91影视

The two combinations of spin and spatial wave.

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

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

(a)

The two combinations of spin and spatial wave function are:

${\mathrm{蠄}}_1({\mathrm{x}}_1,{\mathrm{x}}_2)={\mathrm{蠄}}_{\text{n}}\left({\mathrm{x}}_1\right)鈫�{\mathrm{蠄}}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_2\right)鈫�鈭�{\mathrm{蠄}}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_1\right)鈫�{\mathrm{蠄}}_{\text{n}}\left({\mathrm{x}}_2\right)鈫�\text{鈥�}$鈥︹�︹�� (1)

$鈫�$

${\mathrm{蠄}}_{\text{II}}({\mathrm{x}}_1,{\mathrm{x}}_2)={\mathrm{蠄}}_{\text{n}}\left({\mathrm{x}}_1\right)鈫�{\mathrm{蠄}}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_2\right)鈫�鈭�{\mathrm{蠄}}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_1\right)鈫�{\mathrm{蠄}}_{\text{n}}\left({\mathrm{x}}_2\right)$鈥︹�︹�� (2)

$鈫�$

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

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

${\mathrm{蠄}}_{{\text{I}}^{\text{'}}}({\mathrm{x}}_1,{\mathrm{x}}_2)={\mathrm{蠄}}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_1\right)鈫�{\mathrm{蠄}}_{\text{n}}\left({\mathrm{x}}_2\right)鈫�鈭�{\mathrm{蠄}}_{\text{n}}\left({\mathrm{x}}_1\right)鈫�{\mathrm{蠄}}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_2\right)鈫�$鈥︹�︹��. (3)

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

${\mathrm{蠄}}_{\text{II}}({\mathrm{x}}_1,{\mathrm{x}}_2)={\mathrm{蠄}}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_1\right)鈫�{\mathrm{蠄}}_{\text{n}}\left({\mathrm{x}}_2\right)鈫�鈭�{\mathrm{蠄}}_{\text{n}}\left({\mathrm{x}}_1\right)鈫�{\mathrm{蠄}}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_2\right)鈫�$鈥︹��.. (4)

(b)

The two combinations of spin and spatial wave function are:

${\mathrm{蠄}}_1({\mathrm{x}}_1,{\mathrm{x}}_2)={\mathrm{蠄}}_{\text{n}}\left({\mathrm{x}}_1\right)鈫�{\mathrm{蠄}}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_2\right)鈫�鈭�{\mathrm{蠄}}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_1\right)鈫�{\mathrm{蠄}}_{\text{n}}\left({\mathrm{x}}_2\right)$ 鈥︹�︹�� (1)

$鈫�$

${\mathrm{蠄}}_{\text{II}}({\mathrm{x}}_1,{\mathrm{x}}_2)={\mathrm{蠄}}_{\text{n}}\left({\mathrm{x}}_1\right)鈫�{\mathrm{蠄}}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_2\right)鈫�鈭�{\mathrm{蠄}}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_1\right)鈫�{\mathrm{蠄}}_{\text{n}}\left({\mathrm{x}}_2\right)$鈥︹�︹�� (2)

$鈫�$

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

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

${\mathrm{蠄}}_{{\text{I}}^{\text{'}}}({\mathrm{x}}_1,{\mathrm{x}}_2)={\mathrm{蠄}}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_1\right)鈫�{\mathrm{蠄}}_{\text{n}}\left({\mathrm{x}}_2\right)鈫�鈭�{\mathrm{蠄}}_{\text{n}}\left({\mathrm{x}}_1\right)鈫�{\mathrm{蠄}}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_2\right)鈫�$鈥︹�︹��. (3)

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

${\mathrm{蠄}}_{\text{II}}({\mathrm{x}}_1,{\mathrm{x}}_2)={\mathrm{蠄}}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_1\right)鈫�{\mathrm{蠄}}_{\text{n}}\left({\mathrm{x}}_2\right)鈫�鈭�{\mathrm{蠄}}_{\text{n}}\left({\mathrm{x}}_1\right)鈫�{\mathrm{蠄}}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_2\right)鈫�$鈥︹��.. (4)

(c)

The two combinations of spin and spatial wave function are:

