Singlet vs. Triplet Spin States

Consider the two electrons in a chemical bond. They have spin vectors ${\bf S}_1$ and ${\bf S}_2$. Each of the components of the spin vector is quantized and can take on values $\pm \hbar/2$. We now ask what are the allowed total spin states generated by adding the spins ${\bf S}= {\bf S}_1 + {\bf S}_2$. If the two vectors are completely anti-parallel, i.e. ${\bf S}_1 = -{\bf S}_2$, then the total spin ${\bf S}= 0$, i.e. the magnitude of the total spin is 0. In this case, the only allowed value of the quantum number $m_s$ is 0. However, if any of the components of ${\bf S}_1$ and ${\bf S}_2$ are the same, the total spin will be a nonzero vector. Due to requirements of normalization, the only allowable nonzero value of the total spin is $\hbar$ in magnitude, which leaves three possible values for $m_s$, namely -1,0,1. For this reason, spin states in which the electrons have a total spin of magnitude $\hbar$ are called the triplet states, while the one spin state corresponding to a total spin of 0 is called the singlet state. The two-electron spin wave function of the singlet state is antisymmetric with respect to exchange of the spins:

$$\Psi_{\rm s}(s_1,s_2) = {1 \over \sqrt{2}} \left[\psi_{\upar... ...rrow}(s_2) - \psi_{\uparrow}(s_2)\psi_{\downarrow}(s_1)\right]$$

while there are three possible triplet two-electron spin states corresponding to $m_s=-1,0,1$:

$$\Psi_{{\rm t},m=1} = \psi_{\uparrow}(s_1)\psi_{\uparrow}(s_2)$$

$$\Psi_{{\rm t},m=0} = {1 \over \sqrt{2}} \left[\psi_{\uparrow}(s_1)\psi_{\downarrow}(s_2) + \psi_{\uparrow}(s_2)\psi_{\downarrow}(s_1)\right]$$

$$\Psi_{{\rm t},m=-1} = \psi_{\downarrow}(s_1)\psi_{\downarrow}(s_2)$$

Note that in each of the triplet states, the wave function does not change sign if the electrons are exchanged. Hence, two electrons fixed in space, having only spin wave functions, can never be in a triplet state. However, when there is a spatial part to the wave function, then both singlet and triplet states are possible.

Let us return to the valence bond example of H$_2$ to illustrate how the singlet and triplet states enter into the wave function. According to the valence bond theory, the ground state of H$_2$ could be written approximately as

$$\Psi({\bf r}_1,{\bf r}_2,s_1,s_2) \propto \left[\psi_{1s}({\... ...rrow}(s_2) - \psi_{\uparrow}(s_2)\psi_{\downarrow}(s_1)\right]$$

where ${\bf r}_A$ and ${\bf r}_B$ are the positions of the two protons. We see that, in the ground state, the spin state of the electrons is the singlet state with total spin 0. Recall from our discussion of valence bond theory that the antisymmetric spatial wave function

$$\left[\psi_{1s}({\bf r}_1-{\bf r}_A)\psi_{1s}({\bf r}_2-{\bf... ...{1s}({\bf r}_2-{\bf r}_A)\psi_{1s}({\bf r}_1-{\bf r}_B)\right]$$

has a slighly higher energy than the symmetric state (see Fig. 6.37). Therefore, the two-electron spin state

$$\tilde{\Psi}({\bf r}_1,{\bf r}_2,s_1,s_2) \propto \left[\psi... ...rrow}(s_2) + \psi_{\uparrow}(s_2)\psi_{\downarrow}(s_1)\right]$$

is an excited state in which the electrons are in one of the triplet states. In fact, in the absence of external magnetic fields, the three states:

$$\tilde{\Psi}_1({\bf r}_1,{\bf r}_2,s_1,s_2) \propto \left[\psi_{1s}({\bf r}_1-{\bf r}_A)\psi_{1s}({\bf r}_2-{\bf r}_B... ...)\psi_{1s}({\bf r}_1-{\bf r}_B)\right] \psi_{\uparrow}(s_1)\psi_{\uparrow}(s_2)$$

$$\tilde{\Psi}_2({\bf r}_1,{\bf r}_2,s_1,s_2) \propto \left[\psi_{1s}({\bf r}_1-{\bf r}_A)\psi_{1s}({\bf r}_2-{\bf r}_B... ...(s_1)\psi_{\downarrow}(s_2) + \psi_{\uparrow}(s_2)\psi_{\downarrow}(s_1)\right]$$

$$\tilde{\Psi}_3({\bf r}_1,{\bf r}_2,s_1,s_2) \propto \left[\psi_{1s}({\bf r}_1-{\bf r}_A)\psi_{1s}({\bf r}_2-{\bf r}_B... ...i_{1s}({\bf r}_1-{\bf r}_B)\right] \psi_{\downarrow}(s_1)\psi_{\downarrow}(s_2)$$

have the same energy and are, therefore, threefold degenerate. What if we wanted to generate an excited state in which the spin state is the singlet state? It should be clear from the above analysis that this can only be done by exciting one of the $1s$ states to a $2s$ state, for example, giving a wave function:

$$\Psi({\bf r}_1,{\bf r}_2,s_1,s_2) \propto \left[\psi_{2s}({\... ...rrow}(s_2) - \psi_{\uparrow}(s_2)\psi_{\downarrow}(s_1)\right]$$

The energy needed to promote the $1s$ to the $2s$ orbital is larger than the energy needed to promote the symmetric to the antisymmetric spatial wave function composed of $1s$ orbitals alone. This is why, in general, excited states involving triplet spin states are lower in energy than excited states involving singlet states.

Absorption of a photon by a molecule cannot change the spin state from the singlet to a triplet, hence absorption of a singlet ground state will yield a singlet excited state. Generally, singlet excited states decay quickly back down to the ground state, although the resulting vibrational or rotational state might be different. This process is known as fluoresence. On occasion, however, the singlet excited state can become a triplet state in the course of the quantum-mechanical evolution that results after excitation (see figure below).

Figure: Illustration of how a singlet excited state can convert to a triplet excited state.

Once in the triplet excited state, decay back to the ground state turns out to be quite slow (seconds to hours) due to the orthogonality of the wave functions. The process, therefore, is not spontaneous and proceeds only via the coupling to the environment. This is known as phorphoresence.

It is also possible to excite a molecule to an excited state with overall non-bonding character, leading to a bond dissociation and the production of new molecules known as photofragments. This type of chemistry is known as photochemistry. In some cases, simply exposing a molecule to light will give rise to a photodissociation process. An example is hydrogen peroxide, which has a very weak O-O bond and undergoes the following dissociation reaction:

$${\rm H}_2{\rm O}_2 \longrightarrow 2{\rm OH} \longrightarrow {\rm H}_2{\rm O} + {1 \over 2}{\rm O}_2$$