How much does the uncertainty in position increase with time?

1. Apr 26, 2007

Fredrik

Staff Emeritus

There's a clock at A and a mirror at B. We want to measure the distance L between A and B by measuring the time it takes a light signal to travel from A to B and back. The mass of the clock is m. Its position is represented by a wave function spread out over an interval of length dL. (Pretend that space is one-dimensional). The article claims that in the time that it takes a photon to travel from A to B and back, the clock's wave function will have spread out further, so that the width is now dL+(hbar*L)/(mc*dL).

Appareantly this is something that Wigner proved in 1957. I don't see how to obtain this result. Am I missing something simple?

I assume that we can take the wave function to have a gaussian shape (exp(-ax^2)), but how do we find how much the width has increased in a certain time? I suppose we should have a time evolution operator exp(-iHt) act on this wave function, but I don't see a way to simplify the result. I'm hoping it's just because I haven't been doing this sort of thing in a while.

2. Apr 27, 2007

lalbatros

Expand the wavefunction as a sum of free-space waves.
Observe the spread in momentum and conclude.

The uncertainty principle tells us that: Dx Dt >= hbar/2 .
The equality occurs only for a Gaussian shaped spatial probability distribution.
You can check that the momentum probability distribution is also gaussian in this case.
You can find out the relation between the spatial width and the width of the momentum spectrum.
Since all wave components don't have the same momentum, the wave packet will spread, of course.
You could try to calculate that, this is easy since it is the solution of the SE in free space.

You could also read about the evolution of a wavepacket in many textbooks.

Last edited: Apr 27, 2007
3. Apr 27, 2007

Fredrik

Staff Emeritus
This is what I've got so far:

Suppose the position space wave function is

$$\exp{(-ax^2)}$$

Then the momentum space wave function is

$$\exp{(-p^2/4a)}$$

At least that's what I get when I do the Fourier transform explicitly.

The effect of time evolution is to add $iEt[/tex] to the exponent, with [itex]E=p^2/2m[/tex], but this only changes the phase (of the momentum space wave function), so the width remains the same (right?). This implies that the width of the position space wave function will also remain the same. So the result I get is that a Gaussian wave function doesn't spread at all with time. Can someone verify that this is correct? I still don't have any idea how to obtain Wigner's result. Maybe I have to assume that the wave function has a different shape. Last edited: Apr 27, 2007 4. Apr 28, 2007 lalbatros Each wavelet Exp[i p/hb x]​ evolves as Exp[i p/hb x + i p²/2m t]​ You have to sum that over the initial distribution of wavelets Exp[-p²/4a]​ Note that the phase depends on p². I don't believe this would lead to a constant shape of the wavepacket. Can you try to calculate that? Last edited: Apr 28, 2007 5. Apr 28, 2007 jtbell Staff: Mentor To be more explicit: $$\Psi(x,t) = \int_{-\infty}^{+\infty} { A \exp \left( - \frac{p^2}{4a} \right) \exp \left( \frac{ipx}{\hbar} - \frac{ip^2 t}{2m \hbar} \right) } dp$$ The result of this integral (more precisely, [itex]|\Psi|^2$, the probability distribution) can be put in the form of a Gaussian in x, with a time-varying width. This is a standard topic in quantum-mechanics textbooks.

Last edited: Apr 28, 2007