# Anti-symmetric and Symmetric Helium

## Homework Statement

(a) Find the spatial wavefunction
(b)Show anti-symmetric wavefunctions have larger mean spacing
(c) Discuss the importance of this
(d)Determine the total orbital angular momentum
(e)Hence find the ground state term for Z=15[/B] ## The Attempt at a Solution

Part(a)[/B]
The overall wavefunction must be anti-symmetric due to Pauli's exclusion principle. Since the spin can either be singlet (anti-symmetric) or tripplet (symmetric), the spatial part must be written as a symmetric and anti-symmetric combination of ## u_{A(r_1)} u_{B(r_2)}## and ##u_{A(r_2)}u_{B(r_1)} ##.

Part(b)

$$\phi \phi^* = \frac{1}{2} \left[ | u_{A(r(1)} |^2 |u_{B(r_1)}|^2 + | u_{A(r(2)} |^2 |u_{B(r_2)}|^2 \pm 2 Re \left( u_{A(r_1)}u_{B(r_2)} u^*_{B(r_1)}u^*_{A(r_2)} \right) \right]$$

Hence when spin is aligned (symmetric), the spatial part must be anti-symmetric.

Don't we get ## \phi_{AS} \phi_{AS}^* < \phi_{S} \phi_{S}^* ##?

Which is strange, as I know that spatially anti-symmetric wavefunctions are further away.

Part(c)

Spatially symmetric -> electrons closer -> more shielding -> higher energy (Para-helium)

spatially anti-symmetric -> electrons further -> less shielding -> lower energy (Ortho-helium)

Part (d)

Due to spin-orbit coupling, won't the total angular momentum ##L = l_1 + l_2 + l_3 = 3##?

Part (e)

The ground term is simply ## ^4P_3##.

DrClaude
Mentor
Part(a)
The overall wavefunction must be anti-symmetric due to Pauli's exclusion principle. Since the spin can either be singlet (anti-symmetric) or tripplet (symmetric), the spatial part must be written as a symmetric and anti-symmetric combination of ## u_{A(r_1)} u_{B(r_2)}## and ##u_{A(r_2)}u_{B(r_1)} ##.
What you say is true, but I think that your reasoning is circular. The question could have been "why are the spin states given by...", to which you could answer "because the spatial wave function is either symmetric or anti-symmetric." You can give a better answer by explaining also why the spin and spatial parts of the wave function have to have a definite symmetry.

Part(b)

$$\phi \phi^* = \frac{1}{2} \left[ | u_{A(r(1)} |^2 |u_{B(r_1)}|^2 + | u_{A(r(2)} |^2 |u_{B(r_2)}|^2 \pm 2 Re \left( u_{A(r_1)}u_{B(r_2)} u^*_{B(r_1)}u^*_{A(r_2)} \right) \right]$$

Hence when spin is aligned (symmetric), the spatial part must be anti-symmetric.

Don't we get ## \phi_{AS} \phi_{AS}^* < \phi_{S} \phi_{S}^* ##?

Which is strange, as I know that spatially anti-symmetric wavefunctions are further away.
You haven't inroduced the ##(\vec{r}_1 - \vec{r}_2)^2## yet. Is there a fundamental difference then between the two wave functions?

Part(c)

Spatially symmetric -> electrons closer -> more shielding -> higher energy (Para-helium)

spatially anti-symmetric -> electrons further -> less shielding -> lower energy (Ortho-helium)
I really don't like the use of "closer," which is too classical. I would phrase it in terms of interactions.

Part (d)

Due to spin-orbit coupling, won't the total angular momentum ##L = l_1 + l_2 + l_3 = 3##?
Spin-orbit coupling has nothing to do with it. What does ##S=3/2## tell you? And can you simply apply ##L = \sum l## when the electrons have the same ##nl##? (What would be the ground term of Ne?)

Part (e)

The ground term is simply ## ^4P_3##.
I think you have to be more explicit about how you reach this conclusion.

You haven't inroduced the ##(\vec{r}_1 - \vec{r}_2)^2## yet. Is there a fundamental difference then between the two wave functions?

The difference lies in the ##\pm Re \left( u_{A(r_1)}u_{B(r_2)} u^*_{B(r_1)}u^*_{A(r_2)} \right) ##. The anti-symmetric wavefunction has the ##-## sign, so the overall overlap is smaller? So the overlap becomes:

$$\langle \phi | ((\vec{r}_1 - \vec{r}_2)^2) | \phi \rangle = \langle \phi| r_1^2 | \phi \rangle + \langle \phi | r_2^2|\phi \rangle -2 \langle \phi | \vec r_1 \cdot r_2 | \phi \rangle$$

The only way for the anti-symmetric wave function to be higher is when ##\int Re \left( u_{A(r_1)}u_{B(r_2)} |\vec r_1|^2 u^*_{B(r_1)}u^*_{A(r_2)} \right) d^3\vec r = \left( u_{A(r_1)}u_{B(r_2)} |\vec r_2|^2 u^*_{B(r_1)}u^*_{A(r_2)} \right) d^3 \vec r = 0## so that both ##-## signs coincide at the ## 2 \langle \phi | \vec r_1 \cdot r_2 | \phi \rangle ## to give a positive contribution to the sum.

I really don't like the use of "closer," which is too classical. I would phrase it in terms of interactions.

Spin-orbit coupling has nothing to do with it. What does ##S=3/2## tell you? And can you simply apply ##L = \sum l## when the electrons have the same ##nl##? (What would be the ground term of Ne?)

So if the electrons are further apart in the tripplet state, there is less interaction energy and hence lower energy while in the singlet state they are closer, so higher interaction energy?

