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additive information : $B^0$ and $\bar{B}^0$ don't have the same mass, $K^0$ and $\bar{K}^0$ don't have the same mass. is there energy violation? if yes, is the key explanation of this coming from the relationship $\Delta E*\Delta t \sim \hbar$ :that is it authorized to have energy conservation if this is during a small duration?

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    $\begingroup$ They do have the same mass (or expectation value of the mass, since they are not quite energy eigenstates). This is guaranteed by CPT. $\endgroup$ – Buzz May 8 at 16:25
  • $\begingroup$ Are you sure ? For example : en.wikipedia.org/wiki/B_meson This shows that there is Delta m_d between B0 and B0bar, so there is a difference of mass, so they don't have the same mass. But I'm not sure. Maybe Delta m_d is actually the mass difference between B_L and B_H ?!? Do you mean that B0 and B0 have exactly the same mass ? $\endgroup$ – Mathieu Krisztian May 8 at 19:16
  • $\begingroup$ The mass difference are between the weak eigenstates. Ignoring the small CP violation, the mass eigenstates are symmetric and antisymmetric linear combinations of particle and antiparticle states, like K_S and K_L. $\endgroup$ – Buzz May 8 at 20:06
  • $\begingroup$ @Buzz That seems like an answer, not a comment. $\endgroup$ – rob May 8 at 20:12
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    $\begingroup$ all right. Thank you very much for your kind explanations. So the problem is solved : my question is wrong. $\endgroup$ – Mathieu Krisztian May 8 at 20:36
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Actually, the flavor eigenstates (such as $K^{0}$ and $\bar{K}^{0}$, which are eigenstates of strangeness and isospin) do have the same mass. This is actually guaranteed by CPT invariance. (More precisely, they have the equal expectation values for their the masses, since they are not quite energy eigenstates.)

The actual mass eigenstates are the weak eigenstates. Ignoring the small CP violation, these mass eigenstates are symmetric and antisymmetric linear combinations of particle and antiparticle states, $K_{S}$ and $K_{L}$, which are approximately $\frac{1}{\sqrt{2}}\left(\left|K^{0}\right\rangle\pm\left|\bar{K}^{0}\right\rangle\right)$.

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