Consider any wave packet describing a free particle (so no potential or other forces acting on it). Then it can be shown that $\Delta p$ does not change in time. However, my question is what happens with $\Delta x$ as we go forward in time? Does it have to increase at all times? Or is there a counter-example where the uncertainty in position is decreasing, if even for a short time period?

My initial guess is $\Delta x$ must always increase, because $p \neq 0$, so that $\Delta p \neq0$ and hence $\Delta v = \frac{\Delta p}{m} \neq 0$. But if there's a spread in velocities, then the wave packet must also spread. Is this logic correct? Or could we have a wave packet where the back of it would move forward faster than the front, and for a certain period until that back end catches up with the front one, it would actually be narrower than at the outset, i.e. reducing $\Delta x$? If yes, how would one describe such a free particle (wave packet)?

So it does seem to me that every wave packet describing a free particle will eventually spread, but the question is whether there can be a time period in its evolution when it is actually becoming narrower.

edit: In particular, if it does not have to increase at all times, can this be shown without appealing to time reversal?

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    $\begingroup$ Related, but not the same question: physics.stackexchange.com/q/7231 $\endgroup$
    – David Z
    Feb 20, 2013 at 18:16
  • $\begingroup$ @DavidZaslavsky Yeah, I saw that already, and something similar to my question was discussed in the comments there, as well. However, it was never stated whether $\Delta x$ must increase (or in the ideal case, stay the same) at all times, not just eventually. $\endgroup$
    – Ryker
    Feb 20, 2013 at 18:21

4 Answers 4


If $\Psi(x,t)$ solves the Schrodinger equation, so does $\Psi^*(x,-t)$ , so no, there is nothing at all that must increase.

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    $\begingroup$ Can you elaborate on that? Are you saying that if the uncertainty is increasing on, say, $t \in (t_{1}, t_{2})$, then there is a time period where the uncertainty in position is decreasing, namely $t \in (-t_{2}, -t_{1})$? $\endgroup$
    – Ryker
    Feb 20, 2013 at 19:02
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    $\begingroup$ I am saying that if $\Psi(x,t)$ has increasing uncertainty from t1 to t2, then define $\bar{\Psi}(x,t) = \Psi(x,-t)$, and $\bar{\Psi}$ will have decreasing uncertainty from t1 to t2. $\endgroup$ Feb 20, 2013 at 19:09
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    $\begingroup$ I find it extremely obvious, to be honest. I'm just saying you can time-reverse anything. If something goes up when time goes forward, it goes down when time goes backward, but the Schrodinger equation can't tell the difference between "time goes forward" and "time goes backward", so anything that can increase for one solution can decrease for another solution. $\endgroup$ Feb 20, 2013 at 19:11
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    $\begingroup$ Or, if you want an example easy to visualize, take two Gaussians with a very wide separation but moving towards each other as an example. Then the uncertainty in x is basically just the separation, and the spread of the Gaussians is ignored. Evolve that forward in time some. The Gaussians get closer together, reducing the uncertainty in x, although they individually spread a bit. $\endgroup$ Feb 20, 2013 at 21:32
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    $\begingroup$ Yes, the two Gaussians describe a free particle. If one Gaussian does, then two Gaussians can by superposition. It is easy to adjust the width of the Gaussians and the separations so the statement is correct. $\endgroup$ Feb 20, 2013 at 21:42
  1. We reformulate OP's title question (v1) as follows:

    Show that for all possible wave packets $\psi(p,t)$ of a free particle, the position variance $$ {\rm Var}(\hat{x}) ~=~\langle \hat{x}^2\rangle-\langle \hat{x}\rangle^2\tag{1}$$ decreases in at least some interval $[t_1,t_2]$ of time.

    As Mark Eichenlaub correctly observes in his answer and comments, a solution with decreasing ${\rm Var}(\hat{x})$ in $[t_1,t_2]$ can be mapped by time reversal symmetry to a solution with increasing ${\rm Var}(\hat{x})$ in $[-t_2,-t_1]$. Here it is used that the Hamiltonian $\hat{H}=\frac{\hat{p}^2}{2m}$ for a free particle, and the position operator $\hat{x}$ both commute with the time reversal operator $\hat{T}$. See e.g. my Phys.SE answer here for further details.

    However, time reversal symmetry does logically not rule out alone the possibility that a solution has monotonically increasing ${\rm Var}(\hat{x})$ for all time. (It only implies that if this is the case, then there will also be a solution with monotonically decreasing ${\rm Var}(\hat{x})$ for all time.)

