Orthonormality and the Stark Effect

[BOAS]

Homework Statement

A Hydrogen atom is in a homogeneous electric field. The field's interaction with the atom is described by the Hamiltonian $\hat H = e E_0 r \cos \theta$.

Calculate the energy shift due to the linear stark effect in the following state of Hydrogen.

$\Psi = \frac{1}{\sqrt{2}} (\psi_{200} + \psi_{210})$

Hint: Use the fact that $r \cos \theta = r \sqrt{\frac{4 \pi}{3}} Y_{1,0}$ and the orthonormality of the spherical harmonics.

Homework Equations

The Attempt at a Solution

[/B]
From first order perturbation theory:

$\Delta E = \int_{dv} \Psi^* \hat H \Psi$

$\Delta E = \frac{1}{2} \int^{\infty}_0 \int^{2\pi}_0 \int^{\pi}_0 (\psi_{200}^* + \psi_{210}^*) \hat H (\psi_{200} + \psi_{210})$

Substituting the hint into the Hamiltonian, and using the fact that $\psi_{nlm} = R_{nl} Y_{lm}$ to separate the integral.

$I_{angular} = \int^{2\pi}_0 \int^{\pi}_0 (Y_{00}^* + Y_{10}^*)(Y_{00} + Y_{10}) Y_{10} \sin \theta d\theta d\phi$

I am confused about how to apply the argument of orthonormality to the product of three spherical harmonics. How do I proceed?

Thanks for any help you can give!

 

[Dr Transport]

In general the product of three spherical harmonics is called a Gaunt integral. In this case, we can make the simplification that $Y_{00} = \frac{1}{\sqrt{4\pi}}$. Also you can make the case that since you have $m = 0$ you can convert to Legendre polynomials and use https://en.wikipedia.org/wiki/Associated_Legendre_polynomials to reduce the integrals to product of two only.

 

[BOAS]

In general the product of three spherical harmonics is called a Gaunt integral. In this case, we can make the simplification that $Y_{00} = \frac{1}{\sqrt{4\pi}}$. Also you can make the case that since you have $m = 0$ you can convert to Legendre polynomials and use https://en.wikipedia.org/wiki/Associated_Legendre_polynomials to reduce the integrals to product of two only.

Oh that makes perfect sense, I should have seen that.

Thank you for your help!