Heisenberg Uncertainty

[Ben]

So, I got chatting to a physicist in the bar the other night, and he tried to explain this me. To be truthful, he was rather drunk (is this typical of physicists, one wonders?), so let me see if this makes any sense. He seemed to be saying something like this:

 

To any particle [math]p[/math], one can associate a wave function [math]\psi[/math] with the property that [math]H\psi = \epsilon \psi[/math] as the allowable eigenvalues for the energy of [math]p[/math] (this I recognize as the non-relativistic Schroedinger equation, where [math]H[/math] is the Hamilton operator)

 

Then it seemed that [math]\psi^2[/math] represents the probability of finding [math]p[/math] at any specific place. (I suspect this is probability density)

 

This makes sense; clearly the solution to [math]\psi[/math] is in the real set [math][-1,1][/math], thus the solution to [math]\psi^2 \in [0,1][/math], which is a definite requirement for any sort of probability.

 

So to the meat of the matter. a priori one has no information about our particle's energy, so one must allow all permissible [math]\epsilon[/math] for which [math]H\psi=\epsilon \psi[/math] holds.

 

Fine.

 

Now suppose we actually know where [math]p[/math] is, say at [math]x[/math]. Then here, [math]\psi^2 = 1[/math] and [math] \psi^2 =0[/math] at all other locations. But the only way to generate such a [math]\psi^2[/math] at [math]x[/math] is by allowing all its possible (first order) wave functions to interfere destructively except at one place [math]x[/math].

 

Hence we cannot make any assumptions about its exact energy.

 

Conversely, if we know the exact energy of our particle, there is only one (first order) wave function that will describe this energy, and hence no interference can occur, so there is no non-probabilistic way to describe its position.

 

Question: is this gibberish? Was my friend correct on an intuitional level - it made sense at the time, or was I drunk also?

[Qfwfq]

Although the conversation was indeed a bit tipsy, it is actually on the right track.

 

As you say, it's a probability density when talking about a continuous spectrum, but this brings that it's the integral which must be unity. Of course, probability must be real valued so, strictly, it's the modulus squared of the amplitude for computing it.

 

Also, the uncertainty disequalities hold between position and momentum as well as between time and energy (not between position and energy). The equation you give is not the "real live" Schrödinger equation but instead the eigenvalue equation for the Hamiltonian; the same can be done with any observable and the self-adjoint operator that represents it. That one is, however, often called the "steady state Schrödinger equation". Why?

 

Let's see how you do at sorting the math out... :)