Recall that the position operator is and the momentum operator .

Recalling our abuse of notation that said that the expected value , we find that the expected value of position is . Note: since we can view as a probability density function; hence if is any “reasonable” function of , then . Of course we can calculate the variance and other probability moments in a similar way; e. g. .

Now we turn to momentum; and

So, **back to position**: we can now use the fact that is a valid density function associated with finding the expected value of position and call this the position probability density function. Hence . But we saw that this can change with time so:

This is a great chance to practice putting together: differentiation under the integral sign, Schrödinger’s equation and integration by parts. I recommend that the reader try to show:

The details for the above calculation (students: try this yourself first! 🙂 )

Differentiation under the integral sign:

Schrödinger’s equation (time dependent version) with a little bit of algebra:

Note: is real.

Algebra: eliminate the partial with respect to time terms; multiply the top equation by and the second by . Then add the two to obtain:

Now integrate by parts:

Now the integrals cancel each other and we obtain our result.

It is common to denote by (note the minus sign) and to say (see the reason for the minus sign?)

is called the position probability current at the point at time One can think of this as a "probability flow rate" over the point at time ; the quantity will tell you if the probability of finding the particle between position and is going up (positive sign) or down, and by what rate. But it is important that these are position PROBABILITY current and not PARTICLE current; same for ; this is the position probability density function, not the particle density function.

**NOTE** I haven’t talked about the position and momentum eigenvalues or eigenfuctions. We’ll do that in our next post; we’ll run into some mathematical trouble here. No, it won’t be with the position because we already know what a distribution is; the problem is that we’ll find the momentum eigenvector really isn’t square integrable….or even close.

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