alpha(t) \ \displaystyle\sum_{n \ = \ 0}^{\infty} \ \frac{\alpha^n}{\sqrt{n!}} c_0 e^{-iEt/\hbar} |n\rangle$$
$$\Rightarrow \ \displaystyle\sum_{n \ = \ 0}^{\infty} \ \frac{\alpha^n}{\sqrt{(n \ - \ 1)!}} c_0 e^{-i\Big(n \ + \ \frac{1}{2} \Big) \hbar\omega t/\hbar} |n \ - \ 1\rangle \ = \ \alpha(t) \ \displaystyle\sum_{n \ = \ 0}^{\infty} \ \frac{\alpha^n}{\sqrt{n!}} c_0 e^{-i\Big(n \ + \ \frac{1}{2} \Big) \hbar\omega t/\hbar} |n\rangle$$
We can now match up the coefficients in order to make this equation hold true, by "shifting" the coefficients up by $1$:
$$\frac{\alpha^{n \ + \ 1}}{\sqrt{n!}} c_0 e^{-i\Big(n \ + \ 1 \ + \frac{1}{2} \Big) \hbar\omega t/\hbar} |n\rangle \ = \ \alpha(t) \ \frac{\alpha^n}{\sqrt{n!}} c_0 e^{-i\Big(n \ + \ \frac{1}{2} \Big) \hbar\omega t/\hbar} |n\rangle$$
$$\Rightarrow \ \alpha(t) \ = \ e^{-i\omega t}\alpha$$
And it is evident that when we add the time-dependence to the coherent states, the states remains coherent, as when the states evolve, the state vector is still an eigenvector as all we are doing is multiplying the vector by some scalar coefficient.