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If there's wave packet $\psi(x,t)$ in zero potential and $\psi(x,0)$ is given, what happens when $t<0$? I know that for $t>0$ the wave as a group moves at group velocity $\frac{d\omega}{dk}|_{k_0}$ where we assume that $k$ is strongly peaked around $k_0$. The width of the wave becomes wider when $t$ is large. But what about $t<0$? Does the wave just become narrower?

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    $\begingroup$ Just put a negative $t$ into the equation? $\endgroup$
    – Physiker
    Commented Feb 27 at 5:52

1 Answer 1

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In the Heisenberg picture, the time evolutions of the position operator $X(t)$ and the momentum operator $P(t)$ of a free particle with mass $m$ are given by $$X(t)=X(0)+\frac{P(0) }{m}t, \quad P(t)=P(0). \tag{1} \label{1}$$ Defining $\langle A\rangle_\psi:=\langle \psi |A|\psi \rangle$ and $\left(\Delta_\psi A \right)^2:= \langle \left(A-\langle A \rangle_\psi \right)^2\rangle_\psi =\langle A^2 \rangle_\psi-\langle A\rangle_{\!\psi}^{\,2}$ for an arbitrary (hermitean) operator $A$ and an arbitrary state vector $|\psi\rangle$ (with normalization $\langle \psi |\psi \rangle=1$), we find $$\begin{align}\left(\Delta_\psi X(t)\right)^2&= \langle X(0)^2 \rangle_\psi+\frac{t}{m}\langle \{X(0),P(0)\} \rangle_\psi +\frac{t^2}{m^2}\langle P(0)^2 \rangle_\psi \\[5pt] & -\langle X(0) \rangle_{\! \psi}^{\,2}- \frac{2t}{m} \langle X(0)\rangle_\psi \langle P(0) \rangle_\psi -\frac{t^2}{m^2}\langle P(0) \rangle_{\! \psi}^{\,2} \\[5pt] &=\left( \Delta_\psi X(0) \right)^2 +\frac{t}{m} \left(\langle \{X(0),P(0)\}\rangle_\psi-2\langle X(0)\rangle_\psi\langle P(0)\rangle_\psi \right) +\frac{t^2}{m^2} \left( \Delta_\psi P(0)\right)^2,\end{align} \tag{2}$$ where $\{A,B\}:=AB+BA$ denotes the anticommutator of two operators $A$, $B$. Assuming (without loss of generality) $\langle X(0) \rangle_\psi=0$, the time dependence of the square of $\Delta_\psi X(t)$ assumes the simple form $$\left( \Delta_\psi X(t) \right)^2 = (\left( \Delta_\psi X(0) \right)^2+\frac{t}{m} \langle \{ X(0),P(0) \}\rangle_\psi+\frac{t^2}{m^2}\left( \Delta_\psi P(0) \right)^2 \tag{3} \label{3},$$ a parabola with its minimum at $$t_{\rm min}=-m \langle \{ X(0), P(0) \} \rangle_\psi/2 (\Delta_\psi P(0) )^2 \tag{4} \label{4}$$ with minimum value $$ (\Delta_\psi X(t_{\rm min}))^2 = (\Delta_\psi X(0))^2-\frac{\langle \{X(0),P(0)\}\rangle_{\! \psi}^{\, 2}}{4 (\Delta_\psi P(0))^2} \ge 0 \tag{5} \label{5}$$ and $\Delta_\psi X(t) \to +\infty$ for $t\to \pm \infty$.

Note that for the special case of the Gaussian wave packet $$\langle x |\psi \rangle = \frac{1}{(2\pi)^{1/4}\sigma^{1/2}} e^{-x^2/4\sigma^2} \tag{7} \label{7}$$ with $\Delta_\psi X(0) = \sigma$ and $\Delta_\psi P(0)= \hbar/2 \sigma$, the expectation value of the anticommutator of $X(0)$ and $P(0)$ vanishes and \eqref{3} simplifies to $$(\Delta_\psi X(t) )^2= \sigma^2+\frac{t^2 \hbar^2}{4 m^2 \sigma^2}.\tag{8}$$

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