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hello I have the time dependent Schrödinger equation of a harmonic oscillator $$i \hbar \frac{\partial}{\partial t} \Psi(q,t) = - \frac{\hbar^2}{2M} \frac{\partial^2}{\partial q^2} \Psi(q,t) + \frac{M \Omega^2}{2} q^2 \Psi(q,t)$$ and am supposed to solve it for the initial condition $$\Psi(q,0) = (\frac{A}{\pi})^{1/4} e^{-(A/2)q^2}.$$ Our prof advised us to find an Ansatz for which above differential equation transforms into a riccati differential equation and the only Ansatz I had was to try $$\Psi(q,t) = A e^{\lambda(t)q^2}$$ which then leads to the differential equation $$i \hbar \dot{\lambda}(t) = - \frac{\hbar^2}{Mq^2} \lambda(t) + \frac{2 \hbar^2}{M} \lambda^2(t) + \frac{M \Omega^2}{2}$$ which I still don't know how to solve. Are there any easier methods or does anybody know how to continue from there?

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    $\begingroup$ Notice that you are free to set $A = \frac{\alpha M \Omega}{\hbar}$, based on dimensional analysis, with $\alpha$ a dimensionless quantity. If $\alpha = 1$, your state matches the ground state, and its time evolution is just $|\Psi\rangle = \exp\left(-i \frac{t E_0}{\hbar}\right)|0\rangle = \exp\left(-i \frac{t \Omega}{2}\right)|0\rangle$. $\endgroup$
    – secavara
    Dec 31, 2020 at 15:29
  • $\begingroup$ @TBissinger Sure, I was pointing to the fact that the expansion in your answer will look more natural in terms of $\alpha$, and that, for a specific value, the expansion is trivial. $\endgroup$
    – secavara
    Dec 31, 2020 at 15:40
  • $\begingroup$ Sorry, I misread your statement. You're obviously right. Deleted my comment. $\endgroup$
    – TBissinger
    Dec 31, 2020 at 15:44

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One alternative way to go about it would be to use the already known static solutions of the harmonic oscillator $\phi_n(q)$. They are eigenfunctions, so they evolve in time by picking up a phase factor $e^{-i E_n t/\hbar}$, with $E_n = \hbar \Omega (n + 1/2)$. These functions are a complete basis of the Hilbert space, so you can especially decompose the initial conditions as $$ \Psi_0(q) = \Psi(q,t=0) = \sum_{n\ge 0} \langle\phi_n |\Psi_0\rangle \phi_n(q).$$ You can thus obtain the solution as $$\Psi(q,t) = e^{-iHt/\hbar}\Psi_0(q) = \sum_{n\ge 0} e^{-iE_nt/\hbar} \langle\phi_n |\Psi_0\rangle \phi_n(q).$$ This shifts the task from getting a new solution of the equation to just calculating $\langle \phi_n | \Psi_0\rangle$, which is feasible with Gaussian integrals and using a series expansion for the Hermite polynomials (which should converge).

There isn't much elegance in this solution. It's quite technical and maybe the detour over the Riccati equation yields a more useful result.

The ansatz you use leads to a Riccatti equation, but the anstaz has a flaw: it is not normalized. The norm is time-dependent, $$ \langle \Psi(t) | \Psi(t)\rangle = |A|^2 \sqrt{\frac{\pi}{\lambda(t)}}.$$ This can not yield the correct solution because the Schrödinger equation is unitary, that is the norm of $\Psi$ is necessarily time-independent.

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    $\begingroup$ The time evolution operator should be $\exp\left(- i \frac{t H}{\hbar}\right)$. $\endgroup$
    – secavara
    Dec 31, 2020 at 15:43
  • $\begingroup$ My bad, corrected. $\endgroup$
    – TBissinger
    Dec 31, 2020 at 15:45
  • $\begingroup$ I see what u mean concerning the norm, ty for the quick response. I will try to do what u proposed. $\endgroup$ Dec 31, 2020 at 15:47
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After experimenting and working out quite some algebra with Mathematica and using Hermite polynomials properties, I managed to analytically perform the infinite sum and get a relatively nice answer. This can be fairly time consuming so instead I present first here a simpler approach.

First, I am choosing to write \begin{equation} A = \alpha \frac{M \Omega}{\hbar} \, , \end{equation} where $\alpha$ is a dimensionless positive constant. This does not represent any loss of generality.

Turns out, an ansatz that would actually fit the answer would be \begin{equation} \Psi(x,t) = \frac{A^{1/4}}{\pi^{1/4}} \frac{1}{h(t)^{1/2}} \exp\left[-\frac{1}{2} \frac{M \Omega}{\hbar} x^2 g(t) \right] \, , \end{equation} together with the initial conditions \begin{eqnarray} g(0) &=& \alpha \, , \\ h(0) &=& 1 \, . \end{eqnarray} While in principle normalization necessarily relates $h$ and $g$, it comes as a benefit in this case to ignore this first and leave it as a consistency check for the end of the exercise.

