# Why triplet state |1,0> is considered parallel ?

1. Feb 23, 2012

### harrists

why triplet state |1,0> is considered "parallel"?

let there be a system of two electrons and thus have s1=s2=1/2 and m1=m2=+-1/2.

the singlet state is:
|0,0> = 1/sqrt(2) ( |1/2,-1/2> - |-1/2,1/2> )

and the triplet

|1,1> = |1/2,1/2>
|1,0> = 1/sqrt(2) ( |1/2,-1/2> + |-1/2,1/2> )
|1,-1> = |-1/2,-1/2>

Since they are fermions the total wavefunction has to be antisymmetric.

So, if Stotal = 0 it has to be m1 = -m2 which leads to an antisymmetric spin function so the spacial w/f has to be symmetric.
We consider this spins' state as "antiparallel" and i can totally understand the reason.
(As i also know the spins cannot really be parallel.)
Given that the spins are antiparallel, the two electrons are permitted to occupy the same quantum state. (the rest quantum numbers,except m, to be equal).
Right?

In case Stotal = 1,
i am a bit confused.
All the three triplet states are symmetric, so the spatial w/f should then be antisymmetric.
I have read that, in this case, we consider the spins as parallel, m1 = m2
so the two electrons are not permitted to occupy the same quantum state.

For example, if they both exist in a quantum well, then the one should occupy the basic (n=1) energy state and the other should occupy the next one (n=2).

I wonder why do we consider the state:
|1,0> = 1/sqrt(2) ( |1/2,-1/2> + |-1/2,1/2> )
as "parallel" spin state and therefore enable pauli's exclusion principle ?????

Thank you for your help.

2. Feb 24, 2012

### tom.stoer

Re: why triplet state |1,0> is considered "parallel"?

The symmetry of the wave function is not dictated by spin alone; instead of constructing the two-particle state from the single particle state |sz> you have to construct it from |n,sz> where n labels the energy levels of the quantum well or the harmonic oscillator. The total wave function has to be antisymmetric.

So you can e.g. have

$$|n,s_z\rangle_1 = |n_1,+>$$

$$|n,s_z\rangle_2 = |n_2,+>$$

$$|\psi\rangle = \text{anti-symm}\;|n_1,+\rangle \otimes |n_2,+\rangle = \frac{1}{\sqrt{2}}\left( |n_1,+\rangle \otimes |n_2,+\rangle - |n_2,+\rangle \otimes |n_1,+\rangle\right)$$

Now you see that both electrons can have the same spin state '+' but must differ w.r.t. n; with the same n the state would be 0 trivially.

3. Feb 24, 2012

### The_Duck

Re: why triplet state |1,0> is considered "parallel"?

As you can see it's really not right to call this a "parallel" spin state. We should really call it a "symmetric" spin state. Often it is called a "triplet" spin state since it is one of the three symmetric spin states of 2 spin 1/2 particles.

The precise and general statement of the Pauli principle is "the state of a system of a system of multiple identical fermions must be antisymmetric under exchange of any two of the particles." The behavior of electrons in atoms is a specific instance of this general principle. If two electrons occupy the same orbital in some atom, their two-particle spatial wave function is symmetric; the Pauli principle then requires that their spin state be antisymmetric; i.e., they must be in the singlet spin state. This situation gets simplified in less technical contexts as "electrons in the same orbital must have antiparallel spins," but this wording is not really quite correct: really they must be in the singlet spin state.