Momentum operator

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https://en.wikipedia.org/wiki/Momentum_operator 

In quantum mechanics, the momentum operator is the operator associated with the linear momentum. The momentum operator is, in the position representation, an example of a differential operator. For the case of one particle in one spatial dimension, the definition is:

$${\displaystyle \hat{p}=-i\hslash \frac{\mathrm{\partial }}{\mathrm{\partial }x}}$$

where ħ is Planck's reduced constant, i the imaginary unit, x is the spatial coordinate, and a partial derivative (denoted by $\mathrm{\partial }/\mathrm{\partial }x$) is used instead of a total derivative (d/dx) since the wave function is also a function of time. The "hat" indicates an operator. The "application" of the operator on a differentiable wave function is as follows:

$${\displaystyle \hat{p}\psi =-i\hslash \frac{\mathrm{\partial }\psi }{\mathrm{\partial }x}}$$

In a basis of Hilbert space consisting of momentum eigenstates expressed in the momentum representation, the action of the operator is simply multiplication by p, i.e. it is a multiplication operator, just as the position operator is a multiplication operator in the position representation. Note that the definition above is the canonical momentum, which is not gauge invariant and not a measurable physical quantity for charged particles in an electromagnetic field. In that case, the canonical momentum is not equal to the kinetic momentum.

At the time quantum mechanics was developed in the 1920s, the momentum operator was found by many theoretical physicists, including Niels Bohr, Arnold Sommerfeld, Erwin Schrödinger, and Eugene Wigner. Its existence and form is sometimes taken as one of the foundational postulates of quantum mechanics.

Origin from De Broglie plane waves

The momentum and energy operators can be constructed in the following way.

One dimension

Starting in one dimension, using the plane wave solution to Schrödinger's equation of a single free particle,

$${\displaystyle \psi (x,t)=e^{\frac{i}{\hslash }(px-Et)},}$$

where p is interpreted as momentum in the x-direction and E is the particle energy. The first order partial derivative with respect to space is

$${\displaystyle \frac{\mathrm{\partial }\psi (x,t)}{\mathrm{\partial }x}=\frac{ip}{\hslash }{e}^{\frac{i}{\hslash }(px-Et)}=\frac{ip}{\hslash }\psi .}$$

This suggests the operator equivalence

$${\displaystyle \hat{p}=-i\hslash \frac{\mathrm{\partial }}{\mathrm{\partial }x}}$$

so the momentum of the particle and the value that is measured when a particle is in a plane wave state is the eigenvalue of the above operator.

Since the partial derivative is a linear operator, the momentum operator is also linear, and because any wave function can be expressed as a superposition of other states, when this momentum operator acts on the entire superimposed wave, it yields the momentum eigenvalues for each plane wave component. These new components then superimpose to form the new state, in general not a multiple of the old wave function.

Three dimensions

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

$${\displaystyle \psi =e^{\frac{i}{\hslash }(\mathbf{p}\cdot \mathbf{r}-Et)}}$$

and the gradient is

$${\displaystyle \begin{array}{rl}\mathrm{\nabla }\psi & ={\mathbf{e}}_x\frac{\mathrm{\partial }\psi }{\mathrm{\partial }x}+{\mathbf{e}}_y\frac{\mathrm{\partial }\psi }{\mathrm{\partial }y}+{\mathbf{e}}_z\frac{\mathrm{\partial }\psi }{\mathrm{\partial }z}\\ & =\frac{i}{\hslash }\left(p_x{\mathbf{e}}_x+p_y{\mathbf{e}}_y+p_z{\mathbf{e}}_z\right)\psi \\ & =\frac{i}{\hslash }\mathbf{p}\psi \end{array}}$$

where e_x, e_y, and e_z are the unit vectors for the three spatial dimensions, hence

$${\displaystyle \hat{\mathbf{p}}=-i\hslash \mathrm{\nabla }}$$

This momentum operator is in position space because the partial derivatives were taken with respect to the spatial variables.