  2. Explicit calculation. Consider an arbitrary wave packet $$ \psi(p,t)~=~A(p)e^{-i\theta(p,t)} \tag{2}$$ that represents the general solution to the time-dependent Schrödinger equation in the momentum representation. Here the angle $$ \theta(p,t)~=~\theta_0(p) + \frac{p^2 }{2m}\frac{t}{\hbar},\qquad A(p),\theta_0(p)~\in~\mathbb{R}, \tag{3}$$ is affine in $t$. We assume that the wave packet is normalized $$ 1~=~||\psi(t)||^2~=~ \langle \psi(t)| \psi(t)\rangle~=~\int_{\mathbb{R}} \!dp~A^2 .\tag{4}$$ The position operator in the momentum representation reads $$ \hat{x}~=~i\hbar\frac{\partial }{\partial p}.\tag{5}$$ Therefore the position average $$\langle \hat{x}\rangle~=~ \langle \psi(t)| \hat{x}|\psi(t)\rangle ~=~\hbar\int_{\mathbb{R}} \!dp~A^2\theta^{\prime} \tag{6}$$ is affine in $t$, and the average of the squared position $$ \langle \hat{x}^2\rangle~=~ \langle \psi(t)| \hat{x}^2|\psi(t)\rangle ~=~\hbar^2 \int_{\mathbb{R}} \!dp~(A^{\prime 2} +A^2\theta^{\prime 2} ) \tag{7} $$ is quadratic in $t$. Here primes denote differentiation wrt. $p$.

    Thus we immediately know that the position variance $$ {\rm Var}(\hat{x})~=~at^2+bt+c\tag{8}$$ is quadratic in $t$ as well, where $a$, $b$, and $c$ are constant coefficients. It is straightforward to see from the Cauchy-Schwarz inequality $$\left(\int_{\mathbb{R}} \!dp~A^2\frac{p}{m} \right)^2 ~\leq~ \int_{\mathbb{R}} \!dp~\left(A\frac{p}{m}\right)^2 \cdot \int_{\mathbb{R}} \!dp~A^2 \tag{9}$$ that the second-order coefficient $$ a~=~\int_{\mathbb{R}} \!dp~\left(A\frac{p}{m}\right)^2-\left(\int_{\mathbb{R}} \!dp~A^2\frac{p}{m} \right)^2~\geq ~0, \tag{10}$$ is non-negative, cf. the normalization condition (4). In fact one may show that $a>0$ is strictly positive, and hence the ${\rm Var}(\hat{x})$ is a decreasing parabola for $t$ sufficiently negative, as we wanted to show.

    Sketched indirect proof of $a\neq 0$: The second-order coefficient $a$ becomes zero $~\Leftrightarrow~$ the Cauchy-Schwarz inequality (9) becomes an equality $~\Leftrightarrow~$ $A(p) p$ is proportional to $A(p)$ $~\Leftrightarrow~$ $A(p)$ is proportional to a delta function $\delta(p-p_0)$. But this does not correspond to a normalizable wave packet.

  3. Example. Two mutually approaching wave trains are an easy intuitive example where the position variance ${\rm Var}(\hat{x})$ diminishes in some time-interval $[t_1,t_2]$. But this is, in some sense, a lazy example, which sort of betrays just how universal and omnipresent this behavior is for quantum mechanics.

    For instance, as we know from the general analysis in Section 2, already the simplest possible wave packet, i.e. a single Gaussian wave packet, displays this behavior. However, that is much less intuitive, and thus that much more fascinating/mind boggling to try to understand. Let us for simplicity set $\hbar=1=m$.

