# How can velocity and momentum be in opposite direction for antiparticles as given in the solutions of Klein-Gordon Equation?

This is given in Greiner, Relativistic Quantum Mechanics

For a free particle solution and antiparticle solution with momentum $$\vec{p}$$ the current is given by $$e\frac{c^2\vec{p}}{E_p}$$. The current is same for both particles and antiparticles. But why are the currents for particle and antiparticle with energy $$E_p$$ and momentum $$\vec{p}$$ equal although the charges are different? The reason is that for antiparticles with momentum $$\vec{p}$$ the velocity operator is $$-c^2\vec{p}/E_p$$, i.e. velocity and momentum have opposite directions. One says that the antiparticles move "backwards in time".

My question is: How can velocity and momentum of antiparticles be opposite in direction as given in the paragraph? $$\vec{p}=\gamma m_0\vec{v}$$ where the symbols have their usual meanings. So velocity and momentum should be in the same direction. It also states that the antiparticles move "backward in time". What does it mean? I thought it might be related to Time Reversal but couldn't find anything.

• This does not make much sense. Is this really what Greiner wrote? Aug 19, 2019 at 22:09
• Before this he has proved that the current is equal for both particles and antiparticles when they both have momentum $\vec{p}$. The initial two lines summarize all that. After that I have exactly written as given in the book. Aug 19, 2019 at 22:19
• This is not what the Noether theorem tells you when applied to the Klein-Gordon lagrangian. Aug 19, 2019 at 23:16
• But what he has proven seems correct. I couldn't find any typo or any other mistake. What could be the possible interpretation of what he has proven? Aug 20, 2019 at 7:55
• Oh boy. This is a typical example of why relativistic QM is extremely confusing. I give a treatment of the same problem for Dirac spinors here, maybe you'll find it useful for your scalars. Aug 21, 2019 at 22:34

## 4 Answers

In Klein-Gordon a free particle, for either positive or negative energies, has probability current density $$\vec j = \vec p / E_p$$ ($$\hbar = c = 1$$, natural units). With a normalization that imposes positive and negative charges on positive- and negative-energy solutions, respectively, this is quite surprising. The charge current density is the same regardless of the sign of the charge and energy.

Another difficulty is that while the positive-energy solution is written as $$\Psi_+ \sim exp (-i E_p t + i \vec p \cdot \vec x)$$ for $$E = +E_p$$, the negative-energy solution is written as $$\Psi_- \sim exp (i E_p t + i \vec p \cdot \vec x)$$ for $$E = -E_p$$. The latter expression is not Lorentz invariant.

A way to fix these inconsistencies is to say that the negative-energy particles move backward in time. This not only reverses the sign of $$\vec p$$, but also lets the energy be positive in the exponent of the negative-energy solution.

We can associate the positive-energy solution to a particle and the negative-energy solution to an antiparticle with charges of opposite sign.

Note: If you assume that the antiparticle moves backward in time, velocity and momentum have the same sign.

• The energy is positive for positive and negative frequency alike. Aug 21, 2019 at 23:09
• @my2cts. You misread my post. I stated clearly that the assumption of the move backward in time for the negative-energy solution would cause the energy to be positive also for that solution. I ask you to remove your negative mark. Aug 22, 2019 at 14:39
• The energy is positive, independent of any assumptions. Apart from the normalisation factor N, which should be positive, energy is proportional to $+\omega^2$ and momentum to $+\omega k$, for a plane wave $e^{i\omega t - kx} /N$. For negative $\omega$ the forward moving wave also has momentum proportional to $+\omega k$. Velocity and momentum have the same direction. Aug 27, 2019 at 19:08
• @my2cts. What you say is not how historically the Klein-Gordon was developed. The $E$ which solves the equation is $E = \pm \sqrt{p^2 + m^2}$ in natural units. From that you get an apparently negative energy. Then with some reasoning (refer to my post) you succeed to recover a positive energy. You can find this in any text of relativistic quantum mechanics. Aug 28, 2019 at 7:38
• There are no difficulties in the interpretation of the Klein-Gordon equation. There are only bad textbooks. Aug 28, 2019 at 18:27

Just applying Noether's theorem to the Klein-Gordon lagrangian for a plane wave gives E=$$\omega^2$$, p=$$\omega$$k, q=$$\omega$$ and j=k, apart from a common normalization factor. So E>0 while j and k are parallel for positive and antiparallel for negative charge. Likely, the omission of the factor $$\omega$$ in the Noether momentum is the root cause of the misunderstanding. The issues referred to by the OP do not exist.

