I Antiparticles and Einstein's energy-momentum relation

1. Aug 10, 2016

haushofer

Dear all,

in a lot of undergraduate textbooks you find the claim that antiparticles can be motivated by Einstein's energy-momentum relation $E^2 = p^2 + m^2$, which has both 'negative' and 'positive energy' solutions. In the context of a single wave function this is problematic. In the context of quantum fields however, I thought that these 'negative' energy solutions correspond to those modes of the field which annihilate a particle with that specific energy (which is positive). So how is this related to the concept of antiparticles?

The simplest example I can think of is the real scalar field, as treated in e.g. Peskin&Schroeder. You make the usual expansion in plane waves and find that the 'negative energy solutions' have as a coefficient the operator which annihilates the particle, where all particles have positive energy. No antiparticles here. So what's the precise relation between the existence of antiparticles and the quadratic nature of Einstein's $E^2 = p^2 + m^2$? I'd say this quadratic nature is linked to the creation and annihilation of particles, and antiparticles arise as soon as you consider complex conjugates of the fields.

Last edited: Aug 10, 2016
2. Aug 10, 2016

strangerep

Afaict, the Poincare group alone is not enough. One must also put in causality by hand, which is essentially how Weinberg develops the theory in his QToF vol 1: by constructing causal reps. Electric charge is modeled with the help of complex fields, and one must consider CPT symmetry, i.e., how those discrete symmetry operators act on the various c/a operators...

HTH.

3. Aug 11, 2016

haushofer

Well, I understand that when you complex-conjugate a plane wave with a factor -iEt, you get effectively the other sign for your energy, but the argument as presented ("the quadratic relation has two solutions, hence antiparticles") doesn't make sense to me; if you take a real scalar field or a Majorana spinor you don't have antiparticles in your field excitations but interpret the "negative energy"-terms in your expansion in combination with a destruction operator, hence as "destroying a real particle".

It's basically the first few pages of Peskin&Schroeder, but I guess I have found another undergraduate explanation which is confusing by the route of history.

4. Aug 11, 2016

vanhees71

In addition you also need to assume that you want local realizations of the Poincare group, which leads finally to the characteristic free-field operators as they appear in the interaction representation, consisting of a part with annihilation operators in front of the positive-frequency (not energy!) modes and a part with creation operators in front of the negative-frequency (not energy!) modes. This leads to particles and antiparticles which both have positive energy as it must be for a theory that has a stable ground state, i.e., a Hamiltonian bounded from below. This "Feynman-Stueckelberg trick" of putting a creation operator in front of the negative-energy states and have a "superposition" of both annihilation and creation operators for the quantum field makes the hand-waving argument, which of course in itself doesn't make sense, more reliable. The take-home message is that a first-quantization relativistic QT leads to contradictions, which have to be cured in some way by reinterpreting such an attempt in terms of a many-body theory. This has become clear with Dirac's "hole theory", which is a very complicated version of QED, which is nevertheless equivalent to the modern field-theoretical approach.

5. Aug 12, 2016

haushofer

The confusion I'm having is that already for a real scalar field you have these positive and negative frequency modes and you interpret the negative ones as destroying particles, not antiparticles! So I'd say the reason why one has antiparticles in the first place is complex conjugation, not the quadratic nature of Einstein's energy relation. And then of course in complex conjugation the sign of the frequencies flip.

So what is the logic here?

6. Aug 12, 2016

vanhees71

If you quantize a real scalar field, you have the constraint that the corresponding field operators of the quantized theory should be hermitean. This means that the creation operator (in front of the negative-frequency modes) must be the hermitean adjoint of the annihilation operator (in front of the positive-frequency modes) in the mode decomposition of the (free) field. As it turns out for this field there's no conserved charge, i.e., you describe uncharged particles. In this case the particles and antiparticles are indistinguishable. An example is the neutral pion. Those particles are called "strictly neutral".

Note that only particles which do not carry any charge-like quantum number are strictly neutral. It's not enough to be just electrically uncharged. An example are the neutral kaons, which both carry electric charge 0 but strangeness -1 (kaon) or +1 (anti-kaon). That's an interesting case, because with such a system you have the chance to discover CP violation, and indeed CP is violated by the weak interaction, and it was discovered by Cronin and Fitch in the 1960ies that indeed the $\mathrm{K}^0$-$\overline{\mathrm{K}}^0$ system is not CP symmetric. For more information, see

https://en.wikipedia.org/wiki/CP_violation
https://en.wikipedia.org/wiki/Kaon#Neutral_kaon_mixing

7. Aug 13, 2016

haushofer

So doesn't that indicate that the quadratic nature of Einstein's energy-momentum relation is not the reason for the appearance of antiparticle states?

8. Aug 13, 2016

haushofer

Now I think of it, my logic would also allow for antiparticle states in the non-relativistic case.

