physics geeking
Jul. 21st, 2004 12:59 pmI already looked this up in Griffiths and it was not being helpful. So.
If you know the expected value for x of a one-dimensional, time-independent wave function is x', then is x' the most probable value for x?
My suckitude at probability is one of the reasons why I didn't pursue a career as a physicist. Maybe if I tried retaking probability now, things would be different, but finding out will have to wait until I have time.
If you know the expected value for x of a one-dimensional, time-independent wave function is x', then is x' the most probable value for x?
My suckitude at probability is one of the reasons why I didn't pursue a career as a physicist. Maybe if I tried retaking probability now, things would be different, but finding out will have to wait until I have time.
no subject
Date: 2004-07-21 01:06 pm (UTC)Alternatively, think of a function whose graph is shaped like an "M". The expected value will be near the center of the "M," but the chance that you actually get the expected value as the result is low.
no subject
Date: 2004-07-21 01:08 pm (UTC)no subject
Date: 2004-07-21 01:08 pm (UTC)If the peaks are of identical magnitude but in opposite directions (that is, if -x(t) equals x(-t)), then the expected value, x', would be 0 because the negatives and positives would cancel out. Yet the chance of actually getting a sample with a value of zero is... nothing. There's only a single point where that's even possible. So it's not the most likely value.
no subject
Date: 2004-07-21 02:20 pm (UTC)