What Is the Physical Meaning of a Kind of Probability Spreading Wave of a Free Particle at Rest?

I'm not sure I fully understand the question, but I'll try to explain the physical meaning a bit.The parameter $a$ tells us about how "localized" the wavefunction is in the initial configuration. For large $a$, the particle has a large probability of being observed near the origin (at early times). For small $a$, the particle is more likely to be observed far from the origin.One way to think of this is that the uncertainty on the particle's position is small, so the uncertainty on its velocity must be large. For large $a$, the wavefunction rapidly spreads out, because the particle's velocity is more likely to be large (compared to the small $a$ case). This interpretation fits with the mode of $x$ you posted:

$$x(t) fracsqrta hbar tm . $$For large $a$, the particle is more likely to be observed farther from the origin, than if $a$ were small.

There is an exercise in the Griffiths book on QM, where a wave function:

$$psi (x,0) Ae^-ax^2 $$

is used to calculate $psi (x,t)$, after getting $f(k)$, by Fourier analysis.

$rho psi^2 (x,t)$ is a symmetric bell shaped curve, centred in zero that spreads with time.

My question is about the physical meaning of a kind of wave travelling in both directions (x- and x) as $t$ increases. As the probability density spreads, the curve $rho,$ x $t;$ (for a given $x$ constant) has a local maximum for some $t$.

That is logical because if the probability density decreases near zero, it must increase somewhere.

The expected value of the position $langlepsi|x|psirangle$ is always zero, meaning that the particle is at rest.

But the velocity of that "wave" (except for short times) is exactly what should be expected of a particle:

$x sqrtafrachbarm t$

because $sqrtahbar$ has units of momentum.

·OTHER ANSWER:

There is an exercise in the Griffiths book on QM, where a wave function:

$$psi (x,0) Ae^-ax^2 $$

is used to calculate $psi (x,t)$, after getting $f(k)$, by Fourier analysis.

$rho psi^2 (x,t)$ is a symmetric bell shaped curve, centred in zero that spreads with time.

My question is about the physical meaning of a kind of wave travelling in both directions (x- and x) as $t$ increases. As the probability density spreads, the curve $rho,$ x $t;$ (for a given $x$ constant) has a local maximum for some $t$.

That is logical because if the probability density decreases near zero, it must increase somewhere.

The expected value of the position $langlepsi|x|psirangle$ is always zero, meaning that the particle is at rest.

But the velocity of that "wave" (except for short times) is exactly what should be expected of a particle:

$x sqrtafrachbarm t$

because $sqrtahbar$ has units of momentum.

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