Exercise 45 refers to state I and II and put their algebraic sum in a simple form. (a) Determine algebraic difference of state I and state II.

(b) Determine whether after swapping spatial state and spin state separately, the algebraic difference of state I and state II is symmetric, antisymmetric or neither, and to check whether the algebraic difference becomes antisymmetric after swapping spatial and spin states both.

The algebraic difference of states is given.

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

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

(a)

Let's start by writing down our states I and II.

$$\begin{gathered} {\psi }_{\text{I}}\left(x_1,x_2\right)={\psi }_n\left(x_1\right)\uparrow {\psi }_{n^{\text{'}}}\left(x_2\right)\downarrow -{\psi }_{n^{\text{'}}}\left(x_1\right)\downarrow {\psi }_n\left(x_2\right)\uparrow \\ {\psi }_{\Pi }\left(x_1,x_2\right)={\psi }_n\left(x_1\right)\downarrow {\psi }_{n^{\text{'}}}\left(x_2\right)\uparrow -{\psi }_{n^{\text{'}}}\left(x_1\right)\uparrow {\psi }_n\left(x_2\right)\downarrow \end{gathered}$$

Taking the difference between these states gives us

(b)

Starting with the algebraic difference of part (a), we swap the spatial states-the n and n-alone and show that

$$\begin{gathered} {\psi }_{\text{I}-\text{II}}^{\text{'}}\left(x_1,x_2\right)=\left({\psi }_{n^{\text{'}}}\left(x_1\right){\psi }_n\left(x_2\right)+{\psi }_n\left(x_1\right){\psi }_{n^{\text{'}}}\left(x_2\right)\right)(\uparrow \downarrow -\downarrow \uparrow ) \\ {\psi }_{\text{I}-\text{II}}^{\text{'}}\left(x_1,x_2\right)=\left({\psi }_n\left(x_1\right){\psi }_{n^{\text{'}}}\left(x_2\right)+{\psi }_{n^{\text{'}}}\left(x_1\right){\psi }_n\left(x_2\right)\right)(\uparrow \downarrow -\downarrow \uparrow ) \\ {\psi }_{\text{I}-\text{II}}^{\text{'}}\left(x_1,x_2\right)={\psi }_{\text{I}-\Pi }\left(x_1,x_2\right) \end{gathered}$$

Thus, we find that the algebraic difference is symmetric with respect to swapping spatial states.

Similarly, starting from the algebraic difference again and swapping spin states/arrows alone this time, we have

$$\begin{gathered} {\psi }_{\text{I}-\text{II}}^{\text{'}}\left(x_1,x_2\right)=\left({\psi }_n\left(x_1\right){\psi }_{n^{\text{'}}}\left(x_2\right)+{\psi }_{n^{\text{'}}}\left(x_1\right){\psi }_n\left(x_2\right)\right)(\downarrow \uparrow -\uparrow \downarrow ) \\ {\psi }_{\text{I}-\text{II}}^{\text{'}}\left(x_1,x_2\right)=-\left({\psi }_n\left(x_1\right){\psi }_{n^{\text{'}}}\left(x_2\right)-{\psi }_{n^{\text{'}}}\left(x_1\right){\psi }_n\left(x_2\right)\right)(\uparrow \downarrow -\downarrow \uparrow ) \\ {\psi }_{\text{I}-\text{II}}^{\text{'}}\left(x_1,x_2\right)=-{\psi }_{\text{I}-\text{II}}\left(x_1,x_2\right) \end{gathered}$$

We see that${\psi }_{\text{I}-\Pi \text{I}}$ is anti symmetric with respect to swapping spin states alone.

$$\begin{gathered} {\psi }_{\text{I}-\text{II}}=\left({\psi }_{n^{\text{'}}}\left(x_1\right){\psi }_n\left(x_2\right)+{\psi }_n\left(x_1\right){\psi }_{n^{\text{'}}}\left(x_2\right)\right)(\downarrow \uparrow -\uparrow \downarrow ) \\ {\psi }_{\text{I}-\text{II}}=-\left({\psi }_n\left(x_1\right){\psi }_{n^{\text{'}}}\left(x_2\right)+{\psi }_{n^{\text{'}}}\left(x_1\right){\psi }_n\left(x_2\right)\right)(\uparrow \downarrow -\downarrow \uparrow ) \\ {\psi }_{\text{I}-\text{II}}=-{\psi }_{\text{I}-\text{II}} \end{gathered}$$

which shows that the algebraic difference is still antisymmetric if we swap the particles' total states. This result should be expected for we know that the algebraic difference between two antisymmetric pieces-states I and II-results in an antisymmetric state. Note that the same results apply for the algebraic difference ${\psi }_{\text{II - I.}}$.