9

I am currently reading "Quantum Field Theory for the Gifted Amateur". In chapter 2 Phonons are introduced as solutions (in k-space) of a coupled harmonic oscillator. In real space the oscillator is coupled, but apparently not in k-space (after doing a Fourier-Transform on the x,p Operators). During the solution the author used periodic boundary conditions but I don't see why they should accurately describe a finite crystal that is not shaped like a ring. In a different book the solution was also obtained with periodic boundary conditions.

Are more realistic boundary conditions impossible to solve? I would have guessed that we assume that the wave-function (of the phonons) is supposed to be zero outside of the lattice instead of this infinite periodic behaviour.

A fininte crystal might be very different than a infinite crystal (interpreation of the periodic boundary). They seem like two completely different systems. Why do we use periodic boundary conditions and how accurate is this?

Qmechanic
  • 220,844
Thomas Elliot
  • 839
  • 7
  • 21

1 Answers1

5

In the thermodynamic limit (linear size of the system $L$ to infinity), boundary conditions don't really matter, and most physical observables will be the same for all boundary conditions.

The use of periodic boundary conditions is mostly for practical reasons, in particular, translation symmetry is conserved, which really helps. One could in principle do the calculation with other boundary conditions, such as strict BC as you suggest, but this usually makes the calculation more painful that it has to be.

Of course, if you are interested in the effect of boundaries on the system, you then have to use the proper ones.

Adam
  • 12,192