From Sakurai eq. 6.4.21a we have that $$\langle {\bf k} \vert E,l,m \rangle=\frac{\hbar}{\sqrt{M k}}\delta\left(E-\frac{\hbar^2 k^2 }{2M}\right) Y_l^m({\bf\hat k}),$$ where $M$ is the mass of the particle in question.

Is there an easy way to write down a similar expression in terms of $k=\vert{\bf k}\vert$ instead of $E$? That is, is there a quick and dirty way to rewrite the above expression but with LHS as

$$\langle {\bf k} \vert ~k,l,m \rangle=^?\frac{\hbar}{\sqrt{M k}}~~\delta(?)~~ Y_l^m({\bf\hat k}).$$ Can I simply use this

$$\delta(g(x)) = \sum_i \frac{\delta(x-x_i)}{|g'(x_i)|}$$ Dirac delta formula or must I be more careful eg what happens to the $k$ under the square root?

  • 2
    $\begingroup$ Presumably the energy eigenstates are for a free particle? $\endgroup$ – Emilio Pisanty Apr 15 '14 at 21:24
  • $\begingroup$ Yes! @EmilioPisanty, that is true. $\endgroup$ – Your Majesty Apr 15 '14 at 21:24

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