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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?

trimitri
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1 Answers1

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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 $p$.

Luboš Motl
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