As an Amazon Associate, I earn from qualifying purchases. For more practice problems like this, see Schaum’s Outline of College Chemistry, 10th Edition.

1. What the question is asking (plain language)

You have two one‑electron wavefunctions – usually the ground‑state (state 1) and an excited‑state (state 2) – that are expressed as linear combinations of atomic orbitals (AOs).
From the MO‑coefficient matrices of the two states you want to obtain the transition dipole moment (TDM)

[ \boldsymbol\mu_{12}= \langle\Psi_1|\hat{\boldsymbol\mu}|\Psi_2\rangle ]

for each Cartesian direction ( X, Y, Z ).
The problem is to write down, in a step‑by‑step way, how to construct the transition density matrix from the two coefficient matrices and then contract it with the dipole‑integral matrix (the “multipole matrix” in the OP).

2. Full derivation – every step shown

Below we assume

  • a real Gaussian AO basis (the formulas are the same for complex orbitals, only a complex‑conjugate appears);
  • the AO basis is orthonormal or, if not, the overlap matrix S is known; we will give the expressions for both cases;
  • the calculation is performed in the restricted (closed‑shell) case, i.e. we have separate α‑ and β‑spin blocks but the same spatial orbitals for the two spins. The open‑shell case follows the same algebra, just keep the two spin blocks separate.

2.1 Molecular‑orbital (MO) coefficient matrices

For state 1 (ground) we have

[ \mathbf{C}^{(1)} = \bigl[\,\mathbf{C}^{(1)}{\alpha}\;\; \mathbf{C}^{(1)}{\beta}\,\bigr] ]

where

  • (\mathbf{C}^{(1)}{\alpha}) (size (N{\text{AO}}\times N_{\text{occ}}^{\alpha})) contains the coefficients of the occupied α‑MOs,
  • (\mathbf{C}^{(1)}_{\beta}) contains the occupied β‑MOs.

For state 2 (excited) we have the analogous matrices

[ \mathbf{C}^{(2)} = \bigl[\,\mathbf{C}^{(2)}{\alpha}\;\; \mathbf{C}^{(2)}{\beta}\,\bigr]. ]

(If the excited state is a single‑excitation from the ground state you will usually have the same number of occupied orbitals, only one of them is replaced by a virtual orbital – the formalism does not change.)

2.2 The ordinary (ground‑state) density matrix

For a closed‑shell determinant the (spin‑summed) one‑particle density matrix is

[ \mathbf{P}^{(1)} = \mathbf{P}^{(1)}{\alpha}+\mathbf{P}^{(1)}{\beta} = \mathbf{C}^{(1)}{\alpha}\,(\mathbf{C}^{(1)}{\alpha})^{!T} + \mathbf{C}^{(1)}{\beta}\,(\mathbf{C}^{(1)}{\beta})^{!T}. ]

If the AO basis is not orthonormal the orthonormalised density is

[ \boxed{\; \mathbf{P}^{(1)} = \mathbf{C}^{(1)}{\alpha}\,(\mathbf{C}^{(1)}{\alpha})^{!T}\,\mathbf{S}^{-1} + \mathbf{C}^{(1)}{\beta}\,(\mathbf{C}^{(1)}{\beta})^{!T}\,\mathbf{S}^{-1} \;} \tag{1} ]

where (\mathbf{S}) is the AO overlap matrix ((S_{\mu\nu}= \langle\chi_\mu \chi_\nu\rangle)).

2.3 The transition density matrix

The transition density is the one‑particle operator that connects the two states:

[ \boxed{\; \mathbf{P}^{(12)} = \mathbf{P}^{(12)}{\alpha}+\mathbf{P}^{(12)}{\beta} \;} \tag{2} ]

with

[ \mathbf{P}^{(12)}{\alpha}= \mathbf{C}^{(1)}{\alpha}\,(\mathbf{C}^{(2)}{\alpha})^{!T}, \qquad \mathbf{P}^{(12)}{\beta}= \mathbf{C}^{(1)}{\beta}\,(\mathbf{C}^{(2)}{\beta})^{!T}. ]

If the AO basis is non‑orthogonal one must insert the overlap matrix (or its inverse) once to convert the bra‑side coefficients into the AO metric:

[ \boxed{\; \mathbf{P}^{(12)}{\alpha}= \mathbf{C}^{(1)}{\alpha}\,(\mathbf{C}^{(2)}{\alpha})^{!T}\,\mathbf{S}^{-1}, \qquad \mathbf{P}^{(12)}{\beta}= \mathbf{C}^{(1)}{\beta}\,(\mathbf{C}^{(2)}{\beta})^{!T}\,\mathbf{S}^{-1}. \;} \tag{3} ]

Why only one ( \mathbf{S}^{-1}) ?
The bra‑state coefficient matrix multiplies the AO basis from the left, the ket‑state from the right. The overlap matrix is needed to raise (or lower) one index; it is not needed on both sides because the transition density is a mixed matrix (bra from state 1, ket from state 2).

The total transition density matrix is then

[ \mathbf{P}^{(12)} = \mathbf{C}^{(1)}{\alpha}(\mathbf{C}^{(2)}{\alpha})^{T}\mathbf{S}^{-1} + \mathbf{C}^{(1)}{\beta}(\mathbf{C}^{(2)}{\beta})^{T}\mathbf{S}^{-1}. \tag{4} ]

Note on spin – if the excited state is a pure singlet, the α‑ and β‑contributions are identical and you may simply double the α‑part. For a triplet the two blocks have opposite sign (the transition dipole of a spin‑forbidden transition is zero in the non‑relativistic approximation).

