# Expectation of $r^k$ for a wave function [closed]

So I have a wave function in spherical polars: $$\psi=\left(r\sin(\theta)e^{i\phi}\right)^l\exp\left({\frac{-r}{(l+1)a}}\right)\;,$$ and I have to show $$E[r^k]=\frac{\int_0^\infty r^{2(l+1)+k}\exp\left({\frac{-2r}{(l+1)a}}\right)}{\int_0^\infty r^{2(l+1)}\exp\left({\frac{-2r}{(l+1)a}}\right)}\;.$$ I know the formula for the expectation but can't seem to get the required form, any help will be appreciated!

## closed as off-topic by Emilio Pisanty, sammy gerbil, Kyle Kanos, JMac, Qmechanic♦Dec 22 '17 at 18:20

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• I diddled the math for better typesetting but left one needed change incomplete: neither of your integrals has an indication of the integration variable (i.e. you need a $\,\mathrm{d}r$ (\,\mathrm{d}r) in each before the punctuation). – dmckee Dec 20 '17 at 21:21
• might Mathematics be better suited for your math troubles? – Kyle Kanos Dec 21 '17 at 11:17

The inner product of two wavefunctions is defined as: $$(\phi,\psi)=\int d^3\textbf x\,\phi^*(\textbf x)\psi(\textbf x)$$ Given an operator $\hat O$ and a wavefunction $\psi$, the expectation value of the operator is defined as: $$\mathbb{E}(\hat O)=\frac{(\psi,\hat O \psi)}{(\psi,\psi)}$$ In your case $O=r^k$ and the wavefunction is given in spherical coordinates: $$\psi(r,\theta,\varphi)=\sin^l(\theta)e^{i\varphi l} r^l\exp\left({\frac{-r}{(l+1)a}}\right)$$ Now you have to write down numerator and denominator as integrals and convert the triple integral to spherical polars. You will notice that the angular part of the two integrals will be the same on both the numerator and denominator, and you can therefore simplify.