Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Peak of psi squared and <x> not same?

  1. Oct 25, 2003 #1
    While working on an infinite barrier problem I was surprised to find that the location of |psi|^2 = 0.30 a while <x> is 0.32 a, where a is the width of the well.

    It's often said that |psi|^2 tells one about the probabilty of finding a particle in a location x. Yet the expectation of x, <x> is a slightly different value. How do I reconcile the difference between these two values?

    If <x> is the average of values if I observe a large number of particles prepared in the same way, then what exactly is the peak of |psi|^2 ?
  2. jcsd
  3. Oct 25, 2003 #2
    This isn't anything mysterious about quantum mechanics; it's well known in statistics that the most probable value of a random variable is generally not the the same as the mean (or expected) value of the variable. In fact it rarely is, Gaussian distributions being an exception. I'm not sure what needs "reconciling" here...?
  4. Oct 26, 2003 #3
    That's it, just a heuristic explanation for why the most probable does not equal the mean. Since nearly all the distributions I have ever dealt with are treated as gaussian, I forget those little details like the mean is not always equal to the most probable value.

    I imagine that as the distribution gets broader or more assymmetrical the mean can be further from the most probable value?
  5. Oct 26, 2003 #4
    Yes... imagine, for instance, something that is mostly Gaussian, except with a thin spike far off to the side, that's just slightly higher than the main Gaussian. The most probable value will be the spike, but as far as the mean is concerned, the spike makes almost no difference: the mean will still be at the center of the Gaussian.

    Of course, you can cook up all kinds of screwy functions, and in general, the mean and the maximum will have nothing to do with each other. After all,

    x_max is the x such that dP/dx = 0
    <x> = integral^b_a x P(x) dx / (b-a)

    The two aren't intrinsically related.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook