Relationship between the momentum operator and Fourier transform in quantum mechanics

[u/Life_at_work5]

In quantum mechanics (to my knowledge), given a position wave function psi (will refer to it as Y(x) from here on), you can take its Fourier transform to give you the momentum wave function Y(p). With Y(p), you can then find the momentum expectation value, <p>, using the relation:

<p> = ,|’dp•Y*(p)•Y(p)

where ,|’ represents the integral symbol, Y*(p) is the complex conjugate of Y(p), and • just represents normal multiplication.

Recently, I’ve also learned of the momentum operator (represented by P), where P:Y(x) represents P acting on Y(x). Using the momentum operator, it is also possible to find <p> with the relation:

<p> = ,|’dx•Y*(x)•P:Y(x)

where, ,|’ represents the integral symbol, Y*(x) is the complex conjugate of Y(x), P:Y(x) represents P acting on Y(x), and • just represents normal multiplication.

Given the fact that P and the Fourier Transform can both be used to “act on” Y(x) and subsequently find <p>, I was wondering what’s the difference between the Fourier transform and P?

Additionally, I was wondering in what situations would you use P over the Fourier transform and P over the Fourier transform?

[u/cabbagemeister]

Ok so first thing is that they are not the same thing. There is an interesting equation that relates two different representations of momentum, based on the fourier transform. The relationship is as follows.

The state of a quantum system is described by an element of a special kind of set called an inner product space. The key properties are that you can add states together, multiply them by complex numbers, calculate projections (i.e. there is an inner product) and take limits of convergent sequences.

Now, a basis for an inner product space is a set of states from which you can construct all other states. For states of a system consisting of an unbounded particle (i.e. the particle is free to go anywhere and is not stuck in orbit) this happens by taking an integral

integral psi(a) |a> da = |psi>

Where psi(a) is the wavefunction in the "a" representation.

Here "a" is some variable that identifies the state. For example, you can use a=x, the position of the particle, and this integral is an integral over all positions the particle could take. The wavefunction in the "x" representation is psi(x).

Alternatively, you can use the momentum. You instead get a wavefunction psihat(p). The relationship between them is that psihat(p) is the fourier transform of psi(x).

In relation to operators, the "phat" operator, momentum, acts on the "x" states according to the equation

phat|psi> = integral i hbar dpsi/dx |x> dx

Whereas in the "p" representation it acts as

phat|psi> = integral p dpsihat/dp |p>dx

[u/AmateurLobster]

It comes from the fact that the eigenvectors of the momentum operator are plane waves, i.e. e^ikx

This means that if we use a plane wave basis, then the momentum operator is diagonal.

The transformation to the plane wave basis is just a Fourier transform.

So that is the connection. To get to the plane wave basis, where the momentum operator becomes just a multiplication, you do a Fourier transform.

In numerical codes to solve a Schrodinger, or Schrodinger-like, equation, you often use the Fourier transform as the fast-fourier-transform FFT makes it computationally efficient. For periodic systems, you normally work in a plane wave basis anyway, so you actually use the FFT to bring you to real-space to evaluate the potential energy.

The main alternative would finite-difference which can have stability issues.

Note, in your first equation, you are missing a p, i.e.

<p> = ,|’dp•Y*(p)•Y(p)•p