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Given a valid wave function, like the one shown in the upper-left-hand corner of the animation, how does the form of the wave function affect the average value or expectation value of variables such as position and momentum?  We calculate an expectation value of an observable, A, as

<A> = ∫ Ψ^*(x, t) A Ψ(x, t) dx [integral from −∞ to +∞] (6.10)

which is the weighted average of A; the average weighted by the probability density. If instead we are working with an energy eigenfunction, ψ(x), Eq. (6.10) becomes:

<A> = ∫ ψ^*(x) A ψ(x) dx [integral from −∞ to +∞] (6.11)

Explicitly then, if we have a wave function, we can calculate the expectation values of

<x>,     <x²>,     <p>,     <p²>,    <T>,     <V>, and <H>, (6.12)

where T = p²/2m is the kinetic energy operator, V is the potential energy operator, and H = T + V is the Hamiltonian operator. In general, since the wave function is spread out over space, we expect that there will be some spread (indeterminacy/uncertainty) in our measured values as well. Thus, for position and momentum we can describe this spread using a standard deviation as: Δx = [ <x²> − <x>² ]^1/2  and Δp = [ <p²> − <p>² ]^1/2 and find

ΔxΔp ≥ ħ/2 , (6.13)

which is the Heisenberg uncertainty principle.⁶

In the animation, a wave function for a particle confined between x = -2 and x = 2 is shown with its corresponding probability density. In addition you may create the product of a function of x, f(x), and the probability density, and integrate the result. Such a construction will yield <f(x)>.  To make sure that the wave function is normalized, first integrate the probability density by choosing f(x) = 1.  Now calculate <x> (f(x) = x) and <x²> (f(x) = x*x).  Do your results seem reasonable?  The fact that <x> = 0 follows from the form of the probability density (equal probability for x < 0 and x > 0).  However, it may be surprising that <x²> = 1.2 given <x> = 0.  For <x>, the probability density is weighted by x, which is negative for x < 0.  For <x²>, the probability density is weighted by x², which is always positive. We can also calculate the spread in x (standard deviation) and we find that Δx = (<x²>)^1/2 = 1.095.

⁶A better description of this concept is actually the phrase 'Heisenberg indeterminacy principle' since an inherent indeterminacy associated with the results of measurements better represents the concept that Heisenberg was trying to describe.