Exercise 44 gives an antisymmetric multiparticle state for two particles in a box with opposite spins. Another antisymmetric state with spins opposite and the same quantum numbers is ${\mathbf{\psi }}_{\mathbf{n}}\mathbf{\left(}{\mathbf{x}}_{\mathbf{1}}\mathbf{\right)}{\mathbf{\downarrow }}_{{\mathbf{n}}^{\mathbf{2}}}\mathbf{\left(}{\mathbf{x}}_{\mathbf{2}}\mathbf{\right)}\mathbf{\uparrow }\mathbf{-}{\mathbf{\psi }}_{{\mathbf{n}}^{\mathbf{n}}}\mathbf{\left(}{\mathbf{x}}_{\mathbf{1}}\mathbf{\right)}\mathbf{\uparrow }{\mathbf{\psi }}_{\mathbf{n}}\mathbf{\left(}{\mathbf{x}}_{\mathbf{2}}\mathbf{\right)}\mathbf{\downarrow }$

Refer to these states as 1 and 11. We have tended to characterize exchange symmetry as to whether the state's sign changes when we swap particle labels. but we could achieve the same result by instead swapping the particles' stares, specifically theandin equation (8-22). In this exercise. we look at swapping only parts of the state-spatial or spin.

(a) What is the exchange symmetric-symmetric (unchanged). antisymmetric (switching sign). or neither-of multiparticle states 1 and Itwith respect to swapping spatial states alone?

(b) Answer the same question. but with respect to swapping spin states/arrows alone.

(c) Show that the algebraic sum of states I and II may be written$\mathbf{(}{\mathbf{\psi }}_{\mathbf{n}}\mathbf{\left(}{\mathbf{x}}_{\mathbf{1}}\mathbf{\right)}{\mathbf{\psi }}_{{\mathbf{n}}^{\mathbf{\text{'}}}}\mathbf{\left(}{\mathbf{x}}_{\mathbf{2}}\mathbf{\right)}\mathbf{-}{\mathbf{\psi }}_{{\mathbf{n}}^{\mathbf{\text{'}}}}\mathbf{\left(}{\mathbf{x}}_{\mathbf{1}}\mathbf{\right)}{\mathbf{\psi }}_{\mathbf{n}}\mathbf{\left(}{\mathbf{x}}_{\mathbf{2}}\mathbf{\right)}\mathbf{)}\mathbf{(}\mathbf{\downarrow }\mathbf{\uparrow }\mathbf{+}\mathbf{\uparrow }\mathbf{\downarrow }\mathbf{)}$

Where the left arrow in any couple represents the spin of particle 1 and the right arrow that of particle?

(d) Answer the same questions as in parts (a) and (b), but for this algebraic sum.

(e) ls the sum of states I and 11 still antisymmetric if we swap the particles' total-spatial plus spin-states?

(f) if the two particles repel each other, would any of the three multiparticle states-l. II. and the sum-be preferred?

Explain.

The two combinations of spin and spatial wave.

The energy of electron in ${\mathbf{\text{n}}}^{\mathbf{\text{th}}}$-orbit:$\mathbf{E}\mathbf{=}{\mathbf{Z}}^{\mathbf{2}}\left(-\frac{13.6\mathrm{eV}}{n^2}\right)$

The energy KE which an object of charge q gains by passing through the potential difference$\mathbf{\Delta V}$is $\mathbf{KE}\mathbf{=}\mathbf{q\Delta V}$.

(a)

The two combinations of spin and spatial wave function are:

${\mathrm{\psi }}_1({\mathrm{x}}_1,{\mathrm{x}}_2)={\mathrm{\psi }}_{\text{n}}\left({\mathrm{x}}_1\right)\uparrow {\mathrm{\psi }}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_2\right)\downarrow -{\mathrm{\psi }}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_1\right)\downarrow {\mathrm{\psi }}_{\text{n}}\left({\mathrm{x}}_2\right)\uparrow$……… (1)

$\uparrow$

${\mathrm{\psi }}_{\text{II}}({\mathrm{x}}_1,{\mathrm{x}}_2)={\mathrm{\psi }}_{\text{n}}\left({\mathrm{x}}_1\right)\downarrow {\mathrm{\psi }}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_2\right)\uparrow -{\mathrm{\psi }}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_1\right)\uparrow {\mathrm{\psi }}_{\text{n}}\left({\mathrm{x}}_2\right)$……… (2)

$\downarrow$

Where, ${\mathrm{\psi }}_{\text{n}}\left({\mathrm{x}}_1\right)=\sqrt{\frac{2}{\mathrm{L}}}\sin \frac{{\mathrm{n\pi x}}_1}{\mathrm{L}}$and ${\mathrm{\psi }}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_2\right)=\sqrt{\frac{2}{\mathrm{L}}}\sin \frac{{\mathrm{n\pi x}}_2}{\mathrm{L}}$

