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For a free particle, the solution of the (time independent) Schroedinger equation is

$$\psi = \frac{1}{\sqrt{V}} e^{ip\cdot r /\hbar}$$

Since over infinite space this is undefined (not normalizable) we assume a large finite volume (a cube of size $L_{x}$, $L_{y}$, $L_{z}$) in which the particle must be located. Notably this $\psi$ represents a momentum eigenfunction. In the book "Advanced Quantum Mechanics: A practical guide (2013)" on page 6 periodic boundary conditions are assumed so

$$\psi(x,y,z) = \psi(x+L_x, y, z) = \psi(x, y+L_y, z) = \psi(x,y, z+L_z).$$

This is done to calculate the possible momenta. Why can this be done? Are we not assuming that outside the cube $\psi $ has to be $0$, which would contradict this assumption? Would the boundary condition not have to be that $\psi$ is zero at the boundary for continuity?

Sancol.
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ArcherOfLoafCrosse
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3 Answers3

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Let me start from the beginning. If we are interested in the energy eigenstates of a free particle, we face the problem of their non-normalizability. A simple solution is to confine the particle in a large space where no interaction is present, bounded by hard walls, i.e., infinite energy barriers limiting the accessible configuration space. However, such a solution introduces important modifications of the wavefunctions near the wall, and the resulting expectation values will contain significant surface effects. The vanishing of the wavefunction near the wall implies, for example, that the probability density of finding a particle near the wall vanishes similarly.

Periodic boundary conditions (PBC) minimize such a boundary effect. The probability density remains constant over the whole finite volume independently of the distance from the boundary like in any finite portion of an infinite system. On the other hand, it is possible to prove that the behavior of the observable quantities, when the volume where the particle is free is significant, is independent of the boundary shape and conditions, and this is the rationale for using the PBC (and simple cubes or parallelepiped) in most of the cases.

Notice, however, that independence on boundary conditions for a given shape doesn't imply that every boundary condition is acceptable. Indeed, the choice of a boundary condition has deep implications for the domain of the operators corresponding to important observables, particularly momentum and energy.

Conditions of the wavefunction vanishing at the boundary, or PBC, do not result, as often stated, even in many textbooks, from an inexistent requirement of continuity but stem from the fundamental need that the Hamiltonian and momentum operators are self-adjoint.

GiorgioP-DoomsdayClockIsAt-89
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Are we not assuming that outside the cube psi has to be 0, which would contradict this assumption?

No, we assume that outside of the cube the function continues periodically. This might not necessarily be the case, but which is a trick often used to apply Fourier series to a finite interval - however the function behaves outside this interval, within it the series represent the function faithfully. (Note the distinction between Fourier series and Fourier transform - the latter applied to an infinite interval, and is an alternative to the periodic boundary conditions, although it requires some care.)

Note that if we assumed that $\psi=0$ outside the cube, we would be imposing hard wall boundary conditions, which would kill about half of the solutions (e.g., leaving only sine, but not cosine solutions.)

Finally, in the physical problems where we apply the periodic boundary conditions, we usually intend in the end to take limit $L\rightarrow \infty$, although sometimes it is taken implicitly, as the factors containing cube size cancel out.

Roger V.
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Why can this be done?

The usual explanation is that this can be done because we are interested in bulk properties rather than surface effects. And we assume that the surface boundary conditions don't affect the bulk properties we are interested in.

Alternatively, if you find that unconvincing, just understand that you are solving a pedagogical problem. You might find the solution useful.

Are we not assuming that outside the cube psi has to be 0

No, we are not assuming that. That would be a different boundary condition at the surface.

, which would contradict this assumption?

Right, it would. But we are not applying two different contradictory boundary conditions at once. We are choosing to apply periodic boundary conditions in this case.

Would the boundary condition not have to be that psi is zero at the boundary for continuity?

Would it not have to be? No, it would not have to be. We can choose whatever boundary conditions we want.

hft
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