When we try to construct the relativistic generalization of non-relativistic time dependent Schroedinger equation, there are at least two possible completions - Klein-Gordon equation and Dirac equation. If we keep the single particle interpretation, the Klein-Gordon equation fails because of negative probability density, while Dirac equation does not have this problem, and can be used to describe a relativistic spin 1/2 particle.

Although I understand that single particle approach in relativistic quantum theory is not correct, and when we go to QFT, everything is perfect (probability density is replaced by the charge density, and the later can be positive or negative depending on whether we have particle or antiparticle excitations of the quantum field).

However, still I want to know why in the single particle approach, the case of 1/2 spin works, but the case of 0 spin doesn't. Is there some deep reason, or it is just coincidence and in reality we totally should not think of single particle approach, and using Dirac equation for relativistic quantum mechanics of single electron is meaningless?


  • $\begingroup$ You have the same sorts of problems in single-particle dirac theory: en.wikipedia.org/wiki/Dirac_sea $\endgroup$ – Jerry Schirmer Aug 11 '15 at 17:48
  • 1
    $\begingroup$ @Jerry Schirmer: Dirac theory still suffers from negative energy states, but not negative probability densities. I think the question was why this latter issue can only be mitigated by working with fermions. $\endgroup$ – gj255 Aug 11 '15 at 17:55
  • $\begingroup$ The problem of negative energies appears in both cases, but as I understand it is not really a problem. The modern interpretation of this negative energy states is that they propagate backwards in time, which are equivalent to forward propagating positive energy states - antiparticles. $\endgroup$ – achatrch Aug 11 '15 at 18:06

I don't know whether the OP will be satisfied this answers the original question, but I'd like to offer some context to all this.

A free particle has uniform potential, so without loss of generality $V=0$. The Schrödinger equation then simplifies to $$\dot{\psi}=\frac{i\hbar}{2m}\nabla^2\psi\quad\left(1\right)$$ so $\dot{\rho}=\frac{i\hbar}{2m}\left\{\psi^\ast\nabla^2\psi-\psi\nabla^2\psi^\ast\right\}$. Since probability is conserved, it admits a continuity equation; the probability 3-current $\mathbf{j}$ obeys $\nabla\cdot\mathbf{j}=-\dot{\rho}$. We can choose $\mathbf{j}:=\frac{i\hbar}{2m}\left\{\psi\boldsymbol{\nabla}\psi^\ast-\psi^\ast\boldsymbol{\nabla}\psi\right\}$ (we may add an arbitrary curl to $\mathbf{j}$). If a relativistic 1-particle theory is to work, at the very least a free particle in Minkowski space should be straightforward. In special relativity continuity equations may be written as $\partial_\mu j^\mu=0$. You can check this equation is satisfied by solutions of the mass-$m$ Klein-Gordon equation $$c^{-2}\partial_t^2\psi-\nabla^2\psi+\left(\frac{mc}{\hbar}\right)^2=0,\quad\left(2\right)$$ provided we define $j^\mu\left(\psi\right):=\frac{i\hbar}{2m}\left\{\psi\partial^\mu\psi^\ast-\psi^\ast\partial^\mu\psi\right\}$, which is a natural relativistic upgrade of $\mathbf{j}$. It is therefore natural to suppose a suitable integral of $j^0$ spits out probabilities.

But here we reach a problem. If $\phi,\,\psi$ are equal-mass solutions of the Klein-Gordon equation, we also have a conserved integral called their Klein-Gordon inner product, $$\left\langle\phi,\,\psi\right\rangle_{\text{KG}}:=i\int_{\mathbb{R}^3}\left(\phi^\ast\partial^0\psi-\left(\partial^0\phi^\ast\right)\psi\right)d^3\mathbf{x}.$$ The name is misleading, because this isn't a true inner product; $\left\langle\psi,\,\psi\right\rangle_{\text{KG}}=\frac{2m}{\hbar}\int_{\mathbb{R}^3}j^0\left(\psi\right)d^3\mathbf{x}$ can be negative. Indeed, solutions of Eq. (2) are closed under the operation $\psi\mapsto\psi^\ast$, which multiplies $\left\langle\psi,\,\psi\right\rangle_{\text{KG}}$ by $-1$. Eq. (1) clearly doesn't have an analogous problem (or its usual probability interpretation wouldn't exist). The reason why is that, if we want $\psi\mapsto\psi^\ast$ to send a Schrödinger solution to a Schrödinger solution, we also have to impose $t\mapsto -t$, which also multiplies Klein-Gordon inner products by $-1$.

And the reason why Schrödinger solutions require time reversal and Klein-Gordon solutions don't is because of the parities of the $\partial_t$ exponents. In Schrödinger, the exponent is odd (it's $1$); in Klein-Gordon, the exponent is even (it's $2$). For a classical touchstone, these parities are also why $E=\frac{p^2}{2m}+V$ yields a unique energy but $$E^2=m^2c^4+p^2c^2\,\quad\left(3\right)$$ doesn't.

