# A Probability density of dirac spinors

1. Aug 11, 2017

### Josh1079

The probability density of the dirac spinor is known to be ∑(Ψ)2 and I know how it is derived. However, I'm just wondering why it should be positive definite. Since the lower two components represent antiparticles, so shouldn't the probability density contribution of those two components be negative? Or actually what does that probability density represent? Thanks!

2. Aug 11, 2017

### vanhees71

I'm not sure about your notation. In fact the Dirac-field operators are fermionic operators, and that's why the conserved current due to invariance of the action under global gauge transformations $\psi \rightarrow \exp(\mathrm{i} \alpha)$, $\bar{\psi} \rightarrow \exp(-\mathrm{i} \alpha) \bar{\psi}$, defined by the Lagrangian
$$\mathcal{L}=\bar{\psi}(\mathrm{i} \partial_{\mu} \gamma^{\mu}-m) \psi$$
for the free field, is
$$j^{\mu} = \overline{\psi} \gamma^{\mu} \psi.$$
In order to make this finite in the quantized version you have to normal order this expression, and due to the fermionic nature of the fields you get positive (negative) contributions to the density $\rho=j^{0}$ from particles (antiparticles). Thus, here the conserved Noether current can not be interpreted as a probability current, as in the case of Schrödinger fields, because $\rho$ is not positive definite.

3. Aug 11, 2017

### Josh1079

Hi vanhees, thanks for the reply but sorry I think I don't really get your point. What I mean is essentially this:

I see quite a few textbooks interpret this as the probability density, and I didn't really thought about the lagrangian.

4. Aug 11, 2017

### vanhees71

In which textbook have you found this? It's highly misleading at best!

There is no proper interpretation of a 1st-quantization approach to relativistic QT or only in the non-relativistic limit. The reason is that at relativistic energies you have always the possibility of creating and destroying particles, and that's why relativistic QT is formulated as QFT, which is the most convenient way to describe situations, where the particle number is not conserved, and in my opinion that's the way relativistic QT should be introduced in the 21st century!

5. Aug 11, 2017

### Josh1079

Thanks for the clarification vanhees. So I guess you mean, combined with your previous post, the conserved current represents the density and current of some charge where antiparticles will give negative contribution?

6. Aug 11, 2017

### vanhees71

Yes, the most simple example is QED. You just minimally couple the electromagnetic field to it, and then the above derived conserved current is interpreted as the electromagnetic current:
$$\mathcal{L}=-\frac{1}{4} F_{\mu \nu} F^{\mu \nu} + \bar{\psi}(\mathrm{i} \gamma^{\mu} \partial_{\mu} + e \gamma^{\mu} A_{\mu}) \psi,$$
where
$$F_{\mu \nu}=\partial_{\mu} A_{\nu} -\partial_{\nu} A_{\mu}.$$
Note that now the action is invariant under local gauge transformations!

Last edited: Sep 9, 2018
7. Aug 11, 2017

### Josh1079

Thanks vanhee! Your replies are very insightful!

8. Sep 7, 2018

### gerald V

Recently, I faced the same confusion as Josh1079 did. But, in contrast to him, I haven’t yet got it after reading this post. Please forgive me for the slow-wittedness.

Let’s say the particles are at rest, so the conserved current is just its time component. For charged scalars I have understood the conserved current to measure charge (probability) density, so when integrated over 3-space one arrives at the total charge.

Now, for the Dirac equation this appears to be essentially the same. Bjorken-Drell II, equation (13.49) says that $\int \mbox{d}^3x \psi^\dagger \psi$ is the total charge. But a pile of literature (Ryder „Quantum Field Theory, Holland „The Quantum Theory of Motion", the website https://quantummechanics.ucsd.edu/ph130a/130_notes/node488.html and many more) point out that $\psi^\dagger \psi$ is positive definite. I am aware of negative energy solutions and so on, but this does not lift my confusion. I can neither see that particle-antiparticle generation plays a role, since they appear in pairs as it is necessary to conserve the charge.

Does positivity of $\psi^\dagger \psi$ mean that an electron and a positron both have positive current? How then can the current measure charge (probability) density? If it does not measure charge (probability) density, why is its spatial integral the total charge? Or what else does it measure? It is said „probability“ per se, but probability of what? Or is there simply a confusion w.r.t.\ notation? As I understand from the literature $\psi^\dagger$ is the conjugate transpose (synonym for Hermetian transpose) of $\psi$.

Thank you very much in advance.

9. Sep 9, 2018

### king vitamin

I believe that both @Josh1079 and @gerald V's confusion comes from textbooks which are treating the "Dirac equation" as an equation for a single-particle wave function. This is often done without cautioning issues which this approximation makes. In particular, particle number is not conserved, so a "one-particle" treatment is clearly not possible. Indeed, even if you only deal with energies much less than $mc^2$, and even if you start with a positive-energy single-particle state, if you time-evolve a solution to the Dirac equation long enough you'll get admixtures of states with negative energy, and probability won't stay conserved.

I gave a discussion of how this "one-particle approximation" is constructed, and where it breaks down, here (following the discussion from Merzbacher's textbook).

Last edited: Sep 9, 2018
10. Sep 10, 2018

### gerald V

Thank you very much. It could well be that I mess up aspects of first an second quantization, and some more.

My confusion rather emerges since I do NOT regard the relativistic wave equations as single-particle equations. For a single particle I could understand that the current is semidefinite.

Bjorken-Drell II is on quantum field theory. For complex scalars, its equations 12.63 to 12.64 give the total conserved charge as
$Q = \imath \int \mbox{d}^3x (\varphi^\ast \dot{\varphi} - \varphi \dot{\varphi}^\ast) = \int \mbox{d}^3k [a_+^\dagger(k) a_+(k) - a_-^\dagger(k)a_-(k)] = \int \mbox{d}^3k [N_+(k) - N_-(k)]$, where $N$ are the number operators. So positively charged quanta count positive for the current, while negatively charged quanta count negative. Fairly logic.

I would think that there is a corresponding relation for Fermions. $N_-(k)$, which can take values 0 or 1 is the number of electrons (positive energy solutions of the Dirac equation) with 3-momentum $k$, while $N_+(k)$ is the corresponding number of positrons (negative energy Solutions). Reading above equation backwards, in position space there should be positive and negative contributions to the current, maybe after removing the infinite charge of the Dirac sea.

11. Sep 10, 2018

### gerald V

I've got it!

Itzikson-Zuber puts the facts in didactically perfect words: "In contradistinction to the 'classical' case where we attempted to interpret ….. as a positive square norm it is seen that in the quantum case this is not so. The use of anticommutators has totally reversed the situation; energy is now positive and charge is an indefinite quantity."

I have known fermionic QFT, but obviously did not bother enough about the 'classical' Dirac equation. Only now when reading Holland's book on the deBroglie-Bohm guidance wave theory it crossed my way. If anyone has an opinion on the guidance wave theory in connection with the Dirac equation, I would be glad to hear.