Well, you have noticed, I am confused. :P maybe i should back to Peskin to check the second quantization in full detail.

9. Aug 13, 2016

vanhees71

In non-relativistic theory the free single-particle energy is $E_{\vec{p}}=\frac{\vec{p}^2}{2m}$, and you have no quibbles with the Hamiltonian to be bounded from below. In 2nd quantization the energy of a many-body system of non-interacting particles is given by
$$\hat{H}=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p} \hat{\psi}^{\dagger} \left (-\frac{\Delta}{2m} \right) \hat{\psi}.$$
From the single-particle momentum-spin eigenstates you can define the annihilation (creation) operators $\hat{a}(\vec{p},\sigma)$ ($\hat{a}^{\dagger}(\vec{p},\sigma)$) with the field operator given with annihilation operators only (in the Heisenberg picture):
$$\hat{\psi}(t,\vec{x})=\int_{\mathbb{R}^3} \frac{\mathrm{d}^3 \vec{p}}{(2 \pi)^{3/2}} \sum_{\sigma=-s}^s \hat{a}(\vec{p},\sigma) \exp(-\mathrm{i} E_{\vec{p}}+\mathrm{i} \vec{p} \cdot \vec{x})\chi_{\sigma}.$$
Note that $\hat{\psi}$ is an operator valued spinor with $(2s+1)$ components. The annihilation and creation operators fulfill the commutator (bosons) or anticommutator (fermions) relations,
$$[\hat{a}(\vec{p},\sigma),\hat{a}^{\dagger}(\vec{p}',\sigma')]_{\mp}=\delta^{(3)}(\vec{p}-\vec{p}') \delta_{\sigma,\sigma'}.$$
The total number of particles
$$\hat{N}=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p} \sum_{\sigma=-s}^s \hat{a}^{\dagger}(\vec{p},\sigma)\hat{a}(\vec{p},\sigma)$$
is conserved even for many cases of interacting particles, e.g., if you have only two-body conservative forces. Then you have
$$\hat{H}_{\text{int}}=\frac{1}{2} \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x}_1 \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x}_2 \sum_{\sigma_1,\sigma_2} V(|\vec{x}_1-\vec{x}_2,\sigma_1,\sigma_2) \hat{\psi}^{\dagger}(\vec{x}_2,\sigma_2) \hat{\psi}^{\dagger}(\vec{x}_1,\sigma_1) \hat{\psi}(\vec{x}_1,\sigma_1) \hat{\psi}(\vec{x}_2,\sigma_2).$$
As you see you have as many annilation as creation operators and thus the interaction doesn't change the total number of particles, i.e., if you are initially in a Fock state (i.e., an eigenstates of the number operator) the particle number is fixed at this number for all time. So there is no creation and annihilation of particles going on in such a model.

In the relativistic case, it is impossible to find local interactions that don't change the particle number, i.e., you always can create and annihilate particles in collisions.

10. Aug 13, 2016

haushofer

As a side note, completely off the topic: while I was going through your collection of (excellent looking!) lecture notes on your website, I saw that your "Statistical physics " notes are in English while the link has the comment "(in German)". Thanks again!

11. Aug 13, 2016

vanhees71

Thanks for the hint. I corrected it.

12. Aug 13, 2016

haushofer

Just curious: imagine an undergraduate brings his/her copy of e.g. Perkin's "Particle AstroPhysics", where in section 1.4 it says (after discussing the rel. expresion for energy) "the negative energies are formally connected with the existence of a positive energy anti-particle", and he/she asks you whether that's true. How would you respond?

I would respond "no, it corresponds in th field formalism to the annihilation of a particle (not antiparticle!) state". Am I right?

13. Aug 13, 2016

dextercioby

The concept of antiparticle is well accommodated by Klein-Gordon and Dirac's theories, even in the absence of a quantum field interpretation of the mathematical functions at hand, therefore you don't need quantum fields to speak about the charge conjugation operator acting on specially relativistic wavefunctions.

14. Aug 14, 2016

vanhees71

I would respond that the book is basically right but inaccurate. The point is that the solution of the relativistic wave equations for free particles has, for a given momentum, two frequencies, one positive and one negative $\omega=\pm \sqrt{\vec{p}^2+m^2}$. If you interpret both of these frequencies as energies of particles you have a problem, because your Hamiltonian wouldn't be bounded from below, and that would imply that there's no stable ground state in contradiction to the observation that matter is a pretty stable business.

