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I have some problems understanding the symmetry of spatial wave functions.

In my experimental physics course they tought us that in atoms the total wave function $\Psi_{tot}(\vec{r}_1,\vec{r}_2)=\Psi(\vec{r}_1,\vec{r}_2)\chi(S,M_s)$ has to be anti-symmetric under the exchange of two particles, because that's what the Pauli principle demands for fermions.

So for creating anti-/symmetric spatial wave functions they gave us those two equations:

$$\Psi^a(\vec{r}_1,\vec{r}_2)=\frac{1}{\sqrt{2}}[\Psi_a(\vec{r}_1)\Psi_b(\vec{r}_2)-\Psi_a(\vec{r}_2)\Psi_b(\vec{r}_1)]\\ \Psi^s(\vec{r}_1,\vec{r}_2)=\frac{1}{\sqrt{2}}[\Psi_a(\vec{r}_1)\Psi_b(\vec{r}_2)+\Psi_a(\vec{r}_2)\Psi_b(\vec{r}_1)] $$ Then they said for an antisymmetric spin function you need a symmetric wave function and the other way round. It's clear to me that, if $a$ and $b$ are an identical set of $n,l,m$ the antisymmetric wave function is zero.

On the other hand, I've had an exercise sheet about that and in the video that explained the solution (corona virus yeah!) they stated that the state with $n_1=n_2=3,l_1=l_2=2,m_{l_1}=2,m_{l_2}=0,m_{s_1}=\frac{1}{2},m_{s_2}=\frac{1}{2}$ can not exist because the spatial wave function and the spin function are symmetric. They've been a bit vague but they said something like, if the sum of the $m_l$ is even the spacial wave function is symmetric and antisymmetric if the sum is odd.

Unfortunately I can't understand that. Why can the wave function be symmetric even though $\Psi_{3,2,2}$ and $\Psi_{3,2,0}$ are completely different functions? And why is in this case the wave function not antisymmetric?

I'm going to add a translation of the question on my exercise sheet that caused me to ask this Question:

What states can two electrons inside the 3d-orbital of an atom have and how much?

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  • $\begingroup$ can you give additional info in the system? What is total $L$ of the system? $\endgroup$ Jul 22 '20 at 13:34
  • $\begingroup$ @ZeroTheHero the only information I got is, its a 3d orbital and contains two electrons. So I assume Coulomb potential $\endgroup$
    – Mister00X
    Jul 22 '20 at 13:40
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I think the answer depends on the final value of $L$. The reason behind this is that, when combining systems with angular momenta, the permutation symmetry is dictated in part by the symmetries of the Clebsh-Gordan coefficient coupling the individual angular momenta. To be explicit, when combining systems with angular momenta $\ell_1=2$ and $\ell_2=2$ to final angular momentum $L$, one must use the Clebsch-Gordan coefficients $<2m_1;2m_2\vert L M>$ which, under the interchange of angular momentum, pick up a sign so that \begin{align} <2m_1;2m_2\vert L M>=(-1)^L <2m_2;2m_1\vert LM\rangle\, . \end{align} Thus if $L=2,4$ this part is symmetric under interchange, but if $L=3$ the system is antisymmetric. Note that in your specific case $L=0,1$ cannot appear since $M=m_1+m_2=2$.

Now in your specific case you start with $m_{s_1}=m_{s_2}=1/2$ so the only possible resulting spin state is $S=1$ with $M_S=m_{s_1}+m_{s_2}$ and that's necessarily symmetric. Thus, the spatial part should be antisymmetric.

