## Physics Friday 47

Quantum Mechanics and Momentum
Part 4: The Momentum Operator

Frequently, the quantum mechanical operator for momentum is simply presented to students without justification (that was my experience, at least). However, use of our previous developments in Fourier transforms and observables allow us to see a (non-rigorous) justification of the momentum operator for wavefunctions represented in position space.

Let us consider a one-dimensional problem, with a particle described by the (position domain) wavefunction ψ(x). As previously, we can define a Fourier transform of this to the (angular) wavenumber domain, φ(k):


Now, as we noted previously, the relation  shows that the wavenumber domain represents momentum space.

Now, recall that for an observable with (Hermitian) operator  and (position space) wavefunction ψ(x), the expectation value of the observable is

For example, the expectation value for the position is
,
where * indicates the complex conjugate.

Expectation of a function of momentum g(p) can be found in the wavefunction domain as

and so

(Just as multiplication by x is the position operator is position space, multiplication by  is the momentum operator in momentum space.)

If we wish to write this in terms of ψ(x), we start from our Fourier transform:
.
Taking the complex conjugate,


Now, we know that the Fourier transform of the derivative of a function obeys the rule:

and thus we see that as , then


Combining these, and remembering the delta “function” relation we used in proving Plancherel’s Theorem, we have:
.
Comparing this to our operator expectation value

which implies that the operator  is the momentum operator for wavefunctions in the position domain. Similar arguments can show that the position operator in the frequency domain is .

For three dimensions, we have momentum vector , which gives us vector-valued momentum operator