- #1
mzh
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Dear Physics Forums users
Eq. 2.60 of Ashcroft, Mermin is
[itex] \int \frac{d\vec{k}}{4\pi^3} F(E(\vec{k}) = \int_0^\infty \frac{k^2dk}{\pi^2}F(E(\vec{k})) = \int_{-\infty}^{\infty} dE g(E) F(E) [/itex]
I understand the first transformation is done by introducing spherical coordinates (as written in the text) and to integrate out [itex]\phi[/itex] and [itex]\theta[/itex].
I also get the second transformation, where we insert the expression for the energy [itex]E=\frac{\hbar^2 k^2}{2m} [/itex], but what I don't understand is the new boundaries. How do we arrive at the boundaries from minus to plus infinity? Does E range from -infinity to plus infinity?
Thanks for any hints?
Eq. 2.60 of Ashcroft, Mermin is
[itex] \int \frac{d\vec{k}}{4\pi^3} F(E(\vec{k}) = \int_0^\infty \frac{k^2dk}{\pi^2}F(E(\vec{k})) = \int_{-\infty}^{\infty} dE g(E) F(E) [/itex]
I understand the first transformation is done by introducing spherical coordinates (as written in the text) and to integrate out [itex]\phi[/itex] and [itex]\theta[/itex].
I also get the second transformation, where we insert the expression for the energy [itex]E=\frac{\hbar^2 k^2}{2m} [/itex], but what I don't understand is the new boundaries. How do we arrive at the boundaries from minus to plus infinity? Does E range from -infinity to plus infinity?
Thanks for any hints?