1
$\begingroup$

It is known that Spectrum of a non-hermitian hamiltonian is complex ($E_n-i\gamma_n$) and they represent the dissipative system. eg, Damped harmonic oscillator(DHO), where $E_n$ are the energies of the DHO and $\gamma_n$ describes the decaying rate of its states. It is also known that Spectrum of a hermitian hamiltonian is real and they represent the Non-dissipative system(HO).

In PT-Symmetric hamiltonians: Spectrum is real if PT-symmetry is unbroken and it is complex if it is broken.

Now my ques is: this transition of PT-symmetry can be understood as transition from dissipative system to non-dissipative system ?.

Could anybody give me a simple example ?

$\endgroup$
  • $\begingroup$ PT-symmetric 'quantum mechanics' is, as far as I know, an extremely niche subject mostly advocated by a single person (Carl Bender) and colleagues, and there are some strong objections to it. It might thus be best to email Prof. Bender directly. $\endgroup$ – Al Nejati Sep 18 '18 at 5:48
  • $\begingroup$ @AlNejati, Okay, He is a very busy person, so I wonder whether he will reply to it or not. $\endgroup$ – Epsilon Sep 18 '18 at 11:50
  • $\begingroup$ @Al Nejati No, properly appreciated, the objectors have nuanced their understanding to a mainstream status. At the most conservative, PTQM is a shortcut of plain QM. $\endgroup$ – Cosmas Zachos Jul 12 at 20:03
3
$\begingroup$

Is the breaking of PT symmetry in a quantum mechanical system indicating the transition from a non-dissipative to a dissipative system: no, PT symmetric systems are in between closed (Hermitian) and open (general, non-Hermitian) systems in the following sense.

A quantum mechanical system with a Hermitian Hamiltonian is closed and probability is conserved (and in its classical analog (if it exists), the total energy in the system is conserved). A QM system with non-Hermitian Hamiltonian generally is open, it looses probability and indeed exhibits dissipation in this sense.

A quantum mechanical system with a non-Hermitian, PT-symmetric Hamiltonian may exhibit PT symmetric eigenfunctions and real eigenvalues (unbroken PT symmetry) or complex eigenvalues and non-PT symmetric eigenfunctions.

As a simple example, we consider two subsystems one of which loses probability (and the classical counterpart would lose energy, so there is dissipation) and the other gains exactly the same amount. Such coupled QM system is PT symmetric and exhibits real eigenvalues for sufficiently large coupling strength such that the eigenfunctions are also PT symmetric (unbroken PT symmetry). If the coupling becomes too weak, the eigenvalues cease to be real but become complex, indicating a state of broken PT symmetry, i.e. eigenfunctions are not any longer PT symmetric.

The breaking of PT-symmetry, rather than a transition from an open (dissipative) to a closed (non-dissipative) system, indicates a transition from a balance between loss and gain to a state without such balance.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.