1. May 31, 2008

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

Last edited: May 31, 2008
2. May 31, 2008

### lbrits

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.

3. May 31, 2008

### lbrits

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.

4. Jun 1, 2008

Yep, I had the Schrodinger equation in mind.

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

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?

5. Jun 1, 2008

### lbrits

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.

6. Jun 1, 2008

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 occured 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?

Last edited: Jun 1, 2008
7. Jun 1, 2008

### lbrits

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

8. Jun 1, 2008

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.

9. Jun 1, 2008

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.

Last edited: Jun 1, 2008