Probability 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 (i.e. the probability current density) 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.

The relativistic equivalent of the probability current is known as the probability four-current.

Definition (non-relativistic 3-current)

Free spin-0 particle

In non-relativistic quantum mechanics, the probability current Template:Math of the wave function Template:Math of a particle of mass Template:Mvar in one dimension is defined as $$j=\frac{\hslash }{2mi}({\mathrm{\Psi }}^{*}\frac{\partial \mathrm{\Psi }}{\partial x}-\mathrm{\Psi }\frac{\partial {\mathrm{\Psi }}^{*}}{\partial x})=\frac{\hslash }{m}\mathrm{\Re }\left\{{\mathrm{\Psi }}^{*}\frac{1}{i}\frac{\partial \mathrm{\Psi }}{\partial x}\right\}=\frac{\hslash }{m}\mathrm{\Im }\left\{{\mathrm{\Psi }}^{*}\frac{\partial \mathrm{\Psi }}{\partial x}\right\},$$ where

• $\hslash$ is the reduced Planck constant;
• ${\mathrm{\Psi }}^{*}$ denotes the complex conjugate of the wave function;
• $\mathrm{\Re }$ denotes the real part;
• $\mathrm{\Im }$ denotes the imaginary part.

Note that the probability current is proportional to a Wronskian $W(\mathrm{\Psi },{\mathrm{\Psi }}^{*}).$

In three dimensions, this generalizes to $$𝐣=\frac{\hslash }{2mi}({\mathrm{\Psi }}^{*}\mathrm{\nabla }\mathrm{\Psi }-\mathrm{\Psi }\mathrm{\nabla }{\mathrm{\Psi }}^{*})=\frac{\hslash }{m}\mathrm{\Re }\left\{{\mathrm{\Psi }}^{*}\frac{\mathrm{\nabla }}{i}\mathrm{\Psi }\right\}=\frac{\hslash }{m}\mathrm{\Im }\left\{{\mathrm{\Psi }}^{*}\mathrm{\nabla }\mathrm{\Psi }\right\},$$ where $\mathrm{\nabla }$ denotes the del or gradient operator. This can be simplified in terms of the kinetic momentum operator, $$\widehat{𝐩}=-i\hslash \mathrm{\nabla }$$ to obtain $$𝐣=\frac{1}{2m}({\mathrm{\Psi }}^{*}\widehat{𝐩}\mathrm{\Psi }-\mathrm{\Psi }\widehat{𝐩}{\mathrm{\Psi }}^{*}).$$

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 Template:Mvar and electric charge Template:Mvar includes a term due to the interaction with the electromagnetic field; $$𝐣=\frac{1}{2m}[({\mathrm{\Psi }}^{*}\widehat{𝐩}\mathrm{\Psi }-\mathrm{\Psi }\widehat{𝐩}{\mathrm{\Psi }}^{*})-2q𝐀|\mathrm{\Psi }{|}^2]$$ where Template:Math is the magnetic vector potential. The term Template:Math has dimensions of momentum. Note that $\widehat{𝐩}=-i\hslash \mathrm{\nabla }$ used here is the canonical momentum and is not gauge invariant, unlike the kinetic momentum operator $\widehat{𝐏}=-i\hslash \mathrm{\nabla }-q𝐀$.

In Gaussian units: $$𝐣=\frac{1}{2m}[({\mathrm{\Psi }}^{*}\widehat{𝐩}\mathrm{\Psi }-\mathrm{\Psi }\widehat{𝐩}{\mathrm{\Psi }}^{*})-2\frac{q}{c}𝐀|\mathrm{\Psi }{|}^2]$$ where Template:Mvar 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.