${\mathrm{蠄}}_1({\mathrm{x}}_1,{\mathrm{x}}_2)={\mathrm{蠄}}_{\text{n}}\left({\mathrm{x}}_1\right)鈫�{\mathrm{蠄}}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_2\right)鈫�鈭�{\mathrm{蠄}}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_1\right)鈫�{\mathrm{蠄}}_{\text{n}}\left({\mathrm{x}}_2\right)$ 鈥︹�︹�� (1)

$鈫�$

${\mathrm{蠄}}_{\text{II}}({\mathrm{x}}_1,{\mathrm{x}}_2)={\mathrm{蠄}}_{\text{n}}\left({\mathrm{x}}_1\right)鈫�{\mathrm{蠄}}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_2\right)鈫�鈭�{\mathrm{蠄}}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_1\right)鈫�{\mathrm{蠄}}_{\text{n}}\left({\mathrm{x}}_2\right)$鈥︹�︹�� (2)

$鈫�$

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

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

$\begin{array}{l}{\mathrm{蠄}}_{\text{sum}}={\mathrm{蠄}}_1({\mathrm{x}}_1,{\mathrm{x}}_2)+{\mathrm{蠄}}_{11}({\mathrm{x}}_1,{\mathrm{x}}_2)\\ {\mathrm{蠄}}_{\text{sum}}=\left[\begin{array}{l}{\mathrm{蠄}}_{\text{n}}\left({\mathrm{x}}_1\right)鈫�{\mathrm{蠄}}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_2\right)鈫�鈭�{\mathrm{蠄}}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_1\right)鈫�{\mathrm{蠄}}_{\text{n}}\left({\mathrm{x}}_2\right)鈫�\\ {\mathrm{蠄}}_{\text{n}}\left({\mathrm{x}}_1\right)鈫�{\mathrm{蠄}}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_2\right)鈫�鈭�{\mathrm{蠄}}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_1\right)鈫�{\mathrm{蠄}}_{\text{n}}\left({\mathrm{x}}_2\right)鈫�\end{array}\right]\\ {\mathrm{蠄}}_{\text{sum}}=\left[{\mathrm{蠄}}_{\text{n}}\left({\mathrm{x}}_1\right){\mathrm{蠄}}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_2\right)(鈫�鈫�+鈫�鈫�)鈭�{\mathrm{蠄}}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_1\right){\mathrm{蠄}}_{\text{n}}\left({\mathrm{x}}_2\right)(鈫�鈫�+鈫�鈫�)\right]\\ {\mathrm{蠄}}_{\text{sum}}=[{\mathrm{蠄}}_{\text{n}}\left({\mathrm{x}}_1\right){\mathrm{蠄}}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_2\right)鈭�{\mathrm{蠄}}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_1\right){\mathrm{蠄}}_{\text{n}}\left({\mathrm{x}}_2\right)](鈫�鈫�+鈫�鈫�)\end{array}$

(d)

Algebraic sum of state I and state II is${\mathrm{蠄}}_{\text{sum}}=[{\mathrm{蠄}}_{\text{n}}\left({\mathrm{x}}_1\right){\mathrm{蠄}}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_2\right)鈭�{\mathrm{蠄}}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_1\right){\mathrm{蠄}}_{\text{n}}\left({\mathrm{x}}_2\right)](鈫�鈫�+鈫�鈫�)$

${\mathrm{蠄}}_{\text{sum}}=[{\mathrm{蠄}}_{\text{n}}\left({\mathrm{x}}_1\right){\mathrm{蠄}}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_2\right)鈭�{{\mathrm{蠄}}_{\text{n}}}^{\text{'}}\left({\mathrm{x}}_1\right){\mathrm{蠄}}_{\text{n}}\left({\mathrm{x}}_2\right)](鈫�鈫�+鈫�鈫�)$

Interchanging spatial state alone,

$\begin{array}{l}{\mathrm{蠄}}_{\text{A},\text{sum}}=[{\mathrm{蠄}}_{\text{n}}\left({\mathrm{x}}_2\right){\mathrm{蠄}}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_1\right)鈭�{\mathrm{蠄}}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_2\right){\mathrm{蠄}}_{\text{n}}\left({\mathrm{x}}_1\right)](鈫�鈫�+鈫�鈫�)\\ {\mathrm{蠄}}_{\text{A},\text{sum}}=鈭�[{\mathrm{蠄}}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_2\right){\mathrm{蠄}}_{\text{n}}\left({\mathrm{x}}_1\right)鈭�{\mathrm{蠄}}_{\text{n}}\left({\mathrm{x}}_2\right){\mathrm{蠄}}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_1\right)](鈫�鈫�+鈫�鈫�)\\ {\mathrm{蠄}}_{\text{A},\text{sum}}=鈭�[{\mathrm{蠄}}_{\text{n}}\left({\mathrm{x}}_1\right){\mathrm{蠄}}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_2\right)鈭�{\mathrm{蠄}}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_1\right){\mathrm{蠄}}_{\text{n}}\left({\mathrm{x}}_2\right)](鈫�鈫�+鈫�鈫�)\end{array}$

