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The wave functions of the quantum harmonic oscillator are given by: $$\psi_n(x)=\frac{1}{\sqrt{2^nn!}} \left(\frac{m\omega}{\pi \hbar}\right)^{-1/4} e^{-m\omega x^2/2\hbar}H_n\left(\sqrt{m\omega/ \hbar} x\right)$$ My question is how do these wave functions evolve in time? I could not find any reference concerning this. Meaning, given some $\psi(x,0)$, how does one generally find $\psi(x,t)$ in this case? I was thinking of: $$\left|\psi\left(t\right)\right\rangle =\hat{T}\left|\psi_{0}\right\rangle \implies\left\langle x|\psi\left(t\right)\right\rangle =\psi\left(x,t\right)=\left\langle x\left|e^{-\frac{i}{\hbar}\hat{H}t}\right|\psi_{0}\right\rangle $$ But how does one proceed further? Also, since I'm given $\psi(x,0)$ and not $\left|\psi_{0}\right\rangle $, I'm not sure that's even the way to go.

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  • $\begingroup$ presumably $\psi_0$ is the ground state? Just set $n=0$ in your first expression… $\endgroup$ – ZeroTheHero Jun 11 at 15:39
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    $\begingroup$ These are eigenstates of the Hamiltonian, that is $\psi_n(x,t)=e^{-iE_nt/\hbar}\psi_n(x,0)$. $\endgroup$ – Roger Vadim Jun 11 at 15:51
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In the Dirac notation you use, one can express the initial state vector as

\begin{equation}\left|\Psi_0\right> = \sum_i a_i\left|\psi_i\right>\end{equation}

where the $\left|\psi_i\right>$ are the eigenvector solutions. You then operate on the left of this with $e^{-iHt/\hbar}$, which is an operator (think of the series expansion of this into terms involving products of $iHt/\hbar$). This gives

\begin{align}\left|\Psi(t)\right> &= e^{-iHt/\hbar}\left|\Psi_0\right> \\ &= e^{-iHt/\hbar}\sum_i a_i\left|\psi_i\right> \\ &= \sum_i a_ie^{-iE_it/\hbar}\left|\psi_i\right>.\end{align}

Finally, you take the inner product of this with $\left<x\right|$, which yields the wavefunction

\begin{equation}\left<x|\Psi(t)\right> = \Psi(x,t) = \sum_i a_ie^{-iE_it/\hbar}\psi_i(x),\end{equation}

as in Connor Behan's answer.

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You are introducing $\psi_n(x)$ as the wavefunctions of the harmonic oscillator. But more precisely, they are the energy eigenfunctions having energy $E_n = \hbar \omega \left ( n + \frac{1}{2} \right )$. This means they evolve the way all energy eigenfunctions evolve which is \begin{equation} \psi_n(x, t) = e^{-i E_n t / \hbar} \psi_n(x, 0). \end{equation} A simple mapping like this is the reason why the time independent Schroedinger equation is able to tell us something about solutions to the time dependent Schroedinger equation.

This means that if you prepare your system in an eigenstate (and don't switch on any new Hamiltonian), the time evolution is always a pure phase. If you start off in a non-eigenstate, you need to decompose it and then insert different phases for all terms in the sum. \begin{align} \Psi(x, 0) = \sum_n c_n \psi_n(x) \Rightarrow \\ \Psi(x, t) = \sum_n c_n e^{-i E_n t / \hbar} \psi_n(x) \end{align}

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