Calculating Mean Square Radius with QM Integration: Step-by-Step Guide

[sniffer]

if i want to calculate mean square radius by using:
$$\rho^2=\frac{\int \mid\psi\mid^2 r^2 dr}{\int |\psi|^2 dr}$$
with $\psi=\sum_{k} a_k exp(ikr)$.

How do I solve the numerator part?

 

[CarlB]

Have you considered integrating by parts?

Carl

 

[thepatientmental]

To solve the numerator part, we first need to substitute the given wavefunction \psi into the expression for mean square radius. This gives us:

\rho^2=\frac{\int \mid\sum_{k} a_k exp(ikr)\mid^2 r^2 dr}{\int |\sum_{k} a_k exp(ikr)|^2 dr}

Next, we need to expand the square of the wavefunction using the complex conjugate. This gives us:

\rho^2=\frac{\int \sum_{k} a_k^* exp(-ikr) \sum_{k'} a_{k'} exp(ik'r) r^2 dr}{\int \sum_{k} a_k^* exp(-ikr) \sum_{k'} a_{k'} exp(ik'r) dr}

We can then simplify the numerator by using the product rule for integration and separating the terms with different values of k and k'. This gives us:

\rho^2=\frac{\sum_{k} \sum_{k'} a_k^* a_{k'} \int exp(-ikr) exp(ik'r) r^2 dr}{\sum_{k} \sum_{k'} a_k^* a_{k'} \int exp(-ikr) exp(ik'r) dr}

Now, we can use the fact that \int exp(-ikr) exp(ik'r) dr=2\pi\delta(k-k') to simplify the expression further. This yields:

\rho^2=\frac{\sum_{k} a_k^* a_k \int r^2 dr}{\sum_{k} a_k^* a_k \int dr}

Finally, we can solve the integrals in the numerator and denominator to get:

\rho^2=\frac{\sum_{k} a_k^* a_k \frac{r^3}{3}\Bigg|_0^\infty}{\sum_{k} a_k^* a_k r\Bigg|_0^\infty}

Since the wavefunction is expected to be normalized, i.e. \int |\psi|^2 dr=1, we can simplify the expression further to get:

\rho^2=\frac{\sum_{k} a_k^* a_k}{\sum_{k} a_k^* a_k}=1

Therefore, the mean square radius for the given wave