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I am following a course on condensed matter physics. In our lecture notes, the lecturer has postulated the following formula in the context of Bloch waves: $$| k \rangle = A\sum_{X} e^{ikX} |X \rangle$$ It is either assumed knowledge or bad practice because he does not define his symbols apart from $k$, which is the crystal momentum of the Bloch wave. But from context, it seems that it possibly an expansion of the energy eigenstate with crystal momentum $k$ into a basis indexed by $X$.

  1. What do the symbols mean?
  2. Can you give a proof for the formula?

It would be much appreciated if you are as mathematically precise as possible.

Mikkel Rev
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2 Answers2

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This is actually the fourier transformation. Normally you see $$f(\vec{x})=N\sum_{\vec{k}}e^{i\vec{k}\vec{x}}f(\vec{k})$$ where N is a normalization parameter. This in terms of linear algebra changes your basis from eigenvalues of the position operator to eigenvalues of the momentum operator. The same thing happens here, but instead of functions you transform states. Then you write $$|\vec{k}\rangle=A\sum_{\vec{x}}e^{\pm i\vec{k}\vec{x}}|\vec{x}\rangle\ \ or\ \ |\vec{x}\rangle=A^*\sum_{\vec{k}}e^{\mp i\vec{k}\vec{x}}|\vec{k}\rangle$$ where A is a normalization parameter and the $\pm$ sign can be chosen arbitrarily in only one of the equations. The second is the inverse relation.

I do not know a proof as to why the {$|k>$} forms a basis for all states {$|x\rangle$} (a bit too mathematical for my taste). However if you accept that, all you do is simply change basis.

Maybe prooving the inverse relation I mentioned above will help you get acquainted a bit more. Start with $$|x\rangle=\sum_{\vec{k}}|\vec{k}\rangle\langle\vec{k}|x\rangle$$ then use $$|\vec{k}\rangle=A\sum_{\vec{x}}e^{ i\vec{k}\vec{x}}|\vec{x}\rangle$$ to calculate $$\langle\vec{x'}|\vec{k}\rangle=A\sum_{\vec{x}}e^{\pm i\vec{k}\vec{x}}\langle\vec{x}'|\vec{x}\rangle=A\sum_{\vec{x}}e^{i\vec{k}\vec{{x}}}\delta(\vec{x}'-\vec{x})=Ae^{i\vec{k}\vec{x}'}$$ So $$|x\rangle=\sum_{\vec{k}}(A^*e^{-i\vec{k}\vec{x}})|\vec{k}\rangle=A^*\sum_{\vec{k}}e^{ i\vec{k}\vec{x}}|\vec{k}\rangle$$ Note that some there are several notations about A. Some prefer to put the normalization factor in the delta sum (or integral) and get the second relation without $A^*$. Also, A is usually real. What was mentioned in the comments by Icv and in the begining of this answer, is $\hat{x}|x\rangle=x|x\rangle$ and $\hat{p}|k\rangle=k|k\rangle$. I hope this helps. Cheers.

Quantumight
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Let's start from the basic physical reasoning. The idea is that we want to find the general form of the wavefunction that has the symmetry of the problem. Let's say that we have a one-dimensional lattice with lattice spacing $a$. So, a wavefunction that shares the symmetry of the lattice would be invariant under translations by $na$ where $n\in\mathbb{Z}$.

This means that if the state is $|\psi\rangle$ then $\langle x+na | \psi\rangle=e^{i\phi_n}\langle x|\psi\rangle$ where $\phi_n$ is some phase. Similarly, for some $m\in\mathbb{Z}$, $\langle x+ma | \psi\rangle=e^{i\phi_m}\langle x|\psi\rangle$. Thus, $\langle x+ma | \psi\rangle=e^{i(\phi_m-\phi_n)}\langle x+na|\psi\rangle$. But, we also must have $\langle x+ma | \psi\rangle=e^{i\phi_{m-n}}\langle x+na|\psi\rangle$. Thus, we conclude that $\phi_{m-n}=\phi_m-\phi_n$. This can only hold true for generic $m,n$ if $\phi_n=Kn$.

Thus, $\langle x+na|\psi\rangle=e^{iKn}\langle x|\psi\rangle$.

So, we have a class of states indexed by the parameter $K$. In particular, we can write $$|\psi_K\rangle=\int dx |x\rangle\langle x|\psi_K\rangle=\sum_n\int_{na}^{(n+1)a} dx|x\rangle\langle x|\psi_K\rangle$$$$=\sum_n e^{iKn}\int _0^adx|x+na\rangle\langle x|\psi_K\rangle=\sum_ne^{iKn}|n\rangle$$ where we have defined $|n\rangle=\int _0^adx|x+na\rangle\langle x|\psi_K\rangle$.