I've begun delving into quantum mechanics and encountered a point of confusion. In classical mechanics, we define an initial position and initial momentum, which can take on any values. However, in quantum mechanics, we specify only the initial wave function. I've been exploring ways to create an initial wave function that has a mean coordinate of $x0$, a mean momentum of $p0$, and simultaneously minimizes the Heisenberg uncertainty principle.

I've discovered a function that fits these criteria—the Gaussian wave packet:

$$\psi(x, t=0) = \frac{1}{\sqrt{b\sqrt{\pi}}} \cdot \exp\left(-\frac{1}{2}\left(\frac{ x - x_0}{b}\right)^2\right) \cdot \exp\left(\frac{i}{\hbar} p_0 x\right), \quad b = \sqrt{\frac{\hbar} {{\mu \omega}}}.$$

My main query revolves around whether this initial wave function adheres to the Schrödinger equation and how we construct it. Additionally, what conditions must the initial wave function satisfy?

The only information I've come across regarding this matter is that if $x0 ≠ 0$ and $p0 ≠ 0$, then the state described by such a wave function is non-stationary, and the energy does not possess a specific value.


1 Answer 1


As you are giving simply a specific (possible) initial condition $\psi(x,t=0)$, your question "whether this initial wave function adheres to the Schrödinger equation" does not make sense. Of course, any square-integrable and normalized wave function qualifies as an initial condition. The further steps of computing $\psi(x,t)=\langle x |e^{-iP^2 t/2m\hbar}|\psi(0)\rangle$ for a Gaussian wave packet (your special case) can be found in practically all text books on quantum mechanics.

  • $\begingroup$ Thank you, just as I thought. It just seems incredible that practically any normalized function can be an initial wave function. I still haven't figured out where we derive the specific form for the Gaussian packet. Do you happen to know any literature where I can read about this? $\endgroup$
    – dmitrii
    Commented Dec 27, 2023 at 18:26
  • $\begingroup$ @dmitrii As far as I remember, you can find the calculation e.g. in Cohen Tannoudji's text book on quantum mechanics. One simply computes a Gaussian integral. $\endgroup$
    – Hyperon
    Commented Dec 27, 2023 at 19:37

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