# Free Particle Wavepacket Spreading: Is it True?

In summary, the wavepacket of a free particle generally spreads out as time goes to infinity, due to the linear nature of the Schrodinger equation. This is in contrast to non-linear wave equations that can have "soliton" solutions which do not spread out. It can be shown that the average position of the particle increases with time, and this can be seen by considering the frame where the average position and momentum are both zero. This results in a quadratic increase in the average position with time, while higher order derivatives are zero. However, it is possible for the wavepacket to momentarily squeeze up before it starts to spread out again.

Is it generally true that the wavepacket of a free particle spreads out as time goes to infinity? It seems like it would, since the phase velocities of the component plane waves are different, and therefore the plane waves would get increasingly out of phase with time. A gaussian wave packet spreads with time. I'm just wondering if it is true for arbitrary wavepackets, and if so, how do you show that?

thanks

edit: to be more specific, I'm trying to show that

$$(\Delta x)^2 = \langle x^2 \rangle - \langle x \rangle ^2$$

increases as t -> infinity

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Well, it depends on your wave equation, not so much on the wave packet. The Schrodinger equation is linear and so the wave packets spread out. There are non-linear wave equations, e.g., the Sine-Gordon equation, water waves, etc., which admit "soliton" solutions which don't spread out but which are held together by their non-linearness. Of course I'm talking about "free space" here, since potential wells obviously confine packets.

I'm assuming you're talking about the Schrodinger equation in free space here. You can use Galilean invariance of the SE to go to a frame where $$\left\langle x \right\rangle = 0$$ for all time.

I think it suffices to show that $$\left\langle \psi | x^2 | \phi \right\rangle$$ grows arbitrarily where $$\psi$$ and $$\phi$$ are any two states. Now all you need to do is pick a complete basis, and eigenstates of $$x$$ are well suited.

lbrits said:
I'm assuming you're talking about the Schrodinger equation in free space here.

Yep, I had the Schrodinger equation in mind.

You can use Galilean invariance of the SE to go to a frame where $$\left\langle x \right\rangle = 0$$ for all time.

OK, that is a good suggestion. That should simplify things.

I think it suffices to show that $$\langle \psi |x^2|\phi \rangle$$ grows arbitrarily where $$\psi$$ and $$\phi$$ are any two states.

I don't quite understand this part. Are $$\psi$$ and $$\phi$$ two elements of the complete basis you referred to or just any two arbitrary wavefunctions?

Showing that arbitrary matrix elements of an operator behave in a certain way is equivalent to showing that the matrix elements of an operator behave that way in a certain basis.

Yes, I think I see what you mean now. At any instant the wavefunction can be represented as a superposition of the eigenstates of the position operator, so showing that the matrix element of x^2 between two position eigenstates |x> and |x'> increases with time in the long run is equivalent to showing that the average value w.r.t the wavefunction increases with time. The thing I wasn't seeing was this: since we are dealing with a free particle hamiltonian, a state that is initially (t=0) in a position eigenstate |x> will not remain in a position eigenstate, but will spread out. I was computing the matrix element <x | x^2 | x'> as if a position eigenstate |x> remained a position eigenstate, which gives the not very helpful answer $x^2\delta(x-x')$.

After seeing where I was going wrong, it occurred to me to apply the time-evolution operator directly to the initial wavefunction to compute the average value of x^2:

$$\langle x^2 \rangle(t) = \langle \psi(t)| \hat{x}^2 | \psi(t) \rangle = \langle\psi(0)|e^{it\hat{H}/\hbar}\hat{x}^2e^{-it\hat{H}/\hbar} |\psi(0)\rangle$$

where H = p^2 / 2m.

Then use the commutator relation

$$\left[\hat{x}, e^{-i(t/2m\hbar)\hat{p}^2}\right] = \left( e^{-it\hat{H}/ \hbar}\right) \frac{t\hat{p}}{m}$$

twice to get

$$\langle x^2 \rangle(t) = \langle\psi(0)|\left(\hat{x} + \frac{t\hat{p}}{m}\right)^2|\psi(0)\rangle$$

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

Then, as you suggested, we can consider the frame where <x> = <p> = 0. Since the initial spread in momentum <p^2> is positive, this shows that <x^2> grows like time squared. Does this seem correct?

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Correct, although the term linear in t bothers me a bit.

lbrits said:
Correct, although the term linear in t bothers me a bit.

Yep, I think that means that a wavepacket could possibly spend a finite time squeezing up before it starts to spread out again. But I'll double check to make sure I got the algebra right. Thanks for your help.

lbrits said:
Correct, although the term linear in t bothers me a bit.

Another way of getting the same answer is to write <x^2>(t) as a Taylor expansion around t=0, and compute the derivatives in the following way:
If the operator A is not a function of time,

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

$$\frac{d}{dt}\langle x^2 \rangle = \frac{i}{2m\hbar}\langle[\hat{p}^2 , \hat{x}^2]\rangle$$

But

$$[\hat{p}^2 , \hat{x}^2] = \hat{p}[\hat{p}, \hat{x}^2] + [\hat{p}, \hat{x}^2]\hat{p} = -2i\hbar\left(\hat{x}\hat{p}+\hat{p}\hat{x}\right)$$

So that
$$\frac{d}{dt}\langle x^2 \rangle = \frac{1}{m}\langle \hat{x}\hat{p}+\hat{p}\hat{x}\rangle$$

which is also a function of time. Taking another time derivative gives

$$\frac{d^2}{dt^2}\langle x^2 \rangle = \frac{1}{m}\frac{d}{dt}\langle\hat{x}\hat{p}+\hat{p}\hat{x}\rangle = \frac{i}{2m^2\hbar}\langle[\hat{p}^2, \hat{x}\hat{p}] + [\hat{p}^2, \hat{p}\hat{x}]\rangle = \frac{i}{2m^2\hbar}\langle [\hat{p}^2, \hat{x}]\hat{p} + \hat{p}[\hat{p}^2, \hat{x}] \rangle = \frac{2}{m^2}\langle p^2 \rangle$$

which is constant. Therefore all higher order derivatives are zero. Substituting these results into the taylor expansion

$$\langle x^2 \rangle(t) = \langle x^2 \rangle(0) + \langle x^2 \rangle'(0)t + \langle x^2 \rangle''(0)\frac{t^2}{2}$$

gives the same answer as above. This must put a restriction on the values <xp + px> can take in order to prevent <x^2> from being negative at some time. I don't see how to guarantee that <xp + px> is positive. It is real because the operator is hermitian, but that's all I can conclude about it...weird.

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## 1. What is a free particle wavepacket?

A free particle wavepacket is a mathematical representation of a particle in quantum mechanics, where the particle is not affected by any external forces and can freely move through space.

## 2. What is wavepacket spreading?

Wavepacket spreading refers to the phenomenon in which a free particle's wavepacket, which represents the probability of finding the particle at a given point in space, becomes wider and more spread out over time.

## 3. Is free particle wavepacket spreading true?

Yes, free particle wavepacket spreading is a well-established concept in quantum mechanics and has been experimentally verified through various experiments.

## 4. Why does wavepacket spreading occur?

Wavepacket spreading occurs due to the inherent uncertainty in the position and momentum of a particle in quantum mechanics. As time passes, this uncertainty increases and leads to a wider and more spread out wavepacket.

## 5. Can the rate of wavepacket spreading be controlled?

The rate of wavepacket spreading can be influenced by the initial conditions and properties of the particle, such as its energy and momentum. However, it cannot be fully controlled and is a fundamental aspect of quantum mechanics.

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