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Inquantum mechanics, theprobability current (sometimes calledprobabilityflux) is a mathematical quantity describing the flow ofprobability. Specifically, if one thinks of probability as aheterogeneous fluid, then the probability current is the rate of flow of this fluid. It is arealvector that changes with space and time. Probability currents are analogous tomass currents inhydrodynamics andelectric currents inelectromagnetism. As in those fields, the probability current (i.e. the probability current density) is related to theprobability density function via acontinuity equation. The probability current isinvariant undergauge 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 inBrownian motion and theFokker–Planck equation.^[1]

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

In non-relativistic quantum mechanics, the probability currentj of thewave functionΨ of a particle of massm in one dimension is defined as^[2]$${\displaystyle j=\frac{\hslash }{2mi}\left({\mathrm{\Psi }}^{\ast }\frac{\mathrm{\partial }\mathrm{\Psi }}{\mathrm{\partial }x}-\mathrm{\Psi }\frac{\mathrm{\partial }{\mathrm{\Psi }}^{\ast }}{\mathrm{\partial }x}\right)=\frac{\hslash }{m}\mathrm{\Re }\left\{{\mathrm{\Psi }}^{\ast }\frac{1}{i}\frac{\mathrm{\partial }\mathrm{\Psi }}{\mathrm{\partial }x}\right\}=\frac{\hslash }{m}\mathrm{\Im }\left\{{\mathrm{\Psi }}^{\ast }\frac{\mathrm{\partial }\mathrm{\Psi }}{\mathrm{\partial }x}\right\},}$$where

• ${\displaystyle \hslash }$ is the reducedPlanck constant;
• ${\displaystyle {\mathrm{\Psi }}^{\ast }}$ denotes thecomplex conjugate of thewave function;
• ${\displaystyle \mathrm{\Re }}$ denotes thereal part;
• ${\displaystyle \mathrm{\Im }}$ denotes theimaginary part.

Note that the probability current is proportional to aWronskian${\displaystyle W(\mathrm{\Psi },{\mathrm{\Psi }}^{\ast }).}$

In three dimensions, this generalizes to$${\displaystyle \mathbf{j}=\frac{\hslash }{2mi}\left({\mathrm{\Psi }}^{\ast }\mathrm{\nabla }\mathrm{\Psi }-\mathrm{\Psi }\mathrm{\nabla }{\mathrm{\Psi }}^{\ast }\right)=\frac{\hslash }{m}\mathrm{\Re }\left\{{\mathrm{\Psi }}^{\ast }\frac{\mathrm{\nabla }}{i}\mathrm{\Psi }\right\}=\frac{\hslash }{m}\mathrm{\Im }\left\{{\mathrm{\Psi }}^{\ast }\mathrm{\nabla }\mathrm{\Psi }\right\},}$$where${\displaystyle \mathrm{\nabla }}$ denotes thedel orgradientoperator. This can be simplified in terms of thekinetic momentum operator,$${\displaystyle \hat{\mathbf{p}}=-i\hslash \mathrm{\nabla }}$$to obtain$${\displaystyle \mathbf{j}=\frac{1}{2m}\left({\mathrm{\Psi }}^{\ast }\hat{\mathbf{p}}\mathrm{\Psi }+\mathrm{\Psi }{\left(\hat{\mathbf{p}}\mathrm{\Psi }\right)}^{\ast }\right).}$$

These definitions use the position basis (i.e. for a wavefunction inposition space), butmomentum space is possible. In fact, one can write the probability current operator as

${\displaystyle \hat{\mathbf{j}}(\mathbf{r})=\frac{\hat{\mathbf{p}}|\mathbf{r}⟩⟨\mathbf{r}|+|\mathbf{r}⟩⟨\mathbf{r}|\hat{\mathbf{p}}}{2m}}$

which do not depend on a particular choice of basis. The probability current is then the expectation of this operator,

${\displaystyle \mathbf{j}(\mathbf{r},t)=⟨\mathrm{\Psi }(t)|\hat{\mathbf{j}}(\mathbf{r})|\mathrm{\Psi }(t)⟩.}$