Definition (position space)

See also: Position and momentum space

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

$${\displaystyle \hat{\mathbf{p}}=-i\hslash \mathrm{\nabla }}$$

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

In one spatial dimension, this becomes

$${\displaystyle \hat{p}={\hat{p}}_x=-i\hslash \frac{\mathrm{\partial }}{\mathrm{\partial }x}.}$$

This is the expression for the canonical momentum. For a charged particle q in an electromagnetic field, during a gauge transformation, the position space wave function undergoes a local U(1) group transformation, and $\hat{p}\psi =-i\hslash \frac{\mathrm{\partial }\psi }{\mathrm{\partial }x}$ will change its value. Therefore, the canonical momentum is not gauge invariant, and hence not a measurable physical quantity.

The kinetic momentum, a gauge invariant physical quantity, can be expressed in terms of the canonical momentum, the scalar potential φ and vector potential A:

$${\displaystyle \hat{\mathbf{P}}=-i\hslash \mathrm{\nabla }-q\mathbf{A}}$$

The expression above is called minimal coupling. For electrically neutral particles, the canonical momentum is equal to the kinetic momentum.

Properties

Hermiticity

The momentum operator is always a Hermitian operator (more technically, in math terminology a "self-adjoint operator") when it acts on physical (in particular, normalizable) quantum states.

(In certain artificial situations, such as the quantum states on the semi-infinite interval [0, ∞), there is no way to make the momentum operator Hermitian. This is closely related to the fact that a semi-infinite interval cannot have translational symmetry—more specifically, it does not have unitary translation operators. See below.)

Canonical commutation relation

Further information: Canonical commutation relation

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

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

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.

Fourier transform

The following discussion uses the bra–ket notation. One may write

$${\displaystyle \psi (x)=⟨x|\psi ⟩=\int dp⟨x|p⟩⟨p|\psi ⟩=\int dp\frac{e^{ixp/\hslash }\tilde{\psi }(p)}{\sqrt{2\pi \hslash }},}$$

so the tilde represents the Fourier transform, in converting from coordinate space to momentum space. It then holds that

$${\displaystyle \hat{p}=\int dp|p⟩p⟨p|=-i\hslash \int dx|x⟩\frac{d}{dx}⟨x|,}$$

that is, the momentum acting in coordinate space corresponds to spatial frequency,

$${\displaystyle ⟨x|\hat{p}|\psi ⟩=-i\hslash \frac{d}{dx}\psi (x).}$$

An analogous result applies for the position operator in the momentum basis,

$${\displaystyle ⟨p|\hat{x}|\psi ⟩=i\hslash \frac{d}{dp}\psi (p),}$$

leading to further useful relations,

$${\displaystyle ⟨p|\hat{x}|p^{\prime }⟩=i\hslash \frac{d}{dp}\delta (p-p^{\prime }),}$$

$${\displaystyle ⟨x|\hat{p}|x^{\prime }⟩=-i\hslash \frac{d}{dx}\delta (x-x^{\prime }),}$$

where δ stands for Dirac's delta function.

Derivation from infinitesimal translations

See also: Noether's theorem and Stone's theorem on one-parameter unitary groups

The translation operator is denoted T(ε), where ε represents the length of the translation. It satisfies the following identity:

$${\displaystyle T(\epsilon )|\psi ⟩=\int dxT(\epsilon )|x⟩⟨x|\psi ⟩}$$

that becomes

$${\displaystyle \int dx|x+\epsilon ⟩⟨x|\psi ⟩=\int dx|x⟩⟨x-\epsilon |\psi ⟩=\int dx|x⟩\psi (x-\epsilon )}$$

Assuming the function ψ to be analytic (i.e. differentiable in some domain of the complex plane), one may expand in a Taylor series about x:

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

so for infinitesimal values of ε:

$${\displaystyle T(\epsilon )=1-\epsilon \frac{d}{dx}=1-\frac{i}{\hslash }\epsilon \left(-i\hslash \frac{d}{dx}\right)}$$

