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Usually we say that equality of masses of particle and antiparticle follows from CPT-theorem. But do we need it for showing this equality?

The first method to show that is following.

  1. The equation of free fields of an arbitrary spin $s$: all of them have Klein-Gordon or Dirac form (with some conditions of irreducibility), so the general solution is given in a form $$ \hat {\Psi}_{A}(x) = \sum_{\sigma}\int \frac{d^{3}\mathbf p}{\sqrt{(2 \pi)^{3}2p_{0}}}\left(u^{\sigma}_{A}(\mathbf p)e^{-ipx}\hat {a}_{\sigma}(\mathbf p) + v^{\sigma}_{A}(\mathbf p)e^{ipx}\hat {b}^{\dagger}_{\sigma}(\mathbf p) \right)_{p_{0} = \sqrt{\mathbf p^{2} + m^{2}}} \qquad (1) $$

  2. We must write $\hat {b}^{\dagger}_{\sigma}(\mathbf p)$, not $\hat {a}^{\dagger}_{\sigma}(\mathbf p)$, because of existence of some internal symmetries different from Poincare symmetry. If there exist some symmetry, we have conserving quantity $q$ for field, it means that corresponding operator $\hat {Q}$ commutes with hamiltonian of theory. We construct hamiltonian by having the equation of motion and its physical sense as polynomial of $\hat {\Psi}_{A}(x), \hat {\Psi}^{\dagger}_{B}(x)$ and spinor functions. It leads to the fact that $$ \hat {Q}\hat {a}^{\dagger}(\mathbf p)| \rangle = q\hat {a}^{\dagger}(\mathbf p)| \rangle, \quad \hat {Q}\hat {b}^{\dagger}(\mathbf p)| \rangle = -q\hat {b}^{\dagger}(\mathbf p)| \rangle . $$

The second method is following.

  1. Suppose we don't have the equations for fields. But we know that S-operator must be Lorentz invariant operator or that our theory must be causal theory. S-operator is constructed from hamiltonian. Hamiltonian is constructed from creation and destruction operators and must be poincare scalar. So we must create invariant combination of fock space operators and some non-operator functions. Corresponding object is called creation/destruction field. By having laws of poincare transformation of creation and destruction operators we build the expression $(1)$ for field with one little difference: the first and second summands corresponds, in general, to the different masses. Then we construct the hamiltonian as polynomial of combinations of field functions and some spinor functions.

  2. The statement of paragraph 1 leads us to the conclusion that $$ [\hat {\Psi}_{A}(x), \hat {\Psi}^{\dagger}_{B}(y)]_{\pm} = 0 , \quad (x - y)^{2} < 0. $$ With a bit of derivations we can get $$ [\hat {\Psi}_{A}(x), \hat {\Psi}^{\dagger}_{B}(y)]_{\pm} = P_{AB}\left(i\frac{\partial}{\partial x}\right)\left( D^{m_{1}}_{0}(x - y)\pm (-1)^{s}D^{m_{2}}_{0}(y - x)\right), \qquad (2) $$ where $$ D^{m_{i}}_{0}(x - y) = \int \frac{d^{3}\mathbf p}{(2 \pi )^{3}2p_{0}}e^{-ip(x - y)}, \quad p^{2} = m_{i}^{2}. $$

  3. We conclude that $(3)$ may be equal to sero for spacelike intervals if and only if $m_{1} = m_{2}$.

Are these methods of demonstrating of the identity of particle-antiparticle masses correct?

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  • $\begingroup$ Why do you think they might not be correct? $\endgroup$ – David Z Apr 26 '14 at 20:15
  • $\begingroup$ @DavidZ : because I din't meet them in standart lecture courses or in books. $\endgroup$ – Andrew McAddams Apr 26 '14 at 20:17
  • $\begingroup$ That's not a reason to think they're incorrect, though. Is there some step you don't understand, or some inconsistency between the results, or something like that? That would be a better reason to ask a question. $\endgroup$ – David Z Apr 26 '14 at 20:20
  • $\begingroup$ @DavidZ: they might consist of some incorrect or unstrictly assumptions. I don't know. $\endgroup$ – Andrew McAddams Apr 26 '14 at 20:22
  • $\begingroup$ OK, so what assumptions are made that you think might be incorrect? That's what I'm getting at. $\endgroup$ – David Z Apr 26 '14 at 20:25
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I don't see anything obviously wrong with your derivations either.

My only comment is that you are using i)Lorentz invariance as an axiom and I am pretty sure you are also implicitly using ii)locality and iii)hermitian hamiltonians.

Now i),ii) and iii) are equivalent to CPT invariance so to me it looks that you are not providing an inherently different proof but rather what you are doing is equivalent to the proof that uses the CPT theorem.

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