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When counting $\vec k$-states in a system - in my case, counting the number of states in the first Brillouin Zone - we either use periodic boundary conditions, $$\psi(\vec r) = \psi(\vec r + \vec L)$$ or infinite square well boundary conditions, $$\psi (0) = \psi(L).$$

In the former case, we allow $- \infty < k < \infty $, while for ISW boundary conditions $0<k$. I have been told that these domains simply avoid double-counting the k-states and that they should give the same answer.

My question is:

  • is there a physical argument for the allowed values of k?

  • What does it mean to have a negative k value (is it simply a wave travelling to the left?) and why is this prohibited in the ISW case?

  • Are there circumstances under which these set-ups are inequivalent or one is preferred?

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When counting $k$ states in the context of solid state physics we always use von Neumann boundary conditions (i.e. periodic conditions). The infinite box is not sufficient in the context of solid state physics, we need periodic boundary conditions.

Answering your first questions, there are several points of view.

  1. We know that the energy of an electron becomes quantized when some boundary conditions are imposed. Since the energy is proportional to the momentum, both must be quantized.

  2. Momentum conservation is a consequence of translational symmetry. Inside a crystal the translational symmetry is discontinuous, so we can expect momentum to become so.

  3. Momentum is the Fourier transform of space. When a variable (in our case the space, $x$) is periodic, this implies that its Fourier transform (namely, $k$) it's quantized, and vice versa. We know that the space in a crystal is periodic, so momentum is quantized.

*Proof:

By definition:

$$ |x\rangle = \sum_k{e^{ikx}}|k\rangle $$

Now impose periodicity $|x+L\rangle = |x\rangle$:

$$\sum_k{e^{ik(x+L)}}|k\rangle = \sum_{k'}{e^{ik'x}}|k'\rangle$$ We use orthogonality: $\langle k|k'\rangle = \delta_{kk'}$, this is, we multiply both sides by $\langle k|$ to get rid of the sum:

$$ e^{ik(x+L)} = e^{ikx} \implies e^{ikL}=1 \implies kL = 2\pi n $$

Finally:

$$k=\frac{2\pi}{L}n$$

Now concerning the sign of $k$. For the PBC we can have both positive or negative $k$, there's no constraint on that. For the ISW one can impose either $k>0$ or $k<0$ but this does not affect the states. Why? Because the solutions of the ISW are of the form:

$\psi(x) = C e^{ikx} + C' e^{-ikx}$

This is, the eigenstates are stationary wave functions, so the analogy with electrons moving to the right or to the left used in the case of PBC fails.

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  • $\begingroup$ Thanks for your answer! What are you referring to in the second paragraph, when you say "it"? "It was the very first try to apply..." etc. $\endgroup$
    – Nik
    Commented Jan 4, 2021 at 12:05
  • $\begingroup$ @Nik To the infinite box boundary condition. $\endgroup$ Commented Jan 5, 2021 at 13:11
  • $\begingroup$ To clarify my initial question, what are the allowed values of the integer n? For periodic BCs, I believe it can be negative, while for the ISW case we have k = (pi/L)*n, with n>0. Is this true and why do we impose this restriction on n? $\endgroup$
    – Nik
    Commented Jan 5, 2021 at 13:34
  • $\begingroup$ @Nik First I must correct myself: Even the Free electron model uses PBC (see Aschroft & Mermin, Ch 2 on Sommerfeld's Model). The ISW conditions are not used in solid state physics. Now concerning you're question, I've extended my answer. Hope that helps. $\endgroup$ Commented Jan 5, 2021 at 14:29

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