The singlet and triplet states

The singlet state describes a composite system with a coupled spin eigenstate of $\small \vert j=0,m_j=0\rangle$ , while a triplet state refers to a composite system with coupled spin eigenstates of $\small \vert 1,+1\rangle$ , $\small \vert 1,0\rangle$ and $\small \vert 1,-1\rangle$ .

Consider a system of two spin- $\small \frac{1}{2}$ particles ( $\small s_1=\frac{1}{2}$ and $\small s_2=\frac{1}{2}$ ). According to the Clebsch-Gordan series, the allowed angular momenta of the system are $\small j=1,0$ , which implies that the system is characterised by the spin eigenstates $\small \vert 1,+1\rangle$ , $\small \vert 1,0\rangle$ , $\small \vert 1,-1\rangle$ and $\small \vert 0,0\rangle$ . The basis states forming these four spin eigenstates are given by eq193, with the general spin eigenstate of the system being a linear combination of these basis states (see eq195).

Let’s begin with the determination of the values of the coefficients for the spin eigenstates $\small \vert 1,0\rangle$ and $\small \vert 0,0\rangle$ , which must adhere to the condition: $\small \hat{S}_z^{\; (T)}\vert\psi\rangle=0\hbar\vert\psi\rangle$ . Letting the operator $\small \hat{S}_z^{\; (T)}$ act on eq195 and using eq193 and 197,

$\hat{S}_z^{\; (T)}\vert\psi\rangle=c_1\hbar\begin{pmatrix} 1\\0 \\ 0 \\ 0 \end{pmatrix}-c_4\hbar\begin{pmatrix} 0\\0 \\ 0 \\ 1 \end{pmatrix}+0\hbar\left [c_2\begin{pmatrix} 0\\1 \\ 0 \\ 0 \end{pmatrix}+c_3\begin{pmatrix} 0\\0 \\ 1 \\ 0 \end{pmatrix}\right ]$

For the eigenvalue equation $\hat{S}_z^{\; (T)}\vert\psi\rangle=0\hbar\vert\psi\rangle$ to be valid, $c_1=c_4=0$ , i.e. one possible spin eigenstate when $m_j=0$ is $\vert\psi\rangle=c_2\vert+z\rangle_1\otimes\vert-z\rangle_2+c_3\vert-z\rangle_1\otimes\vert+z\rangle_2$ , while the other possible spin eigenstate is $\vert\psi\rangle=c_2\vert+z\rangle_1\otimes\vert-z\rangle_2-c_3\vert-z\rangle_1\otimes\vert+z\rangle_2$ . These two spin eigenstates must correspond to either $\vert 1,0\rangle$ or $\vert0,0\rangle$ . To distinguish them, we use eq203 and let the operator act on the two spin eigenstates to give:

$\hat{{S}^{2}}^{(T)}\left [ c_2\begin{pmatrix} 0\\1 \\0 \\0 \end{pmatrix}\pm c_3\begin{pmatrix} 0\\0 \\ 1 \\ 0 \end{pmatrix}\right ] =\hbar^{2}\left [ c_2\begin{pmatrix} 0\\1 \\1 \\0 \end{pmatrix}\pm c_3\begin{pmatrix} 0\\1 \\ 1 \\ 0 \end{pmatrix}\right ]$

Since the two eigenstates are characterised by $j=1$ and $j=0$ , the expected results are $\hat{{S}^{2}}^{(T)}\vert\psi\rangle=j(j+1)\hbar^{2}\vert\psi\rangle=2\hbar^{2}\vert\psi\rangle$ and $\hat{{S}^{2}}^{(T)}\vert\psi\rangle=j(j+1)\hbar^{2}\vert\psi\rangle=0\hbar^{2}\vert\psi\rangle$ respectively. Therefore, the above equation is valid only if $c_2=c_3=c$ , such that:

$\small \hat{{S}^{2}}^{(T)}\left [ c\begin{pmatrix} 0\\1 \\0 \\0 \end{pmatrix}+ c\begin{pmatrix} 0\\0 \\ 1 \\ 0 \end{pmatrix}\right ] =\hbar^{2}\left [ c\begin{pmatrix} 0\\1 \\1 \\0 \end{pmatrix}+ c\begin{pmatrix} 0\\1 \\ 1 \\ 0 \end{pmatrix}\right ]=2\hbar^{2}c\begin{pmatrix} 0\\1 \\ 1 \\ 0 \end{pmatrix}=2\hbar^{2}\left [ c\begin{pmatrix} 0\\1 \\0 \\0 \end{pmatrix}+ c\begin{pmatrix} 0\\0 \\ 1 \\ 0 \end{pmatrix}\right ]$

and

$\small \hat{{S}^{2}}^{(T)}\left [ c\begin{pmatrix} 0\\1 \\0 \\0 \end{pmatrix}- c\begin{pmatrix} 0\\0 \\ 1 \\ 0 \end{pmatrix}\right ] =\hbar^{2}\left [ c\begin{pmatrix} 0\\1 \\1 \\0 \end{pmatrix}- c\begin{pmatrix} 0\\1 \\ 1 \\ 0 \end{pmatrix}\right ]=\hbar^{2}c\begin{pmatrix} 0\\0 \\ 0 \\ 0 \end{pmatrix}=0\hbar^{2}\left [ c\begin{pmatrix} 0\\1 \\0 \\0 \end{pmatrix}- c\begin{pmatrix} 0\\0 \\ 1 \\ 0 \end{pmatrix}\right ]$

Therefore, $\small \vert1,0\rangle=c(\vert+z\rangle_1\otimes\vert-z\rangle_2+\vert-z\rangle_1\otimes\vert+z\rangle_2)$ and $\small \vert0,0\rangle=c(\vert+z\rangle_1\otimes\vert-z\rangle_2-\vert-z\rangle_1\otimes\vert+z\rangle_2)$ . The spin state $\small \vert0,0\rangle$ as mentioned earlier is the singlet state, which after normalisation is

$\small \vert0,0\rangle=\frac{1}{\sqrt{2}}(\vert+z\rangle_1\otimes\vert-z\rangle_2-\vert-z\rangle_1\otimes\vert+z\rangle_2)\; \; \; \; \; \; \; \; 220$

Question

Show that the normalisation constant is $\small \frac{1}{\sqrt{2}}$ .