For the spatial part we have \begin{align} &R_{32}(r_1)Y^2_{2}(\theta_1,\phi_1)R_{32}(r_2)Y^2_{0} (\theta_2,\varphi_2)\pm R_{32}(r_2)Y^2_{m_1}(\theta_2,\varphi_2)R_{32}(r_1)Y^2_{m_2}(\theta_1,\varphi_1)\, ,\\ &=R_{32}(r_1)R_{32}(r_2)\left(Y^2_{m_1}(\theta_1,\varphi_1) Y^2_{m_2}(\theta_2,\varphi_2)\pm Y^2_{m_1}(\theta_2,\varphi_2)Y^2_{m_2}(\theta_1,\varphi_1) \right) \end{align} so the symmetry under exchange of particles depends entirely on the angular momentum part. Thus, if $L=4$ \begin{align} [Y^2_{m_1}(\theta_2,\varphi_2)Y^2_{m_2}(\theta_1,\varphi_1)]^{L=4}= [Y^2_{m_1}(\theta_1,\varphi_1)Y^2_{m_2}(\theta_2,\varphi_2)]^{L=4} \end{align} so only the symmetric combination can survive. The same logic applies to the $L=2$ case. However, the argument shows that the $L=3$ states, including the $L=3,M_L=2$ state, are antisymmetric.

In conclusion: it seems to me that it is possible to construct both spatially symmetric and spatially antisymmetric wave functions with the information you have provided. I do not think it's possible to have a complete answer unless you can specify or somehow restrict the possible total $L$.

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Let us simplify the notation. We will represent a state by $|n,l,m\rangle$.

So the states $\Psi_{3,2,2}$ and $\Psi_{3,2,0}$ are represented by $|{3,2,2}\rangle$ and $|{3,2,0}\rangle$.

Now these are individual electron states. However if the system is a two electron one, owing to the indistinguishability of electrons they can be combined in one of the two ways: $$\psi_s=\Big(|{3,2,2}\rangle_1~~|{3,2,0}\rangle_2\Big)+ \Big(|{3,2,0}\rangle_1~~|{3,2,2}\rangle_2\Big) $$ Where the first half of RHS says that electron $1$ is in $m_l=2$ state and electron $2$ is in $m_l=0$ state. And the second half says the other way round. What you may notice is that if you swap the two electrons ie $1\to2$ and $2\to1$ the wavefunction remains the same. Thus we say the wavefunction is symmetric under electron exchange.

The second way to combine is as follows: $$\psi_a=\Big(|{3,2,2}\rangle_1~~|{3,2,0}\rangle_2\Big)- \Big(|{3,2,0}\rangle_1~~|{3,2,2}\rangle_2\Big) $$ Here under electron exchange the wavefunction picked up an overall negative sign. Thus we say the wavefunction is antisymmetric under electron exchange.

As you can see, these are the properties of the wavefunction of the two electrons combined.

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  • $\begingroup$ Yeah I understand that, but in my course they said that for this particular case the wave function is symmetric. So I assume that the antisymmetric wave function has to be zero and that's what I really don't understand. $\endgroup$
    – Mister00X
    Jul 22 '20 at 12:05
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I think I now have understand what my lecturer has done. He used some rule of thumb consisting of the Pauling notation, Hunds Rule and the fact that the Clebsch-Gordan-coefficients disappear for all $M \neq m_{l_1} + m_{l_2}$ and for all $L$s that don't satisfy $|l_1 -l_2| \leq L\leq l_1 +l_2$.

The first step to understand this rule of thumb is to assume that "according" to Hunds rule the first electron in the 3p-orbital has $m_{l_1}=2$ and $m_{s_1}=\frac{1}{2}$, in the picture below that's the blue arrow.

The next step is to define/assume that for those two electrons $L=\sum m_l$ with the $m_l$s written under the boxes in the Pauling notation. This satisfies the $|l_1 -l_2|\leq L \leq l_1 + l_2$ condition.

With this assumptions and the explanation @ZeroTheHero gave about the symmetry of the spatial wave function depending on $L$, it is now possible and allowed to say that the green arrows in the picture below are the allowed states and the red arrows the forbidden states of the second electron because of the Pauli principle.

Thanks everyone for helping me with this problem

The rule of thumb

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