Two Fermions Systems – Energy (2)

Consider two fermions in an infinite square well of width L. On is in the n=1 state and one has been excited to the n=4 state.

b) Using your favorite plotting package, plot the spatial wave function probability density vs. $x_1$ and $x_2$ (a 3D contour plot) for the singlet and triplet spin state.

c) If the fermions were electrons (such that they now interact through the Coulomb force), which spin state (singlet or triplet has a higher energy), or do they all have the same energy? Explain using your results from parts a) and b).


Since all spin states are orthogonal on each other and also normed it follows that the probability density is either

$P^S (x_1,x_2 )=|ψ_{12}^{SA}|^2=$

$=(2/L^2) *[sin⁡(π/L*x_1 )*sin⁡(4π/L*x_2 )+sin⁡(π/L*x_2 )*sin⁡(4π/L*x_1 ) ]^2$

$P^A (x_1,x_2 )=|ψ_{12}^{AS}|^2=$

$=(2/L^2) *[sin⁡(π/L*x_1 )*sin⁡(4π/L*x_2 )-sin⁡(π/L*x_2 )*sin⁡(4π/L*x_1 ) ]^2$

The graphs are below.

$P^S (x_1,x_2 )$ and  $P^A (x_1,x_2 )$:

Show My Homework -Two Fermions Probability Density


The two electrons interact between them. Additional to the “standard” Hamiltonian

$H=(-ℏ^2/2m)*(∇_1^2+∇_2^2)+V_1 (x_1 )+V_2 (x_2)$

There will be an additional term in the Hamiltonian that comes from the interaction between the two electrons $V_{12} (x_2-x_1)$. The perturbation theory says that the energy levels in this case are the same old energy levels $E_{12}$ of non interacting electrons corrected with a term:

$E_{12}’=<ψ_{12} |V_{12} | ψ_{12}>$

For the singlet wave function one has

$E_{12}’ (1)=<ψ_{12}^{AS} |V_{12} | ψ_{12}^{AS}>=<ψ_{12}^A |V_{12} | ψ_{12}^A>*<00|00>=<ψ_{12}^A |V_{12} | ψ_{12}^A>$

The same for the triplet wave function  the correction in energy is:

$E_{12}’ (2)=⋯=<ψ_{12}^S |V_{12} | ψ_{12}^S>$

By defining

$J_{12}=iint (|φ_1 (x_1 ) |^2*V_{12}*|φ_2 (x_2 ) |^2*d x_1 d x_2)$

$K_{12}=iint (φ_1^* (x_1 ) φ_2^* (x_2) V_{12}*φ_1 (x_2 ) φ_2 (x_2 )*d x_1 d x_2)$

The first integral $J_{12}$ is called the coulomb direct integral and the second integral $K_{12}$  is called the coulomb exchange integral. One has for the correction terms of energy (this can be easily seen from the total wave functions defined in a)

$E_{12}’ (1)=J_{12}+K_{12}$  (singlet)  and $E_{12}’ (2)=J_{12}-K_{12}$  (triplet)

Therefore the singlet state will have a higher energy than the triplet states.