I have not read the textbook you describe, but it does seem to have presented this in an unclear way. One thing to note, for example, is that the phrase "moving backwards in time" is rather meaningless, unless of course a text spells out what is meant by it. This is because in spacetime what you have is a worldline. The worldline is a worldline. It is neither "moving forwards in time" nor "moving backwards in time." (I'll return to this, and the arrows on Feynman diagrams, in a moment.) The next reason for the muddle is that historically the Klein-Gordan equation had two incarnations. First it was thought to be a replacement for the Schrodinger equation, before this area became properly understood. But when quantum field theory was developed it turned out that the Schrodinger equation is fine as it stands.

When you look at a vertex on a Feynman diagram, describing, for example, particle--anti-particle annihilation, the lines themselves represent the velocity, via their slope. Regarding momentum, the best advice is to keep clear in your mind that it is conserved at such a vertex, and so is energy, and so is charge. The arrow on the worldline in these diagrams is to help track the conservation of one these quantities: charge. Thus it tells you the charge of that entity, and thus whether we call it a particle or an antiparticle. The momentum is in the same direction as the velocity for both types of particle; the electric current is aligned with the velocity for one type and anti-aligned for the other. The energy is positive for both.

When people talk about negative energy (i.e. negative rest energy) in this context, it seems to be a throw-back to the 'Dirac sea' picture. That picture can remain handy as an alternative way of looking at things, but you don't have to adopt it, and with hindsight perhaps it is better to avoid it in the first instance.

I suspect that if a text is saying that momentum and velocity are in opposite directions for anti-particles they are using the word "velocity" in a non-standard way, trying to make it align with the transport of the conserved quantity (e.g. electric charge, or perhaps baryon number of something like that). If that is what is happening then they should say "current" not "velocity".

The velocity of the antiparticle solution is not in opposite direction to the momentum !!

Take, for instance, the Dirac equation for antiparticles:

$$\gamma^{\mu}p_{\mu} + mc = 0 = i\hslash\gamma^{\mu}\partial_{\mu} + mc$$

or, by simply multiplying all by -1:

$$-\gamma^{\mu}p_{\mu} - mc = 0 = -i\hslash\gamma^{\mu}\partial_{\mu} - mc$$

In fact this last equation is an equation for a positive mass and energy particle in the backwards in time moving universe, since, in the time reversed universe, the total energy and momentum Quantum Mechanical operators are respectively:

$$E_{backInTimeUniverse} = -i\hslash \frac{\partial}{\partial{t}}$$

$$p_{backInTimeUniverse} = i\hslash\nabla$$

Which are symmetrical to the operators in the forward in time moving universe, due to the fact that positive velocities and total energies transform to negative when seen from the perspective of t' = -t, where t is the time parameter for our forward in time universe.

The Dirac equation for antiparticles is exactly like the first equation for particles, when it's seen from the perspective of this backwards in time moving universe.

But then you have the freedom to switch thee sign of this positive mass in the backwards in time universe. When you do that, you have to change the momentum and energy signs in the complex exponential and then change the sign of the momentum again if you want to see the complex exponential from the perspective of the forwards in time moving universe (because momentum operator will switch its sign but the momentum of the observed particle not !, because invariant mass/total energy and velocities will switch signs at the same time when reversing time again!). You don't have to switch the momenta signs in the Dirac spinors, because momenta won't change directions. But you'll have to switch the signs of the invariant mass and total energy in them if you want to express this spinor w.r.t their positive quantities counterparts in our world.

So the momentum of a antiparticle solution is in the same direction of it's velocity because the solution to the second Dirac equation is about a positive energy/invariant mass particle in the backwards in time moving universe. The hability to invert the invariant mass sign gives us the opportunity to have a positive mass and energy object in our universe. The change in sign of the invariant mass and energy and the reversal of time coordinate (changing velocity and invariant mass/total energy signs at the same time) is what gives us velocity and momentum in the same direction when seen from our world.

The same sort of reasoning can be applied to the Klein-Gordon equation for antiparticles, which has as its solution a backwards in time positive mass/total energy particle that can, as well, change its sign of invariant mass and total energy still fulfilling the equation.

I hope I can help you with my answer.

• Hi there! You've made 24 edits to this post over the span of three hours. Every edit bumps a post in the "active" queue and this is distracting for many (especially minor) edits. If you are in the process of revising a post, please try to make each edit substantial instead of submitting each individual correction as its own edit. Jul 14, 2022 at 7:41