I need to calculate the energy of the ground state of a helium athom with the variational method using the wave function:(adsbygoogle = window.adsbygoogle || []).push({});

$$\psi_{Z_e}(r_1,r_2)=u_{1s,Z_e}(r1)u_{1s, Z_e}(r2)=\frac{1}{\pi}\biggr(\frac{Z_e}{a_0}\biggr)^3e^{-\frac{Z_e(r_1+r_2)}{a_0}}$$

with ##Z_e## the effective charge considered as a variational parameter.

The Hamiltonian of the system is:

$$\hat{H}=\hat{H_1}+\hat{H_2} + \frac{e^2}{4\pi\epsilon_0|\vec{x_1}-\vec{x_2}|}$$

The first two terms are idrogen Hamiltonians for the two electrons and the third term is the interaction term.

The expection value of the interaction term is ##\frac{5}{4}Z_eRy## with ##Ry= 13.6 \space eV##.

For the first two term:

$$<\psi_{Z_e}|\hat{H_1}+\hat{H_2}|\psi_{Z_e}>$$

$$<u_{1s,Z_e}(r1)u_{1s,Z_e}(r2)|\hat{H_1}+\hat{H_2}|u_{1s,Z_e}(r1)u_{1s,Z_e}(r2)>$$

$$2<u_{1s,Z_e}(r1)|\hat{H_1}|u_{1s,Z_e}(r1)>$$

Since ##\hat{H_1}=\hat{T}+U## I can calculate:

$$<U> = <u_{Z_e}|U|u_{Z_e}>=-\frac{Ze^2}{4\pi\epsilon_0}<u_{Z_e}|\frac{1}{r}|u_{Z_e}>$$

$$<U>= \frac{Z}{4\pi\epsilon_0a_0} = -2ZZ_{e}Ry$$

Now I have used the virial theorem ##2<\hat{T}> + <U> = 0##, so:

$$<\hat{T}> = ZZ_{e}Ry$$

But this is wrogn because it should be ##Z_{e}^2Ry## and I do not see where is the mistake. Thank you in advance!

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# A Helium atom, variation method and virial theorem

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