After interchanging the spatial state ${\mathrm{\psi }}_{\text{n}}\left({\mathrm{x}}_1\right)$and ${\mathrm{\psi }}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_2\right)$of ${\mathrm{\psi }}_1({\mathrm{x}}_1,{\mathrm{x}}_2)$, it become

${\mathrm{\psi }}_{{\text{I}}^{\text{'}}}({\mathrm{x}}_1,{\mathrm{x}}_2)={\mathrm{\psi }}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_1\right)\uparrow {\mathrm{\psi }}_{\text{n}}\left({\mathrm{x}}_2\right)\downarrow -{\mathrm{\psi }}_{\text{n}}\left({\mathrm{x}}_1\right)\downarrow {\mathrm{\psi }}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_2\right)\uparrow$………. (3)

After Interchange the spatial state ${\mathrm{\psi }}_{\text{n}}\left({\mathrm{x}}_1\right)$and ${\mathrm{\psi }}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_2\right)$of${\mathrm{\psi }}_{\text{II}}({\mathrm{x}}_1,{\mathrm{x}}_2)$ , it become

${\mathrm{\psi }}_{\text{II}}({\mathrm{x}}_1,{\mathrm{x}}_2)={\mathrm{\psi }}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_1\right)\downarrow {\mathrm{\psi }}_{\text{n}}\left({\mathrm{x}}_2\right)\uparrow -{\mathrm{\psi }}_{\text{n}}\left({\mathrm{x}}_1\right)\uparrow {\mathrm{\psi }}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_2\right)\downarrow$…….. (4)

(b)

The two combinations of spin and spatial wave function are:

${\mathrm{\psi }}_1({\mathrm{x}}_1,{\mathrm{x}}_2)={\mathrm{\psi }}_{\text{n}}\left({\mathrm{x}}_1\right)\uparrow {\mathrm{\psi }}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_2\right)\downarrow -{\mathrm{\psi }}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_1\right)\downarrow {\mathrm{\psi }}_{\text{n}}\left({\mathrm{x}}_2\right)$ ……… (1)

$\uparrow$

${\mathrm{\psi }}_{\text{II}}({\mathrm{x}}_1,{\mathrm{x}}_2)={\mathrm{\psi }}_{\text{n}}\left({\mathrm{x}}_1\right)\downarrow {\mathrm{\psi }}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_2\right)\uparrow -{\mathrm{\psi }}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_1\right)\uparrow {\mathrm{\psi }}_{\text{n}}\left({\mathrm{x}}_2\right)$……… (2)

$\downarrow$

Where, ${\mathrm{\psi }}_{\text{n}}\left({\mathrm{x}}_1\right)=\sqrt{\frac{2}{\mathrm{L}}}\sin \frac{{\mathrm{n\pi x}}_1}{\mathrm{L}}$and ${\mathrm{\psi }}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_2\right)=\sqrt{\frac{2}{\mathrm{L}}}\sin \frac{{\mathrm{n\pi x}}_2}{\mathrm{L}}$

After interchanging the spatial state ${\mathrm{\psi }}_{\text{n}}\left({\mathrm{x}}_1\right)$and${\mathrm{\psi }}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_2\right)$ of ${\mathrm{\psi }}_1({\mathrm{x}}_1,{\mathrm{x}}_2)$, it become

${\mathrm{\psi }}_{{\text{I}}^{\text{'}}}({\mathrm{x}}_1,{\mathrm{x}}_2)={\mathrm{\psi }}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_1\right)\uparrow {\mathrm{\psi }}_{\text{n}}\left({\mathrm{x}}_2\right)\downarrow -{\mathrm{\psi }}_{\text{n}}\left({\mathrm{x}}_1\right)\downarrow {\mathrm{\psi }}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_2\right)\uparrow$………. (3)

After Interchange the spatial state${\mathrm{\psi }}_{\text{n}}\left({\mathrm{x}}_1\right)$ and ${\mathrm{\psi }}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_2\right)$of ${\mathrm{\psi }}_{\text{II}}({\mathrm{x}}_1,{\mathrm{x}}_2)$, it become

${\mathrm{\psi }}_{\text{II}}({\mathrm{x}}_1,{\mathrm{x}}_2)={\mathrm{\psi }}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_1\right)\downarrow {\mathrm{\psi }}_{\text{n}}\left({\mathrm{x}}_2\right)\uparrow -{\mathrm{\psi }}_{\text{n}}\left({\mathrm{x}}_1\right)\uparrow {\mathrm{\psi }}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_2\right)\downarrow$…….. (4)

(c)

The two combinations of spin and spatial wave function are:

