Unstable states and imaginary (complex) energy?

Question

I came across the notion of complex energy while studying instanton method to study the unstable state. Unstable states are those which have energy with an imaginary part. But as we know Hamiltonian is a Hermitian operator. So how can the energy be imaginary?

Answer 1

Since nobody is answering this I'll give it a whack.

A Hermitian Hamiltonian gives unitary evolution. The evolution operator is

$$ U(t) = \exp[-i\Omega t] $$

where $\Omega \equiv H/\hbar$. The thing in the exponent is anti-Hermitian because of the $i$ in front of the Hamiltonian. When you exponentiate an anit-Hermitian operator, you get a unitary one, as I'll explain now.

Let $|\Psi'\rangle= \exp[-ih]|\Psi\rangle$ and Let $|\Phi'\rangle= \exp[-ih]|\Phi\rangle$. We compute\begin{equation} \langle \Phi'|\Psi'\rangle = \langle\Phi|e^{ih}e^{-ih}|\Psi\rangle = \langle\Phi|\Psi\rangle. \end{equation} Since the operator $e^{ih}$ preserves the inner product, it is unitary by definition.

So, as you said Hermitian Hamiltonians give unitary evolution. The thing is, if you consider only sub-systems of a complete physical system, the evolution is not unitary. For example, if we're studying a particle in box, there could be a possibility of the particle leaving the box, in which case our assumption that the particle is in the box is wrong. You can account for this by manually inserting a decreasing function into the evolution operator

$$U(t) = e^{-t\Gamma}e^{-i\Omega t}.$$

Here $1/\Gamma$ is an exponential decay time constant which suppresses the magnitude of the wave function (or state vector, whatever). It's convenient to just write this as a single exponential:

$$U(t) = \exp[-i(\Omega - i\Gamma)t].$$

This can be interpreted as $\Omega\rightarrow\Omega-i\Gamma$ which looks like a non-hermitian Hamiltonian. In fact, if you take the point of view that the Hamiltonian is the thing that generates time evolution then the Hamiltonian is not Hermitian for this subsystem. This does not contradict the idea that the Hamiltonian of a physical system must be Hermitian, because we're doing a special case where we're dealing with a subsystem, and we're encapsulating the effect of the outside environment (into which our particle can escape) by using a non-Hermitian Hamiltonian.

of course, in the end, most systems under study are not completely isolated from other degrees of freedom, so in real life the notion of non-Hermitian Hamiltonian is extremely useful.

update

Imagine you have two 2-level systems interacting with one another through the following Hamiltonian (at this point I'm carelessly dropping dimensions from things):

$$ H = \left[ \begin{array}{cccc} 0&0&0&0 \\ 0&0&1&0 \\ 0&1&0&0 \\ 0&0&0&0 \end{array} \right]$$ where the order of basis vectors is $\{|00\rangle, |01\rangle, |10\rangle, |11\rangle\}$. Suppose this interaction is in place for a time $t$. Then the evolution operator is

$$U(t) = \left[ \begin{array}{cccc} 0&0&0&0 \\ 0&\cos(t)&-i\sin(t)&0 \\ 0&-i\sin(t)&\cos(t)&0 \\ 0&0&0&0 \end{array} \right].$$

Suppose we now "forget" that the second of the two 2-level systems exists, thus making it an "environment" of the first 2-level system. In this condition our remaining 2-level system cannot be represented by a state vector; we have to use a density matrix. Suppose the initial state is $$|\Psi(0)\rangle = \left[\begin{array}{c} 0\\1\\0\\0 \end{array}\right].$$ Then after time $t$ the state vector is $$|\Psi(t)\rangle = \left[\begin{array}{c} 0\\\cos(t)\\-i\sin(t)\\0 \end{array}\right].$$ The density matrix of the full system is $$\rho = \left[ \begin{array}{cccc} 0&0&0&0 \\ 0&\cos(t)^2&i\sin(t)\cos(t)&0 \\ 0&-i\cos(t)\sin(t)&\sin(t)^2&0 \\ 0&0&0&0 \end{array} \right].$$ To find the density matrix of the subsystem consisting of just the remaining 2-level system, we trace over the "lost" particle's subspace, which leaves $$\rho_{\text{subsystem}} = \left[ \begin{array}{cc} \sin(t)^2&0 \\ 0&\cos(t)^2 \end{array}\right]$$ where here the states are ordered $\{|0\rangle, |1\rangle\}$.

First of all, note that you cannot get this evolution from a normal Hamiltonian. The state is traveling from one pole of the Bloch sphere to the other, going through the center of the sphere. This is impossible for a pure state under unitary evolution. Also note that the density matrix is always diagonal which means that the information we have about our subsystem is always classical. This is typical of cases where we are missing a piece of an entangled system (in fact it is argued that this is the essence of the quantum to classical transition).

This is a demonstration of how you get non-unitary evolution if you look only at a subsystem of the full quantum system.