# Why do the states of a spin multiplet have to have the same symmetry?

This was said in Prof. Balakrishnan lecture 19 on quantum mechanics for the case of exchange symmetry, but he showed no reason why.

For example, the system corresponding to two spin $\frac{1}{2}$ systems has a singlet state, which is antisymmetric, and three triplet states, which are symmetric. Apparently this has to be always the case. I am also assuming that a multiplet means a set of states with the same total angular momentum (infering from the spin $\frac{1}{2}$ case), although Wikipedia seems to phrase it in more group-theoretic terms with which I'm not yet familiar.

Define $\mathbf{J}=\mathbf{S}_1 + \mathbf{S}_2$ where $\mathbf{J}$ by definition is the total angular momentum to two identical, say spin-half, angular momenta. Define $\sigma$ to be the operation that exchanges the two spins, i.e., it denotes the operation: $1\leftrightarrow 2$. Then, it is easy to see that $\sigma$ commutes with $\mathbf{J}$ as $\mathbf{S}_1 + \mathbf{S}_2=\mathbf{S}_2 + \mathbf{S}_1=\mathbf{J}$. (The argument can be extended to the case of the addition of $N$ identical spins where $\sigma$ gets replaced by any pairwise exchanges which generates the permutation group in $N$ elements.)

Coming back to the case of two spins, it follows that we can simultaneously diagonalise $J^2,J_z$ and $\sigma$. The action of $\sigma$ must necessarily take an eigenstate of $J^2,J_z$ to one with the same $J^2,J_z$ eigenvalues. But the addition of two angular momenta says, that the multiplicity of $J^2$ eigenvalues is one. This is not true when you add more than two spins. It follows that for each given value of $J^2$, the multiplet has a definite $\sigma$ eigenvalue. Since $\sigma^2=1$, its eigenvalues must be $\pm1$.

However, neither $S_{1z}$ or $S_{2z}$ commute with $\sigma$. Thus, we have $$\sigma\ |\ell,m_1,\ell,m_2\rangle = |\ell,m_2,\ell,m_1\rangle\ ,$$ in the notation where $\ell(\ell+1)$ is the $S_i^2$ eigenvalue and $(m_1,m_2)$ are the eigenvalues of $(S_{1z},S_{2z})$ respectively. So in general, the action of sigma is not diagonal, unless $m_1=m_2$. Coming to Prof. Balakrishnan's example. He considered two spin half-particles. The $(j=1,m=\pm1)$ states only arise when $m_1=m_2=\pm\tfrac12$ and from the above argument must be symmetric.However, the $(j=1,m=0)$ and $(j=0,m=0)$ states arises as linear combinations of $m_1=-m_2$ state which are not diagonal. However, since the $(j=1,m=0)$ state can be obtained by using the lowering operator $J_-$ (which commutes with $\sigma$), it will have the same $\sigma$ eigenvalue as the state it was constructed from, i.e., $(j=m=1)$, which is symmetric. The singlet state must be (i) orthogonal to the triplet $m=0$ state and must necessarily be an eigenstate of $\sigma$ (I can give the precise argument if you wish) (ii) antisymmetric.

For the sum of N identical spins, all one can say is that there exists a basis of multiplets organised/labelled by the representations of the permutation group (given by Young diagrams) in addition to the $J^2$ eigenvalue.

• I like the argument with the lowering operator (which commutes with $\sigma$ because $J_{x}$ and $J_{y}$ do, doesn't it?) Just one thing, you say that the multiplicity of $J^{2}$ eigenvalues is one for two spins. Is this so? Isn't it one for the singlet, and three for the triplet? Or am I misunderstanding 'multiplicity'? – guillefix May 8 '14 at 11:58
• I am talking about the multiplicity of multiplets. So I count the triplet and singlet as two multiplets. – suresh May 8 '14 at 15:22
• Multiplets are indexed only by the $J^2$ eigenvalue. Of course, states within a multiplet have different $J_z$ eigenvalue. – suresh May 9 '14 at 0:59
• So what you are saying is that in the case of three spins for example, we have two multiplets corresponding to the same $J^{2}$? – guillefix May 9 '14 at 13:05
• Yes. For instance, $\frac12 \otimes \frac12 \otimes \frac12=\frac32\oplus\frac12 \oplus \frac12$. – suresh May 9 '14 at 13:11

One of the symmetries that's discussed in physics is isotropy of space, against translations and rotations. If you and I are doing the same experiment, but I'm doing the experiment in a rotated reference frame where I'm standing on my head, we expect to get the same result. It's invariance under rotations that leads us to conservation of angular momentum.

(Indeed, here on Earth's surface standing on my head is quite different from standing on my feet, and as a consequence angular momentum in this non-inertial frame is not conserved: tilt a gyroscope and its axis will precess.)

