Probability four-current

In quantum mechanics, the probability current (sometimes called probability flux) is a mathematical quantity describing the flow of probability. Specifically, if one thinks of probability as a heterogeneous fluid, then the probability current is the rate of flow of this fluid. It is a real vector that changes with space and time. Probability currents are analogous to mass currents in hydrodynamics and electric currents in electromagnetism. As in those fields, the probability current is related to the probability density function via a continuity equation. The probability current is invariant under gauge transformation.

The concept of probability current is also used outside of quantum mechanics, when dealing with probability density functions that change over time, for instance in Brownian motion and the Fokker–Planck equation.

Definition (non-relativistic 3-current)

Free spin-0 particle

In non-relativistic quantum mechanics, the probability current j of the wave function $\Psi$ of a particle of mass $m$ in one dimension is defined as
$j = \frac \left(\Psi^* \frac- \Psi \frac \right) = \frac \Re\left\ = \frac \Im\left\,$
where $\Psi^*$ denotes the complex conjugate of the wave function, $\Re$ denotes the real part, and $\Im$ denotes the imaginary part. Note that the probability current is proportional to a Wronskian $W(\Psi,\Psi^*)$.

In three dimensions, this generalizes to
$\mathbf j = \frac \left( \Psi^* \mathbf \nabla \Psi - \Psi \mathbf \nabla \Psi^ \right) = \frac \Re\left\ = \frac\Im\left\ \,,$
where $\hbar$ is the reduced Planck constant, $m$ is the particle's mass, $\Psi$ is the wavefunction, and $\nabla$ denotes the del or gradient operator.

This can be simplified in terms of the kinetic momentum operator,
$\mathbf = -i\hbar\nabla$
to obtain
$\mathbf j = \frac \left(\Psi^* \mathbf \Psi - \Psi \mathbf \Psi^*\right)\,.$

These definitions use the position basis (i.e. for a wavefunction in position space), but momentum space is possible.

Spin-0 particle in an electromagnetic field

The above definition should be modified for a system in an external electromagnetic field. In SI units, a charged particle of mass ''m'' and electric charge ''q'' includes a term due to the interaction with the electromagnetic field;
\mathbf j = \frac\left \Psi, ^2 \right/math>
where is the magnetic potential (aka "A-field"). The term ''q''A has dimensions of momentum. Note that $\mathbf = -i\hbar\nabla$ used here is the canonical momentum and is not gauge invariant, unlike the kinetic momentum operator $\mathbf = -i\hbar\nabla-q\mathbf$.

In Gaussian units:
\mathbf j = \frac\left \Psi, ^2 \right/math>
where ''c'' is the speed of light.

Spin-''s'' particle in an electromagnetic field

If the particle has spin, it has a corresponding magnetic moment, so an extra term needs to be added incorporating the spin interaction with the electromagnetic field. In SI units:
\mathbf j = \frac\left \Psi, ^2 \right+ \frac\nabla\times(\Psi^* \mathbf\Psi)
where S is the spin vector of the particle with corresponding spin magnetic moment μ_''S'' and spin quantum number ''s''. In Gaussian units:
\mathbf j = \frac\left \Psi, ^2 \right+ \frac\nabla\times(\Psi^* \mathbf\Psi)

Connection with classical mechanics

The wave function can also be written in the complex exponential ( polar) form:
$\Psi = R e^$
where ''R'' and ''S'' are real functions of r and ''t''.

Written this way, the probability density is $\rho = \Psi^* \Psi = R^2$ and the probability current is:
\begin
\mathbf & = \frac\left(\Psi^ \mathbf \Psi - \Psi \mathbf\Psi^ \right) \\ pt & = \frac\left(R e^ \mathbfR e^ - R e^ \mathbfR e^\right) \\ pt & = \frac\left R e^ \left( e^ \mathbfR + \fracR e^ \mathbfS \right) - R e^ \left( e^ \mathbfR - \frac R e^ \mathbf S \right)\right
\end

The exponentials and ''R''∇''R'' terms cancel:
\mathbf = \frac\left frac R^2 \mathbf S + \frac R^2 \mathbf S \right

Finally, combining and cancelling the constants, and replacing ''R''² with ''ρ'',
$\mathbf = \rho \frac.$

If we take the familiar formula for the mass flux in hydrodynamics:
$\mathbf = \rho \mathbf,$
where $\rho$ is the mass density of the fluid and v is its velocity (also the group velocity of the wave), we can associate the velocity with ∇''S''/''m'', which is the same as equating ∇''S'' with the classical momentum p = ''m''v. This interpretation fits with Hamilton–Jacobi theory, in which
$\mathbf = \nabla S$
in Cartesian coordinates is given by ∇''S'', where ''S'' is Hamilton's principal function.