So, it asymmetric

Interchanging spin state alone,

$\begin{array}{c}{\mathrm{蠄}}_{\text{s},\text{sum}}=[{\mathrm{蠄}}_{\text{n}}\left({\mathrm{x}}_1\right){\mathrm{蠄}}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_2\right)鈭�{\mathrm{蠄}}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_1\right){\mathrm{蠄}}_{\text{n}}\left({\mathrm{x}}_2\right)](鈫�鈫�+鈫�鈫�)\\ [{\mathrm{蠄}}_{\text{n}}\left({\mathrm{x}}_1\right){\mathrm{蠄}}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_2\right)鈭�{\mathrm{蠄}}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_1\right){\mathrm{蠄}}_{\text{n}}\left({\mathrm{x}}_2\right)](鈫�鈫�+鈫�鈫�)\end{array}$

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

(e)

Algebraic sum of state I and state II is${\mathrm{蠄}}_{\text{sum}}=[{\mathrm{蠄}}_{\text{n}}\left({\mathrm{x}}_1\right){\mathrm{蠄}}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_2\right)鈭�{\mathrm{蠄}}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_1\right){\mathrm{蠄}}_{\text{n}}\left({\mathrm{x}}_2\right)](鈫�鈫�+鈫�鈫�)$

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.

${\mathrm{蠄}}_{\text{sum}}=[{\mathrm{蠄}}_{\text{n}}\left({\mathrm{x}}_1\right){\mathrm{蠄}}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_2\right)鈭�{\mathrm{蠄}}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_1\right){\mathrm{蠄}}_{\text{n}}\left({\mathrm{x}}_2\right)](鈫�鈫�+鈫�鈫�)$

So, first swapping spatial state

${\mathrm{蠄}}_{\text{sum}}^{\text{'}}=[{\mathrm{蠄}}_{\text{n}}\left({\mathrm{x}}_2\right){\mathrm{蠄}}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_1\right)鈭�{\mathrm{蠄}}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_2\right){\mathrm{蠄}}_{\text{n}}\left({\mathrm{x}}_1\right)](鈫�鈫�+鈫�鈫�)$

Secondly swap spin state

${\mathrm{蠄}}_{\text{sum}}^{\text{'}}=[{\mathrm{蠄}}_{\text{n}}\left({\mathrm{x}}_2\right){\mathrm{蠄}}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_1\right)鈭�{\mathrm{蠄}}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_2\right){\mathrm{蠄}}_{\text{n}}\left({\mathrm{x}}_1\right)](鈫�鈫�+鈫�鈫�)$

Rearranging the above function

$\begin{array}{l}{\mathrm{蠄}}_{\text{sum}}^{\text{'}}=鈭�[{\mathrm{蠄}}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_2\right){\mathrm{蠄}}_{\text{n}}\left({\mathrm{x}}_1\right)鈭�{\mathrm{蠄}}_{\text{n}}\left({\mathrm{x}}_2\right){\mathrm{蠄}}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_1\right)](鈫�鈫�+鈫�鈫�)\\ {\mathrm{蠄}}_{\text{sum}}^{\text{'}}=鈭�[{\mathrm{蠄}}_{\text{n}}\left({\mathrm{x}}_1\right){\mathrm{蠄}}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_2\right)鈭�{\mathrm{蠄}}_{{\text{n}}^{\text{'}}}\left({\mathrm{x}}_1\right){\mathrm{蠄}}_{\text{n}}\left({\mathrm{x}}_2\right)](鈫�鈫�+鈫�鈫�)\\ {\mathrm{蠄}}_{\text{sum}}^{\text{'}}=鈭�{\mathrm{蠄}}_{\text{sum}}\end{array}$

(f)

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