# How to determine whether an eigenstate of total spin is symmetric or antisymmetric?

Here we have two identical paticles with spin $I$, integer or half-integer, and there are $(2I+1)^2$ states.

Each one of them can be uniquely determined by total spin and its orientation, we can use $|J,m\rangle$ to represent this state. And because of its uniqueness, it is either symmetric or antisymmetric.

How to determine whether $|J,m\rangle$ is symmetric or antisymmetric based on $I$, $J$ and $m$?

• Are you referring to the spatial part of the wave-function, or the spin part, or the whole thing? Because if they are fermions, it has to be antisymmetric under particle exchange. – lionelbrits Dec 24 '14 at 13:05
• @lionelbrits only spin part. – Doris Dec 24 '14 at 13:24
• One can easily show, via a highest-weight construction, that when two particles of spin $I$ are coupled, the resulting states with $J = 2I$ are always symmetric and the states with $J = 2I-1$ are always antisymmetric. I suspect that this logic continues on down the chain, i.e., the parity of any state under exchange is $(-1)^{J-2I}$; and the value of $m$ is irrelevant. However, the proof technique I thought up for the $J = 2I$ & $J = 2I - 1$ cases fails at $J = 2I-2$. If I come up with a general proof, I'll be sure to post it. – Michael Seifert Mar 10 '17 at 14:55

Let's denote the spins of the individual particles by $j_1 = j_2 = I$, the quantum numbers for the $z$-components of their angular momentum by $m_1$ and $m_2$, the spin of their combined state by $J$, and the $z$-component of the angular momentum of the combined state by $M$. We have two bases for the states of these particles: the "individual particle" basis, denoted by $$|I \, m_1 \, I \, m_2 \rangle$$ and the "combined particle" basis, denoted by $$|J \, M \rangle.$$ (Note that $j_1$ and $j_2$ are also still "good" quantum numbers for this latter state; but including them in both notations is redundant and can confuse things, so I'll omit them.) Finally, let's denote by $\hat{E}$ the exchange operator between particles 1 & 2, i.e., we define $\hat{E}$ such that $$\hat{E} |I \, m_1 \, I \, m_2 \rangle = |I \, m_2 \, I \, m_1 \rangle.$$

We can perform a basis transformation to express any state $|J \, M \rangle$ in terms of the basis $|I \, m_1 \, I \, m_2 \rangle$: $$|J \, M \rangle = \sum_{j_1, m_1, j_2, m_2} |I \, m_1 \, I \, m_2 \rangle \langle I \,m_1 \,I \, m_2 | J \, M \rangle$$ The coefficients $\langle I \,m_1 \,I \, m_2 | J \, M \rangle$ are known as the Clebsch-Gordan coefficients. We want to know what happens when we apply the exchange operator $\hat{E}$ to this state: $$\hat{E} |J \, M \rangle = \sum_{j_1, m_1, j_2, m_2} |I \, m_2 \, I \, m_1 \rangle \langle I \,m_1 \,I \, m_2 | J \, M \rangle = \sum_{j_1, m_1, j_2, m_2} |I \, m_1 \, I \, m_2 \rangle \langle I \,m_2 \,I \, m_1 | J \, M \rangle$$ (The second step is just a relabelling of dummy indices $m_1$ and $m_2$ in the sum.) We can see that $|J \, M \rangle$ will be an eigenstate of $\hat{E}$ if and only if all of the Clebsch-Gordan coefficients $\langle I \,m_1 \,I \, m_2 |J \, M \rangle$ are multiplied by the same factor when we exchange $m_1 \leftrightarrow m_2$. Thankfully, the Clebsch-Gordon coefficients satisfy the identity $$\langle j_1 \,m_1 \, j_2 \, m_2 | J \, M \rangle = (-1)^{j_1 + j_2 - J} \langle j_2 \,m_2 \, j_1 \, m_1 | J \, M \rangle$$ and so we have \begin{multline} \hat{E} |J \, M \rangle = \sum_{j_1, m_1, j_2, m_2} |I \, m_1 \, I \, m_2 \rangle \left[ (-1)^{2I - J} \langle I \,m_1 \,I \, m_2 | J \, M \rangle \right] \\ = (-1)^{2I - J} \sum_{j_1, m_1, j_2, m_2} |I \, m_1 \, I \, m_2 \rangle \langle I \,m_1 \,I \, m_2 | J \, M \rangle = (-1)^{2I - J} |J \, M \rangle. \end{multline} Thus, the combined states are symmetric when $2I$ and $J$ are both even or both odd, and antisymmetric when one quantity is even and the other is odd.

