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conantee
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« on: September 21, 2010, 12:57:56 PM »

59. The ground state of the helium atom is a spin
(A) singlet
(B) doublet
(C) triplet
(D) quartet
(E) quintuplet
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« Reply #1 on: September 22, 2010, 07:55:37 PM »

อันนี้ตอบ Singlet หรือเปล่าครับ เพราะ spin มันจะชี้ทิศตรงข้ามกัน สงสัยครับอยากรู้
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conantee
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« Reply #2 on: September 25, 2010, 02:32:17 PM »

The correct answer is (A) singlet krub, but the reason is beyond the fact that spins of two electrons are just opposite.

First, let's do this problem without knowledge about symmetric and antisymmetric wave function.
If you have been through the issue of helium atom, you might heard that wave function of two electrons (with spin 1/2) has four basis. They split into one and three degenerate states. Singlet and triplet are only two reasonable choices.
Then, by your experience, as the energy level increases, the number of degeneracy increases. Think of 2 dimensional infinite square well as an example. Hence, the answer is singlet.

Now, if you want more elaborate reason, you need notion of symmetric and antisymmetric wave function.
Electrons are fermions - they have antisymmetric wave function. The wave function describing particle A and B is antisymmetric if and only if, when the labels A and B are swapped, the wave function is negative of the former wave function.
Then, antisymmetric wave function can be obtained from antisymmetric wave function = (symmetric spatial wave function) x (antisymmetric spin wave function)
or from antisymmetric wave function = (antisymmetric spatial wave function) x (symmetric spin wave function)

For 2 fermions, the familiar set of basis for spin wave function is { |\uparrow>_A|\uparrow>_B, |\downarrow>_A|\downarrow>_B, |\uparrow>_A|\downarrow>_B|\downarrow>_A|\uparrow>_B}
However, we can form another set of basis according to property of symmetry/antisymmetry as
 { |\uparrow>_A|\uparrow>_B, |\downarrow>_A|\downarrow>_B, \frac{1}{\sqrt{2}}|\uparrow>_A|\downarrow>_B+\frac{1}{\sqrt{2}}|\downarrow>_A|\uparrow>_B\frac{1}{\sqrt{2}}|\uparrow>_A|\downarrow>_B-\frac{1}{\sqrt{2}}|\downarrow>_A|\uparrow>_B}
The last basis is the only one representing antisymmetry. That is singlet state.
Since now the spatial wave function of the singlet sate is symmetric, both electrons can be same at the first orbital (orbital is spatial quantity) yielding lowest energy. On the contrary, the triplet states correspond to antisymmetric spatial wave function. Therefore, both electrons cannot be the at the first orbital simultaneously. Otherwise, the whole wave function vanishes due to negative sign in wave function.

The correct answer is therefore (A). 29 of 100 people answer the question correctly.
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