I'm reading the Wikipedia page for the Dirac equation:



$J = -\frac{i\hbar}{2m}(\phi^*\nabla\phi - \phi\nabla\phi^*)$

with the conservation of probability current and density following from the Schrödinger equation:

$\nabla\cdot J + \frac{\partial\rho}{\partial t} = 0.$

The fact that the density is positive definite and convected according to this continuity equation, implies that we may integrate the density over a certain domain and set the total to 1, and this condition will be maintained by the conservation law. A proper relativistic theory with a probability density current must also share this feature. Now, if we wish to maintain the notion of a convected density, then we must generalize the Schrödinger expression of the density and current so that the space and time derivatives again enter symmetrically in relation to the scalar wave function. We are allowed to keep the Schrödinger expression for the current, but must replace by probability density by the symmetrically formed expression

$\rho = \frac{i\hbar}{2m}(\psi^*\partial_t\psi - \psi\partial_t\psi^*).$

which now becomes the 4th component of a space-time vector, and the entire 4-current density has the relativistically covariant expression

$J^\mu = \frac{i\hbar}{2m}(\psi^*\partial^\mu\psi - \psi\partial^\mu\psi^*)$

The continuity equation is as before. Everything is compatible with relativity now, but we see immediately that the expression for the density is no longer positive definite - the initial values of both ψ and ∂tψ may be freely chosen, and the density may thus become negative, something that is impossible for a legitimate probability density. Thus we cannot get a simple generalization of the Schrödinger equation under the naive assumption that the wave function is a relativistic scalar, and the equation it satisfies, second order in time.

I am not sure how one gets a new $\rho$ and $J^\mu$. How does one do to derive these two? And can anyone show me why the expression for density not positive definite?

  • $\begingroup$ any comment...? $\endgroup$ Oct 7, 2012 at 6:41
  • $\begingroup$ please see below, hope that helps $\endgroup$
    – user11547
    Oct 7, 2012 at 17:17

1 Answer 1



This particular writing of the problem in the article I have always thought was sloppy as well. The most confusing part of the discussion is the statement "The continuity equation is as before". At first one writes the continuity equation as:

$$\nabla \cdot J + \dfrac{\partial\rho}{\partial t} = 0$$

Although the del operator can be defined to be infinite dimensional, it is frequently reserved for three dimensions and so the construction of the sentence does not provide a clear interpretation. If you look up conserved current you find the 4-vector version of the continuity equation:

$$\partial_\mu j^\mu = 0$$

What is important about the derivation in the wikipedia article is the conversion of the non time dependent density to a time dependent density, or rather:

$$\rho = \phi^*\phi$$


$$\rho = \dfrac{i\hbar}{2m}(\psi^*\partial_t\psi - \psi\partial_t\psi^*)$$

the intent is clear, the want to make the time component have the same form as the space components. The equation of the current is now:

$$J^\mu = \dfrac{i\hbar}{2m}(\psi^*\partial^\mu\psi - \psi\partial^\mu\psi^*)$$

which now contains the time component. So the continuity equation that should be used is:

$$\partial_\mu J^\mu = 0$$

where the capitalization of $J$ appears to be arbitrary choice in the derivation.

One can verify that this is the intent by referring to the article on probability current.

From the above I can see that the sudden insertion of the statement that one can arbitrarily pick $$\psi$$ and $$\dfrac{\partial \psi}{\partial t}$$ isn't well explained. This part the article was a source of confusion for me as well until one realized that the author was trying to get to a discussion about the Klein Gordon equation

A quick search of web for "probability current and klein gordan equation" finds good links, including a good one from the physics department at UC Davis. If you follow the discussion in the paper you can see it confirms that the argument is really trying to get to a discussion about the Klein Gordon equation and make the connection to probability density.

Now, if one does another quick search for "negative solutions to the klein gordan equation" one can find a nice paper from the physics department of the Ohio University. There we get some good discussion around equation 3.13 in the paper which reiterates that, when we redefined the density we introduced some additional variability. So the equation:

$$\rho = \dfrac{i\hbar}{2mc^2}(\psi^*\partial_t\psi - \psi\partial_t\psi^*)$$

(where in the orginal, c was set at 1) really is at the root of the problem (confirming the intent in the original article). However, it probably still doesn't satisfy the question,

"can anyone show me why the expression for density not positive definite?",

but if one goes on a little shopping spree you can find the book Quantum Field Theory Demystified by David McMahon (and there are some free downloads out there, but I won't link to them out of respect for the author), and if you go to pg 116 you will find the discussion:

Remembering the free particle solution $$\varphi(\vec{x},t) = e^{-ip\cdot x} = e^{-i(Et- px)}$$ the time derivatives are $$\dfrac{\partial\varphi}{\partial t} = -iEe^{-i(Et- px)}$$ $$\dfrac{\partial\varphi^*}{\partial t} = iEe^{i(Et- px)}$$ We have $$\varphi^*\dfrac{\partial\varphi}{\partial t} = e^{i(Et- px)}[-iEe^{-i(Et- px)}] = -iE$$ $$\varphi\dfrac{\partial\varphi^*}{\partial t} = e^{-i(Et- px)}[iEe^{i(Et- px)}] = iE$$ So the probability density is $$\rho = i(\varphi^*\dfrac{\partial\varphi}{\partial t} - \varphi\dfrac{\partial\varphi^*}{\partial t}) = i(-iE-iE) = 2E$$ Looks good so far-except for those pesky negative energy solutions. Remember that $$E = \pm\sqrt{p^2+m^2}$$ In the case of the negative energy solution $$\rho = 2E =-2\sqrt{p^2+m^2}<0$$ which is a negative probability density, something which simply does not make sense.

Hopefully that helps, the notion of a negative probability does not make sense because we define probability on the interval [0,1], so by definition negative probabilities have no meaning. This point is sometimes lost on people when they try to make sense of things, but logically any discussion of negative probabilities is non-sense. This is why QFT ended up reinterpreting the Klein Gordan equation and re purposing it for an equation that governs creation and annihilation operators.

  • $\begingroup$ With respect to McMahon's books, please see the cooperative effort to make errata sheets here $\endgroup$ Jan 31, 2013 at 16:24

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.