Motivation

Continuity equation for quantum mechanics

The definition of probability current and Schrödinger's equation can be used to derive the continuity equation, which has ''exactly'' the same forms as those for hydrodynamics and electromagnetism:Quantum Mechanics, E. Abers, Pearson Ed., Addison Wesley, Prentice Hall Inc, 2004,
$\frac + \mathbf \nabla \cdot \mathbf j = 0$
where the probability density $\rho\,$ is defined as
$\rho(\mathbf,t) = , \Psi, ^2 = \Psi^*(\mathbf,t)\Psi(\mathbf,t) .$

If one were to integrate both sides of the continuity equation with respect to volume, so that
$\int_V \left( \frac \right) \mathrmV + \int_V \left( \mathbf \nabla \cdot \mathbf j \right) \mathrmV = 0$

then the divergence theorem implies the continuity equation is equivalent to the integral equation

:

where the ''V'' is any volume and ''S'' is the boundary of ''V''. This is the conservation law for probability in quantum mechanics.

In particular, if ''Ψ'' is a wavefunction describing a single particle, the integral in the first term of the preceding equation, sans time derivative, is the probability of obtaining a value within ''V'' when the position of the particle is measured. The second term is then the rate at which probability is flowing out of the volume ''V''. Altogether the equation states that the time derivative of the probability of the particle being measured in ''V'' is equal to the rate at which probability flows into ''V''.

Transmission and reflection through potentials

In regions where a step potential or potential barrier occurs, the probability current is related to the transmission and reflection coefficients, respectively ''T'' and ''R''; they measure the extent the particles reflect from the potential barrier or are transmitted through it. Both satisfy:
$T + R = 1\,,$
where ''T'' and ''R'' can be defined by:
$T= \frac \, , \quad R = \frac \, ,$
where ''j''_inc, ''j''_ref and ''j''_trans are the incident, reflected and transmitted probability currents respectively, and the vertical bars indicate the magnitudes of the current vectors. The relation between ''T'' and ''R'' can be obtained from probability conservation:
$\mathbf_\mathrm + \mathbf_\mathrm=\mathbf_\mathrm\,.$

In terms of a unit vector n normal to the barrier, these are equivalently:
$T= \left, \frac\\,, \qquad R= \left, \frac \ \,,$
where the absolute values are required to prevent ''T'' and ''R'' being negative.

Examples

Plane wave

For a plane wave propagating in space:
$\Psi(\mathbf,t) = \, A e^$
the probability density is constant everywhere;
$\rho(\mathbf,t) = , A, ^2 \rightarrow \frac = 0$
(that is, plane waves are stationary states) but the probability current is nonzero – the square of the absolute amplitude of the wave times the particle's speed;
$\mathbf\left(\mathbf,t\right) = \left, A\^2 = \rho \frac = \rho \mathbf$

illustrating that the particle may be in motion even if its spatial probability density has no explicit time dependence.

Particle in a box

For a particle in a box, in one spatial dimension and of length ''L'', confined to the region $0 < x < L$, the energy eigenstates are
$\Psi_n = \sqrt \sin \left( \frac x \right)$
and zero elsewhere. The associated probability currents are
$j_n = \frac\left( \Psi_n^* \frac - \Psi_n \frac \right) = 0$
since $\Psi_n = \Psi_n^*$

Discrete definition

For a particle in one dimension on $\ell^2\left(\mathbb\right)$, we have the Hamiltonian $H = -\Delta + V$ where $-\Delta \equiv 2 I - S - S^\ast$ is the discrete Laplacian, with $S$ being the right shift operator on $\ell^2\left(\mathbb\right)$. Then the probability current is defined as $j \equiv 2 \Im\left\$, with $v$ the velocity operator, equal to v \equiv -i ,\, H/math> and $X$ is the position operator on $\ell^2\left(\mathbb\right)$. Since $V$ is usually a multiplication operator on $\ell^2\left(\mathbb\right)$, we get to safely write -i ,\, H= -i ,\, -\Delta= -i\left ,\, -S - S^\right= i S - i S^.

As a result, we find: $j\left(x\right) \equiv 2 \Im\left\ = 2 \Im\left\ = 2 \Im\left\$

References

Further reading

*{{cite book , title=Quantum Physics of Atoms, Molecules, Solids, Nuclei and Particles , edition=2nd , first=R. , last=Resnick , first2=R. , last2=Eisberg , publisher=John Wiley & Sons , year=1985 , isbn=0-471-87373-X

Quantum mechanics