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.

 

 

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