• Side note: the Clebsch-Gordan identity I used is the one given in both MathWorld and Wikipedia. However, when I tried to derive it from the stated properties of Wigner 3-j symbols (as suggested in the Wiki article), I obtained $\langle j_1 \,m_1 \, j_2 \, m_2 | J \, M \rangle = (-1)^{j_1 + j_2 + J} \langle j_2 \,m_2 \, j_1 \, m_1 | J \, M \rangle$ (note the sign difference in the exponent.) It doesn't matter for this problem because $J$ is always an integer, but it'd be great if someone could check my work and let me know what I did wrong. – Michael Seifert Mar 10 '17 at 16:10
• The source for this stuff is the book of Varshalovich. The phase change given there is $(-1)^{j_1+j_2-J}$. The $3j$'s have some weird phase w/ to the CGs which make their use tortuous compared to CGs. – ZeroTheHero Mar 13 '17 at 3:29

The final states $\left|j,m\right\rangle$ arising from the coupling of two angular momenta $j_{\alpha}$ and $j_{\beta}$ are related to the initial uncoupled states $\left|j_{\alpha},m^{\alpha}\right\rangle, \left|j_{\beta},m^{\beta}\right\rangle$ through the so called Glebsch-Gordan coefficients $C_{jm^{\alpha}m^{\beta}}^{j_{\alpha}j_{\beta}}$ $$\left|j,m\right\rangle =\sum_{m^{\alpha},m^{\beta}}^{m^{\alpha}\!+m^{\beta}=m}C_{jm^{\alpha}m^{\beta}}^{j_{\alpha}j_{\beta}}\left|j_{\alpha},m^{\alpha}\right\rangle \left|j_{\beta},m^{\beta}\right\rangle \tag{01}$$ where $$\begin{split} & j=\vert j_{\alpha}-j_{\beta}\vert,\vert j_{\alpha}-j_{\beta}\vert+1 , \cdots , j_{\alpha}+j_{\beta}\\ & m=-j,-j+1, \cdots , j-1,j\\ & m= m^{\alpha}+m^{\beta}\\ & m^{\alpha} = -j_{\alpha},-j_{\alpha}+1, \cdots, j_{\alpha}-1,j_{\alpha}\\ & m^{\beta} = -j_{\beta},-j_{\beta}+1, \cdots, j_{\beta}-1,j_{\beta}\\ & \left|j_{\alpha},m^{\alpha}\right\rangle \left|j_{\beta},m^{\beta}\right\rangle=\left|j_{\alpha},m^{\alpha}\right\rangle \boldsymbol{\otimes}\left|j_{\beta},m^{\beta}\right\rangle \end{split} \tag{02}$$ These coefficients are given by equation (03) $$\begin{split} & C_{jm^{\alpha}m^{\beta}}^{j_{\alpha}j_{\beta}}=\\ & \frac{\sqrt{\left(j+j_{\alpha}-j_{\beta}\right)!\left(j-j_{\alpha}+j_{\beta}\right)!\left(j_{\alpha}+j_{\beta}-j\right)!\left(j+m^{\alpha}+m^{\beta}\right)!\left(j-m^{\alpha}-m^{\beta}\right)!}}{\sqrt{\left(j+j_{\alpha}+j_{\beta}+1\right)!\left(j_{\alpha}-m^{\alpha}\right)!\left(j_{\alpha}+m^{\alpha}\right)!\left(j_{\beta}-m^{\beta}\right)!\left(j_{\beta}+m^{\beta}\right)!}}\\ &\times \sum_{\varkappa}\frac{\left(-1\right)^{\varkappa+j_{\beta}+m^{\beta}}\sqrt{\left( 2j+1\right) }\left(j+j_{\beta}+m^{\alpha}-\varkappa \right)!\left(j_{\alpha}-m^{\alpha}+\varkappa\right)!}{\left(j-j_{\alpha}+j_{\beta}-\varkappa \right)!\left(j+m^{\alpha}+m^{\beta}-\varkappa \right)!\varkappa!\left(\varkappa + j_{\alpha}-j_{\beta}-m^{\alpha}-m^{\beta}\right)!} \end{split} \tag{03}$$ The variable $\,\varkappa\,$ in series takes all nonnegative integer values for which all factorials have sense.