${\mathrm{\psi }}_1({\mathrm{x}}_1,{\mathrm{x}}_2)={\mathrm{\psi }}_{\text{n}}\left({\mathrm{x}}_1\right)\uparrow {\mathrm{\psi }}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_2\right)\downarrow -{\mathrm{\psi }}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_1\right)\downarrow {\mathrm{\psi }}_{\text{n}}\left({\mathrm{x}}_2\right)$ ……… (1)

$\uparrow$

${\mathrm{\psi }}_{\text{II}}({\mathrm{x}}_1,{\mathrm{x}}_2)={\mathrm{\psi }}_{\text{n}}\left({\mathrm{x}}_1\right)\downarrow {\mathrm{\psi }}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_2\right)\uparrow -{\mathrm{\psi }}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_1\right)\uparrow {\mathrm{\psi }}_{\text{n}}\left({\mathrm{x}}_2\right)$……… (2)

$\downarrow$

Where, ${\mathrm{\psi }}_{\text{n}}\left({\mathrm{x}}_1\right)=\sqrt{\frac{2}{\mathrm{L}}}\sin \frac{{\mathrm{n\pi x}}_1}{\mathrm{L}}$ and ${\mathrm{\psi }}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_2\right)=\sqrt{\frac{2}{\mathrm{L}}}\sin \frac{{\mathrm{n\pi x}}_2}{\mathrm{L}}$

Add states (I) and (II) together to obtain

$\begin{array}{l}{\mathrm{\psi }}_{\text{sum}}={\mathrm{\psi }}_1({\mathrm{x}}_1,{\mathrm{x}}_2)+{\mathrm{\psi }}_{11}({\mathrm{x}}_1,{\mathrm{x}}_2)\\ {\mathrm{\psi }}_{\text{sum}}=\left[\begin{array}{l}{\mathrm{\psi }}_{\text{n}}\left({\mathrm{x}}_1\right)\uparrow {\mathrm{\psi }}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_2\right)\downarrow -{\mathrm{\psi }}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_1\right)\downarrow {\mathrm{\psi }}_{\text{n}}\left({\mathrm{x}}_2\right)\uparrow \\ {\mathrm{\psi }}_{\text{n}}\left({\mathrm{x}}_1\right)\downarrow {\mathrm{\psi }}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_2\right)\uparrow -{\mathrm{\psi }}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_1\right)\uparrow {\mathrm{\psi }}_{\text{n}}\left({\mathrm{x}}_2\right)\downarrow \end{array}\right]\\ {\mathrm{\psi }}_{\text{sum}}=\left[{\mathrm{\psi }}_{\text{n}}\left({\mathrm{x}}_1\right){\mathrm{\psi }}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_2\right)(\uparrow \downarrow +\downarrow \uparrow )-{\mathrm{\psi }}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_1\right){\mathrm{\psi }}_{\text{n}}\left({\mathrm{x}}_2\right)(\uparrow \downarrow +\downarrow \uparrow )\right]\\ {\mathrm{\psi }}_{\text{sum}}=[{\mathrm{\psi }}_{\text{n}}\left({\mathrm{x}}_1\right){\mathrm{\psi }}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_2\right)-{\mathrm{\psi }}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_1\right){\mathrm{\psi }}_{\text{n}}\left({\mathrm{x}}_2\right)](\uparrow \downarrow +\downarrow \uparrow )\end{array}$

(d)

Algebraic sum of state I and state II is${\mathrm{\psi }}_{\text{sum}}=[{\mathrm{\psi }}_{\text{n}}\left({\mathrm{x}}_1\right){\mathrm{\psi }}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_2\right)-{\mathrm{\psi }}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_1\right){\mathrm{\psi }}_{\text{n}}\left({\mathrm{x}}_2\right)](\uparrow \downarrow +\downarrow \uparrow )$

${\mathrm{\psi }}_{\text{sum}}=[{\mathrm{\psi }}_{\text{n}}\left({\mathrm{x}}_1\right){\mathrm{\psi }}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_2\right)-{{\mathrm{\psi }}_{\text{n}}}^{\text{'}}\left({\mathrm{x}}_1\right){\mathrm{\psi }}_{\text{n}}\left({\mathrm{x}}_2\right)](\uparrow \downarrow +\downarrow \uparrow )$

