# Free particle propagator, wavefunction not moving problem

I'm learning the QM propagator and the first example is of course the free particle: $$\hat{H}=\frac{p^2}{2m}$$, then the new wavefunction is found by: $$\psi(x,t)=\int dx_0\;K(x,t;x_0,t_0)\;\psi(x_0,t_0)$$ and $$K=\langle x|U(t,t_0)|x_0\rangle;$$ Thus evaluating the time-evolution operator and etc we find: $$\psi(x,t)=\int dx_0\;\frac{A}{\sqrt{t}}\exp\bigg({\frac{im(x-x_0)^2}{2t}}\bigg)\psi(x_0,t_0)$$ with $$A$$ being a constant.

So the problem for me is this: giving a gaussian like $$\psi(x_0,t_0)$$ centered at $$x_0$$. I supposed that the latter-time wavefunction $$\psi(x,t)$$ would have a maximum at $$\bar x=(p/m)(t-t_0)$$, but it seems that the wavefunction keeps centered at $$x_0$$ and just goes spreading.

Am I in the proper frame of the particle? How can I see the particle moving at all?

• what is the $p$ in your formula for $\bar{x}$ and how did you derive that formula? Why do you think a particle with initially Gaußian wavefunction should move? Commented Apr 22, 2023 at 13:42

If you start with a particle having a definite position,e.g., $$\psi(x_0,t_0)=\delta(x_0-y)$$, then its momentum ic completely uncertain - it can go either right or left with equal probability. Thus we do not expect directed motion.
If we start with a particle having a definite momentum, $$\psi(x_0,t_0)\propto e^{ikx_0}$$, it will remain in this state, since it is the eigenstate of the Hamiltonian - nothing changes.
Now, you can experiment with Gaussian wave packets - having finite position and momentum uncertainty - these will indeed behave as moving with speed, which is the mean speed of the initial packet. But this information is encoded in $$\psi(x_0,t_0)$$, not in the propagator.
• To really tie this home: you can multiply a Gaussian wavepacket with $e^{ikx_0}$ to get a wavepacket that is moving with wavenumber $k$. Commented Apr 22, 2023 at 14:58