The above definition should be modified for a system in an externalelectromagnetic field. InSI units, acharged particle of massm andelectric chargeq includes a term due to the interaction with the electromagnetic field;^[3]$${\displaystyle \mathbf{j}=\frac{1}{2m}\left[\left({\mathrm{\Psi }}^{\ast }\hat{\mathbf{p}}\mathrm{\Psi }-\mathrm{\Psi }\hat{\mathbf{p}}{\mathrm{\Psi }}^{\ast }\right)-2q\mathbf{A}|\mathrm{\Psi }{|}^2\right]}$$whereA =A(r,t) is themagnetic vector potential. The termqA has dimensions of momentum. Note that${\displaystyle \hat{\mathbf{p}}=-i\hslash \mathrm{\nabla }}$ used here is thecanonical momentum and is notgauge invariant, unlike thekinetic momentum operator${\displaystyle \hat{\mathbf{P}}=-i\hslash \mathrm{\nabla }-q\mathbf{A}}$.

InGaussian units:$${\displaystyle \mathbf{j}=\frac{1}{2m}\left[\left({\mathrm{\Psi }}^{\ast }\hat{\mathbf{p}}\mathrm{\Psi }-\mathrm{\Psi }\hat{\mathbf{p}}{\mathrm{\Psi }}^{\ast }\right)-2\frac{q}{c}\mathbf{A}|\mathrm{\Psi }{|}^2\right]}$$wherec is thespeed of light.

If the particle hasspin, it has a correspondingmagnetic moment, so an extra term needs to be added incorporating the spin interaction with the electromagnetic field.

According to Landau-Lifschitz'sCourse of Theoretical Physics the electric current density is in Gaussian units:^[4]$${\displaystyle {\mathbf{j}}_e=\frac{q}{2m}\left[\left({\mathrm{\Psi }}^{\ast }\hat{\mathbf{p}}\mathrm{\Psi }-\mathrm{\Psi }\hat{\mathbf{p}}{\mathrm{\Psi }}^{\ast }\right)-\frac{2q}{c}\mathbf{A}|\mathrm{\Psi }{|}^2\right]+\frac{{\mu }_{S}c}{s\hslash }\mathrm{\nabla }\times ({\mathrm{\Psi }}^{\ast }\mathbf{S}\mathrm{\Psi })}$$

And in SI units:$${\displaystyle {\mathbf{j}}_e=\frac{q}{2m}\left[\left({\mathrm{\Psi }}^{\ast }\hat{\mathbf{p}}\mathrm{\Psi }-\mathrm{\Psi }\hat{\mathbf{p}}{\mathrm{\Psi }}^{\ast }\right)-2q\mathbf{A}|\mathrm{\Psi }{|}^2\right]+\frac{{\mu }_S}{s\hslash }\mathrm{\nabla }\times ({\mathrm{\Psi }}^{\ast }\mathbf{S}\mathrm{\Psi })}$$

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

whereS is thespin vector of the particle with correspondingspin magnetic momentμ_S andspin quantum numbers.

It is doubtful if this formula is valid for particles with an interior structure.^{[citation needed]} Theneutron has zero charge but non-zero magnetic moment, so${\displaystyle \frac{{\mu }_S}{qs\hslash }}$ would be impossible (except${\displaystyle \mathrm{\nabla }\times ({\mathrm{\Psi }}^{\ast }\mathbf{S}\mathrm{\Psi })}$ would also be zero in this case). For composite particles with a non-zero charge – like theproton which has spin quantum number s=1/2 and μ_S= 2.7927·μ_N or thedeuteron (H-2 nucleus) which has s=1 and μ_S=0.8574·μ_N^[5] – it is mathematically possible but doubtful.

The wave function can also be written in thecomplexexponential (polar) form:$${\displaystyle \mathrm{\Psi }=R{e}^{iS/\hslash }}$$whereR, S are real functions ofr andt.

