I'm reconsidering the problem of resonance states.(adsbygoogle = window.adsbygoogle || []).push({});

We know that the resonances in QM are described as the complex energy poles in the scattering amplitude. In the version of QFT, the resonances are described by the complex mass poles of the scattering matrix.

In QFT, I can understand that the masses of intermediate particles develops imaginary masses from loop corrections.

But in the case of QM, I don't quite understand the situation. I read from a book that because the wavefunction of the unstable state extends to infinity. Hence the boundary condition changes (different from bound state's), that's why the complex energy. (The complex energy is still the eigenvalue of the Hamiltonian.)

But I remember that we can prove that the eigenvalues of a Hermitian operators are always real. Like the following proof in the braket language,

from

[tex] A|a'\rangle = a'|a'\rangle [/tex] and [tex] \langle a''|A = a''^*\langle a''| [/tex]

where [tex] A [/tex] is an Hermitian operator and [tex] a',a'' [/tex] are its eigenvalues.

We times the first equation with [tex] \langle a''| [/tex], the second equation with [tex] |a'\rangle [/tex], then substract,

[tex] \Rightarrow (a' - a''^*)\langle a''|a'\rangle = 0 [/tex]

now we select [tex] a' = a'' [/tex], then we conclude that [tex] a' [/tex] is real.

So, eigenvalues of a Hermitian operator must be real.

In short, my question is,

(1) is the complex energy of resonance in QM the eigenvalue of Hamiltonian?

(2) If (1) is true, then how to explain the breakdown of the proof I wrote above?

Thanks for any ideas.

Sincerely

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# Resonance states and complex energies

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