Probability current

In non-relativistic quantum mechanics, the probability current j of the wave function Ψ of a particle of mass m in one dimension is defined as[2]

$${\displaystyle j={\frac {\hbar }{2mi}}\left(\Psi ^{*}{\frac {\partial \Psi }{\partial x}}-\Psi {\frac {\partial \Psi ^{*}}{\partial x}}\right)={\frac {\hbar }{m}}\Re \left\{\Psi ^{*}{\frac {1}{i}}{\frac {\partial \Psi }{\partial x}}\right\}={\frac {\hbar }{m}}\Im \left\{\Psi ^{*}{\frac {\partial \Psi }{\partial x}}\right\},}$$

where

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

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

In three dimensions, this generalizes to

$${\displaystyle \mathbf {j} ={\frac {\hbar }{2mi}}\left(\Psi ^{*}\mathbf {\nabla } \Psi -\Psi \mathbf {\nabla } \Psi ^{*}\right)={\frac {\hbar }{m}}\Re \left\{\Psi ^{*}{\frac {\nabla }{i}}\Psi \right\}={\frac {\hbar }{m}}\Im \left\{\Psi ^{*}\nabla \Psi \right\}\,,}$$

where ${\displaystyle \nabla }$ denotes the del or gradient operator. This can be simplified in terms of the kinetic momentum operator,

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

to obtain

$${\displaystyle \mathbf {j} ={\frac {1}{2m}}\left(\Psi ^{*}\mathbf {\hat {p}} \Psi -\Psi \mathbf {\hat {p}} \Psi ^{*}\right)\,.}$$

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

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;[3]

$${\displaystyle \mathbf {j} ={\frac {1}{2m}}\left[\left(\Psi ^{*}\mathbf {\hat {p}} \Psi -\Psi \mathbf {\hat {p}} \Psi ^{*}\right)-2q\mathbf {A} |\Psi |^{2}\right]}$$

where A = A(r, t) is the magnetic vector potential. The term qA has dimensions of momentum. Note that ${\displaystyle \mathbf {\hat {p}} =-i\hbar \nabla }$ used here is the canonical momentum and is not gauge invariant, unlike the kinetic momentum operator ${\displaystyle \mathbf {\hat {P}} =-i\hbar \nabla -q\mathbf {A} }$.

In Gaussian units:

$${\displaystyle \mathbf {j} ={\frac {1}{2m}}\left[\left(\Psi ^{*}\mathbf {\hat {p}} \Psi -\Psi \mathbf {\hat {p}} \Psi ^{*}\right)-2{\frac {q}{c}}\mathbf {A} |\Psi |^{2}\right]}$$

where c is the speed of light.

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:[4]

$${\displaystyle \mathbf {j} _{e}={\frac {q}{2m}}\left[\left(\Psi ^{*}\mathbf {\hat {p}} \Psi -\Psi \mathbf {\hat {p}} \Psi ^{*}\right)-{\frac {2q}{c}}\mathbf {A} |\Psi |^{2}\right]+{\frac {\mu _{S}c}{s\hbar }}\nabla \times (\Psi ^{*}\mathbf {S} \Psi )}$$

And in SI units:

$${\displaystyle \mathbf {j} _{e}={\frac {q}{2m}}\left[\left(\Psi ^{*}\mathbf {\hat {p}} \Psi -\Psi \mathbf {\hat {p}} \Psi ^{*}\right)-2q\mathbf {A} |\Psi |^{2}\right]+{\frac {\mu _{S}}{s\hbar }}\nabla \times (\Psi ^{*}\mathbf {S} \Psi )}$$

Hence the probability current (density) is in SI units:

$${\displaystyle \mathbf {j} =\mathbf {j} _{e}/q={\frac {1}{2m}}\left[\left(\Psi ^{*}\mathbf {\hat {p}} \Psi -\Psi \mathbf {\hat {p}} \Psi ^{*}\right)-2q\mathbf {A} |\Psi |^{2}\right]+{\frac {\mu _{S}}{qs\hbar }}\nabla \times (\Psi ^{*}\mathbf {S} \Psi )}$$

where S is the spin vector of the particle with corresponding spin magnetic moment μ_S and spin quantum number s.

