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.

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. 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 \hbar is the reduced Planck constant; \Psi^* denotes the complex conjugate of the wave function; \Re denotes the real part; \Im denotes the imaginary part. Note that the probability current is proportional to a Wronskian W(\Psi,\Psi^*). In three dimensions, this generalizes to \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 \nabla denotes the del or gradient operator. This can be simplified in terms of the kinetic momentum operator, \mathbf{\hat{p}} = -i\hbar\nabla to obtain \mathbf j = \frac{1}{2m} \left(\Psi^* \mathbf{\hat{p}} \Psi + \Psi \left( \mathbf{\hat{p}} \Psi \right)^*\right)\,. These definitions use the position basis (i.e. for a wavefunction in position space), but momentum space is possible. In fact, one can write the probability current operator as \mathbf{\hat{j}}(\mathbf{r}) = \frac{\mathbf{\hat{p}} |\mathbf{r}\rangle\langle\mathbf{r}|+|\mathbf{r}\rangle\langle\mathbf{r}|\mathbf{\hat{p}}}{2m} which do not depend on a particular choice of basis. The probability current is then the expectation of this operator, \mathbf{j}(\mathbf{r},t) = \langle \Psi(t)|\hat{\mathbf{j}}(\mathbf{r})| \Psi(t)\rangle. \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 is the magnetic vector potential. The term has dimensions of momentum. Note that \mathbf{\hat{p}} = -i\hbar\nabla used here is the canonical momentum and is not gauge invariant, unlike the kinetic momentum operator \mathbf{\hat{P}} = -i\hbar\nabla-q\mathbf{A} . In Gaussian units: \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 is the speed of light. \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: \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: \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}{q s\hbar}\nabla\times(\Psi^* \mathbf{S}\Psi) where is the spin vector of the particle with corresponding spin magnetic moment and spin quantum number . It is doubtful if this formula is valid for particles with an interior structure. The neutron has zero charge but non-zero magnetic moment, so \frac{\mu_S}{q s\hbar} would be impossible (except \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 – it is mathematically possible but doubtful. \Psi = R e^{i S / \hbar } where are real functions of and . Written this way, the probability density is \rho = \Psi^* \Psi = R^2 and the probability current is: \begin{align} \mathbf{j} & = \frac{\hbar}{2mi}\left(\Psi^{*} \mathbf{\nabla} \Psi - \Psi \mathbf{\nabla}\Psi^{*} \right) \\[5pt] & = \frac{\hbar}{2mi}\left(R e^{-i S / \hbar} \mathbf{\nabla}R e^{i S / \hbar} - R e^{i S / \hbar} \mathbf{\nabla}R e^{-i S / \hbar}\right) \\[5pt] & = \frac{\hbar}{2mi}\left[ R e^{-i S / \hbar} \left( e^{i S / \hbar} \mathbf{\nabla}R + \frac{i}{\hbar}R e^{i S / \hbar} \mathbf{\nabla}S \right) - R e^{i S / \hbar} \left( e^{-i S / \hbar} \mathbf{\nabla}R - \frac{i}{\hbar} R e^{-i S / \hbar} \mathbf{\nabla} S \right)\right]. \end{align} The exponentials and terms cancel: \mathbf{j} = \frac{\hbar}{2mi}\left[\frac {i}{\hbar} R^2 \mathbf{\nabla} S + \frac {i}{\hbar} R^2 \mathbf{\nabla} S \right]. Finally, combining and cancelling the constants, and replacing with , \mathbf{j} = \rho \frac{\mathbf{\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: \mathbf{j} = \rho \mathbf{v}, where \rho is the mass density of the fluid and is its velocity (also the group velocity of the wave). In the classical limit, we can associate the velocity with \tfrac{\nabla S}{m}, which is the same as equating with the classical momentum 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 \mathbf{p} = \nabla S in Cartesian coordinates is given by , where is Hamilton's principal function. The de Broglie-Bohm theory equates the velocity with \tfrac{\nabla S}{m} in general (not only in the classical limit) so it is always well defined. It is an interpretation of quantum mechanics. \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, \begin{align} \frac{d}{dt}\int_\mathcal{V}dV\, \rho &= \int_\mathcal{V}dV\, \left(\frac {\partial \psi} {\partial t}\psi^*+ \psi \frac {\partial \psi^*} {\partial t}\right) \\ &=\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} \left[\left(\nabla^2\psi\right) \psi^* - \psi \left(\nabla^2 \psi^*\right)\right]\\ &=\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{align} where is any volume and is the boundary of . This is the conservation law for probability in quantum mechanics. The integral form is stated as: \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 \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: \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: In particular, if 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 when the position of the particle is measured. The second term is then the rate at which probability is flowing out of the volume . Altogether the equation states that the time derivative of the probability of the particle being measured in is equal to the rate at which probability flows into . 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. \mathcal{L}=\partial_\mu\phi^*\,\partial^\mu\phi +V(\phi^*\,\phi) of the complex scalar field \phi:\mathbb{R}^{n+1}\mapsto\mathbb{C} . This is invariant under the symmetry transformation \phi\mapsto\phi'=\phi\,e^{i\alpha}\, . Defining \delta\phi=\phi'-\phi we find the Noether current j^\mu:= \frac{d\mathcal{L}}{d \dot{\mathbf{q}}} \cdot \mathbf{Q}_r = \frac{d\mathcal{L}}{d(\partial_\mu \phi)}\,\frac{d(\delta\phi)}{d\alpha}\bigg|_{\alpha=0}+\frac{d\mathcal{L}}{d(\partial_\mu \phi^*)}\,\frac{d(\delta\phi^*)}{d\alpha}\bigg|_{\alpha=0}= i\,\phi\,(\partial^\mu\phi^*)-i\,\phi^*\,(\partial^\mu\phi) which satisfies the continuity equation. Here \mathbf{Q}_r is the generator of the symmetry, which is \frac{d (\delta \mathbf{q})}{d \alpha_r} in the case of a single parameter \alpha . The continuity equation \partial_\mu j^\mu = 0 is satisfied. However, note that now, the analog of the probability density is not \phi \phi^* but rather \phi^* \partial_t \phi - \phi \partial_t \phi^* . As this quantity can now be negative, we must interpret it as a charge density, with an associated current density and 4-current. T + R = 1\,, where and can be defined by: T= \frac \, , \quad R = \frac \, , where are the incident, reflected and transmitted probability currents respectively, and the vertical bars indicate the magnitudes of the current vectors. The relation between and can be obtained from probability conservation: \mathbf{j}_\mathrm{trans} + \mathbf{j}_\mathrm{ref}=\mathbf{j}_\mathrm{inc}\,. In terms of a unit vector normal to the barrier, these are equivalently: 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 and being negative. \Psi(\mathbf{r},t) = \, A e^{ i (\mathbf{k}\cdot{\mathbf{r}} - \omega t)} the probability density is constant everywhere; \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; \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. \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\hbar}{2m}\left( \Psi_n^* \frac{\partial \Psi_n}{\partial x} - \Psi_n \frac{\partial \Psi_n^*}{\partial x} \right) = 0 since \Psi_n = \Psi_n^* For a particle in one dimension on \ell^2(\Z), we have the Hamiltonian H = -\Delta + V where -\Delta \equiv 2 I - S - S^\ast is the discrete Laplacian, with being the right shift operator on \ell^2(\Z). Then the probability current is defined as j \equiv 2 \Im\left\{\bar{\Psi} i v \Psi\right\}, with the velocity operator, equal to v \equiv -i[X,\, H] and is the position operator on \ell^2\left(\mathbb{Z}\right). Since is usually a multiplication operator on \ell^2(\Z), we get to safely write -i[X,\, H] = -i[X,\, -\Delta] = -i\left[X,\, -S - S^{\ast}\right] = i S - i S^{\ast}. As a result, we find: \begin{align} j\left(x\right) \equiv 2 \Im\left\{\bar{\Psi}(x) i v \Psi(x)\right\} &= 2 \Im\left\{\bar{\Psi}(x) \left((-S\Psi)(x) + \left(S^\ast \Psi\right)(x)\right)\right\}\\ &= 2 \Im\left\{\bar{\Psi}(x) \left(-\Psi(x-1) + \Psi(x + 1)\right)\right\} \end{align}

Text derived from https://en.wikipedia.org/w/index.php?oldid=1371638992 under CC BY-SA 4.0; modifications have been made.