I'm currently reading Quantum Optics by Scully and Zubairy and come across a derivation in which I am stuck as to what to do next.
Starting with a general solution to the harmonic oscillator Hamiltonian:
$$i\hbar\frac{\partial\psi}{\partial t} = \left(-\frac{\hbar}{2}\frac{\partial^2}{\partial q^2}+\frac{\nu^2q^2}{2}\right)\psi,\qquad \psi(q,t)=\sum_{n=0}^{\infty}a_n\phi_n(q)e^{-iE_nt/\hbar}.$$
Where $\phi_n(q)$ is the coordinate representations of the oscillator number state $|{n}\rangle$.
$\phi_n$ are orthonormal, i.e. $\int_{-\infty}^{\infty}\phi_n^*\phi_mdq=\delta_{nm}.$
Also given is the initial wavefunction at time $t = 0$:
$$\psi(q,0) = \left(\frac{\nu}{\pi\hbar}\right)^{1/4}\exp\left[\frac{-\nu}{2\hbar}(q-q_0)^2\right].$$
Where $q_0>0$ is the initial displacement of the particle in the oscillator potential centred at $q=0$.
I want to find the coefficients $a_n$, so considering the expansion of $\psi$ at $t=0$ leads to the following equality:
$$\psi(q,0) =\sum_{n=0}^{\infty}a_n\phi_n(q) = \left(\frac{\nu}{\pi\hbar}\right)^{1/4}\exp\left[\frac{-\nu}{2\hbar}(q-q_0)^2\right].$$
Then multiplying by $\phi_m^*$, integrating over all of $q$ and using the orthonormality property leads to:
$$a_m = \left(\frac{\nu}{\pi\hbar}\right)^{1/4}\int_{-\infty}^{\infty}\phi_m^*(q)\exp\left[\frac{-\nu}{2\hbar}(q-q_0)^2\right]dq.$$
It is here I am at a loss as what to do to simplify this integral as I want to calculate the probability density at any future time $t$. I have considered using the definition of $\phi_n$:
$$\phi_n(q)= \frac{1}{\sqrt{2^nn!}}H_n\left(\sqrt{\frac{\nu}{\hbar}}\,q\right)\!\phi_0(q),\qquad
\phi_0(q) = \left(\frac{\nu}{\pi\hbar}\right)^{1/4}
\exp\left(\!-\frac{\nu q^2}{2\hbar}\right).$$
With $H_n$ the Hermite polynomials, but this leads to an even more complex integral.
Any help would be greatly appreciated, thanks for reading.
Best Answer
You can notice that $\psi(q)$ corresponds to the spatial translation of the vacuum state: $$ \psi(q) = \phi_0(q-q_0) = e^{-q_0\frac{d}{dq}}\phi_0(q). $$ It is convenient to use creation and annihilation operators in further computations: $$ \hat{a}^+ = \sqrt{\frac{\hbar}{2\nu}}\left(\frac{\nu}{\hbar}q-\frac{d}{dq}\right),\quad \hat{a}^- = \sqrt{\frac{\hbar}{2\nu}}\left(\frac{\nu}{\hbar}q+\frac{d}{dq}\right). $$ These operators possess the following remarkable properties: $$ [\hat{a}^-,\hat{a}^+] = 1,\quad \hat{a}^-\phi_0(q) = 0, \quad \frac1{\sqrt{n!}}(\hat{a}^+)^n \phi_0(q) = \phi_n(q), \quad e^{\alpha\hat{a}^+ + \beta\hat{a}^-} = e^{\alpha\hat{a}^+}e^{\beta\hat{a}^-}e^{\alpha\beta/2}. $$ Now, it is easy to derive the following chain of equalities: $$ \psi(q) = \exp\left(\sqrt{\frac{\nu q_0^2}{2\hbar}}(\hat{a}^+ - \hat{a}^-)\right)\phi_0(q) = $$ $$ =\exp\left(-\frac{\nu q_0^2}{4\hbar}\right) \exp\left(\sqrt{\frac{\nu q_0^2}{2\hbar}}\hat{a}^+\right) \exp\left(-\sqrt{\frac{\nu q_0^2}{2\hbar}}\hat{a}^-\right)\phi_0(q) = $$ $$ =\exp\left(-\frac{\nu q_0^2}{4\hbar}\right)\sum_{n=0}^\infty \frac1{\sqrt{n!}}\left(\frac{\nu q_0^2}{2\hbar}\right)^{n/2}\phi_n(q). $$ From the last equality, you obtain: $$ a_n = \exp\left(-\frac{\nu q_0^2}{4\hbar}\right) \frac1{\sqrt{n!}}\left(\frac{\nu q_0^2}{2\hbar}\right)^{n/2}. $$
Update. Equivalent solution without $\hat{a}^\pm$ operators. For the Hermite polynomials, it is known the generating function: $$ e^{2xt-t^2} = \sum_{n=0}^\infty H_n(x)\frac{t^n}{n!}. $$ Let's define $$ t = \frac{q_0}2\sqrt{\frac{\nu}{\hbar}},\quad x = q\sqrt{\frac{\nu}{\hbar}} $$ and rewrite $\psi(q)$ in the following manner: $$ \psi(q) = \left(\frac{\nu}{\pi\hbar}\right)^{1/4} e^{-\nu q^2/2\hbar}e^{\nu qq_0/\hbar}e^{-\nu q^2_0/2\hbar} = \phi_0(q) e^{2xt-t^2} e^{-\nu q^2_0/4\hbar} = $$ $$ = \phi_0(q) \sum_{n=0}^\infty H_n\left(q\sqrt{\frac{\nu}{\hbar}}\right)\frac1{n!}\left(\frac{q_0}2\sqrt{\frac{\nu}{\hbar}}\right)^n e^{-\nu q^2_0/4\hbar} = \sum_{n=0}^\infty \phi_n(q) \frac1{\sqrt{n!}}\left(\frac{\nu q_0^2}{2\hbar}\right)^{n/2} e^{-\nu q^2_0/4\hbar}. $$ This representation of $\psi(q)$ coincides with the one obtained earlier and gives the same coefficients $a_n$. This solution also doesn't involve taking integrals. The explicit expression of the Hermite polynomials is known. I think that direct computation of integrals for $a_n$ is possible but tedious.