Can Hamiltonian gain/lost some symmetry through boundary conditions?

Initial expression may not have symmetry, but after specyfing boundary conditions and predicted form of the wavefunction it will actually acquire this symmetry. Interesting things may happen with time-reversal symmetry and magnetic field in curved geometries I suppose.

For example, $$H = -i\hbar v_{F}\left[ \sigma_{x}\partial_{x} + \frac{\sigma_{y}}{R}\left(\partial_{\phi} + i\eta \right)\right]$$ It is not time-reversal symmetric because of non-zero value of $\eta$. If I know tell that this Hamiltonian describes massles Dirac fermion on the surface of a topological nanowire with radius $R$ where wavefunctions has to satisfy antiperiodic boundary conditions in $\phi$ and transaltion in $x$-direction, then You will solve Schrodinger equations through ansatz, $$\Psi = \Phi e^{ikx}e^{il_{n}\phi}$$ with $l_{n} = n - 1/2$, quantized with integer numbers $n$. You can rewritte Hamiltonian, $$H(k,n) = \hbar v_{F}\left[k\sigma_{x} + \frac{\sigma_{y}}{R}\left(l_{n} + \eta\right)\right]$$ Now for $\mathcal{T} = i\sigma_{y}\mathcal{K}$, with complex conjugation operator $\mathcal{K}$ one gets, $$\mathcal{T}H(k,n)\mathcal{T}^{-1} = -H(k,n)$$ but for $\eta=1/2$ one can actually write, $$\mathcal{T}H(k,n)\mathcal{T}^{-1} = H(-k,-n)$$ It implies that the time reversed wavefunction $\mathcal{T}\Psi$ is a solution of the Hamiltonian $H(-k,-n)$ with the same energy $E$, $$H(k,n) \Psi = E \Psi \\ H(-k,-n) \mathcal{T}\Psi = E \mathcal{T}\Psi$$ You can look at this as a pseudo "particle-hole symmetry", cause for every state $\Psi$ with energy $E$ exists time-reversed state $\mathcal{T}\Psi$ with energy $-E$.

  • $\begingroup$ What kind of boundary conditions do you mean? Since when do Hamiltonian need such things, and why should the "predicted form of the wavefunction" change anything about the symmetries of the Hamiltonian (from which this predicted form has presumably been derived)? $\endgroup$ – ACuriousMind Aug 27 '14 at 21:06

I think that in order to get a proper answer, the question should be made more precise mathematically. Since the problem is not really specified without any boundary conditions, it's not obvious what the "initial expression" is. The two following examples of interplay between BCs, definition of the Hamiltonian and symmetry might be relevant.

  1. It's very common in statistical/condensed matter physics to impose periodic boundary conditions on finite size systems. Usually there is then translation invariance, which is obviously lost with non-periodic BCs.
  2. Quantum fields on star graphs (arXiv:1110.5713) - this involves time-reversal invariance (the theory is developed in earlier papers which contain more details, but the first pages of this one are a concise introduction).

Consider several one-dimensional systems, joined at a vertex. A wave function for this system should be a collection of wave functions on each edge $\psi=(\psi_1\dots\psi_m)$. In the bulk, we would like the Hamiltonian to be proportional to the second derivative wrt the coordinate. Requiring that the Hamiltonian be self-adjoint involves studying its domain; they find that this gives constraints on the BCs at the vertex, and a way to parametrise them.

Long story short, the physics is described by a scattering matrix, which may not be symmetric. If it is, rime-reversal symmetry is present, otherwise it's not. So, boundary conditions at the vertex determine time-reversal invariance.

Further discussion of time-reversal and relation to magnetic fields in this setting is found in arXiv:0907.4221 and arXiv:1202.4270.


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.