Momentum operator

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.

Contents

[hide]

Definition

For a single particle with no electric charge and no spin, the momentum operator can be written in the position basis as: [1]

\bold{\hat{p}}=-i\hbar\nabla

where ∇ is the gradient operator, ħ is the reduced Planck constant, and i is the imaginary unit.

In one spatial dimension this becomes:

\hat{p}=\hat{p}_x=-i\hbar{\partial \over \partial x}.

This is a commonly encountered form of the momentum operator, though not the most general one. For a charged particle q in an electromagnetic field, described by the scalar potential φ and vector potential A, the momentum operator must be replaced by the kinetic momentum operator, which includes a contribution from the A field: [2]

\bold{\hat{\Pi}} = \bold{\hat{P}} - q\bold{A}

where \bold{\hat{P}} is the canonical momentum operator given as the usual momentum operator:

\bold{\hat{P}} = -i\hbar\nabla

so the momentum operator generalizes to

\bold{\hat{\Pi}} = -i\hbar\nabla - q\bold{A}

This is of course true for electrically charged or neutral particles, since the second term vanishes if q is zero and the original operator appears.

Properties

Hermiticity

The momentum operator is always a Hermitian operator when it acts on physical (in particular, normalizable) quantum states.[3]

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.

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 \delta stands for Dirac's delta function.

Derivations

De Broglie plane waves

The momentum and energy operators can easily be constructed in the following way. [4] Starting in one dimension, using the plane wave solution to Schrödinger's equation:

 \Psi = e^{i(kx-\omega t)} \,\!

The first order partial derivative with respect to space is

 \frac{\partial \Psi}{\partial x} = i k e^{i(kx-\omega t)} = i k \Psi \,\!

by the De Broglie relation:

 p=\hbar k\,\!

we have

 \frac{\partial \Psi}{\partial x} = i \frac{p}{\hbar} \Psi \,\!

The momentum value p is a scalar factor, the momentum of the particle and the value that is mesured. Cancellation of Ψ leads to

 p = -i\hbar \frac{\partial \Psi}{\partial x} \,\!

The partial derivative is a linear operator so this expression can only be the momentum operator:

 \hat{p} = -i\hbar \frac{\partial }{\partial x} \,\!

So it can be concluded that the scalar p is the eigenvalue of the operator, while  \hat{p} \,\! is the operator. Summarizing these results:

 \hat{p}\Psi = -i\hbar \frac{\partial \Psi}{\partial x} = p\Psi\,\!

The derivation in three dimensions is the same, except using the gradient operator instead of one partial derivative. In three dimensions, the plane wave solution to Schrödinger's equation is:

 \Psi = e^{i(\bold{k}\cdot\bold{r}-\omega t)} \,\!

The gradeint is

 \begin{align} \nabla \Psi & = \bold{e}_{\rm x}\frac{\partial \Psi}{\partial x} %2B \bold{e}_{\rm y}\frac{\partial \Psi}{\partial y} %2B \bold{e}_{\rm z}\frac{\partial \Psi}{\partial z} \\ 
& = i k_x\Psi\bold{e}_{\rm x} %2B i k_y\Psi\bold{e}_{\rm y}%2B i k_z\Psi\bold{e}_{\rm z} \\
& = \frac{i}{\hbar} \left ( p_x\bold{e}_{\rm x} %2B p_y\bold{e}_{\rm y}%2B p_z\bold{e}_{\rm z} \right)\Psi \\
& = \frac{i}{\hbar} \bold{\hat{p}}\Psi \\
\end{align} \,\!

hence

 \bold{\hat{p}} = -i \hbar \nabla \,\!

Since the operators are linear, they are valid for any linear combination of plane waves, and so they can act on any wavefunction without affecting the properties of the wavefunction or operators.

Infinitesimal translations

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 %2B \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 \psi 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 }

See also

References

  1. ^ Quantum Mechanics Demystified, D. McMahon, Mc Graw Hill (USA), 2006, ISBN(10) 0 07 145546 9
  2. ^ Quantum Physics of Atoms, Molecules, Solids, Nuclei and Particles (2nd Edition), R. Resnick, R. Eisberg, John Wiley & Sons, 1985, ISBN 978-0-471-873730
  3. ^ 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.
  4. ^ Quantum Physics of Atoms, Molecules, Solids, Nuclei and Particles (2nd Edition), R. Resnick, R. Eisberg, John Wiley & Sons, 1985, ISBN 978-0-471-873730