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In DJ Griffiths' ''Introduction to Quantum Mechanics" to describe an unstable particle that spontaneously disintegrates he assumed an imaginary part in the potential. What does that signify? What is meant by an imaginary potential?

I have added the image of the problem.

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    $\begingroup$ You should also mention the context of the question: from memory, it's one of those cases where you can't derive the continuity equation from the Schrodinger Equation. What do you think that means? $\endgroup$
    – Philip
    Jan 21, 2021 at 8:02
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    $\begingroup$ Related: physics.stackexchange.com/questions/60185 $\endgroup$
    – kaylimekay
    Jan 21, 2021 at 8:10
  • $\begingroup$ I have solved the given problem but had a conceptual doubt. $\endgroup$ Jan 21, 2021 at 8:11

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Recall that a state evolves as $|{\psi(t)}\rangle = e^{-i E t/\hbar} |{\psi(0)}\rangle$. If $E = E_R + i E_I$, then $e^{-i E t/\hbar} = e^{-i E_R t/\hbar} e^{- E_I t/\hbar}$. The imaginary part of the energy causes exponential decay.

We can use an imaginary energy as a phenomenological method to describe unstable, decaying states, and the imaginary part of the energy is related to the average lifetime of the state, as you calculated while solving the problem.

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Imaginary potential is an ad-hoc way of including dissipation in the Schrödinger equation. As noted in the comments, doing it in a straightforward way is a rather crude approach, as it immediately breaks particle conservation, unitarity, etc.

A cohent way to include dissipation is by coupling the system of interest to a bath, e.g., a bath of harmonic oscillators, and than tracing this bath out. This is the method behind the Caldeira-Legget model, Master equation formalism and many Green's function approaches. But dissipative Schrödinger equation still has its advantages, particularly in numerical domain, and more sophisticated techniques to tjsi end have been developed, see, e.g., here.

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