Probability current

From Wikipedia, the free encyclopedia

In quantum mechanics, the probability current (sometimes called probability flux) is a concept describing the flow of probability density. In particular, if one pictures the probability density as an inhomogeneous fluid, then the probability current is the rate of flow of this fluid (the density times the velocity).

1 Definition
2 Examples
- 2.1 Plane wave
- 2.2 Particle in a box
3 Derivation of continuity equation
4 Definition in an external field

[edit] Definition

The probability current, $\vec j$ , in non-relativistic quantum mechanics is defined as

$\vec j = \frac{\hbar}{2mi}\left(\Psi^* \vec \nabla \Psi - \Psi \vec \nabla \Psi^*\right) = \frac\hbar m \mbox{Im}(\Psi^*\vec\nabla\Psi)=\mbox{Re}(\Psi^* \frac{\hbar}{im} \vec\nabla \Psi)$

in the position basis and satisfies the quantum mechanical continuity equation

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

with the probability density $\rho\,$ defined as

$\rho = |\Psi|^2 \,$ .

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

$\frac{\partial}{\partial t} \int_V |\Psi|^2 dV + \int_S \vec j \cdot \vec {dA} = 0$

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

In particular, if $\Psi\,$ is a wavefunction describing a single particle, the integral in the first term of the preceding equation (without the 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 change of the probability of the particle being measured in $V$ is equal to the rate at which probability flows into $V$ .

[edit] Examples

[edit] Plane wave

The probability current associated with the (three dimensional) plane wave

$\Psi = A e^{i\vec k \cdot \vec r} e^{i \omega t}$

$\vec j = \frac{\hbar}{2mi} |A|^2 \left( e^{-i\vec k \cdot \vec r} \vec \nabla e^{i\vec k \cdot \vec r} - e^{i\vec k \cdot \vec r} \vec \nabla e^{-i\vec k \cdot \vec r} \right) = |A|^2 \frac{\hbar \vec k}{m}.$

This is just the square of the amplitude of the wave times the particle's velocity,

$\vec v = \frac{\vec p}{m} = \frac{\hbar \vec k}{m}$ .

Note that the probability current is nonzero despite the fact that plane waves are stationary states and hence

$\frac{d|\Psi|^2}{dt} = 0\,$

everywhere. This demonstrates that a particle may be in motion even if its spatial probability density has no explicit time dependence.

[edit] Particle in a box

The energy eigenstates of a particle in a box of one spatial dimension and of length $L$ are, for $0 < x < L$ ,

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

and zero elsewhere. The associated probability currents are

$j_n = \frac{\hbar}{2mi}\left( \Psi_n^* \frac{\partial \Psi_n}{\partial x} - \Psi_n \frac{\partial \Psi_n^*}{\partial x} \right) = 0$

since $\Psi_n = \Psi_n^*.$

[edit] Derivation of continuity equation

In this section the continuity equation is derived from the definition of probability current and the basic principles of quantum mechanics.

Suppose $\Psi \,$ is the wavefunction for a single particle in the position basis (i.e. $\Psi \,$ is a function of $x$ , $y$ , and $z$ ). Then

$P = \int_V |\Psi|^2 dV \,$

is the probability that a measurement of the particle's position will yield a value within V. The time derivative of this is

$\frac{dP}{dt} = \frac{\partial}{\partial t} \int_V |\Psi|^2 dV = \int_V \left( \frac{\partial \Psi}{\partial t}\Psi^* + \Psi \frac{\partial \Psi^*}{\partial t} \right) dV$

where the last equality follows from the product rule and the fact that the shape of $V$ is presumed to be independent of time (i.e. the time derivative can be moved through the integral). In order to simplify this further consider the time dependent Schrödinger equation

$i\hbar \frac{\partial \Psi}{\partial t} = \frac{-\hbar^2}{2m} \nabla^2 \Psi + U\Psi$

and use it to solve for the time derivative of $\Psi\,$ :

$\frac{\partial \Psi}{\partial t} = \frac{i \hbar}{2m} \nabla^2 \Psi - \frac{i}{\hbar} U \Psi$

When substituted back into the preceding equation for $\frac{dP}{dt}$ this gives

$\frac{dP}{dt} = - \int_V \frac{\hbar}{2mi} \left(\Psi^* \nabla^2 \Psi - \Psi \nabla^2 \Psi^* \right) dV$ .

Now from the product rule for the divergence operator

$\vec \nabla \cdot \left(\Psi^* \vec \nabla \Psi - \Psi \vec \nabla \Psi^* \right) = \vec \nabla \Psi^* \cdot \vec \nabla \Psi + \Psi^* \nabla^2 \Psi - \vec \nabla \Psi \cdot \vec \nabla \Psi^* - \Psi \nabla^2 \Psi^*$

and since the first and third terms cancel:

$\frac{dP}{dt} = - \int_V \vec \nabla \cdot \frac{\hbar}{2mi} \left(\Psi^* \vec \nabla \Psi - \Psi \vec \nabla \Psi^* \right) dV$

If we now recall the expression for $P$ and note that the argument of the divergence operator is just $\vec j$ this becomes

$\int_V \left( \frac{\partial |\Psi|^2}{\partial t} \right) dV = - \int_V \left( \vec \nabla \cdot \vec j \right) dV$

$\int_V \left( \frac{\partial |\Psi|^2}{\partial t} + \vec \nabla \cdot \vec j \right) dV = 0$

which is the integral form of the continuity equation. The differential form follows from the fact that the preceding equation holds for all $V$ , and as the integrand is a continuous function of space, it must vanish everywhere:

$\frac{\partial |\Psi|^2}{\partial t} + \vec \nabla \cdot \vec j = 0.$

[edit] Definition in an external field

The standard definition should be modified for a system in an external electromagnetic field. E.g. for a particle with a charge $q$ the time-dependent Schrödinger equation that satisfies local gauge invariance is given in coordinate representation with minimal-coupling Hamiltonian by

$i\hbar \frac{\partial \Psi}{\partial t} =- \frac{\hbar^2}{2m} \nabla^2 \Psi + i (1 - \beta) \frac{\hbar q}{m} \vec{A}(t)\cdot\vec{\nabla} \Psi +\frac{q^2}{2m}(1-\beta)^2 A^2(t) \Psi - \beta q \vec{F}(t) \cdot\vec{r}\,\Psi + U\Psi$

where $\vec F(t)$ is an electric field, $\vec A(t)$ is a vector potential, and $β$ is a parameter specifying the gauge. For example, $β = 0$ corresponds to the velocity gauge, whereas $β = 1$ corresponds to the length gauge. Rederiving the continuity equation one obtains