As it is known from classical mechanics, the momentum is the generator of translation, so the relation between translation and momentum operators is:

$${\displaystyle T(\epsilon )=1-\frac{i}{\hslash }\epsilon \hat{p}}$$

thus

$${\displaystyle \hat{p}=-i\hslash \frac{d}{dx}.}$$

4-momentum operator

Inserting the 3d momentum operator above and the energy operator into the 4-momentum (as a 1-form with (+ − − −) metric signature):

$${\displaystyle P_{\mu }=\left(\frac{E}{c},-\mathbf{p}\right)}$$

obtains the 4-momentum operator:

$${\displaystyle {\hat{P}}_{\mu }=\left(\frac{1}{c}\hat{E},-\hat{\mathbf{p}}\right)=i\hslash \left(\frac{1}{c}\frac{\mathrm{\partial }}{\mathrm{\partial }t},\mathrm{\nabla }\right)=i\hslash {\mathrm{\partial }}_{\mu }}$$

where ∂_μ is the 4-gradient, and the −iħ  becomes +iħ  preceding the 3-momentum operator. This operator occurs in relativistic quantum field theory, such as the Dirac equation and other relativistic wave equations, since energy and momentum combine into the 4-momentum vector above, momentum and energy operators correspond to space and time derivatives, and they need to be first order partial derivatives for Lorentz covariance.

The Dirac operator and Dirac slash of the 4-momentum is given by contracting with the gamma matrices:

$${\displaystyle {\gamma }^{\mu }{\hat{P}}_{\mu }=i\hslash {\gamma }^{\mu }{\mathrm{\partial }}_{\mu }=\hat{P}=i\hslash \mathrm{\partial }/}$$

If the signature was (− + + +), the operator would be

$${\displaystyle {\hat{P}}_{\mu }=\left(-\frac{1}{c}\hat{E},\hat{\mathbf{p}}\right)=-i\hslash \left(\frac{1}{c}\frac{\mathrm{\partial }}{\mathrm{\partial }t},\mathrm{\nabla }\right)=-i\hslash {\mathrm{\partial }}_{\mu }}$$

instead.

Energy operator

From Wikipedia, the free encyclopedia

https://en.wikipedia.org/wiki/Energy_operator

In quantum mechanics, energy is defined in terms of the energy operator, acting on the wave function of the system as a consequence of time translation symmetry.

Definition

It is given by:

$${\displaystyle \hat{E}=i\hslash \frac{\mathrm{\partial }}{\mathrm{\partial }t}}$$

It acts on the wave function (the probability amplitude for different configurations of the system)

$${\displaystyle \mathrm{\Psi }\left(\mathbf{r},t\right)}$$

Application

The energy operator corresponds to the full energy of a system. The Schrödinger equation describes the space- and time-dependence of the slow changing (non-relativistic) wave function of a quantum system. The solution of this equation for a bound system is discrete (a set of permitted states, each characterized by an energy level) which results in the concept of quanta.

Schrödinger equation

Using the energy operator to the Schrödinger equation:

$${\displaystyle i\hslash \frac{\mathrm{\partial }}{\mathrm{\partial }t}\mathrm{\Psi }(\mathbf{r},t)=\hat{H}\mathrm{\Psi }(\mathbf{r},t)}$$

can be obtained:

$${\displaystyle \hat{E}\mathrm{\Psi }(\mathbf{r},t)=\hat{H}\mathrm{\Psi }(\mathbf{r},t)}$$

where i is the imaginary unit, ħ is the reduced Planck constant, and $\hat{H}$ is the Hamiltonian operator.

