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.

d) What is, in units of L, the expectation value of the square of the separation between the two particles $<(x_1-x_2 )^2>$ for the singlet and triplet spin states. Does this agree with your answer in part c)?

e) If the fermions were replaced with two bosons that had zero spin explain/show how or if your answers to a), b) and c) would change

d)

This is the hardest part. In general

$<(x_1-x_2 )^2>=<x_1^2>+<x_2^2>-2<x_1 x_2>$

We will show first how to arrive at the simple expression for the $<x_1^2>$ and $<x_2^2>:$

$<x_1^2>=<x_2^2>=1/2*[\int [φ_1^* (x_1 ) x_1^2 φ_1 (x_1 )d x_1+ \int [φ_2^* (x_2 ) x_2^2 φ_2 (x_2 )d x_1]]$

The computations are similar for singlet and triplet states. Take for example the triplet state

$<x_1^2>=<ψ_12^{AS} |x_1^2 | ψ_{12}^{AS}>=<ψ_{12}^A |x_1^2 | ψ_{12}^A>*<1M|1M>=<ψ_{12}^A |x_1^2 | ψ_{12}^A>$

We write again the total function we found above for this state $ψ_{12}^A$:

$ψ_{12}^A (x_1,x_2 )=1/√2 [φ_1 (1) φ_2 (2)-φ_1 (2) φ_2 (1)]$

$<x_1^2>=(1/2)*<[φ_1 (1) φ_2 (2)-φ_1 (2) φ_2 (1)]*|x_1^2 |*[φ_1 (1) φ_2 (2)-φ_1 (2) φ_2 (1) ]>$

$<x_1^2>=(1/2)*{<φ_1 (1)|x_1^2 | φ_1 (1)><φ_2 (2)|φ_2 (2)>- <φ_1 (1) | x_1^2 |φ_2 (1)>*<φ_2 (2) | φ_1 (2)>-<φ_2 (1)|x_1^2 | φ_1 (1)><φ_1 (2)|φ_2 (2)> +<φ_2 (1) | x_1^2 |φ_2 (1)><φ_1 (2) | φ_1 (2)>}$

we know that $<φ_n (2)|φ_m (2)> =δ_{m n}$

So that

$<x_1^2>=1/2*{<φ_1 (1)|x_1^2 | φ_1 (1)>+<φ_2 (1)|x_1^2 | φ_2 (1)>}$

Will say again that this result is independent of which singlet or triplet states we use in the computations. However for the last average value $<x_1,x_2>$ one has for singlet states

$<x_1 x_2>=<ψ_{12}^S |x_1 x_2 | ψ_12^S>=$

$=(2/2) [\iint φ_1^2 (x_1 ) φ_2^2 (x_2 ) x_1 x_2 d x_1 d x_2+\iint φ_1 (x_1 ) φ_2 (x_2 )*φ_1 (x_2 ) φ_2 (x_1 ) x_1 x_2 d x_1 d x_2 ]$

And for triplet states:

$<x_1 x_2>=<ψ_{12}^A |x_1 x_2 | ψ_{12}^A>=(2/2) [\iint φ_1^2 (x_1 ) φ_2^2 (x_2 ) x_1 x_2 d x_1 d x_2-$

$-\iintφ_1 (x_1 ) φ_2 (x_2 )*φ_1 (x_2 ) φ_2 (x_1 ) x_1 x_2 d x_1 d x_2]$

Next we compute the values of the integrals:

$(2/L) ∫_0^L x^2 \sin^2 ((π/L)*x)d x=(2/L)*0.141L^3==0.282L^2$

and $(2/L)*∫_0^L x^2 \sin^2 ((4π/L) x)d x=2/L*0.165L^3=0.33L^2$

Therefore the first two average terms are:

$2<x^2>=(0.282+0.33) L^2=0.612L^2$

For the last average term $2<x_1,x_2>$ we have:

$(4/L^2) ∫_0^L ∫_0^L x_1 x_2*\sin^2 ((π/L) x_1 ) \sin^2 ((4π/L)*x_2 )d x_1 d x_2=$

$=(4/L^2) *(L^2/4)*(L^2/4)=(L^2/4)=0.25L^2$

and

$(4/L^2) *∫_0^L∫_0^L x_1 x_2*\sin ((π/L) x_1 )*\sin ((π/L x_2 )*\sin ((4π/L) x_2 )*\sin ((4π/L) x_1 )d x_1 d x_2=$

$= 4*0.0072^2*L^2=2*10^{-4}*L^2$

Therefore

$2<x_1 x_2>=(0.25+0.0002) L^2=0.2502L^2$ for singlet states

$2<x_1 x_2>=(0.25-0.0002) L^2=0.2498L^2$ for triplet states

And the total separation is

$<(x_1-x_2 )^2>=(0.612+0.2502) L^2$ for singlet states

$<(x_1-x_2 )^2>=(0.612-0.2498) L^2$ for triplet states

The explanation for getting a different average distance between the two electrons is simple. Since in the singlet state the electrons are closer they will have a higher energy (it is necessary more energy to keep two negative charges closer) At point c) above we have seen that the energy of the singlet state is higher.

$E_{singlet}=E_{12}^0+J_{12}+K_{12}$ and $E_{triplet}=E_{12}^0+J_{12}-K_{12}$

e)

Bosons are particles of integer spin. For bosons the total wave function (spatial and spin parts together) need to by symmetrical. Since the spin part is always symmetrical and equal to |00> the total wave function is

$ψ_{12} (x_1,x_2 )=|ψ_{12}^{SS}>=(1/\sqrt{2})*[φ_1 (x_1 ) φ_2 (x_2 )+φ_1 (x_2 ) φ_2 (x_1 )]*|00>$

Since there is a single state possible always the energy will be corrected just in the positive “sense”, that is

$E_{12}=E_{12}^0+J_{12}+K_{12}=(E_1+E_4)+J_{12}+K_{12}$

Therefore all bosons will share the same energy level.