    A single Gaussian wave packet at $t=0$ is of the form $$ \psi(p,t=0)~=~N \exp\left(-ipc - \frac{p^2}{2}\tau\right) ,\tag{11}$$ where $\tau>0$ and $c=a+ib\in\mathbb{C}$ are constants. Here $N>0$ is a normalization constant. The normalization condition (4) implies that $$ N~=~\left(\frac{\tau}{\pi}\right)^{\frac{1}{4}}\exp\left(-\frac{b^2}{2\tau}\right).\tag{12}$$ Thus for arbitrary time $t$, the Gaussian wave packet reads $$ \begin{align}\psi(p,t)~=~&\psi(p,0)\exp\left(-i\frac{p^2}{2}t\right)\cr ~=~& N \exp\left(-ipc - \frac{p^2}{2}(\tau+it)\right) .\end{align}\tag{13}$$ In the position representation, the Gaussian wave packet becomes $$ \begin{align}\psi(x,t) ~=~&\int_{\mathbb{R}} \frac{dp}{\sqrt{2\pi}} \exp\left(ipx\right) \psi(p,t)\cr ~=~&\frac{N}{\sqrt{\tau+it}} \exp\left(-\frac{(x-c)^2}{2(\tau+it)}\right).\end{align} \tag{14}$$ The position probability distributions becomes $$\begin{align}|\psi&(x,t)|^2\cr ~=~&\frac{N^2}{|\tau+it|} \exp\left(-{\rm Re}\frac{(x-c)^2}{\tau+it}\right) \cr ~=~&\frac{N^2}{\sqrt{\tau^2+t^2}} \exp\left(-\frac{[(x-a)^2-b^2]\tau-2(x-a)bt}{\tau^2+t^2}\right) \cr ~=~&\sqrt{\frac{\tau}{\pi(\tau^2+t^2)}} \exp\left(-\frac{(x-a-\frac{bt}{\tau})^2\tau }{\tau^2+t^2} \right). \end{align}\tag{15}$$ Therefore the position average $$ \langle \hat{x}\rangle~=~\int_{\mathbb{R}} \!dx~x|\psi(x,t)|^2 ~=~a+\frac{bt}{\tau}\tag{16}$$ is an affine function of $t$, while the position variance $$ {\rm Var}(\hat{x}) ~=~\int_{\mathbb{R}} \!dx~(x-\langle \hat{x}\rangle)^2|\psi(x,t)|^2 ~=~\frac{\tau^2+t^2}{2\tau}\tag{17}$$ is a quadratic function of $t$.

    The time symmetric profile (17) of the position variance ${\rm Var}(\hat{x})$ of a single Gaussian wave packet is probably somewhat surprising to anyone who borrows his intuition from classical physics.


I don't know the answer to this yet, but here is a calculation that others might find useful in determining the answer. Let's compute the time derivative of $\sigma_X$. First note that $$ \frac{d}{dt}\sigma_X^2 = 2\sigma_X\dot\sigma_X, \qquad \sigma_X = \langle X^2\rangle - \langle X\rangle^2 $$ so \begin{align} \dot\sigma_X &= \frac{1}{2\sigma_X}\frac{d}{dt}\left(\langle X^2\rangle - \langle X\rangle^2\right) \\ &= \frac{1}{2\sigma_X}\left(\frac{d}{dt}\langle X^2\rangle - 2\langle X\rangle\frac{d}{dt}\langle X\rangle \right)\\ \end{align} Now use the general relation $$ \frac{d}{dt}\langle O\rangle = \frac{i}{\hbar}\langle[H,O]\rangle $$ for an operator with no explicit time-dependence. It follows that \begin{align} \frac{d}{dt}\langle X\rangle = \frac{i}{\hbar}\langle[P^2/2m, X]\rangle =\frac{i}{2m\hbar}\langle -2i\hbar P\rangle = \frac{\langle P\rangle}{m} \end{align} so that also \begin{align} \frac{d}{dt}\langle X^2\rangle &= \frac{i}{\hbar} \left\langle\left[\frac{P^2}{2m}, X^2\right]\right\rangle = \frac{i}{2m\hbar}(-2i\hbar)\langle\{P,X\}\rangle = \frac{\langle \{P,X\}\rangle}{m} \end{align} where $\{P, X\}=PX+XP$, and therefore $$ \dot\sigma_X = \frac{1}{2m\sigma_X}\Big(\langle \{P,X\}\rangle - 2\langle P \rangle \langle X \rangle\Big) $$ We want to know if there is a state for which at some time, $\dot\sigma_X<0$. Since $\sigma_X>0$, the expression just derived leads us to an equivalent question; is there a state for which the following inequality holds at some time? $$ \langle\{P,X\}\rangle < 2\langle P\rangle \langle X \rangle\qquad ? $$ Let me know if you find any errors in the math as I did it quickly.

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    $\begingroup$ Looks fine to me. Apply Mark Eichenlaub's approach to this. Suppose a state exists such that $\dot{\sigma}_x>0$. Then by your calculation $2\left\langle P\right\rangle \left\langle X\right\rangle > \left\langle \{P,X\}\right\rangle $. Now, under time reverse $P$ changes sign but not $X$, so under T: $2\left\langle -P\right\rangle \left\langle X\right\rangle > \left\langle \{-P,X\}\right\rangle $ Now multiply through by a minus and you get the desired result. $\endgroup$
    – Michael
    Feb 20, 2013 at 21:29
  • $\begingroup$ @MichaelBrown Cool thanks. Somehow I would still be more satisfied with a calculation for a particular state in which the bound were explicitly verified. It would be nice to see how it works out. $\endgroup$ Feb 20, 2013 at 21:56
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    $\begingroup$ You could take an example state of two Gaussians that are coming towards each other, with x=0 in between. Then <p> = <x> = 0, but the quadratic operator doesn't get killed by the mirror-image symmetry. $\endgroup$ Feb 21, 2013 at 0:20
  • $\begingroup$ If my brain hasn't broken (not enough coffee today) then $|0>-|2>$ (in harmonic oscillator eigenfunctions) should be another example. Not too confident it works atm, but if it doesn't it shouldn't be hard to tweak into an example. :) $\endgroup$
    – Michael
    Feb 21, 2013 at 1:54
  • $\begingroup$ Derp derp doesn't work. Shows I need to make a coffee and have a lay down. $\endgroup$
    – Michael
    Feb 21, 2013 at 3:21