It is doubtful if this formula is vaild for particles with an interior structure. The neutron has zero charge but non-zero magnetic moment, so ${\displaystyle {\frac {\mu _{S}}{qs\hbar }}}$ would be impossible (except ${\displaystyle \nabla \times (\Psi ^{*}\mathbf {S} \Psi )}$ 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 [5] – it is mathematically possible but doubtful.

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:[7]

$${\displaystyle {\frac {\partial \rho }{\partial t}}+\mathbf {\nabla } \cdot \mathbf {j} =0}$$

where the probability density ${\displaystyle \rho \,}$ is defined as

$${\displaystyle \rho (\mathbf {r} ,t)=|\Psi |^{2}=\Psi ^{*}(\mathbf {r} ,t)\Psi (\mathbf {r} ,t).}$$

If one were to integrate both sides of the continuity equation with respect to volume, so that

$${\displaystyle \int _{V}\left({\frac {\partial |\Psi |^{2}}{\partial t}}\right)\mathrm {d} V+\int _{V}\left(\mathbf {\nabla } \cdot \mathbf {j} \right)\mathrm {d} V=0}$$

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

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

where 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.

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:

$${\displaystyle T+R=1\,,}$$

where T and R can be defined by:

$${\displaystyle T={\frac {|\mathbf {j} _{\mathrm {trans} }|}{|\mathbf {j} _{\mathrm {inc} }|}}\,,\quad R={\frac {|\mathbf {j} _{\mathrm {ref} }|}{|\mathbf {j} _{\mathrm {inc} }|}}\,,}$$

where j_inc, j_ref, 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:

$${\displaystyle \mathbf {j} _{\mathrm {trans} }+\mathbf {j} _{\mathrm {ref} }=\mathbf {j} _{\mathrm {inc} }\,.}$$

In terms of a unit vector n normal to the barrier, these are equivalently:

$${\displaystyle T=\left|{\frac {\mathbf {j} _{\mathrm {trans} }\cdot \mathbf {n} }{\mathbf {j} _{\mathrm {inc} }\cdot \mathbf {n} }}\right|\,,\qquad R=\left|{\frac {\mathbf {j} _{\mathrm {ref} }\cdot \mathbf {n} }{\mathbf {j} _{\mathrm {inc} }\cdot \mathbf {n} }}\right|\,,}$$

where the absolute values are required to prevent T and R being negative.

For a plane wave propagating in space:

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

the probability density is constant everywhere;

$${\displaystyle \rho (\mathbf {r} ,t)=|A|^{2}\rightarrow {\frac {\partial |\Psi |^{2}}{\partial t}}=0}$$

(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;

$${\displaystyle \mathbf {j} \left(\mathbf {r} ,t\right)=\left|A\right|^{2}{\hbar \mathbf {k} \over m}=\rho {\frac {\mathbf {p} }{m}}=\rho \mathbf {v} }$$

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

For a particle in a box, in one spatial dimension and of length L, confined to the region ${\displaystyle 0<x<L}$, the energy eigenstates are

$${\displaystyle \Psi _{n}={\sqrt {\frac {2}{L}}}\sin \left({\frac {n\pi }{L}}x\right)}$$

and zero elsewhere. The associated probability currents are

$${\displaystyle j_{n}={\frac {i\hbar }{2m}}\left(\Psi _{n}^{*}{\frac {\partial \Psi _{n}}{\partial x}}-\Psi _{n}{\frac {\partial \Psi _{n}^{*}}{\partial x}}\right)=0}$$

since

$${\displaystyle \Psi _{n}=\Psi _{n}^{*}}$$