According to Landau-Lifschitz's Course of Theoretical Physics the electric current density is in Gaussian units: $${𝐣}_e=\frac{q}{2m}[({\mathrm{\Psi }}^{*}\widehat{𝐩}\mathrm{\Psi }-\mathrm{\Psi }\widehat{𝐩}{\mathrm{\Psi }}^{*})-\frac{2q}{c}𝐀|\mathrm{\Psi }{|}^2]+\frac{{\mu }_{S}c}{s\hslash }\mathrm{\nabla }\times ({\mathrm{\Psi }}^{*}𝐒\mathrm{\Psi })$$

And in SI units: $${𝐣}_e=\frac{q}{2m}[({\mathrm{\Psi }}^{*}\widehat{𝐩}\mathrm{\Psi }-\mathrm{\Psi }\widehat{𝐩}{\mathrm{\Psi }}^{*})-2q𝐀|\mathrm{\Psi }{|}^2]+\frac{{\mu }_S}{s\hslash }\mathrm{\nabla }\times ({\mathrm{\Psi }}^{*}𝐒\mathrm{\Psi })$$

Hence the probability current (density) is in SI units: $$𝐣={𝐣}_e/q=\frac{1}{2m}[({\mathrm{\Psi }}^{*}\widehat{𝐩}\mathrm{\Psi }-\mathrm{\Psi }\widehat{𝐩}{\mathrm{\Psi }}^{*})-2q𝐀|\mathrm{\Psi }{|}^2]+\frac{{\mu }_S}{qs\hslash }\mathrm{\nabla }\times ({\mathrm{\Psi }}^{*}𝐒\mathrm{\Psi })$$

where Template:Math is the spin vector of the particle with corresponding spin magnetic moment Template:Math and spin quantum number Template:Mvar.

It is doubtful if this formula is valid for particles with an interior structure.Template:Citation needed The neutron has zero charge but non-zero magnetic moment, so $\frac{{\mu }_S}{qs\hslash }$ would be impossible (except $\mathrm{\nabla }\times ({\mathrm{\Psi }}^{*}𝐒\mathrm{\Psi })$ would also be zero in this case). For composite particles with a non-zero charge – like the proton which has spin quantum number s=1/2 and μ_S= 2.7927·μ_N or the deuteron (H-2 nucleus) which has s=1 and μ_S=0.8574·μ_N – it is mathematically possible but doubtful.

Connection with classical mechanics

The wave function can also be written in the complex exponential (polar) form: $$\mathrm{\Psi }=R{e}^{iS/\hslash }$$ where Template:Mvar are real functions of Template:Math and Template:Math.

Written this way, the probability density is $$\rho ={\mathrm{\Psi }}^{*}\mathrm{\Psi }=R^2$$ and the probability current is: $$\begin{array}{cc}𝐣& =\frac{\hslash }{2mi}({\mathrm{\Psi }}^{*}\mathrm{\nabla }\mathrm{\Psi }-\mathrm{\Psi }\mathrm{\nabla }{\mathrm{\Psi }}^{*})\\ & =\frac{\hslash }{2mi}(R{e}^{-iS/\hslash }\mathrm{\nabla }R{e}^{iS/\hslash }-R{e}^{iS/\hslash }\mathrm{\nabla }R{e}^{-iS/\hslash })\\ & =\frac{\hslash }{2mi}[R{e}^{-iS/\hslash }(e^{iS/\hslash }\mathrm{\nabla }R+\frac{i}{\hslash }R{e}^{iS/\hslash }\mathrm{\nabla }S)-R{e}^{iS/\hslash }(e^{-iS/\hslash }\mathrm{\nabla }R-\frac{i}{\hslash }R{e}^{-iS/\hslash }\mathrm{\nabla }S)].\end{array}$$

The exponentials and Template:Math terms cancel: $$𝐣=\frac{\hslash }{2mi}[\frac{i}{\hslash }{R}^2\mathrm{\nabla }S+\frac{i}{\hslash }{R}^2\mathrm{\nabla }S].$$

Finally, combining and cancelling the constants, and replacing Template:Math with Template:Mvar, $$𝐣=\rho \frac{\mathrm{\nabla }S}{m}.$$Hence, the spatial variation of the phase of a wavefunction is said to characterize the probability flux of the wavefunction. If we take the familiar formula for the mass flux in hydrodynamics: $$𝐣=\rho 𝐯,$$

where $\rho$ is the mass density of the fluid and Template:Math is its velocity (also the group velocity of the wave). In the classical limit, we can associate the velocity with $\frac{\mathrm{\nabla }S}{m},$ which is the same as equating Template:Math with the classical momentum Template:Math however, it does not represent a physical velocity or momentum at a point since simultaneous measurement of position and velocity violates uncertainty principle. This interpretation fits with Hamilton–Jacobi theory, in which $$𝐩=\mathrm{\nabla }S$$ in Cartesian coordinates is given by Template:Math, where Template:Mvar is Hamilton's principal function.

