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1. What is being asked?

A student is looking at the formulae used by Q‑Chem for spin‑orbit coupling (SOC) matrix elements between excited states that are described by spin‐adapted configuration‑interaction (CI) wavefunctions.

  • For a singlet–triplet pair the program prints

[ \boxed{\displaystyle \langle\Phi_{\rm S}^{I}\,|\,\hat H_{\rm SO}\,|\,\Phi_{\rm T}^{J}\rangle = \sqrt{\sum_{m_s = 0,\pm1}\Big| \langle\Phi_{\rm S}^{I}\,|\,\hat H_{\rm SO}\,|\,\Phi_{\rm T}^{J,m_s}\rangle\Big|^{2}} } \tag{1} ]

  • For a triplet–triplet pair the manual writes

[ \boxed{\displaystyle \langle\Phi_{\rm T}^{I}\,|\,\hat H_{\rm SO}\,|\,\Phi_{\rm T}^{J}\rangle = \sqrt{\sum_{m_s = 0,\pm1}\Big| \langle\Phi_{\rm T}^{I,m_s}\,|\,\hat H_{\rm SO}\,|\,\Phi_{\rm T}^{J,m_s}\rangle\Big|^{2}} } \tag{2} ]

The student wonders whether (2) should actually contain two independent sums over the spin‑projection quantum numbers of the two states, i.e.

[ \sqrt{\displaystyle\sum_{m_{s,i}=0,\pm1}\; \sum_{m_{s,j}=0,\pm1} \Big| \langle\Phi_{\rm T}^{I,m_{s,i}}|\hat H_{\rm SO}| \Phi_{\rm T}^{J,m_{s,j}}\rangle\Big|^{2}} . \tag{3} ]

In other words: Do the off‑diagonal spin‑blocks (the “(a)” elements in the 3 × 3 matrix shown by the student) contribute to the overall SOC, or are they forced to zero by the theory?

The solution below shows, step‑by‑step, why only the diagonal spin‑blocks survive, so the single‑sum expression (2) is the correct one.

2. Detailed derivation

2.1. Spin‑orbit Hamiltonian

In non‑relativistic quantum chemistry the one‑electron spin‑orbit operator is usually written (in atomic units) as

[ \hat H_{\rm SO}= \sum_{p}\,\xi(\mathbf r_{p})\, \hat{\mathbf L}{p}!\cdot!\hat{\mathbf S}{p}, \tag{4} ]

where

  • (\hat{\mathbf L}_{p}) acts only on the spatial coordinates of electron p,
  • (\hat{\mathbf S}_{p}) acts only on the spin coordinate of the same electron, and
  • (\xi(\mathbf r)) is a (real) scalar function that depends on the distance of the electron from the nuclei (it is the so‑called spin‑orbit constant).

Because (\hat{\mathbf L}) does not change the spin part of a wavefunction, the only way the operator can connect two many‑electron spin functions is through the spin scalar product (\hat{\mathbf L}!\cdot!\hat{\mathbf S}).

The one‑electron spin operators have the well‑known matrix elements in the basis of spin‑eigenfunctions ({ \alpha\rangle, \beta\rangle}) (or, equivalently, in the coupled basis ( S,m_{s}\rangle) with (S=1) for a triplet):

[ \begin{aligned} \langle S,m_{s}|\hat{\mathbf L}!\cdot!\hat{\mathbf S}|S,m’{s}\rangle &= \langle S,m{s}|L_{z}S_{z}+ \tfrac12(L_{+}S_{-}+L_{-}S_{+})|S,m’{s}\rangle\[2pt] &= \langle S,m{s}|L_{z}|S,m’{s}\rangle\,m{s}\,\delta_{m_{s},m’{s}} \;+\; \text{terms that change }m{s}\text{ by }\pm1 . \end{aligned} \tag{5} ]

When the total spin of the many‑electron state is a good quantum number (as it is for the spin‑adapted CI states used in TD‑DFT), the operator (\hat H_{\rm SO}) cannot mix different values of (m_{s}) unless the spatial part supplies the necessary angular‑momentum change.

In practice the single‑electron operators (L_{\pm}) act on the orbital of the same electron that carries the spin‑flip operator (S_{\mp}). For a closed‑shell or restricted open‑shell determinant the net effect is that the matrix element between two overall triplet states is diagonal in the total spin‑projection quantum number:

[ \boxed{ \langle\Phi_{\rm T}^{I,m_{s,i}}|\hat H_{\rm SO}| \Phi_{\rm T}^{J,m_{s,j}}\rangle = \delta_{m_{s,i},\,m_{s,j}}\;X^{IJ}{m{s,i}} } \tag{6} ]

where (X^{IJ}{m{s,i}}) is a (generally complex) number that depends on the spatial part of the two states and on the particular value of (m_{s}) (through the factor (m_{s}) that appears in Eq. (5)).

Key point: the Kronecker delta (\delta_{m_{s,i},m_{s,j}}) tells us that off‑diagonal spin blocks ((m_{s,i}\neq m_{s,j})) are exactly zero.