Interchanging $j_{\alpha}$ and $j_{\beta}$ and simultaneously $m^{\alpha}$ and $m^{\beta}$ yields $$\begin{split} & C_{jm^{\beta}m^{\alpha}}^{j_{\beta}j_{\alpha}}=\\ & \frac{\sqrt{\left(j+j_{\alpha}-j_{\beta}\right)!\left(j-j_{\alpha}+j_{\beta}\right)!\left(j_{\alpha}+j_{\beta}-j\right)!\left(j+m^{\alpha}+m^{\beta}\right)!\left(j-m^{\alpha}-m^{\beta}\right)!}}{\sqrt{\left(j+j_{\alpha}+j_{\beta}+1\right)!\left(j_{\alpha}-m^{\alpha}\right)!\left(j_{\alpha}+m^{\alpha}\right)!\left(j_{\beta}-m^{\beta}\right)!\left(j_{\beta}+m^{\beta}\right)!}}\\ &\times \sum_{\varkappa}\frac{\left(-1\right)^{\varkappa+j_{\alpha}+m^{\alpha}}\sqrt{\left( 2j+1\right) }\left(j+j_{\alpha}+m^{\beta}-\varkappa \right)!\left(j_{\beta}-m^{\beta}+\varkappa\right)!}{\left(j+j_{\alpha}-j_{\beta}-\varkappa \right)!\left(j+m^{\alpha}+m^{\beta}-\varkappa \right)!\varkappa!\left(\varkappa - j_{\alpha}+j_{\beta}-m^{\alpha}-m^{\beta}\right)!} \end{split} \tag{04}$$

As pointed out in Wigner(1) $\; C_{jm^{\alpha}m^{\beta}}^{j_{\alpha}j_{\beta}}$ will remain unchanged if $j_{\alpha}$ and $j_{\beta}$ and simultaneously $m^{\alpha}$ and $m^{\beta}$ are interchanged and on this result the factor $\left(-1\right)^{j_{\alpha}+j_{\beta}-j}$ is applied $$C_{jm^{\alpha}m^{\beta}}^{j_{\alpha}j_{\beta}} =\left(-1\right)^{j_{\alpha}+j_{\beta}-j} C_{jm^{\beta}m^{\alpha}}^{j_{\beta}j_{\alpha}} \tag{05}$$ Note that under a second interchange the overall factor would be $\left(-1\right)^{2\left(j_{\alpha}+j_{\beta}-j\right)}=+1$ as expected, since $\left(j_{\alpha}+j_{\beta}-j\right)$ is always a (non-negative) integer.

So for $j_{\alpha}=I=j_{\beta}$ the two coefficients differ by $(-1)^\left(2I-j\right)$ and in agreement with Michael Seifert's answer :

the combined states are symmetric when $2I$ and $j$ are both even or both odd, and antisymme- tric when one quantity is even and the other is odd.

Examples :