The de Broglie-Bohm theory equates the velocity with $\frac{\mathrm{\nabla }S}{m}$ in general (not only in the classical limit) so it is always well defined. It is an interpretation of quantum mechanics.

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.

For some wave function Template:Math, let:

$$\rho (𝐫,t)=|\mathrm{\Psi }{|}^2={\mathrm{\Psi }}^{*}(𝐫,t)\mathrm{\Psi }(𝐫,t).$$be the probability density (probability per unit volume, Template:Math denotes complex conjugate). Then,

$\begin{array}{cc}\frac{d}{dt}\underset{𝒱}{\int }dV\rho & =\underset{𝒱}{\int }dV(\frac{\partial \psi }{\partial t}{\psi }^{*}+\psi \frac{\partial {\psi }^{*}}{\partial t})\\ & =\underset{𝒱}{\int }dV(-\frac{i}{\hslash }(-\frac{{\hslash }^2}{2m}{\mathrm{\nabla }}^2\psi +V\psi ){\psi }^{*}+\frac{i}{\hslash }(-\frac{{\hslash }^2}{2m}{\mathrm{\nabla }}^2{\psi }^{*}+V{\psi }^{*})\psi )\\ & =\underset{𝒱}{\int }dV\frac{i\hslash }{2m}[({\mathrm{\nabla }}^2\psi ){\psi }^{*}-\psi ({\mathrm{\nabla }}^2{\psi }^{*})]\\ & =\underset{𝒱}{\int }dV\mathrm{\nabla }\cdot (\frac{i\hslash }{2m}({\psi }^{*}\mathrm{\nabla }\psi -\psi \mathrm{\nabla }{\psi }^{*}))\\ & =\underset{𝒮}{\int }d𝐚\cdot (\frac{i\hslash }{2m}({\psi }^{*}\mathrm{\nabla }\psi -\psi \mathrm{\nabla }{\psi }^{*}))\\ \end{array}$

where Template:Mvar is any volume and Template:Mvar is the boundary of Template:Mvar.

This is the conservation law for probability in quantum mechanics. The integral form is stated as:

$$\underset{V}{\int }\left(\frac{\partial |\mathrm{\Psi }{|}^2}{\partial t}\right)\mathrm{d}V+\underset{V}{\int }(\mathrm{\nabla }\cdot 𝐣)\mathrm{d}V=0$$where$$𝐣=\frac{1}{2m}({\mathrm{\Psi }}^{*}\widehat{𝐩}\mathrm{\Psi }-\mathrm{\Psi }\widehat{𝐩}{\mathrm{\Psi }}^{*})=-\frac{i\hslash }{2m}({\psi }^{*}\mathrm{\nabla }\psi -\psi \mathrm{\nabla }{\psi }^{*})=\frac{\hslash }{m}\mathrm{Im}({\psi }^{*}\mathrm{\nabla }\psi )$$is the probability current or probability flux (flow per unit area).

Here, equating the terms inside the integral gives the continuity equation for probability:$$\frac{\partial }{\partial t}\rho (𝐫,t)+\mathrm{\nabla }\cdot 𝐣=0,$$and the integral equation can also be restated using the divergence theorem as:

Template:Oiint.

In particular, if Template:Mvar 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 Template:Mvar when the position of the particle is measured. The second term is then the rate at which probability is flowing out of the volume Template:Mvar. Altogether the equation states that the time derivative of the probability of the particle being measured in Template:Mvar is equal to the rate at which probability flows into Template:Mvar.

By taking the limit of volume integral to include all regions of space, a well-behaved wavefunction that goes to zero at infinities in the surface integral term implies that the time derivative of total probability is zero ie. the normalization condition is conserved. This result is in agreement with the unitary nature of time evolution operators which preserve length of the vector by definition.

