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Non-Hermitian Hamiltonians

  1. Feb 20, 2009 #1
    Can someone explain how non-Hermitian Hamiltonians are used to account for particle decay?
     
  2. jcsd
  3. Feb 20, 2009 #2
    Hermitian operators always have real eigenvalues. Non-Hermitian operators can have complex eigenvalues. The evolution of an eigenstate of the Hamiltonian is ~exp(-iEt) where E is the energy. If E is complex, there could be a decay.
     
  4. Feb 20, 2009 #3
    I infer from you terse reply that the decay is seen in the imaginary part of the eigen-value. It becomes an exponential decay of the state amplitudes. Yes?
     
  5. Feb 20, 2009 #4
    Yes, imaginary eigenvalues mean that the probability of finding the particle decreases exponentially with time (decay). However, you should keep in mind that this is an approximate way to study decays. In this approach the probability is not conserved and the evolution is non-unitary, which contradicts basic postulates of quantum mechanics.

    If you want to have a rigorous model of decays, you'll need to form a bigger Hilbert space, which includes states of both unstable particle and its decay products. In this Hilbert space the decay can be described by a Hermitian Hamiltonian and unitary time evolution operator. The total probability will be conserved, as required. The decreasing probability of finding the particle will be compensated by increasing probability of finding decay products.
     
  6. Feb 20, 2009 #5
    very cool. Thanks
     
  7. Feb 21, 2009 #6
    You could also consider cases where particles are not bound perfectly, but within a finite barrier. In such case you could find so called meta stable eigen states, which have imaginary components. The Hamiltonian is Hermitian though, but the boundary conditions are wave like at the boundary. Here the decay of the wave function is compensated by the fact that a current is produced outwards (or inwards) at the quasi bound domain, i.e.,

    [tex]\frac{d\mid\Psi\mid^2}{dt}+\nabla\cdot\vec{j}=0[/tex]

    where the quantum current j is non vanishing. There are some simple cases where you could solve this exactly, like V=V0 for a<x<a+d and V=infinite at x<0 and else V=0. Eigenstates are normally degenerated symmetrically so that E=E0+/-i*Ej which means yo have both decaying and growing solution in time.
     
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