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A Helium atom, variation method and virial theorem

  1. May 14, 2018 #1
    I need to calculate the energy of the ground state of a helium athom with the variational method using the wave function:
    $$\psi_{Z_e}(r_1,r_2)=u_{1s,Z_e}(r1)u_{1s, Z_e}(r2)=\frac{1}{\pi}\biggr(\frac{Z_e}{a_0}\biggr)^3e^{-\frac{Z_e(r_1+r_2)}{a_0}}$$
    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!
     
  2. jcsd
  3. May 14, 2018 #2
    It's a know problem. You can find a cool explanation/solution in "Introduction to Quantum Mechanics" by Griffiths Chap 7 (p. 264). Here's an extract:
     

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