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I have heard that the eigenvalue of Hamiltonian in an unstable system can contain an imaginary part corresponding the tunneling. Is that true? If it is the case, then I am very confused about it.

Let consider the quantum mechanics case. Suppose a particle move in an potential shown as following (replace the field variable with x). enter image description here

Then the Hamiltonian is $$H=\frac{\hat{p}^2}{2m}+U(x).$$

We can simply obtain the eigenvalue of Hamiltonian by solving the eigen equation $$\left(\frac{\hat{p}^2}{2m}+U(x)-E\right)\psi(x)=0.$$ Now since every term inside the bracket is hermitian, how can the Hamiltonian get a complex eigenvalue?

Wein Eld
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1 Answers1

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The hamiltonian you're considering is not unstable, it is metastable.

If you place a particle at the bottom of the right-hand well at $+v$ it will stay there, even if it is not at the global minimum. Quantum mechanically, you get two closely spaced ground states, separated by ($h$ times) the frequency for the particule to tunnel back and forth.

An unstable system looks more like this:

Mathematica graphics

Classically, this system is metastable, too, but if it makes it past the barrier it is completely gone. Quantum mechanically, the particle can 'sit' in the well, in the sense that the wavefunction can reflect back and forth and form a resonance. However, the resonance isn't perfect, because the right-hand-side wall is not perfectly reflective - since the particle can tunnel out - so eventually you will lose all your population, and you cannot have a stationary state.

In terms of how you phrased the Schrödinger equation, it is true that the hamiltonian's eigenvalue equation looks hermitian, $$\left(\frac{\hat{p}^2}{2m}+U(x)-E\right)\psi(x)=0,$$ and indeed it is; you will normally have a full orthogonal set of continuum eigenfunctions with real eigenvalues. In addition to them, you will also have a resonance, which is exactly the wave I talked about above: it is a state with the property that $$ |\psi(t)⟩=e^{-iE_0t/\hbar}e^{-\Gamma t/2}|\psi(0)⟩, $$ so it continuously loses population (but it is otherwise stationary), and it may also have uglier decay properties at infinity than the regular eigenstates. This is often thought of as a state with a complex eigenvalue, but you need some caution before you interpret it as an eigenvalue equation since the state might not fall in your normal Hilbert spaces.

The best place I know about how to formalize all of this is at

Decay theory of unstable quantum systems. L Fonda, G C Ghirardi and A Rimini. Rep. Prog. Phys. 41, 587 (1978).

which is well worth a read.

Emilio Pisanty
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  • Thank you very much, Emilio. So it seems that the picture in my mind that all states are evolved unitarily form eigenstates of Hamiltonian is not correct always. Because the time-dependent state above can not be obtained that way. – Wein Eld Dec 15 '16 at 19:58
  • @WeinEld That is not necessarily the case - don't give up on an expression for the resonance as a linear combination of actual eigenstates. It looks like it breaches unitarity because norm isn't conserved, but the norm is infinite to begin with (so it is always infinite anyway). These are pretty subtle waters so I'll refrain from categoric statements - that Fonda paper is a good place to start reading. – Emilio Pisanty Dec 15 '16 at 20:29