- #1
wc2351
- 6
- 0
Homework Statement
The problem is from Ashcroft&Mermin, Ch32, #2(a). (This is for self-study, not coursework.)
The mean energy of a two-electron system with Hamiltonian
[tex]\mathcal{H} = -\frac{\hbar^2}{2m}(∇_1^2 + ∇_2^2) + V(r_1, r_2)[/tex]
in the state ψ can be written (after an integration by parts) in the form:
[tex]
E = \int d{\bf r}_1 d{\bf r}_2 \left[\frac{\hbar^2}{2m}\{|∇_1\psi|^2 + |∇_2\psi|^2\} + V({\bf r}_1,{\bf r}_2)|\psi|^2 \right]
[/tex]
Show that the lowest value of the above equation assumes over all normalized antisymmetric differentiable wavefunctions ψ that vanish at infinity is the triplet ground-state energy E_t, and that when symmetric functions are used the lowest value is the singlet ground-state energy E_s.
Homework Equations
Just basic knowledge of quantum mechanics? The proof should be elementary it seems...
The Attempt at a Solution
Before proving this theorem, it seemed natural to me that the ground state should be singlet because if the spin state is triplet, it is symmetric so the spatial wavefunction is anti-symmetric, and thus the existence of a node indicates that it has higher energy than the singlet wavefunction.
But I don't know how to proceed from the statement of the problem. Naive applicatin of the Euler-lagrange equation gives me Schrodinger's equation with zero energy. I don't know how I should use the parity of wavefunction since it is enclosed by the absolute signs.
Thank you for your help in advance.