# Possible spin states?

Given a system of two particles with spin up and down, I have troubles to understand the possible states of this system.

I would have normally thought, that the possible states are the tensor products of the states of each particle with one of the other. This would have given us the states that are listed in Wikipedia under: substituting in the four basis states. So why do we define these new states(three with total spin $1$ and one with total spin $0$) that are given by triplet and singlet representations? So maybe you don't know what I am asking so I repeat my question: What is the meaning of these triplet and singlet states, because I would have thought that the four states listed above are the possible states of a system? The reference to what I am talking about: wikipedia reference

• @ Lipschitz It would be better if you can specify your motivation of your question or present some specific example. – Kai Li Jan 16 '14 at 19:38
• @K-boy I am talking about the calculation done in the wikipedia article. I thought the states of a system of two spin 1/2 particles would be the states written down there after the phrase ' substituting in the four basis states', but apparently there is some need to look at so called triplet and singulett states. my question is, why this is so? – Xin Wang Jan 16 '14 at 19:41
• @ Lipschitz From the last sentence "The result is that a combination of two spin-1/2 particles can carry a total spin of 1 or 0, depending on whether they occupy a triplet or singlet state.", I guess that the wiki want to introduce the concept of total spin?... – Kai Li Jan 16 '14 at 19:58

If we consider two spin $1/2$ particles with spin up and spin down states, then there are four possibilities in total: $$\begin{array}{cccc} |++\rangle \; ,& |+-\rangle \; ,& |-+\rangle \; ,& |--\rangle \end{array}$$ where this notation means for instance: $$|+-\rangle = |s_1=1/2,m_1=1/2;s_2=1/2,m_2=-1/2\rangle$$ We will suppose that the system consisting of both particle has zero orbital angular momentum and let $S_z$ denote that operator acting on the system. Then it is very easy to evaluate the following eigenvalue equations: \begin{align} S_z|++\rangle &=\hbar|++\rangle \\ S_z|+-\rangle &=0|+-\rangle\\ S_z|-+\rangle &=0|-+\rangle \\ S_z|--\rangle &=-\hbar|--\rangle \\ \end{align} and so $S_z$ can be written as: $$S_z = \hbar \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix}$$ Now, if $s=1$ (remember $s$ denotes the quantum number for the total system), then the three states of the system are: \begin{align} |s=1,m=1\rangle & = |++\rangle \\ |s=1,m=0\rangle & = \sqrt{\frac{1}{2}}(|+-\rangle+|-+\rangle) \\ |s=1,m=-1\rangle & = |--\rangle \end{align} with eigenvalues: \begin{align} S_z |1,1\rangle & = \hbar |1,1\rangle \\ S_z |1,0\rangle & = 0|1,0\rangle \\ S_z |1,-1\rangle & = -\hbar |1,-1\rangle \end{align} where I have switched to a more compact notation: $$|s=1,m=1\rangle \equiv |1,1\rangle$$ However, we can also get an eigenstate with $s=0$: $$|0,0\rangle = \sqrt{\frac{1}{2}}(|+-\rangle-|-+\rangle)$$ with eigenvalue: $$S_z |0,0\rangle = 0|0,0\rangle$$ Now, we can also verify that: $$S^2 \equiv S_x^2 + S_y^2 + S_z^2$$ satisfies: \begin{align} S^2 |1,1\rangle & = 2 \hbar^2 |1,1\rangle \\ S^2 |1,0\rangle & = 2 \hbar^2 |1,0\rangle \\ S^2 |1,-1\rangle & = 2 \hbar^2 |1,-1\rangle \\ S^2 |0,0\rangle & =0|0,0\rangle \end{align} Therefore, we see that the following two important equations are satisfied: $$\begin{array}{cc} S^2 |s , m \rangle = \hbar^2 s (s+1) |s , m \rangle \; ,& S_z |s , m \rangle = \hbar m |s , m \rangle \end{array}$$ To sum up, we have found the possible eigenvalues for the magnitude and $z$-component of the system and the eigenstates corresponding to these values: the allowed values for total spin are $s=1$ and $s=0$, while the allowed value of $s_z$ are $\hbar$, $0$, and $-\hbar$ and the corresponding eigenstates in the product basis $|1,1\rangle$, $|1,0\rangle$, $|1,-1\rangle$ and $|0,0\rangle$. Thus, the meaning of these triplet and singlet states is that they are the possible states of the system consisting of the two aforementioned particles. This is often written as: $$\frac{1}{2} \otimes \frac{1}{2} = 1 \oplus 0$$ which means that the tensor product of two spin-$1/2$ Hilbert spaces is a direct sum of a spin-$1$ space and a spin-$0$ space.

• @Lipschitz no problems. Happy to hear that. – Hunter Jan 16 '14 at 20:20

A representation of $SU(2)$, indiced by the spin $s$, has $2s+1$ states. The projection of the spin under some axis $z$, $s_z$, for a state belonging to the representation $s$, verify: $-s \leq s_z \leq s$, and is an integer or a semi-integer, depending on $s$ ($s_z = -s, -s+1, -s+2,....+s$).

When you take the product of $2$ representations, the result is a sum of representations. Note that this means that all the states of the initial and final representations must be taken in account (you cannot consider only one part of the states of a representation).

If you take the product of $2$ representations of spin $s=\frac{1}{2}$, the projections of the spin under some axis $z$ add : $s_z = (s_1)_z+ (s_2)_z$, so the maximum possible value, for $|s_z|$, is $1$ . So the only possibilities for the decomposition of the product of representations $\frac{1}{2} \otimes \frac{1}{2}$, are the representations $s=0$ ($1$ state), and $s=1$ ($3$ states)

For instance, we cannot take the representation $s=2$, because there is a state with $s_z = +2$, and this state cannot be obtained, and we know that all the states of a representation must be taken in account.

Because you must find $4$ states for the total final number of states (each representation $\frac{1}{2}$ has $2$ states), you have to use both the representations $s=0$ and $s=1$, so you may write $\frac{1}{2} \otimes \frac{1}{2} = 1 \oplus 0$ (this is an equation between representations)

So the triplet states correspond to the representation $s=1$, and the singlet state corresponds to the representation $s=0$