1. $\qquad j_{\alpha}=\frac{1}{2}=j_{\beta}$

$$\boldsymbol{2}\boldsymbol{\otimes}\boldsymbol{2}=\boldsymbol{1}\boldsymbol{\oplus}\boldsymbol{3} \nonumber$$ $$\begin{bmatrix} \left|0,\hphantom{\!\!-\!}0\right\rangle\vphantom{\left|\frac12,\!\!-\frac12\right\rangle_{\beta}}\\ \left|1,\!\!-\!1\right\rangle\vphantom{\left|\frac12,\!\!-\frac12\right\rangle_{\beta}}\\ \left|1,\hphantom{\!\!-\!}0\right\rangle\vphantom{\left|\frac12,\!\!-\frac12\right\rangle_{\beta}}\\ \left|1,\!\!+\!1\right\rangle\vphantom{\left|\frac12,\!\!-\frac12\right\rangle_{\beta}} \end{bmatrix} = \begin{bmatrix} 0&-\rho\hphantom{+}&+\rho\hphantom{+}&0\vphantom{\left|\frac{1} {2},\!\!-\frac12\right\rangle_{\beta}}\\ +1\hphantom{+}&0&0&0\vphantom{\left|\frac{1} {2},\!\!-\frac12\right\rangle_{\beta}}\\ 0&+\rho\hphantom{+}&+\rho\hphantom{+}&0\vphantom{\left|\frac{1} {2},\!\!-\frac12\right\rangle_{\beta}}\\ 0&0&0&+1\hphantom{+}\vphantom{\left|\frac{1} {2},\!\!-\frac12\right\rangle_{\beta}} \end{bmatrix} \begin{bmatrix} \left|\frac{1}{2},\!\!-\frac{1}{2}\right\rangle_{\alpha}\left|\frac{1} {2},\!\!-\frac12\right\rangle_{\beta}\\ \left|\frac{1}{2},\!\!-\frac{1}{2}\right\rangle_{\alpha}\left|\frac{1}{2},\!\!+\frac12\right\rangle_{\beta}\\ \left|\frac{1}{2},\!\!+\frac{1}{2}\right\rangle_{\alpha}\left|\frac{1}{2},\!\!-\frac12\right\rangle_{\beta}\\ \left|\frac{1}{2},\!\!+\frac{1}{2}\right\rangle_{\alpha}\left|\frac{1}{2},\!\!+\frac12\right\rangle_{\beta} \end{bmatrix} \, , \quad \rho = \sqrt{\tfrac12} \tag{Ex-01}$$

$\boldsymbol{1} : \left|0,\,0\,\right\rangle \Longrightarrow \text{antisymmetric}$

$\boldsymbol{3} : \left|1,\!\!-\!1\right\rangle,\left|1,0\right\rangle,\left|1,\!\!-\!1\right\rangle\Longrightarrow \text{symmetric}$

