# Why are periodic boundary conditions used for the derivation of phonons? [duplicate]

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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?

## marked as duplicate by Ruslan, AccidentalFourierTransform, Norbert Schuch, John Rennie quantum-mechanics StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Apr 18 '16 at 8:28

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.
• @ThomasElliot - Take a $1 m^{3}$ cube of silicon crystal - how many atoms are there? How many atoms are there on the interfaces? How about a $1 mm^{3}$ crystal? OK, how about a $1 \mu m^{3}$ crystal? When dealing with a billion atoms, you are still well into the thermodynamic limit and away from significant edge effects. – Jon Custer Apr 18 '16 at 13:48