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In the LHC they have found a meson(the charm meson) which oscillates between itself and its antiparticle and they have also found there is a difference in mass between the charm meson and the anticharm meson. Where does that difference in mass come from?

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In a neutral meson system, the mass eigenstates are not necessarily the same as the flavor eigenstates. Since both the $D^0$ and the $\bar D^0$ -- flavor eigenstates, as each has well defined valence quark content (e.g. $c \bar u$, $\bar c u$) -- are neutral, a typical such particle exists in a superposition of the two. There is a mass matrix (i.e. operator) in this space, and the crucial point is that is it not diagonal in the flavor basis. Instead, it works something like the following: $$ \left( \begin{array}{cc} M & m \\ m^* & M \end{array} \right) \left( \begin{array}{c}\alpha_{D^0} \\ \alpha_{D^1} \end{array} \right) $$ where the column vector contains the amplitudes of the superposition in the flavor basis. The mass matrix is symmetric are invariant under permutations of the flavor labels, which is required by approximate $CP$ symmetry, and must have real eigenvalues. But $D^0 = (1 0)$ and $\bar D^0 = (01)$ are not the eigenstates. The eigenvalues are given by $$ (M-\lambda)^2 - \left| m \right|^2 = 0 \Rightarrow \lambda = M \pm \left| m \right| $$ The corresponding eigenvectors are denoted $D_H$ and $D_L$. A neutral meson is not generally created in one of these states, though, because the forces that lead to their creation respect various symmetries and instead create flavor eigenstates. But the flavor eigenstates can be written in the mass eigenbasis, e.g. $$ D^0 = \beta D_H + \gamma D_L $$ As this state evolves in time, the unitary evolution operator $e^{-iHt}$ leads to different coefficients for the mass eigenstates, $$ e^{-iHt} D^0 = \beta e^{-i(M+\left|m\right|t)} D_H + \gamma e^{-i(M-\left|m\right|t)} D_L. $$ And of course, that right hand term can be rewritten as a superposition of $D^0$ and $\bar D^0$ states with oscillating, time-dependent amplitudes, which is the genesis of neutral meson oscillation (and which is how these mass differences get measured).

jwimberley
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  • Is it a coincidence all particles (composite or fundamental) which have baryon number of 0 and are electrically neutral have this kind of behavior? – Jun Seo-He Aug 02 '21 at 14:15
  • Comment got deleted accidentally -- it's not true that all such particles have this behavior (e.g. $\gamma$, $J/\psi$), but those that aren't their own antiparticles do. The rule that applies here is "everything that isn't forbidden is permitted" -- there's no symmetry forcing the off-diagonal matrix elements to be zero, so they should have some value, though they could be a lot smaller in magnitude than the diagonal elements. – jwimberley Aug 02 '21 at 14:39
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    This is a nice answer which however does not exactly answer the question itself. I think the question was "why is the mass difference not exactly zero from the physics point of view" rather than how to describe this effect mathematically. Some hint: excited $D^{0}$ and $\overline{D^{0}}$ are flavour eigenstates, but are they mass eigenstates? The role of the weak interaction should be explained. – Martino Aug 05 '21 at 11:25
  • @Martino Ah, that's a good point, I interpreted it probably in a different way than was intended (i.e., why mass differences are permitted without violating symmetries). – jwimberley Aug 05 '21 at 17:10