Momentum operator

See also: Momentum

In quantum mechanics, momentum is defined as an operator on the wave function. The Heisenberg uncertainty principle defines limits on how accurately the momentum and position of a single observable system can be known at once. In quantum mechanics, position and momentum are conjugate variables.

For a single particle with no electric charge and no spin, the momentum operator can be written in the position basis as

$\mathbf{p}={\hbar\over i}\nabla=-i\hbar\nabla$

where:

• $\nabla$ is the gradient operator;
• $\hbar$ is the reduced Planck constant;
• i is the imaginary unit.

In one space dimension this becomes:

$\mathbf{p}=p_{x}={\hbar\over i}{\partial \over \partial x}=-i\hbar{\partial \over \partial x}.$

This is a commonly encountered form of the momentum operator, though not the most general one.

The momentum operator is always a Hermitian operator when it acts on physical (in particular, normalizable) quantum states.^[1]

Fourier transform

One can show that the Fourier transform of the momentum in quantum mechanics is the position operator. The Fourier transform turns the momentum-basis into the position-basis.

$\langle x | \hat{p} | \psi \rangle = - i \hbar {d \over dx} \psi ( x )$

The same applies for the Position operator in the momentum basis:

$\langle p | \hat{x} | \psi \rangle = i \hbar {d \over dp} \psi ( p )$

and other useful relations:

$\langle p | \hat{x} | p' \rangle = i \hbar {d \over dp} \delta (p - p')$

$\langle x | \hat{p} | x' \rangle = -i \hbar {d \over dx} \delta (x - x')$

where δ stands for Dirac's delta function.

Derivation

Suppose we have an infinitesimal translation operator $T(\epsilon)$, where $\epsilon$ represents the length of the infinitesimal translation, then

$T(\epsilon) | \psi \rangle = \int dx T(\epsilon) | x \rangle \langle x | \psi \rangle$

that becomes

$\int dx | x + \epsilon \rangle \langle x | \psi \rangle = \int dx | x \rangle \langle x - \epsilon | \psi \rangle = \int dx | x \rangle \psi(x - \epsilon)$

We assume the function ψ to be analytic (or simply differentiable, for simplicity), so we may write:

$\psi(x-\epsilon) = \psi(x) - \epsilon {d \psi \over dx}$

so we have for infinitesimal values of epsilon:

$T(\epsilon) = 1 - \epsilon {d \over dx} = 1 - {i \over \hbar} \epsilon \left ( - i \hbar{ d \over dx} \right )$

we know from classical mechanics that momentum is the generator of translation, so we know that

$T(\epsilon) = 1 - {i \over \hbar} \epsilon \hat{p}$

thus

$\hat{p} = - i \hbar { d \over dx }$

Canonical commutation relation

Further information: Canonical commutation relation

One can easily show that by appropriately using the momentum basis and the position basis:

$\left [ \hat{ x }, \hat{ p } \right ] = \hat{x} \hat{p} - \hat{p} \hat{x} = i \hbar.$

References

1. ^ See Lecture notes 1 by Robert Littlejohn for a specific mathematical discussion and proof for the case of a single, uncharged, spin-zero particle. See Lecture notes 4 by Robert Littlejohn for the general case.