In David Tong's notes on the quantum hall effect, when calculating the current due to edge modes, he writes it's expression as $$I_y = -e \int \frac{dk}{2 \pi} v_y(k)$$ where $v_y$ is the drift velocity of the state with momentum $\hbar k$. But I fail to understand how that gives the current. And why is there a factor of $2 \pi$ in there?
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The $2\pi$is there because when you have a an edge with lenth $L$ the allowed values of $k$ are $k_n= 2\pi n/L$. So as $L$ becomes large we have
$$
dk = \frac{2\pi}{L} dn
$$
and so
$$
\sum_{n} f(k_n) \approx \int f(k_n) dn \to L \int \frac{dk}{2\pi} f(k)
$$
The $L$ drops out because to get the current you need to multiply the drift velocity $v$ by the charge density which is
$$
\rho = \frac 1 L \sum_n \theta(k_f-k_n)\to \int_{k<k_f} \frac{dk}{2\pi}
$$
mike stone
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Current in a crystalline solid can be expressed as $$ \mathbf{j}\propto \int_{\text{BZ}}\mathbf{v}_g(\mathbf{k})n(\mathbf{k})d^3\mathbf{k}, $$ i.e., a product of electron group velocity in a state, times the occupation number of this state, summed over all the states (the proof can be found in solid state physics texts, but see also this answer for more discussion).
At zero temperature the distribution function is a step function, i.e., it merely defines the limit of integration: $$ \mathbf{j}\propto \int_0^{k_F} v_g(k)dk = \int_0^{\epsilon_F} v_g(E)\rho(E)dE, $$ Further it can be shown that in 1D the density of states is inversely proportional to the group velocity and the two cancel out. This is the core of the integer quantum Hall effect as well as conductance quantization in 1D structures.
See Landauer formula, Conductance quantization, and Intuition on why quantum hall effect?.
Roger V.
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