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Two-electron Ground State of a Spin-Independent Hamiltonian is a singlet

  1. Jun 8, 2012 #1
    1. The problem statement, all variables and given/known data
    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.

    2. Relevant equations
    Just basic knowledge of quantum mechanics? The proof should be elementary it seems...

    3. 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.
     
  2. jcsd
  3. Mar 16, 2015 #2
    Hi wc2351,

    it just so happens that I just solved this problem for the class I am TAing.

    You have the right idea for the proof, but let me lay down the final steps in order for the benefit of anyone who might read it. The key is that the two-electrons Schrödinger equation for the orbital wavefunction [itex]\psi(\mathbf{r}_1,\mathbf{r}_2)[/itex] can be viewed as the single-particle Schrödinger equation in 6 dimensions. Therefore, all we know about the single particle wavefunction also applies here.

    I) As you point out, the orbital ground state cannot have any nodes because a node costs more kinetic energy than it can decrease the potential energy.

    II) Because the Schrödinger equation is real, the eigenfunctions can always be chosen real. More precisely, if [itex]\psi[/itex] is a solution to the time-independent Schrödinger equation, then so is [itex]\psi + \psi^*[/itex]. Because the orbital ground state wavefunction has no nodes, it can then always be chosen positive.

    III) As a bonus, this shows that the orbital ground state wavefunction is unique, because two positive wavefunctions cannot be orthogonal.

    IV) Because the potential is symmetric under interchange of particles, we must have:

    [itex]|\psi(\mathbf{r}_1,\mathbf{r}_2)|^2 = |\psi(\mathbf{r}_2,\mathbf{r}_1)|^2[/itex]

    But the wavefunction is real so:

    [itex]\psi(\mathbf{r}_1,\mathbf{r}_2)^2= \psi(\mathbf{r}_2,\mathbf{r}_1)^2[/itex]

    and it can be chosen positive so we can take the square root:

    [itex]\psi(\mathbf{r}_1,\mathbf{r}_2)= \psi(\mathbf{r}_2,\mathbf{r}_1)[/itex]

    i.e. the orbital ground state is symmetric.

    V) By Pauli exclusion, the spin ground state must therefore be antisymmetric. But for two electrons, the only antisymmetric spin state is the singlet. That completes the proof.

    B.D.
     
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