Nowadays, we know that the way to handle "wrong-sign" solutions of the Klein-Gordon equation is to (i) write solutions as sums of "positive-frequency" and "negative-frequency" parts which interchange under complex conjugation (so the spaces thereof have bases which conjugate to each other's bases) and (ii) say that our space integrals compute differences between numbers of particles and antiparticles.

Let's think now about the Dirac equation. Dirac hoped he could bin negative-energy solutions of Eq. (3) with an equation which, like Schrödinger, was first-order in time. That's how we ended up with $\gamma^\mu\partial_\mu\psi=-im\psi$. This time, to close solutions under $\psi\mapsto\psi^\ast$ we have to append the transformation $x^\mu\mapsto -x^\mu$, which is more than enough to explain the this-time-it-works finding. This time we have not only time reversal but also the spatial equivalent, the parity reversal $\mathbf{x}\mapsto -\mathbf{x}$.

Let's briefly discuss what happens to plane-wave solutions of all three equations. When I conjugate an $\exp i\left(\mathbf{k}\cdot\mathbf{x}-\omega t\right)$ solution (which suffices for the KGE) $\mathbf{k},\,\omega$ change sign. When I then reverse time $\omega$ changes back to its old sign (the Schrödinger requirement), so the only overall change is the sign of $\mathbf{k}$. If I apply a parity transformation as well at the end (Dirac needs this), even this sign change in $\mathbf{k}$ is lost. So actually, our "symmetry" for Dirac solutions does nothing at all!

The last question is what all this has to do with spin and bosons and fermions. The anticommutators $\left\{\gamma^\mu,\,\gamma^\nu\right\}=2\eta^{\mu\nu}$ in $4$-dimensional spacetime require the gamma matrices to be at least $4\times 4$, so the Dirac spinor $\psi$ has at least $4$ components. Dirac realised that the symmetries of the Dirac equation's solutions (not the above "symmetry", some proper ones!) relate these components with a combination of a $2S+1$ spin degeneracy and a matter-antimatter factor of $2$, so $4S+2=4$ and $S=\frac{1}{2}$. Dirac's theory was vindicated not only by predicting the positron, but also by finally explaining spin as a consequence of relativity, whereas before that it was just an empirical fact you had to add to the axioms of quantum mechanics for no apparent reason. This set the stage for later findings concerning spin, such as the spin-statistics theorem. For now, we'll note that a spin-$\tfrac{1}{2}$ Dirac spinor has to be a fermion.

  • $\begingroup$ +1 for the bare $d^3$. It feels, though, that you can go further in addressing the OP. It's it not the case that to get a single-electron Dirac equation you need to discard the position half of the solution space? The question then perhaps becomes less mysterious. I'm any case, it would help to go deeper into the Dirac bilinears. $\endgroup$ – Emilio Pisanty Jan 24 '16 at 21:58

Your questions are answered in detail in the 1st chapter of this book: W. Greiner, Relativistic Quantum Mechanics, http://iate.oac.uncor.edu/~manuel/libros/Modern%20Physics/Quantum%20Mechanics/Relativistic%20Quantum%20Mechanics.%20Wave%20Equations,%203rd%20ed.%20-%20W.%20Greiner.pdf

It treats the Klein-Gordon equation (KGE), its current interpretation, and several related advanced topics in impressive depth.

Short answer to the question: Let $\rho$ be the probability density initially associated to the KGE. The KGE was reinstated as a valid equation for relativistic spin-0 particles when it was realized that $e\rho$ should be identified instead as a charge density, while the negative energy solutions correspond to antiparticles, as for the Dirac equations. This reinterpretation made the KGE instrumental in describing both charged and neutral spin-0 particles. The chapter cited provides several examples concerning the pion triplet, $\pi^0$, $\pi^±$, and much more.

  • $\begingroup$ I don't see how this answers the question - the question is what makes it possible to interpret the solution to the Dirac equation as a probability, in contrast to the KG equation. This answer seems to only say how one removes the apparent problem for the KG equation. $\endgroup$ – ACuriousMind Aug 12 '15 at 1:57
  • $\begingroup$ The actual question stated reads "However, still I want to know why in the single particle approach, the case of 1/2 spin works, but the case of 0 spin doesn't." The answer is: "the case of spin 0 works too, provided it is correctly interpreted". $\endgroup$ – udrv Aug 12 '15 at 2:24
  • $\begingroup$ @ACuriousMind: To answer your question, a commonly cited fundamental reason why the Dirac eq admits a probability density while the KGE does not, is simply that the Dirac eq is 1st order in time, whereas the KGE is 2nd order. Same holds for equations for higher-spin massive particles. Note however that each spinor component in the Dirac eq (and the higher-spin eqs) also satisfies the KGE, while the KGE itself can be cast in 1st-order-in-time "Schroedinger representation". The latter also admits, at least formally, a positive definite conserved quantity, but I've never seen it discussed. $\endgroup$ – udrv Aug 12 '15 at 4:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.