Interchanging spatial state alone,

$\begin{array}{l}{\mathrm{\psi }}_{\text{A},\text{sum}}=[{\mathrm{\psi }}_{\text{n}}\left({\mathrm{x}}_2\right){\mathrm{\psi }}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_1\right)-{\mathrm{\psi }}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_2\right){\mathrm{\psi }}_{\text{n}}\left({\mathrm{x}}_1\right)](\uparrow \downarrow +\downarrow \uparrow )\\ {\mathrm{\psi }}_{\text{A},\text{sum}}=-[{\mathrm{\psi }}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_2\right){\mathrm{\psi }}_{\text{n}}\left({\mathrm{x}}_1\right)-{\mathrm{\psi }}_{\text{n}}\left({\mathrm{x}}_2\right){\mathrm{\psi }}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_1\right)](\uparrow \downarrow +\downarrow \uparrow )\\ {\mathrm{\psi }}_{\text{A},\text{sum}}=-[{\mathrm{\psi }}_{\text{n}}\left({\mathrm{x}}_1\right){\mathrm{\psi }}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_2\right)-{\mathrm{\psi }}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_1\right){\mathrm{\psi }}_{\text{n}}\left({\mathrm{x}}_2\right)](\uparrow \downarrow +\downarrow \uparrow )\end{array}$

So, it asymmetric

Interchanging spin state alone,

$\begin{array}{c}{\mathrm{\psi }}_{\text{s},\text{sum}}=[{\mathrm{\psi }}_{\text{n}}\left({\mathrm{x}}_1\right){\mathrm{\psi }}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_2\right)-{\mathrm{\psi }}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_1\right){\mathrm{\psi }}_{\text{n}}\left({\mathrm{x}}_2\right)](\downarrow \uparrow +\uparrow \downarrow )\\ [{\mathrm{\psi }}_{\text{n}}\left({\mathrm{x}}_1\right){\mathrm{\psi }}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_2\right)-{\mathrm{\psi }}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_1\right){\mathrm{\psi }}_{\text{n}}\left({\mathrm{x}}_2\right)](\uparrow \downarrow +\downarrow \uparrow )\end{array}$

Interchanging spin doesn't change the sign, so it is symmetric.

(e)

Algebraic sum of state I and state II is${\mathrm{\psi }}_{\text{sum}}=[{\mathrm{\psi }}_{\text{n}}\left({\mathrm{x}}_1\right){\mathrm{\psi }}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_2\right)-{\mathrm{\psi }}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_1\right){\mathrm{\psi }}_{\text{n}}\left({\mathrm{x}}_2\right)](\uparrow \downarrow +\downarrow \uparrow )$

Swapping spatial state would change the sign of the algebraic sum, so it become asymmetric. But swapping spin state would not able to the sign of the algebraic sum, so it become symmetric.

${\mathrm{\psi }}_{\text{sum}}=[{\mathrm{\psi }}_{\text{n}}\left({\mathrm{x}}_1\right){\mathrm{\psi }}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_2\right)-{\mathrm{\psi }}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_1\right){\mathrm{\psi }}_{\text{n}}\left({\mathrm{x}}_2\right)](\uparrow \downarrow +\downarrow \uparrow )$

So, first swapping spatial state

${\mathrm{\psi }}_{\text{sum}}^{\text{'}}=[{\mathrm{\psi }}_{\text{n}}\left({\mathrm{x}}_2\right){\mathrm{\psi }}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_1\right)-{\mathrm{\psi }}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_2\right){\mathrm{\psi }}_{\text{n}}\left({\mathrm{x}}_1\right)](\uparrow \downarrow +\downarrow \uparrow )$

Secondly swap spin state

${\mathrm{\psi }}_{\text{sum}}^{\text{'}}=[{\mathrm{\psi }}_{\text{n}}\left({\mathrm{x}}_2\right){\mathrm{\psi }}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_1\right)-{\mathrm{\psi }}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_2\right){\mathrm{\psi }}_{\text{n}}\left({\mathrm{x}}_1\right)](\downarrow \uparrow +\uparrow \downarrow )$

Rearranging the above function

$\begin{array}{l}{\mathrm{\psi }}_{\text{sum}}^{\text{'}}=-[{\mathrm{\psi }}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_2\right){\mathrm{\psi }}_{\text{n}}\left({\mathrm{x}}_1\right)-{\mathrm{\psi }}_{\text{n}}\left({\mathrm{x}}_2\right){\mathrm{\psi }}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_1\right)](\uparrow \downarrow +\downarrow \uparrow )\\ {\mathrm{\psi }}_{\text{sum}}^{\text{'}}=-[{\mathrm{\psi }}_{\text{n}}\left({\mathrm{x}}_1\right){\mathrm{\psi }}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_2\right)-{\mathrm{\psi }}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_1\right){\mathrm{\psi }}_{\text{n}}\left({\mathrm{x}}_2\right)](\uparrow \downarrow +\downarrow \uparrow )\\ {\mathrm{\psi }}_{\text{sum}}^{\text{'}}=-{\mathrm{\psi }}_{\text{sum}}\end{array}$

(f)

Since, multiple of particle states I and II are neither symmetric nor antisymmetric but the sum of the state ${\mathrm{\psi }}_{\text{sum}}$is asymmetric so it would be preferred to keep the repelling particles farther apart, to keep retain the lower energy.