2.4 Dipole‑integral (multipole) matrix

For each Cartesian direction (k\in{x,y,z}) you have an AO matrix

[ \boldsymbol{\mu}^{(k)}{\mu\nu}= -\langle\chi\mu|\hat{k}|\chi_\nu\rangle = -\int \chi_\mu(\mathbf{r})\, k \,\chi_\nu(\mathbf{r})\,\mathrm{d}\mathbf{r}, ]

where the minus sign is the electronic charge convention (the nuclear contribution is added later).

These three matrices are often supplied by quantum‑chemistry programs under the name dipole integrals or multipole matrix.

2.5 Contracting the transition density with the dipole matrix

The electronic part of the transition dipole vector is simply the trace of the product of the transition density with each dipole matrix:

[ \boxed{\; \mu^{\text{el}}k = \operatorname{Tr}!\bigl[\,\mathbf{P}^{(12)}\,\boldsymbol{\mu}^{(k)}\,\bigr] = \sum{\mu\nu} P^{(12)}{\mu\nu}\,\mu^{(k)}{\nu\mu}. \;} \tag{5} ]

Because (\mathbf{P}^{(12)}) is not symmetric the order of indices matters, but the trace eliminates the ordering issue (the product is a square matrix, then you sum the diagonal).

2.6 Adding the nuclear term

The nuclear contribution to the dipole moment is the same for both electronic states because the nuclei do not move during an electronic transition (the Born–Oppenheimer approximation). Therefore, when you evaluate a transition moment you may omit the nuclear term – it cancels out when the matrix element is taken between two electronic wavefunctions at the same geometry.

If you still wish to write the full expression, it is

[ \boxed{\; \mu_{k}^{(12)} = \mu^{\text{el}}{k} + \mu^{\text{nuc}}{k}, \qquad \mu^{\text{nuc}}{k}= \sum{A} Z_A\,R_{A,k}, \;} \tag{6} ]

where (Z_A) is the atomic number of nucleus (A) and (R_{A,k}) its Cartesian coordinate.

Since (\mu^{\text{nuc}}_{k}) is identical for state 1 and state 2, the transition dipole that appears in spectroscopy is simply the electronic part, Eq. (5).

2.7 Summary – algorithm in matrix notation

  1. Read the AO overlap matrix (\mathbf{S}) (identity if the basis is orthonormal).
  2. Read the MO‑coefficient blocks (\mathbf{C}^{(1)}{\alpha},\mathbf{C}^{(1)}{\beta},\mathbf{C}^{(2)}{\alpha},\mathbf{C}^{(2)}{\beta}).

  3. Form the transition density (Eq. 4)

    [ \mathbf{P}^{(12)} = \bigl[\mathbf{C}^{(1)}{\alpha}(\mathbf{C}^{(2)}{\alpha})^{!T} +\mathbf{C}^{(1)}{\beta}(\mathbf{C}^{(2)}{\beta})^{!T}\bigr]\, \mathbf{S}^{-1}. ]

  4. Load the three dipole‑integral matrices (\boldsymbol{\mu}^{(x)},\boldsymbol{\mu}^{(y)},\boldsymbol{\mu}^{(z)}).

  5. Contract for each Cartesian component

    [ \mu^{\text{el}}_k = \operatorname{Tr}!\bigl[\mathbf{P}^{(12)}\,\boldsymbol{\mu}^{(k)}\bigr], \qquad k=x,y,z . ]

  6. (Optional) add the nuclear term of Eq. (6).

  7. Result: the transition dipole vector

    [ \boxed{\;\boldsymbol\mu_{12}= (\mu^{\text{el}}_x,\mu^{\text{el}}_y,\mu^{\text{el}}_z)\;} ]

    whose magnitude ( \boldsymbol\mu_{12} ) is the quantity that enters the Einstein‑A coefficient, oscillator strength, etc.

3. Final answer (concise form)

For a closed‑shell system, the transition dipole moment between electronic states 1 and 2 is

[ \boxed{ \boldsymbol\mu_{12}= -!\operatorname{Tr} \Bigl[ \bigl( \mathbf{C}^{(1)}{\alpha}(\mathbf{C}^{(2)}{\alpha})^{!T} +\mathbf{C}^{(1)}{\beta}(\mathbf{C}^{(2)}{\beta})^{!T} \bigr)\mathbf{S}^{-1} \, \boldsymbol{\mu} \Bigr]

  • \sum_A Z_A\mathbf{R}_A } ]

where

  • (\boldsymbol{\mu}) is the 3‑component dipole‑integral matrix (one matrix per Cartesian direction),
  • (\mathbf{S}) is the AO overlap matrix (use the identity if the basis is orthonormal),
  • the nuclear term is geometry‑dependent and cancels for a pure electronic transition, so the spectroscopic TDM is usually just the trace term.

If the basis is orthonormal ((\mathbf{S}= \mathbf{I})), the expression simplifies to

[ \boldsymbol\mu_{12

Original question: How to calculate transition dipole moment from two known wavefunctions on Chemistry Stack Exchange, licensed CC BY-SA.