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 Template:Mvar and Template:Mvar; they measure the extent the particles reflect from the potential barrier or are transmitted through it. Both satisfy: $$T+R=1,$$ where Template:Mvar and Template:Mvar can be defined by: $$T=\frac{|{𝐣}_{\mathrm{t}\mathrm{r}\mathrm{a}\mathrm{n}\mathrm{s}}|}{|{𝐣}_{\mathrm{i}\mathrm{n}\mathrm{c}}|},R=\frac{|{𝐣}_{\mathrm{r}\mathrm{e}\mathrm{f}}|}{|{𝐣}_{\mathrm{i}\mathrm{n}\mathrm{c}}|},$$ where Template:Math are the incident, reflected and transmitted probability currents respectively, and the vertical bars indicate the magnitudes of the current vectors. The relation between Template:Mvar and Template:Mvar can be obtained from probability conservation: $${𝐣}_{\mathrm{t}\mathrm{r}\mathrm{a}\mathrm{n}\mathrm{s}}+{𝐣}_{\mathrm{r}\mathrm{e}\mathrm{f}}={𝐣}_{\mathrm{i}\mathrm{n}\mathrm{c}}.$$

In terms of a unit vector Template:Math normal to the barrier, these are equivalently: $$T=\left|\frac{{𝐣}_{\mathrm{t}\mathrm{r}\mathrm{a}\mathrm{n}\mathrm{s}}\cdot 𝐧}{{𝐣}_{\mathrm{i}\mathrm{n}\mathrm{c}}\cdot 𝐧}\right|,R=\left|\frac{{𝐣}_{\mathrm{r}\mathrm{e}\mathrm{f}}\cdot 𝐧}{{𝐣}_{\mathrm{i}\mathrm{n}\mathrm{c}}\cdot 𝐧}\right|,$$ where the absolute values are required to prevent Template:Mvar and Template:Mvar being negative.

Examples

Plane wave

For a plane wave propagating in space: $$\mathrm{\Psi }(𝐫,t)=A{e}^{i(𝐤\cdot 𝐫-\omega t)}$$ the probability density is constant everywhere; $$\rho (𝐫,t)=|A{|}^2\to \frac{\partial |\mathrm{\Psi }{|}^2}{\partial t}=0$$ (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; $$𝐣(𝐫,t)={\left|A\right|}^2\frac{\hslash 𝐤}{m}=\rho \frac{𝐩}{m}=\rho 𝐯$$

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 Template:Mvar, confined to the region $0<x<L$, the energy eigenstates are $${\mathrm{\Psi }}_n=\sqrt{\frac{2}{L}}\sin \left(\frac{n\pi }{L}x\right)$$ and zero elsewhere. The associated probability currents are $$j_n=\frac{i\hslash }{2m}({\mathrm{\Psi }}_n^{*}\frac{\partial {\mathrm{\Psi }}_n}{\partial x}-{\mathrm{\Psi }}_n\frac{\partial {\mathrm{\Psi }}_n^{*}}{\partial x})=0$$ since $${\mathrm{\Psi }}_n={\mathrm{\Psi }}_n^{*}$$

Discrete definition

For a particle in one dimension on ${\ell }^2(\mathbb{Z}),$ we have the Hamiltonian $H=-\mathrm{\Delta }+V$ where $-\mathrm{\Delta }\equiv 2I-S-S^{\ast }$ is the discrete Laplacian, with Template:Mvar being the right shift operator on ${\ell }^2(\mathbb{Z}).$ Then the probability current is defined as $j\equiv 2\mathrm{\Im }\left\{\overline{\mathrm{\Psi }}iv\mathrm{\Psi }\right\},$ with Template:Mvar the velocity operator, equal to $v\equiv -i[X,H]$ and Template:Mvar is the position operator on ${\ell }^2\left(\mathbb{Z}\right).$ Since Template:Mvar is usually a multiplication operator on ${\ell }^2(\mathbb{Z}),$ we get to safely write $$-i[X,H]=-i[X,-\mathrm{\Delta }]=-i[X,-S-S^{\ast }]=iS-i{S}^{\ast }.$$

As a result, we find: $$\begin{array}{cc}j\left(x\right)\equiv 2\mathrm{\Im }\{\overline{\mathrm{\Psi }}(x)iv\mathrm{\Psi }(x)\}& =2\mathrm{\Im }\{\overline{\mathrm{\Psi }}(x)((-S\mathrm{\Psi })(x)+\left(S^{\ast }\mathrm{\Psi }\right)(x))\}\\ & =2\mathrm{\Im }\{\overline{\mathrm{\Psi }}(x)(-\mathrm{\Psi }(x-1)+\mathrm{\Psi }(x+1))\}\end{array}$$

Further reading

• Template:Cite book