# Helium singlet/triplet state with total spin of 1

1. Feb 7, 2016

### misko

Say we have a Helium atom with the first electron in 1s orbital and second electron in 2s orbital, and if these electrons have parallel spin so that the total spin of the system is S=1, then is this a triplet or a singlet state?

According to LS coupling, multiplicity is defined as:
2S+1 (if L>=S)
and
2L+1 (if L<S)

Since in our case we have L=0 (both electrons in S orbitals) then multiplicity should be 1 (according to the second case) which means we have a singlet. However, I find on many places that states with total spin S=1 are triplets without mentioning the L (total orbital angular moment).
Even for Orthohelium is defined as the Helium state atom in which the spins of the two electrons are parallel. Here http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/helium.html and here https://en.wiktionary.org/wiki/orthohelium.

2. Feb 7, 2016

### blue_leaf77

Those relations are for determining the number of possible total angular momenta $J$ when you take the spin-orbit coupling into account. This problem deals with He atom which does not exhibit prominent relativistic effect, therefore the fine splitting due to spin-orbit coupling is negligible here.
The multiplicity is just $2S+1$, which equals three for $S=1$. That's why it's called triplet.

3. Feb 7, 2016

### misko

Thank you for the reply. I am trying to wrap my head around multiplicity concept. We are not learning it through strict QM formalism and we just barely mentioned spin-orbit coupling without much details like relativistic effects etc... My course is based on old quantum theory (https://en.wikipedia.org/wiki/Old_quantum_theory) with some explanations from modern QM so it is pretty unfortunate mix with a lot of holes in my knowledge.

Anyway, how can I think about multiplicity in a intuitive way rather than strict mathematical? When should I take total orbital angular momentum L into account and when I should just calculate it as 2S+1? I am pretty confused and trying to find some basic logic for this concept and be able to understand what practically it means when I see some Term symbol with multiplicity in the upper left part. I think it is related to splitting of lines in magnetic field but I am not sure exactly how it works.

4. Feb 8, 2016

### blue_leaf77

In atoms with low $Z$ (usually below $Z=25$), it is enough to consider L-S coupling only without taking the fine splitting into account. In this case, the energy levels of the atoms are specified, among others, by two quantum numbers: $L$ and $S$. So, both of them are important as specifiers of energy level - you cannot ignore even one of them.
Terms symbol is a way to specify energy levels of multielectron atoms, as I have put forward above. For example, a hypothetical atom has two unpaired, outer electrons. Here, "two outer electrons" means that the other electrons in the atom has already formed filled subshells and only two are left. Suppose that the last two electrons occupy different shell, with the first one in subshell $p$ the second one in subshell $d$: $np\hspace{2mm}n'd$. Since both electrons have spin 1/2, the possible total spin are
$$S = 0,1$$
For the orbital angular momenta, one electron is in $L_1=1$ the other one in $L_2=2$. So the total orbital angular momenta are
$$L = 1,2,3$$
corresponding to the letters $P,D,F$ respectively. From those possibilities of spin and orbital angular momenta, one can form 6 terms obtained by pairing each $L$ and $S$ (in this simple example, the two electrons occupy different shells, if they occupy the same shell Pauli principle must also be taken into account in forming this pairing between $L$ and $S$). The terms are
$$^1P,^3P,^1D,^3D,^1F,^3F$$
Each of these terms actually has different energies. The way in which those terms are ordered based their energies is called Hund's rules.
No, it has nothing to do with external magnetic field, although it may influence the splitting when an external magnetic field is present.

5. Feb 8, 2016

### misko

If that is so then why for He atom (with Z=2) we calculate multiplicity as 2S+1 without considering L anywhere?

If we have $L_1=1$ $L_2=2$ then why total L is not always 3? In what cases it will be 1 or 2?

So Term is associated with the state of the atom as a whole and not to the specific electron? In our course we introduced Term concept with the hydrogen atom so I associated Term to the energy level of a single electron in the atom. Was this a wrong thinking?

6. Feb 8, 2016

### blue_leaf77

As far as I know, in atomic physics, for a given $L$, the multiplicity is defined as $2S+1$. If one wants to specify energies, however, both $L$ and $S$ matter.
Here, we are talking about the addition of angular momentum vector operators, $\mathbf{L} = \mathbf{L}_1+\mathbf{L}_2$, instead of the addition of the quantum numbers $L_1+L_2$. The operator nature of $\mathbf{L}_1$ and $\mathbf{L}_2$ makes the addition between them not as easy to carry out as the addition of simple numbers. The magnitude of the total angular momentum in this case is
$$L = |L_1-L_2|,|L_1-L_2|+1,\ldots,L_1+L_2-1,L_1+L_2$$
The possible total angular momenta above along with their respective z components actually form a different way of writing the composite states formed by two angular momenta ($L_1$ and $L_2$). If you write these composite states in terms of $|m_1,m_2\rangle$ where $m_1$ and $m_2$ are the z component of $L_1$ and $L_2$ respectively, you will get $(2L_1+1)(2L_2+1)$ different states. For the present example with $L_1=1$ and $L_2=2$, there will be $3\times 5=15$ states of the form $|m_1,m_2\rangle$. An equivalent way of writing the composite states are by using the total angular momentum and its z component as specifiers: $|L,m\rangle$. Using the same example, we have $L=1,2,3$. $L=1$ has three z components, $L=2$ five z components, and $L=3$ has seven z components, summing all these you obtain 15 possible states of the form $|L,m\rangle$ which is the same number as that with $|m_1,m_2\rangle$ way of writing. In terms of linear algebra, writing state as $|m_1,m_2\rangle$ or $|L,m\rangle$ corresponds to using different bases (basis transformation) in the same vector space, which is the vector space of composite angular momentum. In atoms with low $Z$ however, the individual orbital and spin angular momenta do not commute with the Hamiltonian, therefore they cannot be used as the good quantum numbers. Instead, it is the total orbital and spin angular momenta which commute with the Hamiltonian, that's why the term symbol is written using the total orbital and spin angular momenta, instead of the individual ones.

Yes, it's associated with the entire electron configuration in the atom.

Last edited: Feb 8, 2016