Written this way, the probability density is$${\displaystyle \rho ={\mathrm{\Psi }}^{\ast }\mathrm{\Psi }=R^2}$$ and the probability current is:

The exponentials andR∇R terms cancel:$${\displaystyle \mathbf{j}=\frac{\hslash }{2mi}\left[\frac{i}{\hslash }{R}^2\mathrm{\nabla }S+\frac{i}{\hslash }{R}^2\mathrm{\nabla }S\right].}$$

Finally, combining and cancelling the constants, and replacingR² withρ,$${\displaystyle \mathbf{j}=\rho \frac{\mathrm{\nabla }S}{m}.}$$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:$${\displaystyle \mathbf{j}=\rho \mathbf{v},}$$

where${\displaystyle \rho }$ is the mass density of the fluid andv is its velocity (also thegroup velocity of the wave). In the classical limit, we can associate the velocity with${\displaystyle \frac{\mathrm{\nabla }S}{m},}$ which is the same as equating∇S with the classical momentump =mv however, it does not represent a physical velocity or momentum at a point since simultaneous measurement of position and velocity violatesuncertainty principle. This interpretation fits withHamilton–Jacobi theory, in which$${\displaystyle \mathbf{p}=\mathrm{\nabla }S}$$in Cartesian coordinates is given by∇S, whereS isHamilton's principal function.

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

The definition of probability current and Schrödinger's equation can be used to derive thecontinuity equation, which hasexactly the same forms as those forhydrodynamics andelectromagnetism.^[6]

For somewave functionΨ, let:

$${\displaystyle \rho (\mathbf{r},t)=|\mathrm{\Psi }{|}^2={\mathrm{\Psi }}^{\ast }(\mathbf{r},t)\mathrm{\Psi }(\mathbf{r},t).}$$be theprobability density (probability per unit volume,* denotescomplex conjugate). Then,

whereV is any volume andS is the boundary ofV.

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

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

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

${\displaystyle \frac{\mathrm{\partial }}{\mathrm{\partial }t}{\int }_V|\mathrm{\Psi }{|}^2\mathrm{d}V+}$${\displaystyle S}$${\displaystyle \mathbf{j}\cdot \mathrm{d}\mathbf{S}=0}$.

In particular, ifΨ 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 withinV when the position of the particle is measured. The second term is then the rate at which probability is flowing out of the volumeV. Altogether the equation states that the time derivative of the probability of the particle being measured inV is equal to the rate at which probability flows intoV.

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.^[7] This result is in agreement with the unitary nature of time evolution operators which preserve length of the vector by definition.

The probability (4-)current arises fromNoether's theorem as applied to theLagrangian theKlein-Gordon Lagrangian density

$${\displaystyle \mathcal{L}={\mathrm{\partial }}_{\mu }{\varphi }^{\ast }{\mathrm{\partial }}^{\mu }\varphi +V({\varphi }^{\ast }\varphi )}$$of the complex scalar field$\varphi :{\mathbb{R}}^{n+1}\mapsto \mathbb{C}$. This is invariant under the symmetry transformation$${\displaystyle \varphi \mapsto {\varphi }^{\prime }=\varphi e^{i\alpha }.}$$ Defining$\delta \varphi ={\varphi }^{\prime }-\varphi$ we find the Noether current$${\displaystyle j^{\mu }:=\frac{d\mathcal{L}}{d\dot{\mathbf{q}}}\cdot {\mathbf{Q}}_r=\frac{d\mathcal{L}}{d({\mathrm{\partial }}_{\mu })\varphi }\frac{d(\delta \varphi )}{d\alpha }{|}_{\alpha =0}+\frac{d\mathcal{L}}{d({\mathrm{\partial }}_{\mu }){\varphi }^{\ast }}\frac{d(\delta {\varphi }^{\ast })}{d\alpha }{|}_{\alpha =0}=i\varphi ({\mathrm{\partial }}^{\mu }{\varphi }^{\ast })-i{\varphi }^{\ast }({\mathrm{\partial }}^{\mu }\varphi )}$$which satisfies the continuity equation. Here${\displaystyle {\mathbf{Q}}_r}$ is the generator of the symmetry, which is${\displaystyle \frac{d(\delta \mathbf{q})}{d{\alpha }_r}}$ in the case of a single parameter${\displaystyle \alpha }$.

The continuity equation${\displaystyle {\mathrm{\partial }}_{\mu }{j}^{\mu }=0}$ is satisfied. However, note that now, the analog of the probability density is not${\displaystyle \varphi {\varphi }^{\ast }}$ but rather${\displaystyle {\varphi }^{\ast }{\mathrm{\partial }}_t\varphi -\varphi {\mathrm{\partial }}_t{\varphi }^{\ast }}$. As this quantity can now be negative, we must interpret it as a charge density, with an associated current density and4-current.

