In condensed matter physics, I heard that if chern number of a band $n$ is non zero, it is impossible to choose a gauge such that $\psi_{nk}$ is smooth in the whole brillouin zone.

However, it is know that $\psi_{nk}$ can be written as:

$$\psi_{nk}(x)=\sum_R e^{ikR}a_R(x)$$

where $a_R(x)$ is wannier function.

It seems that the above equation gives a continuous $\psi_{nk}$ over $k$ and contradicts the fact such gauge does not exist in some cases.

So how to resolve such a contradiction?

  • 2
    $\begingroup$ It is also the obstruction to the existence of localized Wannier functions. $\endgroup$
    – Meng Cheng
    Mar 5, 2016 at 7:02
  • $\begingroup$ @MengCheng, Really? But even if $a_R(x)$ is not localized, the equation in the original question still seems to provide a smooth gauge. $\endgroup$
    – atbug
    Mar 5, 2016 at 7:17
  • $\begingroup$ No, that's wrong. You have to be more careful about the analytical properties. For example, if the bands touch, you can still formally have Wannier functions, but then they have power-law tails. $\endgroup$
    – Meng Cheng
    Mar 5, 2016 at 7:25
  • $\begingroup$ @MengCheng, what happens when Wannier functions are not localized, does that $\sum_R$ blows up? $\endgroup$
    – atbug
    Mar 5, 2016 at 7:28

1 Answer 1


The basic resolution is the following: the property of smoothness doesn't nicely correspond under the Fourier transform (which your formula is an example of). In particular, usually the Wannier function is indeed smooth, but this does not mean $\psi_k$ is!

As a simple example, consider the Dirac-delta function $\delta(k)$. We know we (at least if we are physicists and not mathematicians) can write $\delta(k) = \sum_R e^{ikR}$. In other words: the Fourier transform of the delta function is the constant function $1$. Clearly `$1$' is smooth, but its Fourier transform is not (or in more physical terms: if your wave function is a delta function, the corresponding Wannier function is a constant). In general, the Fourier transform of a function is smooth if that function is integrable. Clearly $1$ is not integrable on the real line, in fact it's a severe form of not being integrable, corresponding to the delta function being a severe form of not being smooth.


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