Free spin-0 particle

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.

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;^[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.

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:^[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.

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.^[7]

For some wave function Ψ, let:

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

be the probability density (probability per unit volume, * denotes complex conjugate). Then,

${\displaystyle {\begin{aligned}{\frac {d}{dt}}\int _{\mathcal {V}}dV\,\rho &=\int _{\mathcal {V}}dV\,(\psi '^{*}\psi +\psi ^{*}\psi ')\\&=\int _{\mathcal {V}}dV\,\left(-{\frac {i}{\hbar }}\left(-{\frac {\hbar ^{2}}{2m}}\nabla ^{2}\psi +V\psi \right)\psi ^{*}+{\frac {i}{\hbar }}\left(-{\frac {\hbar ^{2}}{2m}}\nabla ^{2}\psi ^{*}+V\psi ^{*}\right)\psi \right)\\&=\int _{\mathcal {V}}dV\,{\frac {i\hbar }{2m}}(\nabla ^{2}\psi \psi ^{*}-\psi \nabla ^{2}\psi ^{*})\\&=\int _{\mathcal {V}}dV\,\nabla \cdot \left({\frac {i\hbar }{2m}}(\psi ^{*}\nabla \psi -\psi \nabla \psi ^{*})\right)\\&=\int _{\mathcal {S}}d\mathbf {a} \cdot \left({\frac {i\hbar }{2m}}(\psi ^{*}\nabla \psi -\psi \nabla \psi ^{*})\right)\\\end{aligned}}}$

where V is any volume and S is the boundary of V.

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

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

where

$${\displaystyle \mathbf {j} ={\frac {1}{2m}}\left(\Psi ^{*}{\hat {\mathbf {p} }}\Psi -\Psi {\hat {\mathbf {p} }}\Psi ^{*}\right)=-{\frac {i\hbar }{2m}}(\psi ^{*}\nabla \psi -\psi \nabla \psi ^{*})={\frac {\hbar }{m}}\operatorname {Im} (\psi ^{*}\nabla \psi )}$$

is the probability current or probability flux (flow per unit area). Here, equating the terms inside the integral gives the continuity equation for probability:

$${\displaystyle {\frac {\partial }{\partial t}}\rho \left(\mathbf {r} ,t\right)+\nabla \cdot \mathbf {j} =0,}$$

and the integral equation can also be restated using the divergence theorem as:

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

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.

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.^[8] 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 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.

Plane wave

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.

Particle in a box

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

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