W

Spin-orbit coupling has nothing to do with it. What does ##S=3/2## tell you? And can you simply apply ##L = \sum l## when the electrons have the same ##nl##? (What would be the ground term of Ne?)

##S=3/2## means that all the spins are aligned (+++), so the spin state is symmetric. Thus the spatial part must be anti-symmetric. Neon is represented as ## 1s^2 2s^2 2p^6 ##

DrClaude
Mentor
The difference lies in the ##\pm Re \left( u_{A(r_1)}u_{B(r_2)} u^*_{B(r_1)}u^*_{A(r_2)} \right) ##. The anti-symmetric wavefunction has the ##-## sign, so the overall overlap is smaller? So the overlap becomes:

$$\langle \phi | ((\vec{r}_1 - \vec{r}_2)^2) | \phi \rangle = \langle \phi| r_1^2 | \phi \rangle + \langle \phi | r_2^2|\phi \rangle -2 \langle \phi | \vec r_1 \cdot r_2 | \phi \rangle$$
But what happens to the wave functions at ##\vec{r}_1 = \vec{r}_2##?

So if the electrons are further apart in the tripplet state, there is less interaction energy and hence lower energy while in the singlet state they are closer, so higher interaction energy?
If the mean square separation is less, then the Coulomb interaction is reduced.

##S=3/2## means that all the spins are aligned (+++), so the spin state is symmetric. Thus the spatial part must be anti-symmetric.
Yes, but you have to include orbital angular momentum somehow in the picture.

Neon is represented as ## 1s^2 2s^2 2p^6 ##
So what would ##L## be?

But what happens to the wave functions at ##\vec{r}_1 = \vec{r}_2##?

If the mean square separation is less, then the Coulomb interaction is reduced.

Yes, but you have to include orbital angular momentum somehow in the picture.

So what would ##L## be?

When ##\vec{r}_1 = \vec{r}_2##, the anti-symmetric spatial wavefunction disappears while the symmetric one doubles. (Also, why are we equating ##\vec r_1 = \vec r_2##?)

If the condition that spatial part must be anti-symmetric, what does it say about the orbital angular momentum?

For neon, since the outermost is ##2p^6##, it means that ##L = 1##.

DrClaude
Mentor
When ##\vec{r}_1 = \vec{r}_2##, the anti-symmetric spatial wavefunction disappears while the symmetric one doubles. (Also, why are we equating ##\vec r_1 = \vec r_2##?)
You now have to relate that to ##\langle (\vec r_1 - \vec r_2)^2 \rangle##.

If the condition that spatial part must be anti-symmetric, what does it say about the orbital angular momentum?
Nothing in itself. That was my point. The fact that all spins are aligned should tell you something about the total orbital angular momentum. This is also the point for me asking about neon.

For neon, since the outermost is ##2p^6##, it means that ##L = 1##.
That's not correct. How did you arrive at that answer?

You now have to relate that to ##\langle (\vec r_1 - \vec r_2)^2 \rangle##.

Nothing in itself. That was my point. The fact that all spins are aligned should tell you something about the total orbital angular momentum. This is also the point for me asking about neon.

That's not correct. How did you arrive at that answer?

I'm thinking that ##\langle r_1^2 \rangle = \langle r_2^2 \rangle ##, so the expectation becomes ## 2\langle \phi | r_1^2| \phi \rangle - 2\langle \phi | \vec r_1 \cdot \vec r_2 | \phi \rangle##.

If all spins are aligned, total orbital angular momentum is zero? I'm not sure why though.

Since neon is a noble gas, the overall shell is symmetrical and hence ##L=0##?

DrClaude
Mentor
If all spins are aligned, total orbital angular momentum is zero? I'm not sure why though.

Since neon is a noble gas, the overall shell is symmetrical and hence ##L=0##?
Yes. You have to think in terms of ##m_l## to understand why in both cases.

Yes. You have to think in terms of ##m_l## to understand why in both cases.

I suppose a qualitative explanation would be the lowest energy levels would be when there is most overlap between the orbitals, i.e. sphere-sphere-sphere overlap. This implies that ##l_1 = l_2 = l_3 = 0##.

For neon, I would explain that that it is a full octet -> spherically symmetrical and stable so overall ##L=0##.

Do you think there is a better explanation?

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DrClaude
Mentor
I suppose a qualitative explanation would be the lowest energy levels would be when there is most overlap between the orbitals, i.e. sphere-sphere-sphere overlap. This implies that ##l_1 = l_2 = l_3 = 0##.

For neon, I would explain that that it is a full octet -> spherically symmetrical and stable so overall ##L=0##.
But you are considering p electrons, so ##l=1## for each of them.

The thing you have to consider is that you can apply ##L = \sum l## naively only for electrons with different principal quantum numbers n. Otherwise, the Pauli exclusion principle will forbid some combinations (the actual equation is for operators, ##\hat{L} = \sum \hat{l}##, and the sum applies to allowed wave functions). What always applies is ##M_L = \sum m_l##, so you need to sum up ##m_l##'s for the electrons to get which ##M_L## states are present, from which you can deduce what is the corresponding ##L##. (For example, if you find ##M_L = 2## for a certain electronic configuration, then there is a D term associated with that configuration.)

There is a systematic way to go about it, where you list all the microstates (each electron in all possible orbitals with all possible spins) allowed by Pauli, figure out ##M_L## and ##M_S## for each microstate, and deduce the term symbols. For instance, if you do that with an np2 configuration, you will find three possible terms, 1D, 3P, and 1S. Your case is simpler.