Unless you have a term in your Hamiltonian that explicitly defines a preferred direction (such as coupling to a magnetic field), any property of your system must be invariant against an arbitrary choice of coordinates. This includes the projection of angular momentum against an arbitrary axis.

For instance, let's suppose that I've prepared a two spin-half particles in a spin-one state with a large spin energy, such as the two protons in orthohydrogen. I prepare some of them polarized along one direction and send them to you. But we miscommunicate and I give them to you polarized along the $x$-axis instead of the $z$-axis. In the $S_z$ basis, the eigenvectors of $S_x$ are $$\left|\rightarrow\right> = \frac{\left|\uparrow\right>+\left|\downarrow\right>}{\sqrt2} \quad\quad \left|\leftarrow\right> = \frac{\left|\uparrow\right>-\left|\downarrow\right>}{\sqrt2}$$ and so you analyze my two-spin state as \begin{align*} \left|\rightarrow\rightarrow\right> &= \frac{\left|\uparrow\right>+\left|\downarrow\right>}{\sqrt2} \otimes \frac{\left|\uparrow\right>+\left|\downarrow\right>}{\sqrt2} \\ &= \frac{\left|\uparrow\uparrow\right>}{2} + \frac{\left|\uparrow\downarrow\right> + \left|\uparrow\downarrow\right>}{2} + \frac{\left|\downarrow\downarrow\right>}{2}. \end{align*}

Notice how this symmetric state in one basis overlaps with all the symmetric states in the other basis, but has no overlap with the antisymmetric state. But the more compelling argument to me is the thermal physics. Liquifying orthohydrogen requires you to remove about twice as much heat as liquifying parahydrogen, because the ortho converts to para after the density jump, and the heat of that transformation is comparable to the heat of vaporization. If the symmetric, $m_s=0$ state corresponds to spinless parahydrogen, these lines of algebra suggest that the heat required to liquify polarized orthohydrogen is different depending on the spin axis used by your refrigerator. That wouldn't make any sense at all!

You can verify for yourself that the projection of $\left|\uparrow\uparrow\right>$ onto the $y$-axis also involves only symmetrical combinations (eigenvectors are $(1,±i)$), that the spin singlet is antisymmetric in every basis, and so on. In particular, it's easy to show that a symmetric $m=0$ projection on the $x$-axis contains only projections with $m=±1$ on the $z$-axis: \begin{align*} \frac{ \left| \rightarrow\leftarrow \right> + \left| \leftarrow\rightarrow \right> }{\sqrt2} &= \frac{1}{\sqrt2} \left( \frac{ \left| \uparrow \right> + \left| \downarrow \right> }{\sqrt2} \otimes \frac{ \left| \uparrow \right> - \left| \downarrow \right> }{\sqrt2} + \frac{ \left| \uparrow \right> - \left| \downarrow \right> }{\sqrt2} \otimes \frac{ \left| \uparrow \right> + \left| \downarrow \right> }{\sqrt2} \right) \\ &= \frac{ \left| \uparrow\uparrow \right> - \left| \uparrow\downarrow \right> + \left| \downarrow\uparrow \right> - \left| \downarrow\downarrow \right> }{2\sqrt2} + \frac{ \left| \uparrow\uparrow \right> + \left| \uparrow\downarrow \right> - \left| \downarrow\uparrow \right> - \left| \downarrow\downarrow \right> }{2\sqrt2} \\ &= \frac{ \left| \uparrow\uparrow \right> - \left| \downarrow\downarrow \right> }{\sqrt2} \end{align*}

• Not sure if this answers the initial question, which I interpret to be: why is the |up,down> - |down,up> anti-symmetric state the S=0, Sz=0 and the |up,down> + |down,up> symmetric state the S=1, Sz=0. The logic used in Balakrishnan's lecture was that because the other two states in the S=1 group |up,up> and |down,down> are ALSO symmetric, then we group the symmetric combination in there as well. There wasn't a "deeper" description of why that is. – Kent May 7 '14 at 21:42
• For me, this naturally leads to questions about a system of three spin-1/2 particles: what are symmetry properties of the S=3/2 and S=1/2 groups? – Kent May 7 '14 at 21:45
• Yes, the reason this doesn't answer the question is that the state S=1, Sz=0 is not really a rotated version of S=1, Sz=1 say, as far as I know. – guillefix May 7 '14 at 23:46
• @Kent I prefer symmetry arguments to algebraic arguments, but here you go. – rob May 8 '14 at 0:06
• But again the issue is in knowing that the symmetric state you used there (|right,left>+|left,right>) corresponds to total angular momentum 1 and not 0. However, I think I can see why this is now: the RHS has total angular momentum 1 with probability 1, so the LHS must as well. A similar argument can be used for your first derivation. – guillefix May 8 '14 at 12:02