The way out of this dilemma is to reinterpret the plane-wave modes in terms of a quantum field theory taking into account that in the relativistic realm it is always possible that particles get produced and/or destroyed in collisions. Since I further want the field operators to transform under Lorentz transformations as their classical pendants (i.e., in a local way), on the other hand one necessarily needs both modes, with positive and negative frequencies. For quantum fields, it's however no problem to write a creation operator in front of the negative-frequency modes and flip the momentum in the corresponding integral over momenta. This means that you have now reinterpreted the negative-frequency modes as one-particle states with positive energy, and the Hamiltonian gets bounded from below (after normal ordering!), provided you quantize half-integer-spin particles as fermions and integer-spin particles as bosons. This is the famous spin-statistics theorem.

15. Aug 14, 2016

haushofer

Ok, so I am wrong then. You say you interpret the neg.freq. modes as creating positive energy states, which correspond to the Antiparticles? I thought these modes destroy ordinary particles. But clearly I'm still missing something basic. As i said, i need to take a closer look at the details, but i really appreciate your help with this.

16. Aug 14, 2016

haushofer

So that suggests it is intimately connected to the energy relation.

17. Aug 14, 2016

vanhees71

Take again the free scalar field (Klein-Gordon field). The plane-wave solutions are
$$u_{\vec{p}}^{(\pm)}(x)=\frac{1}{\sqrt{(2 \pi)^3 2 E_{\vec{p}}}} \exp(\mp \mathrm{i} E_{\vec{p}} t+\mathrm{i} \vec{x} \cdot \vec{p}).$$
The general solution of the classical field is thus
$$\phi(t,\vec{x})=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p} [A(\vec{p})u_{\vec{p}}^{(+)}(x)+B(\vec{p}) u_{\vec{p}}^{(-)}(x)].$$
Now when quantizing the field, you can write for $A$ an annihilation and for $B$ a creation operator. In order to make the solution also easier to handle concerning Lorentz covariance you also flip the sign of $\vec{p}$ in the negative-frequency modes. Then you have
$$\hat{\phi}(t,\vec{x}) = \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p} [\hat{a}(\vec{p})u_{\vec{p}}^{(+)}(x)+\hat{b}(\vec{p}) u_{\vec{p}}^{(+)*}(x)].$$
Doing the analysis with the Poincare generators (Noether's theorem leading to the expressions for energy, momentum, and angular momentum) and the operators for the U(1) global gauge symmetry (invariance under multiplication with a phase factor), leading to a conserved current leads to the conclusion that the KG field must be quantized as bosons in order to get a positive Hamiltonian (after normal ordering to get rid of all irrelevant divergences).

The $\hat{a}(\vec{p})$ and $\hat{b}(\vec{p})$ annihilate two different kinds of particles, which are however closely related, because they both are necessary to create a theory, where the field operator behaves under Poincare transformations as the corresponding classical field, i.e., in a local way. It turns out that both particles have the same mass but opposite charge (of the conserved U(1) current). One calls such pairs of particles particle and antiparticle. You can also define a new kind of discrete symmetry (on top of the parity (spatial reflections) and time-reversal (time reflections) of the full Poincare group), namely you can exchange particles and anti particles by exchanging $\hat{a}$ and $\hat{b}$. That's called charge-conjugation invariance, and as has been shown by Pauli and Lüders for the first time, one can show that for all relativistic QFTs of this kind (local relativistic QFT) is not only invariant under the proper orthochronous Poincare group but necessarily also under the transformation $\hat{C} \hat{P} \hat{T}$. There's no necessity for any of the other discrete transformations to be a symmetry, and indeed the weak interaction breaks parity, T, and CP (all directly proven experimentally in various experiments concerning cases as the neutral-kaon and -B-meson system).

Of course, you can also have real classical and hermitean quantum KG fields. Then you have $\hat{a}=\hat{b}$. Then you have no conserved current anymore, because you have no U(1) symmetry. This means that such a field describes strictly neutral particles, and obviously here the particle is identical with its anti-particle. An example is the photon as the quantum of the electromagnetic field (which has its own mathematical quibbles since it's a massless spin-1 field).

18. Aug 15, 2016

haushofer

Thank you for your patience. I think a keypoint is the flipping of the momentum, because I see that in your expansion of phi it allows you to get annihilation operators of particle and antiparticle states, whereas I'm considering phi as an expansion of creation and annihilation of just one type of particle.

19. Aug 15, 2016

vanhees71

As I said, if you consider it as creation and annihilation of one type of particle, you have the special case of a strictly neutral particle, for which by definition particle and antiparticle are the same. This implies that there's no conserved charge-like quantum number, i.e., all possible charges within a larger theory with particles which have some are vanishing. An example is the photon. Take simple QED with just a Dirac particle (which has distinguishable particles and antiparticles, e.g., electrons and positrons) and photons. The photon is strictly neutral and thus does not carry electric charge.

20. Aug 15, 2016

haushofer

So in general e.g. phi creates particles and annihilates antiparticles, while its complex conjugate creates antiparticles and annihilates particles?