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Exchange term introduced by Felix Bloch is given by

$$ E_x = c_2 \int n(\vec{r})^{4 / 3} d\vec{r} $$

In the example 1.6.7 of the attached reference, the power of $n(\vec{r})$ is derived as following

$$ E_x \propto \int d\vec{r} e^2 n(\vec{r})^d $$

And the argument is that, that the dimensions of integrand is of energy density, so

$$ \text{energy density} = \left[ \frac{M}{LT^2} \right] \implies e^2 n^d = \left[ \frac{ML^3}{T^2} L^{-3d} \right] $$

I did not understand why there is a $L^3$ term in the second fraction above, because the dimensions of energy is

$$ \left[ E \right] = \left[ ML^2T^{-2} \right] $$

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1 Answers1

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The left hand side of the equation has the dimension energy, while on the right hand side there is a $L^3$ (from the integral) times $e^2\,n^d$. The dimension of $n^d$ is $L^{-3d}$ and the dimension of $e^2$ is given by $M\,L^3\, T^{-2}$. Hence we obtain the dimension of $e^2\,n^d$ as

$$ M\, L^3\, T^{-2}\, L^{-3d} \quad.$$

Finally, we see that the right hand side has thus a dimension of $$M\, L^6\,T^{-2}\,L^{-3d}$$ and in order for both sides to have the same dimension, we require $d=4/3$.

Edit: The dimensions of $e^2$ as given here can be found in cgs units, see for example Physical dimensions for the Coulomb force in SI and CGS. Also in this reference, on page 12, the same argument is used: https://www.aimspress.com/fileOther/PDF/Materials/matersci-04-01372.pdf.

Tobias Fünke
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