# I Triplet State Symmetric Wavefunction

1. May 28, 2016

### looseleaf

Hi everybody. I was reading about the singlet and triplet states. It makes sense that we use an antisymmetric wavefunction for the singlet state, as we are talking about two fermions. But why are we using a symmetric wavefunction for the Sz = 0 triplet state? Doesn't this go against the electrons fermion-ness?

Thanks!

2. May 28, 2016

### vanhees71

Only the spin part of the triplet state is symmetric (for all three states, which are given by $|1/2,1/2 \rangle$, $(|1/2,-1/2 \rangle+|-1/2,1/2 \rangle)/\sqrt{2}$, and $|-1/2,-1/2 \rangle$). The spatial part of the wave function then must of course be antisymmetric, so that the overall two-particle wave function is anti-symmetric under exchange of the two fermions. For the singlet state, which is the antisymmetric combination of the spin state, the spatial part must be symmetric.

3. May 28, 2016

### PeroK

The overall state of two electrons must be anti-symmetric. The triplet spin state is symmetric, so that must be combined with an anti-symmetric spatial wavefunction. The singlet spin state, on the other hand, is anti-symmetric, so it must be combined with a symmetric spatial wavefunction in order for the overall state of the system to be anti-symmetric.

In both cases, the state of the system is the product of one symmetric and one anti-symmetric component, giving overall an anti-symmetric state.

4. May 29, 2016

### looseleaf

Ok thanks, that makes sense. Would you mind telling me then what the symmetric and antisymmetric spatial wf's look like in this case?

5. May 30, 2016

### blue_leaf77

That depends on the Hamiltonian operator of the system.

6. May 30, 2016

### looseleaf

Sure, but isn't the Hamiltonian specified by looking at singlet / triplet states?

Last edited: May 30, 2016
7. May 30, 2016

### blue_leaf77

No, the Hamiltonian does not depend on any particular state. You start everything from the Schroedinger equation, in which in turn one needs to know the Hamiltonian operator.

By the way, the separation between the spatial and spin wavefunctions only applies for non-relativistic atoms, i.e. atoms with $Z<20$ because in this situation the spin-orbit coupling can be neglected. For heavy atoms, the Hamiltonian contains terms involving spins, which is due to the spin-orbit effect, and therefore in general $[H,S_z]\neq 0$. In this case the wavefunctions cannot be separated into spatial and spin parts.

A rather intricate aspect about the spin wavefunctions of many electrons is that one cannot form an anti-symmetrical wavefunction when there are more than two electrons. This means for atoms having more than two electrons, even if the spin orbit effect is not significant, the total wavefunction does not separate into spatial and spin parts. In general, for such systems, the eigenfunctions of spin and of Hamiltonian are combined to form the so-called Slater determinant and the total wavefunction is equal to a Slater determinant (or a linear combination of it - since the eigenfunctions of Hamiltonian is infinitely numerable, there are also infinite possible Slater determinants). It follows from the interchange nature of a determinant, this kind of wavefunction is guaranteed to be antisymmetric.

Last edited: May 30, 2016