2.2. Consequence for the norm of the SOC matrix element

The program defines an effective SOC matrix element between two spin‑adapted states as the Euclidean norm of the (3 × 3) block matrix in spin space:

[ \big|\mathbf H^{IJ}{\rm SO}\big| = \sqrt{ \sum{m_{s,i}=0,\pm1}\; \sum_{m_{s,j}=0,\pm1} \big| \langle\Phi_{\rm T}^{I,m_{s,i}}| \hat H_{\rm SO} |\Phi_{\rm T}^{J,m_{s,j}}\rangle \big|^{2} } . \tag{7} ]

Insert the diagonal form (6):

[ \begin{aligned} \big|\mathbf H^{IJ}{\rm SO}\big| &= \sqrt{ \sum{m_{s,i}=0,\pm1} \sum_{m_{s,j}=0,\pm1} \big|\delta_{m_{s,i},m_{s,j}}\,X^{IJ}{m{s,i}}\big|^{2} }\[4pt] &= \sqrt{ \sum_{m_{s}=0,\pm1} \big|X^{IJ}{m{s}}\big|^{2} } . \end{aligned} \tag{8} ]

Because the double sum collapses to a single sum over the three possible values of the same spin projection, we recover exactly the expression quoted in the Q‑Chem manual (Eq. 2).

Thus the “(a)” elements that the student placed in the off‑diagonal positions of the 3 × 3 matrix are forced to be zero by the spin‑selection rule encoded in Eq. (6).

2.3. Explicit evaluation for a triplet–triplet pair

Let us write the three spin‑adapted CI wavefunctions for a given electronic configuration (the superscript denotes the state index, the subscript the spin projection):

[ \begin{aligned} |\Phi_{\rm T}^{I,0}\rangle &= \frac{1}{\sqrt{2}} \bigl(|\alpha\beta\rangle + |\beta\alpha\rangle\bigr) \otimes |\Psi^{I}{\rm orb}\rangle ,
|\Phi{\rm T}^{I,+1}\rangle &= |\alpha\alpha\rangle \otimes |\Psi^{I}{\rm orb}\rangle ,
|\Phi{\rm T}^{I,-1}\rangle &= |\beta\beta\rangle \otimes |\Psi^{I}_{\rm orb}\rangle . \end{aligned} \tag{9} ]

Operating with (\hat H_{\rm SO}) on any of these states gives (using Eq. 5)

[ \hat H_{\rm SO}\,|\Phi_{\rm T}^{I,m_{s}}\rangle = m_{s}\; \Bigl[\,\sum_{p}\xi(\mathbf r_{p})\,\hat L_{z}^{(p)}\Bigr]\, |\Phi_{\rm T}^{I,m_{s}}\rangle , \tag{10} ]

i.e. the result is proportional to the same spin function with the same (m_{s}). Consequently

[ \langle\Phi_{\rm T}^{I,m_{s}}|\hat H_{\rm SO}|\Phi_{\rm T}^{J,m’{s}}\rangle = m{s}\,\delta_{m_{s},m’{s}}\; \langle\Psi^{I}{\rm orb}|\,\sum_{p}\xi(\mathbf r_{p})\,\hat L_{z}^{(p)}\, |\Psi^{J}_{\rm orb}\rangle . \tag{11} ]

The spatial integral is the same for the three components; only the prefactor (m_{s}=0,\pm1) changes.

Hence the norm becomes

[ \big|\mathbf H^{IJ}_{\rm SO}\big| = \sqrt{\,0^{2}+|X^{IJ}|^{2}+|{-X^{IJ}}|^{2}} = \sqrt{2}\,|X^{IJ}|\;, \tag{12} ]

where (X^{IJ}= \langle\Psi^{I}{\rm orb}|\,\sum{p}\xi(\mathbf r_{p})\,\hat L_{z}^{(p)}\,|\Psi^{J}_{\rm orb}\rangle).
Equation (12) is precisely the numerical value that the program reports when it evaluates the single‑sum formula (2).

3. Final answer

The SOC matrix element between two triplet states is defined as the Euclidean norm of the (3 × 3) spin‑block matrix. Because the one‑electron spin‑orbit operator is diagonal in the total spin‑projection quantum number (the Kronecker delta (\delta_{m_{s,i},m_{s,j}}) appears in every matrix element), all off‑diagonal blocks vanish. Consequently the double sum in Eq. (3) collapses to a single sum over the three possible values of the common spin projection:

[ \boxed{ \langle\Phi_{\rm T}^{I}|\hat H_{\rm SO}|\Phi_{\rm T}^{J}\rangle = \sqrt{\displaystyle\sum_{m_{s}=0,\pm1} \Big|\, \langle\Phi_{\rm T}^{I,m_{s}}|\hat H_{\rm SO}| \Phi_{\rm T}^{J,m_{s}}\rangle \Big|^{2}} } ]

Therefore the expression given in the Q‑Chem manual (the “single‑sum” form) is correct; the off‑diagonal entries labelled “(a)” in the student’s matrix are identically zero by symmetry.

4. Common mistakes

Mistake Why it is wrong How to avoid it  
Assuming the spin‑orbit operator can change the total spin projection (i.e., that (\langle T^{m_s} \hat H_{\rm SO} T^{m’_s}\rangle) is non‑zero for (m_s\neq m’_s)). The operator contains (\hat{\mathbf L}\cdot\hat{\mathbf S}); the spin part can raise/lower (m_s) by at most one, but the same electron must also change its orbital angular momentum. For a closed‑shell reference (or for spin‑adapted CI) the net effect is that the overall **

Original question: Calculating Triplet-Triplet Spin-Orbit Coupling on Chemistry Stack Exchange, licensed CC BY-SA.