Answer

$\langle\psi\vert\psi\rangle=1\; \; \; \; \Rightarrow \; \; \; \; c^{*}\left [ \begin{pmatrix} 0\\1 \\ 0 \\ 0 \end{pmatrix}-\begin{pmatrix} 0\\0 \\ 1 \\ 0 \end{pmatrix} \right ] ^{\dagger}c\left [ \begin{pmatrix} 0\\1 \\ 0 \\ 0 \end{pmatrix}-\begin{pmatrix} 0\\0 \\ 1 \\ 0 \end{pmatrix} \right ]=1$

$\vert c\vert^{2}\begin{pmatrix} 0 &1 &-1 &0 \end{pmatrix}\begin{pmatrix} 0\\1 \\ -1 \\ 0 \end{pmatrix}=1\; \; \; \; \Rightarrow \; \; \; \; c=\frac{1}{\sqrt{2}}$

Likewise, the state $\vert1,0\rangle$ after normalisation is

$\vert1,0\rangle=\frac{1}{\sqrt{2}}(\vert+z\rangle_1\otimes\vert-z\rangle_2+\vert-z\rangle_1\otimes\vert+z\rangle_2)\; \; \; \; \; \; \; \; 221$

We have two more spin eigenstates of the system to consider: $\vert1,+1\rangle$ and $\vert1,-1\rangle$ . Since the allowed values of the total magnetic quantum number $m_j$ are the sum of the allowed values of the two contributing magnetic quantum numbers (see eq207), $\vert1,+1\rangle=\vert+z\rangle_1\otimes\vert+z\rangle_2$ and $\vert1,-1\rangle=\vert-z\rangle_1\otimes\vert-z\rangle_2$ ., both of which are already normalised because they are basis states.

Question

Verify that $\hat{{S}^{^{2}}}^{{T}}\vert1,\pm1\rangle=2\hbar^{2}\vert1,\pm1\rangle$ and $\hat{S}_z^{\; {T}}\vert1,\pm1\rangle=\pm\hbar\vert1,\pm1\rangle$ .

Answer

Using eq203

$\hat{{S}^{2}}^{(T)}\vert1,+1\rangle=\hbar^{2}\begin{pmatrix} 2 &0 &0 &0 \\ 0 &1 &1 &0 \\ 0& 1 & 1 & 0\\ 0& 0& 0& 2 \end{pmatrix}\begin{pmatrix} 1\\0 \\ 0 \\ 0 \end{pmatrix}=2\hbar^{2}\begin{pmatrix} 1\\0 \\ 0 \\ 0 \end{pmatrix}$

Similarly, $\hat{{S}^{2}}^{(T)}\vert1,-1\rangle=2\hbar^{2}\vert1,-1\rangle$ . Using eq197,

$\hat{S}_z^{\; {T}}\vert1,+1\rangle=\hbar\begin{pmatrix} 1 &0 &0 &0 \\ 0 &0 &0 &0 \\ 0& 0 & 0 & 0\\ 0& 0& 0& -1 \end{pmatrix}\begin{pmatrix} 1\\0 \\ 0 \\ 0 \end{pmatrix}=\hbar\begin{pmatrix} 1\\0 \\ 0 \\ 0 \end{pmatrix}$

Similarly, $\hat{S}_z^{\; (T)}\vert1,-1\rangle=-\hbar\vert1,-1\rangle$ .

Therefore, we have:

$Triplet\: state=\left\{\begin{matrix} \vert1,+1\rangle&=\vert+z\rangle_1\otimes\vert+z\rangle_2 \\ \vert1,0\rangle &=\frac{1}{\sqrt{2}}(\vert+z\rangle_1\otimes\vert-z\rangle_2+\vert-z\rangle_1\otimes\vert+z\rangle_2)\\ \vert1,-1\rangle&=\vert-z\rangle_1\otimes\vert-z\rangle_2 \end{matrix}\right.$

$Singlet\: state=\vert0,0\rangle=\frac{1}{\sqrt{2}}[\vert+z\rangle_1\otimes\vert-z\rangle_2-\vert-z\rangle_1\otimes\vert+z\rangle_2]$

or in the wave function notation:

$Triplet\: state=\left\{\begin{matrix} \sigma_a&=\alpha(1)\alpha(2)\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; 222 \\ \sigma_b &=\frac{1}{\sqrt{2}}[\alpha(1)\beta(2)+\beta(1)\alpha(2)]\; \; \; \; \; \; 223\\ \sigma_c&=\beta(1)\beta(2)\; \; \; \; \; \; \; \;\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; 224 \end{matrix}\right.$

$Singlet\: state=\sigma_d=\frac{1}{\sqrt{2}}[\alpha(1)\beta(2)-\beta(1)\alpha(2)]\; \; \; \; \; \; \; \; 225$

The helium atom with the excited state configuration of 1s¹2s¹ is commonly used to illustrate the singlet and triplet states. If the electrons have anti-parallel spins, the excited atom is characterised by the singlet state. If the electrons have parallel spins, the excited helium is in the triplet state.

The singlet and triplet states

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