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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?

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marked as duplicate by Ruslan, AccidentalFourierTransform, Norbert Schuch, John Rennie quantum-mechanics Apr 18 '16 at 8:28

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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.

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  • $\begingroup$ So if they (different conditions) only agree in the thermodynamic limit how does one justify using it? $\endgroup$ – Thomas Elliot Apr 18 '16 at 7:58
  • $\begingroup$ @ThomasElliot: we are usually interested in the thermodynamic limit, and we don't really care about finite size effects, that will be subdominant. If this is the case, you can then use whatever BC you prefer, which is usually the one that makes the calculation the easiest, that is PBC. $\endgroup$ – Adam Apr 18 '16 at 8:01
  • $\begingroup$ @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. $\endgroup$ – Jon Custer Apr 18 '16 at 13:48
  • $\begingroup$ Ok. Well, I knew that there are a lot atoms in one crystal but in mathematics infinity often gives weird results so I was just wondering how certain a result is that is obtained by using infinity without it being really present. $\endgroup$ – Thomas Elliot Apr 18 '16 at 18:30
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    $\begingroup$ @ThomasElliot: in physics, infinity just means large enough. Typically, these finite size corrections will vanish as the inverse of the volume, i.e. very fast, and will usually not be the main error coming from your modelization (read, almost never be). There are very few cases (such as the BKT transition), where the finite size of the system really changes the physics. (In that case, it has been argued that one would need system of the size of Texas to see properly (vanishing magnetization) the BKT phase). But that's quite an advanced topic compare to plain vanilla phonons. $\endgroup$ – Adam Apr 18 '16 at 19:29

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