Intuition behind the $k$-indexing of wavefunctions

In solid state physics we can write the wavefunctions of a crystal electron according to Bloch's theorem:

$$\psi_k(x)=\mathrm{e}^{\mathrm{i}kx}u_k(x)$$

where $$u_k(x)$$ is a lattice periodic function.

We always index the wavefunctions with the wavevector $$k$$. Sometimes this is even written as

$$\psi(k,x)=\mathrm{e}^{\mathrm{i}kx}u(k,x)$$

which makes it looks like the function depends on real space and reciprocal space simultanously. Are we looking at a wavefunction in real space or in reciprocal space? Can we transform between both representations using a fourier transform?

For some reason, e.g. when reading textbooks this indexing always confuses me. I somehow find it hard to get intuition on why the wavefunction is indexed with the wavevector.

What I do understand however is that when looking at a finite crystal with periodic boundaries, there is only a finite number of allowed $$k$$ values, which becomes continuous in the limit of an infinite crystal.

Edit

How are this $$k$$ and reciprocal space connected? I think this is what is really confusing me. And how would the fourier transform of $$\psi(k,x)$$ look like? Would that be $$\tilde{\psi}(x,k)=\tilde{\psi}_x(k)$$?

2 Answers

We always index the wavefunctions with the wavevector $$k$$. Sometimes this is even written as

$$\psi(k,x)=\mathrm{e}^{\mathrm{i}kx}u(k,x)$$

which makes it looks like the function depends on real space and reciprocal space simultanously.

That would be weird, but that's not what the notation is saying. For example, consider the eigenstates $$|n \rangle$$ of a particle in a harmonic potential. Each eigenstate has a wavefunction, which is usually denoted $$\psi_n(x) \equiv \langle x | n \rangle.$$ If you want, you can also define $$\psi(n, x) \equiv \psi_n(x)$$, but this is just a trivial change of notation. The point is that the energy eigenstates are indexed by $$n$$, so the argument $$n$$ just tells you which energy eigenstate's wavefunction you're looking at.

Similarly, you can consider the energy eigenstates of electrons in a periodic potential. The lattice has a discrete translational symmetry, so the eigenstates can be indexed by the phase picked up upon such a translation, which itself is determined by the quantity $$k$$, the crystal momentum. The notation $$\psi(k, x)$$ means the wavefunction of an energy eigenstate with crystal momentum $$k$$.

How are this $$k$$ and reciprocal space connected?

Great question! Crystal momentum $$k$$ and ordinary reciprocal space (which I'll index with $$p$$ to avoid confusion) are closely connected, though of course not exactly the same thing.

The wavefunction $$\psi_k(x)$$ has the Fourier transform $$\tilde{\psi}_k(p)$$. Now consider the case $$k = 0$$. Then $$\psi_0(x) = u_0(x)$$ is periodic with period $$L$$, where $$L$$ is the lattice spacing. So by the ordinary logic of Fourier series, $$\tilde{\psi}_0(p) \text{ has support at } p = \frac{n}{L}, n \in \mathbb{Z}.$$ For the case $$k \neq 0$$, we note that $$\psi_k(x)$$ is just $$e^{ikx}$$ times such a periodic function, which shifts everything by $$k$$. So $$\tilde{\psi}_k(p) \text{ has support at } p = \frac{n}{L} + k, n \in \mathbb{Z}.$$ The physical intuition is that, if you didn't have the lattice around, the energy eigenstates would just be plane waves $$e^{ipx}$$. In a lattice, the quantity $$p$$ is no longer conserved, because you can pick it up from the lattice itself in multiples of $$1/L$$. The crystal momemtum $$k$$ indicates the "base" momentum that gets modified by multiples of $$1/L$$.

• I have made an edit to my question. Jan 13, 2020 at 14:23

This arises from the discrete translational symmetry of the lattice (Bloch's theorem). Since the potential is periodic, the Schrodinger's equation is invariant to discrete translations by lattice vectors. Suppose, $$\left|0\right>$$ is the original state we start with, and $$\left|a\right>$$ represent the state after a translation by a. $$H(0)\left|0\right> = E \left|0\right>$$, and $$H(a)\left|a\right> = E \left|a\right>$$.

From the translational symmetry, we have : $$H(0) = H(a)$$, hence $$H(0)\left|a\right> = E \left|a\right>$$. Hence, $$\left|a\right>$$ is also a eigenstate of $$H(0)$$. $$\left|a\right>$$ and $$\left|0\right>$$ are degenerate in energy.

With a bit of mathematics, it can be proved that this is possible only when $$\left|a\right> = e^{ika}\left|0\right>$$, i.e the states differ by a phase factor. Quoting Ziman, "For any wave-function that satisfies the Schrodinger equation, there exists a vector $$\vec{k}$$ such that translation by a lattice vector $$\vec{l}$$ is equivalent to multiplying by the phase factor $$e^{i\vec{k}.\vec{l}}$$. Each different wave function may have a different wave-vector $$\vec{j}$$, but each one must satisfy this condition."

Hence, we use k to index the different wavefunctions that are possible which satisfy the translation symmetry of the lattice. The wavefunction is clearly in real space.