I came across this question in Griffiths QM, which asked to show that this equation

$$\Psi(x,t)=\left(\frac{m\omega}{\pi \hbar}\right)^{1/4} \exp\left[-\frac{m\omega}{2\hbar} \left(x^2+\frac{a^2}{2}(1+e^{-2i\omega t})+\frac{i\hbar t}{m}-2axe^{-i\omega t}\right)\right]$$

$a$ is any real constant

satisfies the time-dependent Schrodinger's equation for a harmonic potential

I was able to show this (simple substitution), I know the ladder and the brute-force solutions to the SE for a harmonic potential, but I'm interested in knowing how the equation was obtained, I tried Google but didn't find any resource related to this.

Any hints or concepts on how this equation is derived would be helpful

  • $\begingroup$ Wow, there's no typo in that equation? Can you give the page number and edition number? $\endgroup$
    – user12029
    Commented Oct 29, 2015 at 4:10
  • $\begingroup$ Page - 70, Chapter 2, problem-2.45, Its a 1994 edition. (It also said that the equation was discovered by Schrodinger himself) $\endgroup$
    – Courage
    Commented Oct 29, 2015 at 4:17
  • $\begingroup$ how it was obtained... if he didn't list a reference for the wave function, I imagine he might've just come up with a mathematical expression that did (or didn't) satisfy the TDSE. $\endgroup$
    – Kyle Kanos
    Commented Oct 29, 2015 at 11:40
  • $\begingroup$ Possible, But the equation doesn't look easy and definitely not obvious. $\endgroup$
    – Courage
    Commented Oct 29, 2015 at 11:46

2 Answers 2


First note the following: If $\psi(x,0)$ is any normalizable wave-function in your Hilbert space, then $\psi(x,t) = e^{-iHt} \psi(x,0)$ (I've set $\hbar = 1$) satisfies the time-dependent Schrodinger equation (we call this "time evolution"). This can be seen by using the fact that the stationary states form a basis for your Hilbert space and so $\psi(x,0)$ can be expanded in terms of them. One can then check by substitution (after plugging in the expanded form of $\psi(x,0)$ into $\psi(x,t)$) that the TDSE is satisfied.

Now I haven't checked this explicitly, but to me this seems like the time evolved wave function for the ground state shifted to the right by $a$ (this has the right properties at $a=0$ and $t=0$). Namely, we simply apply the time evolution operator to $\psi(x,0) = \psi_0(x-a)$ where $\psi_0(x)$ is the ground state wave function. To actually verify that $\psi(x,t) = e^{-iHt}\psi(x,0)$ coincides with the wave function you write, one can proceed as follows:

Use $$\psi(x,t) = e^{-iHt}\psi(x,0) = \int dy \,K(x,y,t)\psi(y,0),$$ where $K(x,y,t)$ is the quantum mechanical propagator for the harmonic oscillator for which the explicit expression can be found on this Wikipedia link. Perform the subsequent Gaussian integral. If I'm right you'll end up with exactly the wave function that you want.

In short, finding the most general time-dependent solution boils down to finding the propagator. One can then use that to time evolve any stationary wave function that one wishes.


This is the wavefunction of a coherent state; it is a well-known solution of the quantum harmonic oscillator and it has a large number of nice properties. There is a good number of independent ways to derive it so it is pretty pointless to try and guess which one Griffiths was thinking of.


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