There is yet another solution (maybe more elementary)$^1$, with some components of the answers from Qmechanic and JoshPhysics (Currently I'm taking my first QM course and I don't quite understand the solution of Qmechanic, and this answer complement JoshPhysics's answer) the solution uses the Heisenberg Equation:

The time evolution of an operator $\hat{A}$ in the Heisenberg Picture of Quantum mechanics is given by:

$$\frac{d\hat{A}}{dt} = \frac{1}{i\hbar} [\hat{A},\hat{H}]$$

For a free particle with $\hat{H} = \frac{\hat{p}^2}{2m} $, the time evolution of the operators $\hat{x}$ and $\hat{p}$ are :

$$\frac{d\hat{p}(t)}{dt}=0 \:\:\:\:\:\:\: \frac{d\hat{x}(t)}{dt}=\frac{\hat{p}}{m}$$

These are the equations of motion in the Heisenberg picture of a free particle, now the solutions yield the same that for a classical free particle:

$$\hat{p}(t) = \hat{p}(0) \:\:\:\:\:\:\: \hat{x}(t) =\hat{x}(0)+\frac{\hat{p}(0)}{m}t$$

With this expressions is easy to see that (or use Ehrenfest theorem ),

$$\langle \hat{x}(t) \rangle^2 = \langle \hat{x}(0) \rangle^2 + \frac{2t}{m} \langle \hat{x}(0) \rangle \langle \hat{p}(0) \rangle +\frac{t^2}{m^2} \langle \hat{p}(0) \rangle^2$$ $$\langle \hat{x}(t)^2 \rangle = \langle \hat{x}(0)^2 \rangle + \frac{t}{m}\langle \{\hat{x}(0),\hat{p}(0)\} \rangle +\langle \hat{x}(0) \rangle +\frac{t^2}{m^2} \langle \hat{p}(0)^2 \rangle $$ $$\langle \hat{p}(t) \rangle^2 = \langle \hat{p}(0) \rangle^2$$ $$\langle \hat{p}(t)^2 \rangle = \langle \hat{p}(0)^2 \rangle $$

Where we defined the anticonmutator as$ \{\hat{x}(0),\hat{p}(0)\} = \hat{x}(0)\hat{p}(0) +\hat{p}(0) \hat{x}(0)$, with this we can compute the standart deviation of the momentum and position operators of the particle, now for the position operator:

\begin{align} (\Delta \hat{x}(t) )^2 & = \langle \hat{x}(t)^2 \rangle - \langle \hat{x}(t) \rangle^2 \\\\ & = (\Delta \hat{x}(0) )^2 +\frac{t}{m} \Big( \langle \{\hat{x}(0),\hat{p}(0)\} \rangle- 2\langle \hat{x}(0) \rangle \langle \hat{p}(0) \rangle \Big) + \frac{t^2}{m^2} (\Delta \hat{p}(0) )^2 \end{align}

We can see that it's an expression of the form, $at^2+bt+c$ with

$$a= (\Delta \hat{p}(0) )^2 \:\:\:\:\:\:\: b=\Big( \langle \{\hat{x}(0),\hat{p}(0)\} \rangle- 2\langle \hat{x}(0) \rangle \langle \hat{p}(0) \rangle \Big) \:\:\:\:\:\:\: c= (\Delta \hat{x}(0) )^2 $$

We can see that $ a,c \geq 0 $ , however there are no restrictions for $b$, it in principle can take any value, one can imagine an initial state where $b$ is negative and very big for small $t$, the standard deviation starts decreasing, but eventually for a big enough $t$ it start increasing as one can see in the section 5 of this article Wave packet spreading: Temperature and squeezing effects with applications to quantum measurement and decoherence

Its interesting that we arrive to the same result that Joshphysics with somehow different approaches.

$^1$ Again, I post this answer to complement the already given ones, personally when I first encountered this problem I was confused with this thread, I read it but I didin't quite grasp the time reversal symetry and the other solutions where confusing or to advanced for my level. I hope this gives a light to an student searching for this in the future. A good complement for starters can be found here Heisenberg Picture: U Colorado Advanced Quantum Mechanics


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