In regions where astep potential orpotential barrier occurs, the probability current is related to the transmission and reflection coefficients, respectivelyT andR; they measure the extent the particles reflect from the potential barrier or are transmitted through it. Both satisfy:$${\displaystyle T+R=1,}$$whereT andR can be defined by:$${\displaystyle T=\frac{|{\mathbf{j}}_{\mathrm{t}\mathrm{r}\mathrm{a}\mathrm{n}\mathrm{s}}|}{|{\mathbf{j}}_{\mathrm{i}\mathrm{n}\mathrm{c}}|},R=\frac{|{\mathbf{j}}_{\mathrm{r}\mathrm{e}\mathrm{f}}|}{|{\mathbf{j}}_{\mathrm{i}\mathrm{n}\mathrm{c}}|},}$$wherej_inc,j_ref,j_trans are the incident, reflected and transmitted probability currents respectively, and the vertical bars indicate themagnitudes of the current vectors. The relation betweenT andR can be obtained from probability conservation:$${\displaystyle {\mathbf{j}}_{\mathrm{t}\mathrm{r}\mathrm{a}\mathrm{n}\mathrm{s}}+{\mathbf{j}}_{\mathrm{r}\mathrm{e}\mathrm{f}}={\mathbf{j}}_{\mathrm{i}\mathrm{n}\mathrm{c}}.}$$

In terms of aunit vectornnormal to the barrier, these are equivalently:$${\displaystyle T=\left|\frac{{\mathbf{j}}_{\mathrm{t}\mathrm{r}\mathrm{a}\mathrm{n}\mathrm{s}}\cdot \mathbf{n}}{{\mathbf{j}}_{\mathrm{i}\mathrm{n}\mathrm{c}}\cdot \mathbf{n}}\right|,R=\left|\frac{{\mathbf{j}}_{\mathrm{r}\mathrm{e}\mathrm{f}}\cdot \mathbf{n}}{{\mathbf{j}}_{\mathrm{i}\mathrm{n}\mathrm{c}}\cdot \mathbf{n}}\right|,}$$where the absolute values are required to preventT andR being negative.

For aplane wave propagating in space:$${\displaystyle \mathrm{\Psi }(\mathbf{r},t)=A{e}^{i(\mathbf{k}\cdot \mathbf{r}-\omega t)}}$$the probability density is constant everywhere;$${\displaystyle \rho (\mathbf{r},t)=|A{|}^2\to \frac{\mathrm{\partial }|\mathrm{\Psi }{|}^2}{\mathrm{\partial }t}=0}$$(that is, plane waves arestationary states) but the probability current is nonzero – the square of the absolute amplitude of the wave times the particle's speed;$${\displaystyle \mathbf{j}\left(\mathbf{r},t\right)={\left|A\right|}^2\frac{\hslash \mathbf{k}}{m}=\rho \frac{\mathbf{p}}{m}=\rho \mathbf{v}}$$

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

For aparticle in a box, in one spatial dimension and of lengthL, confined to the region, the energy eigenstates are$${\displaystyle {\mathrm{\Psi }}_n=\sqrt{\frac{2}{L}}\sin \left(\frac{n\pi }{L}x\right)}$$and zero elsewhere. The associated probability currents are$${\displaystyle j_n=\frac{i\hslash }{2m}\left({\mathrm{\Psi }}_n^{\ast }\frac{\mathrm{\partial }{\mathrm{\Psi }}_n}{\mathrm{\partial }x}-{\mathrm{\Psi }}_n\frac{\mathrm{\partial }{\mathrm{\Psi }}_n^{\ast }}{\mathrm{\partial }x}\right)=0}$$since$${\displaystyle {\mathrm{\Psi }}_n={\mathrm{\Psi }}_n^{\ast }}$$

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

As a result, we find:

• Resnick, R.; Eisberg, R. (1985).Quantum Physics of Atoms, Molecules, Solids, Nuclei and Particles (2nd ed.). John Wiley & Sons.ISBN 0-471-87373-X.

• Quantum mechanics

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