# Definition of probablity current in dirac space not including spatial dimension?

I'm currently reviewing (basic) relativistic quantum mechanics and stumbled upon the probability current in "dirac space", defined as

$j^μ = (j^0,\vec j)^\mathrm T$ with $j^0 = c\,ρ = c\,ψ^+ψ$ and $\vec j = cψ^+\,\hat{\vec α}\,ψ$,

where $\hat{\vec α}$ is the set of Dirac α-Matrices, composed of the pauli spin matrices like

$\hat α_i = \begin{pmatrix} 0 & \hat σ_i\\ \hat σ_i & 0\end{pmatrix}$.

Why is there no spatial derivative in the probability current, as there is for non-relativistic quantum mechanics? Or am I just failing to see it?

• Why should there be such a derivative in there? Aug 8 '15 at 12:49
• Because there is in the definition of the non-relativistic probability current, defined (for a free spin-0 particle in this case) as $\mathbf j = \frac{\hbar}{2mi}\left(\Psi^* \mathbf \nabla \Psi - \Psi \mathbf \nabla \Psi^{*} \right)$ and because to me the way to derive this as $\partial_t\,ρ$ makes a lot of sense... Aug 8 '15 at 12:55
• You don't define the probability current as that, it's just the conserved current for the $\mathrm{U}(1)$ phase symmetry of quantum mechanics. If you look at the Lagrangian expressions (or the equations of motion) for the scalar and the Dirac spinor, surely you'll notice the difference (second order vs. first order) that leads to the difference in probability current. Aug 8 '15 at 12:55
• So do I get it correctly, that this is not the "full" (spacial) probability current, but just some special prob. current we defined to get a nice continuity equation to work? Aug 8 '15 at 12:58
• I don't know what you mean by "full". The current $j^\mu = \bar\psi \gamma^\mu \psi$ is the probability current (i.e. the conserved current under the phase symmetry of quantum mechanics) for a Dirac spinor. There's nothing special about it. Aug 8 '15 at 13:00

Unlike its non-relativistic counterpart, the Dirac current doesn't need any derivatives in its definition because the Dirac equation is more "beautiful". It represents the charge density and obeys the continuity equation $\partial_\mu j^\mu=0$, anyway.
There is no contradiction with the non-relativistic limit because the Dirac equation basically says that if the Dirac 4-spinor is divided to two 2-spinor parts, $C$ and $D$, then (in a certain basis optimized for non-relativistic physics) the Dirac equation reduces to $$(i\vec \sigma\cdot \nabla) C = (m+i\partial / \partial t) D$$ and a similar equation with $C,D$ basically reverted. You may reconstruct the precise signs and coefficients if you fix a convention. In the non-relativistic limit, the right hand side also basically reduces to $2m\cdot D$ plus small corrections that increase with the velocity.
Consequently, the 2-spinor $D$ is equal to the spatial derivatives of $C$ which is why the bilinear expressions constructed both from $C$ and $D$ – from the whole 4-spinor – de facto depend on the spatial derivatives of $C$, anyway (if you eliminate $D$). This $C$ may then be identified with the 2-spinor that appears in the non-relativistic Pauli equation.
The Dirac equation's ability to reduce the number of derivatives may surprise in other contexts, too. For example, it is a first-order equation – linear in the derivatives or $p_\mu$ – even though it replaces either the Klein-Gordon equation or the non-relativistic Schrödinger's equation, both of which are second-order differential equations (at least second-order in spatial derivatives). Again, this is no contradiction because the Dirac spinor has (at least) twice as many components and one-half of the components end up being approximately equal to the derivatives of the other half. In this way, a first-order equation may become equivalent to a second-order equation. The trick is similar as the trick to rewrite the second-order equations in mechanics, $m\ddot x = F(x)$, as a "Hamiltonian formalism" set of first-order equations for $x$ and $p$.