I Antiproton and positron annihilation

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To make antihydrogen (an antiproton bound to a positron) requires supplies of positrons and antiprotons that are moving very slowly (enough to combine into atoms). I am wondering does someone have knowledge how much energy is released if antiproton annihilate itself with positron? What would be the outcome of this annihilation?
 
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Sasho Andonov said:
I am wondering does someone have knowledge how much energy is released if antiproton annihilate itself with positron? What would be the outcome of this annihilation?
An antiproton can't annihilate itself with a positron. They are not each other's antiparticle. An antiproton can "capture" a positron to form a neutral antihydrogen anti-atom. The energy released is exactly the same as for a proton "capturing" an electron: the ionization energy (absolute value) plus the free electron's excess (depends on the coordinate frame of reference).

In most cases, particle-antiparticle annihilation results in EM radiation (photons). The total energy of the photons is twice the energy associated with a single particle's rest mass (##~E_0=2mc^2~##) plus the excess due to the "kinetic" corrections.
 
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Sasho Andonov said:
I am wondering does someone have knowledge how much energy is released if antiproton annihilate itself with positron?
Generally annhiliation is between a particle and its antiparticle. The interaction between an antiproton and a positron would be equivalent to the interaction between a proton and an electron.
 
For the quantum state ##|l,m\rangle= |2,0\rangle## the z-component of angular momentum is zero and ##|L^2|=6 \hbar^2##. According to uncertainty it is impossible to determine the values of ##L_x, L_y, L_z## simultaneously. However, we know that ##L_x## and ## L_y##, like ##L_z##, get the values ##(-2,-1,0,1,2) \hbar##. In other words, for the state ##|2,0\rangle## we have ##\vec{L}=(L_x, L_y,0)## with ##L_x## and ## L_y## one of the values ##(-2,-1,0,1,2) \hbar##. But none of these...

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