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I'm trying to understand the meaning of the group velocity for Bloch electrons given by $$ \mathbf{v}=\frac{1}{\hbar}\frac{\partial E(\mathbf{k})}{\partial \mathbf{k}} $$ where $E(\mathbf{k})$ is the energy of the band, and $\mathbf{k}$ is the crystal momentum. So defined, this velocity is only applicable for periodic boundary conditions.

However, in the thermodynamic limit, the bulk properties should be independent of the boundary conditions, so $\mathbf{v}$ should describe the velocity of bulk eigenstates of the Hamiltonian. Generally, $\mathbf{v}$ depends on $\mathbf{k}$, and under open boundary conditions $\mathbf{k}$ is not a good quantum number. This confuses me:

What is the relation between the Bloch velocity and the (local) velocity of the bulk electrons under open boundary conditions?

Tobias Fünke
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NessunDorma
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1 Answers1

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The short answer is, in the thermodynamic limit, k could always be defined as a good quantum number for bulk states despite of different boundary conditions.

This can be shown by considering a finite potential of several equally distributed wells and infinite high walls as boundary.

The solution of wave function is then planewave-like and has some phase shift if we move from one well to another. And, we can see that at the thermodynamic limit (# of wells -> infinite), the phase shift is nothing but exp(-ikR), where the definition of k is more complecated than the perodic case. However the difference eventually goes to zero if we take the thermodynamic limit. Therefore we can still construct wave packets with these newly defined "wave vectors", and thus the group veolocity.

Siqi Wu
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