# Equality of masses of particle and antiparticle

+ 0 like - 0 dislike
269 views

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?

This post imported from StackExchange Physics at 2014-05-04 11:35 (UCT), posted by SE-user Andrew McAddams
retagged May 4, 2014
Why do you think they might not be correct?

This post imported from StackExchange Physics at 2014-05-04 11:35 (UCT), posted by SE-user David Z
@DavidZ : because I din't meet them in standart lecture courses or in books.

This post imported from StackExchange Physics at 2014-05-04 11:35 (UCT), posted by SE-user Andrew McAddams
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.

This post imported from StackExchange Physics at 2014-05-04 11:35 (UCT), posted by SE-user David Z
@DavidZ: they might consist of some incorrect or unstrictly assumptions. I don't know.

This post imported from StackExchange Physics at 2014-05-04 11:35 (UCT), posted by SE-user Andrew McAddams
OK, so what assumptions are made that you think might be incorrect? That's what I'm getting at.

This post imported from StackExchange Physics at 2014-05-04 11:35 (UCT), posted by SE-user David Z

+ 3 like - 0 dislike

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.

This post imported from StackExchange Physics at 2014-05-04 11:35 (UCT), posted by SE-user Heterotic
answered Apr 27, 2014 by (525 points)

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysics$\varnothing$verflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.