Plugging this in the Schrödinger equation gives you a system of differential equations that can be decoupled and solved without too much effort. You should find something like this \begin{equation} \Psi(x,t) = \frac{A^{1/4}}{\pi^{1/4}} \frac{1}{\left[ \cos (t \Omega )+ i \alpha \sin (t \Omega ) \right]^{1/2}} \exp\left[-\frac{1}{2} \frac{M \Omega}{\hbar} x^2 \frac{\alpha \cos (t \Omega )+i \sin (t \Omega )}{\cos (t \Omega )+i \alpha \sin (t \Omega )}\right] \, . \end{equation}

Other options:

In case there is interest in finding this result through the sum over the eigenstates, I present here the short version of how to do so. We have \begin{eqnarray} |\Psi(t)\rangle &=& \exp\left[-i \frac{t H}{\hbar}\right] |\Psi(0)\rangle \\ &=& \exp\left[-i \frac{t H}{\hbar}\right] \sum_{n=0}^\infty | n \rangle \langle n |\Psi(0)\rangle \\ &=& \mathrm{e}^{- \frac{i t \Omega}{2}} \sum_{n=0}^\infty \left( \mathrm{e}^{-i t \Omega} \right)^n | n \rangle \langle n |\Psi(0)\rangle \, . \end{eqnarray} The wave function can then be found with \begin{eqnarray} \Psi(x,t) &=& \langle x |\Psi(t)\rangle \\ &=& \mathrm{e}^{- \frac{i t \Omega}{2}} \sum_{n=0}^\infty \left( \mathrm{e}^{-i t \Omega} \right)^n \langle x | n \rangle \langle n |\Psi(0)\rangle \\ &=& \mathrm{e}^{- \frac{i t \Omega}{2}} \int_{- \infty}^\infty \mathrm{d} x' \sum_{n=0}^\infty \left( \mathrm{e}^{-i t \Omega} \right)^n \langle x | n \rangle \langle n | x' \rangle \langle x' |\Psi(0)\rangle \\ &=& \mathrm{e}^{- \frac{i t \Omega}{2}} \int_{- \infty}^\infty \mathrm{d} x' \sum_{n=0}^\infty \left( \mathrm{e}^{-i t \Omega} \right)^n \langle x | n \rangle \langle n | x' \rangle \Psi(x', 0) \, . \end{eqnarray}

At this point it is possible to approach this expression by either performing the $x'$ integral first and then the sum (Option 1) or viceversa (Option 2).

Option 1: Integration gives you \begin{equation} \int_{- \infty}^\infty \mathrm{d} x' \langle n | x' \rangle \Psi(x', 0) = \left(1 + (-1)^n \right) \frac{\alpha^{1/4}}{\sqrt{2} \left( \alpha + 1 \right)^{1/2}} \frac{(n!)^{1/2}}{\left(\frac{n}{2}\right)!} \left[\frac{i \left( \alpha - 1 \right)^{1/2}}{\sqrt{2}\left( \alpha + 1 \right)^{1/2}}\right]^n \, . \end{equation} Then you can use the explicit expression for the eigenfunction $\langle x | n \rangle$ and you will find yourself with sums of the form \begin{equation} \sum_{n=0}^{\infty} \frac{k^n}{\left(\frac{n}{2}\right)!} H_n(y) = \frac{4 k^2+2 k y+1}{\left(4 k^2+1\right)^{3/2}} \exp\left[ \frac{4 k^2 y^2}{4 k^2+1} \right] \, . \end{equation} I have to thank Mathematica for both results. Putting that together and a bit of algebra gives you the final result.

Option 2: Performing the sum first gives you an intermediate result of greater generality. In this case, we have to deal with the expression \begin{equation} \sum_{n=0}^\infty \mathrm{e}^{- \frac{i t \Omega}{2}} \left( \mathrm{e}^{-i t \Omega} \right)^n \langle x | n \rangle \langle n | x' \rangle \,. \end{equation} This happens to be the propagator of the quantum harmonic oscillator, also known as the Mehler kernel. This is \begin{equation} \langle x | \exp \left[ - i \frac{t H}{ \hbar} \right] | x' \rangle = \sum_{n = 0}^\infty \sum_{m = 0}^\infty \langle x | n \rangle \langle n | \exp \left[ - i \frac{t H}{ \hbar} \right] | m \rangle \langle m | x' \rangle = \sum_{n=0}^\infty \mathrm{e}^{- \frac{i t \Omega}{2}} \left( \mathrm{e}^{-i t \Omega} \right)^n \langle x | n \rangle \langle n | x' \rangle \,. \end{equation} Looking for this in references you will find \begin{equation} \langle x | \exp \left[ - i \frac{t H}{ \hbar} \right] | x' \rangle = \sqrt{\frac{M \Omega}{\hbar}} \frac{1}{\sqrt{2 \pi i \sin(t \Omega)}} \exp \left\{ \frac{M \Omega }{\hbar } \frac{i}{2 \sin (t \Omega )} \left[ \left(x^2+x'^2\right) \cos (t \Omega )-2 x x' \right] \right\} \, . \end{equation} From here, it only remains to perform the $x'$ integral, which looks messy, but in reality it is only a shifted version of a Gaussian integral: \begin{equation} \Psi(x,t) = \int_{- \infty}^\infty \mathrm{d} x' \langle x | \exp \left[ - i \frac{t H}{ \hbar} \right] | x' \rangle \Psi(x',0) \, . \end{equation}

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  • $\begingroup$ Wow, this is very thorough! Great work, this answer should be upvoted. There is an article on this on the math stackexchange: math.stackexchange.com/questions/873916/… $\endgroup$
    – TBissinger
    Jan 1, 2021 at 21:30
  • $\begingroup$ Yea I also have to add thank you very much for making the effort! $\endgroup$ Jan 2, 2021 at 16:55

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