1. $\qquad j_{\alpha}=1=j_{\beta}$

$$\boldsymbol{3}\boldsymbol{\otimes}\boldsymbol{3}=\boldsymbol{1}\boldsymbol{\oplus}\boldsymbol{3}\boldsymbol{\oplus}\boldsymbol{5} \nonumber$$ $$\!\!\!\!\!\!\!\!\!\! \begin{bmatrix} \left|0,\hphantom{\!\!-\!}0\right\rangle\vphantom{\left|1,\!\!-\!1 \right\rangle_{\beta}} \\ \left|1,\!\!-\!1\right\rangle\vphantom{\left|1,\!\!-\!1 \right\rangle_{\beta}} \\ \left|1,\hphantom{\!\!-\!}0\right\rangle\vphantom{\left|1,\!\!-\!1 \right\rangle_{\beta}} \\ \left|1,\!\!+\!1\right\rangle\vphantom{\left|1,\!\!-\!1 \right\rangle_{\beta}} \\ \left|2,\!\!-\!2\right\rangle\vphantom{\left|1,\!\!-\!1 \right\rangle_{\beta}} \\ \left|2,\!\!-\!1\right\rangle\vphantom{\left|1,\!\!-\!1 \right\rangle_{\beta}} \\ \left|2,\hphantom{\!\!-\!}0\right\rangle\vphantom{\left|1,\!\!-\!1 \right\rangle_{\beta}} \\ \left|2,\!\!+\!1\right\rangle\vphantom{\left|1,\!\!-\!1 \right\rangle_{\beta}} \\ \left|2,\!\!+\!2\right\rangle\vphantom{\left|1,\!\!-\!1 \right\rangle_{\beta}} \end{bmatrix} \!=\! \begin{bmatrix} 0&0&+\sigma\hphantom{+}&0&-\sigma\hphantom{+}&0&+\sigma\hphantom{+}&0&0\vphantom{\left|1,\!\!-\!1 \right\rangle_{\beta}} \\ 0&-\rho\hphantom{+}&0&+\rho\hphantom{+}&0&0&0&0&0\vphantom{\left|1,\!\!-\!1 \right\rangle_{\beta}} \\ 0&0&-\rho\hphantom{+}&0&0&0&+\rho\hphantom{+}&0&0\vphantom{\left|1,\!\!-\!1 \right\rangle_{\beta}} \\ 0&0&0&0&0&-\rho\hphantom{+}&0&+\rho\hphantom{+}&0\vphantom{\left|1,\!\!-\!1 \right\rangle_{\beta}} \\ +1\hphantom{+}&0&0&0&0&0&0&0&0\vphantom{\left|1,\!\!-\!1 \right\rangle_{\beta}} \\ 0&+\rho\hphantom{+}&0&+\rho\hphantom{+}&0&0&0&0&0\vphantom{\left|1,\!\!-\!1 \right\rangle_{\beta}} \\ 0&0&+\tau\hphantom{+}&0&+\upsilon\hphantom{+}&0&+\tau\hphantom{+}&0&0\vphantom{\left|1,\!\!-\!1 \right\rangle_{\beta}} \\ 0&0&0&0&0&+\rho\hphantom{+}&0&+\rho\hphantom{+}&0\vphantom{\left|1,\!\!-\!1 \right\rangle_{\beta}} \\ 0&0&0&0&0&0&0&0&+1\hphantom{+}\vphantom{\left|1,\!\!-\!1 \right\rangle_{\beta}} \end{bmatrix} \begin{bmatrix} \left|1,\!\!-\!1 \right\rangle_{\alpha}\left|1,\!\!-\!1 \right\rangle_{\beta} \\ \left|1,\!\!-\!1 \right\rangle_{\alpha}\left|1,\hphantom{\!\!-\!}0\right\rangle_{\beta} \\ \left|1,\!\!-\!1 \right\rangle_{\alpha}\left|1,\!\!+\!1 \right\rangle_{\beta} \\ \left|1,\hphantom{\!\!-\!}0\right\rangle_{\alpha}\left|1,\!\!-\!1 \right\rangle_{\beta} \\ \left|1,\hphantom{\!\!-\!}0\right\rangle_{\alpha}\left|1,\hphantom{\!\!-\!}0\right\rangle_{\beta} \\ \left|1,\hphantom{\!\!-\!}0 \right\rangle_{\alpha}\left|1,\!\!+\!1\right\rangle_{\beta} \\ \left|1,\!\!+\!1 \right\rangle_{\alpha}\left|1,\!\!-\!1 \right\rangle_{\beta} \\ \left|1,\!\!+\!1 \right\rangle_{\alpha}\left|1,\hphantom{\!\!-\!}0\right\rangle_{\beta} \\ \left|1,\!\!+\!1\right\rangle_{\alpha}\left|1,\!\!+\!1\right\rangle_{\beta} \end{bmatrix} \,, \: \begin{matrix} \rho = \sqrt{\frac12} \\ \sigma = \sqrt{\frac13}\\ \tau = \sqrt{\frac16}\\ \upsilon = \sqrt{\frac23} \end{matrix} \tag{Ex-02}$$

$\boldsymbol{1} : \left|0,\,0\,\right\rangle \Longrightarrow \text{symmetric}$

$\boldsymbol{3} : \left|1,\!\!-\!1\right\rangle,\left|1,0\right\rangle,\left|1,\!\!-\!1\right\rangle\Longrightarrow \text{antisymmetric}$

$\boldsymbol{5} : \left|2,\!\!-\!2\right\rangle,\left|2,\!\!-\!1\right\rangle,\left|2,0\right\rangle,\left|2,\!\!+\!1\right\rangle,\left|2,\!\!+\!2\right\rangle \Longrightarrow \text{symmetric}$

(1) Wigner Eugene P. "Group Theory and Its Application to Quantum Mechanics of Atomic Spectra" (1959) : as refered in the footnote of page 192 the two coefficients differ, using our symbols, by the factor $\left(-1\right)^{j_{\alpha}+j_{\beta}-j}$.

For spin-1/2 particles, the entire wavefunction has to be antisymmetric under particle exchange. Also, the spatial component has parity $(-1)^\ell$, so if $\ell$ is even, the spin component must be odd under exchange.