Constant energy

Working from the definition, a partial solution for a wavefunction of a particle with a constant energy can be constructed. If the wavefunction is assumed to be separable, then the time dependence can be stated as ${\displaystyle e^{-iEt/\hslash }}$, where E is the constant energy. In full,

$${\displaystyle \mathrm{\Psi }(\mathbf{r},t)=\psi (\mathbf{r})e^{-iEt/\hslash }}$$

where $\psi (\mathbf{r})$ is the partial solution of the wavefunction dependent on position. Applying the energy operator, we have

$${\displaystyle \hat{E}\mathrm{\Psi }(\mathbf{r},t)=i\hslash \frac{\mathrm{\partial }}{\mathrm{\partial }t}\psi (\mathbf{r})e^{-iEt/\hslash }=i\hslash \left(\frac{-iE}{\hslash }\right)\psi (\mathbf{r})e^{-iEt/\hslash }=E\psi (\mathbf{r})e^{-iEt/\hslash }=E\mathrm{\Psi }(\mathbf{r},t).}$$

This is also known as the stationary state, and can be used to analyse the time-independent Schrödinger equation:

$${\displaystyle E\mathrm{\Psi }(\mathbf{r},t)=\hat{H}\mathrm{\Psi }(\mathbf{r},t)}$$

where E is an eigenvalue of energy.

Klein–Gordon equation

The relativistic mass-energy relation:

$${\displaystyle E^2=(pc{)}^2+(m{c}^2{)}^2}$$

where again E = total energy, p = total 3-momentum of the particle, m = invariant mass, and c = speed of light, can similarly yield the Klein–Gordon equation:

$${\displaystyle \begin{array}{rl}& {\hat{E}}^2=c^2{\hat{p}}^2+(m{c}^2{)}^2\\ & {\hat{E}}^2\mathrm{\Psi }=c^2{\hat{p}}^2\mathrm{\Psi }+(m{c}^2{)}^2\mathrm{\Psi }\end{array}}$$

where $\hat{p}$ is the momentum operator. That is:

$${\displaystyle \frac{{\mathrm{\partial }}^2\mathrm{\Psi }}{\mathrm{\partial }{t}^2}=c^2{\mathrm{\nabla }}^2\mathrm{\Psi }-{\left(\frac{m{c}^2}{\hslash }\right)}^2\mathrm{\Psi }}$$

Derivation

The energy operator is easily derived from using the free particle wave function (plane wave solution to Schrödinger's equation). Starting in one dimension the wave function is

$${\displaystyle \mathrm{\Psi }=e^{i(kx-\omega t)}}$$

The time derivative of Ψ is

$${\displaystyle \frac{\mathrm{\partial }\mathrm{\Psi }}{\mathrm{\partial }t}=-i\omega e^{i(kx-\omega t)}=-i\omega \mathrm{\Psi }.}$$

By the De Broglie relation:

$${\displaystyle E=\hslash \omega ,}$$

we have

$${\displaystyle \frac{\mathrm{\partial }\mathrm{\Psi }}{\mathrm{\partial }t}=-i\frac{E}{\hslash }\mathrm{\Psi }.}$$

Re-arranging the equation leads to

$${\displaystyle E\mathrm{\Psi }=i\hslash \frac{\mathrm{\partial }\mathrm{\Psi }}{\mathrm{\partial }t},}$$

where the energy factor E is a scalar value, the energy the particle has and the value that is measured. The partial derivative is a linear operator so this expression is the operator for energy:

$${\displaystyle \hat{E}=i\hslash \frac{\mathrm{\partial }}{\mathrm{\partial }t}.}$$

It can be concluded that the scalar E is the eigenvalue of the operator, while ${\displaystyle \hat{E}}$ is the operator. Summarizing these results:

$${\displaystyle \hat{E}\mathrm{\Psi }=i\hslash \frac{\mathrm{\partial }}{\mathrm{\partial }t}\mathrm{\Psi }=E\mathrm{\Psi }}$$

For a 3-d plane wave

$${\displaystyle \mathrm{\Psi }=e^{i(\mathbf{k}\cdot \mathbf{r}-\omega t)}}$$

the derivation is exactly identical, as no change is made to the term including time and therefore the time derivative. Since the operator is linear, they are valid for any linear combination of plane waves, and so they can act on any wave function without affecting the properties of the wave function or operators. Hence this must be true for any wave function. It turns out to work even in relativistic quantum mechanics, such as the Klein–Gordon equation above.