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Chemistry 352

Direct Products

In symmetry analysis, a direct product describes how two symmetry-adapted quantities behave when they are combined. Direct products are used to determine whether interactions, couplings, or transitions are symmetry-allowed.

Examples include:

Mathematically, the direct product of two representations is formed by multiplying their characters element-by-element.

Direct products in \(C_{2v}\)

In \(C_{2v}\), all irreducible representations are one-dimensional. As a result, direct products are especially simple: characters are just multiplied together, and the result corresponds directly to another irreducible representation.

Example 1: \(A_1 \times B_2\)

Using the characters:

\[ A_1 = (1,\;1,\;1,\;1), \qquad B_2 = (1,\;-1,\;-1,\;1) \]

Element-by-element multiplication gives:

\[ (1,\;1,\;1,\;1) \times (1,\;-1,\;-1,\;1) = (1,\;-1,\;-1,\;1) \]

which corresponds to \(B_2\).

Therefore: \(A_1 \times B_2 = B_2\).

Example 2: \(A_2 \times B_2\)

Using the characters:

\[ A_2 = (1,\;1,\;-1,\;-1), \qquad B_2 = (1,\;-1,\;-1,\;1) \]

Multiplying element-by-element:

\[ (1,\;1,\;-1,\;-1) \times (1,\;-1,\;-1,\;1) = (1,\;-1,\;1,\;-1) \]

which corresponds to \(B_1\).

Therefore: \(A_2 \times B_2 = B_1\).

Direct products in \(C_{3v}\)

In \(C_{3v}\), not all irreducible representations are one-dimensional. Direct products involving the two-dimensional representation \(E\) generally produce reducible representations that must be decomposed.

Example 3: \(A_2 \times A_2\)

The characters for \(A_2\) are

\[ A_2 = (1,\;1,\;-1) \quad \text{for } (E,\;2C_3,\;3\sigma_v) \]

Multiplying the characters by themselves:

\[ (1,\;1,\;-1) \times (1,\;1,\;-1) = (1,\;1,\;1) \]

which corresponds to \(A_1\).

Therefore: \(A_2 \times A_2 = A_1\).

Example 4: \(E \times E\)

The characters of \(E\) in \(C_{3v}\) are

\[ E = (2,\;-1,\;0) \]

Forming the direct product by multiplying characters:

\[ (2,\;-1,\;0) \times (2,\;-1,\;0) = (4,\;1,\;0) \]

This result is a reducible representation, which must be decomposed into irreducible representations.

Decomposing \(E\times E\) by dot products (using orthogonality)

For \(C_{3v}\), the classes are \(E\), \(2C_3\), and \(3\sigma_v\), and the group order is \(h=6\).

The irreducible characters are: \(A_1=(1,1,1)\), \(A_2=(1,1,-1)\), \(E=(2,-1,0)\).

First form the direct product characters:

\[ E\times E = (2,-1,0)\times(2,-1,0)=(4,1,0). \]

To find how many times an irrep \(\alpha\) appears in a reducible representation \(\Gamma\), take the weighted dot product over classes:

\[ a_\alpha=\frac{1}{h}\sum_{\text{classes}} n_R\,\chi^{(\alpha)}(R)\,\chi^{(\Gamma)}(R). \]

Dot product with \(A_1\)

\[ \sum n_R\,\chi^{(A_1)}\chi^{(E\times E)} = 1(1)(4)+2(1)(1)+3(1)(0) = 4+2+0 = 6. \]

\[ a_{A_1}=\frac{6}{h}=\frac{6}{6}=1 \quad\Rightarrow\quad A_1 \text{ appears once in } E\times E. \]

Dot product with \(A_2\)

\[ \sum n_R\,\chi^{(A_2)}\chi^{(E\times E)} = 1(1)(4)+2(1)(1)+3(-1)(0) = 4+2+0 = 6. \]

\[ a_{A_2}=\frac{6}{6}=1 \quad\Rightarrow\quad A_2 \text{ appears once in } E\times E. \]

Dot product with \(E\)

\[ \sum n_R\,\chi^{(E)}\chi^{(E\times E)} = 1(2)(4)+2(-1)(1)+3(0)(0) = 8-2+0 = 6. \]

\[ a_{E}=\frac{6}{6}=1 \quad\Rightarrow\quad E \text{ appears once in } E\times E. \]

Therefore the decomposition is: \(E\times E = A_1 + A_2 + E\).

This result is extremely important in spectroscopy and bonding, as it determines which combinations of functions can produce totally symmetric components.

Big idea: direct products reveal how symmetry properties combine. One-dimensional representations multiply directly, while higher-dimensional products often produce reducible representations that must be decomposed using the Great Orthogonality Theorem.

A powerful consequence of symmetry is the following rule:

An integral over all space is zero unless the integrand contains at least one component that transforms as the \(A_1\) (totally symmetric) irreducible representation.

This rule underlies selection rules in spectroscopy, orbital overlap arguments, and the use of direct products in symmetry analysis.

Why symmetry forces integrals to vanish

Consider a familiar example from calculus. An odd function (antisymmetric with respect to inversion) satisfies

\[ f(-x) = -f(x). \]

If we integrate an odd function over a symmetric interval,

\[ \int_{-a}^{a} f(x)\,dx, \]

the result is zero. Contributions from positive \(x\) cancel exactly with contributions from negative \(x\). The integral vanishes by symmetry, not because the function is small.

The symmetry generalization

In molecular symmetry, the same idea applies. An integral such as

\[ \int \psi_i\,\hat{O}\,\psi_j\,d\tau \]

will be nonzero only if the product \( \psi_i\,\hat{O}\,\psi_j \) transforms as the totally symmetric representation of the point group (or has a component that does).

If the integrand changes sign or otherwise fails to map onto itself under any symmetry operation of the group, positive and negative contributions cancel over the full integration domain, and the integral is zero.

Connection to direct products

This is why direct products are so useful. By examining the symmetry of

\[ \Gamma(\psi_i)\times\Gamma(\hat{O})\times\Gamma(\psi_j), \]

we can immediately determine whether the totally symmetric irreducible representation is present.

Big idea: symmetry does not just simplify calculations—it can tell you whether a quantity is exactly zero before you ever evaluate an integral.

Worked example (C2v): is an integral symmetry-allowed?

Suppose a molecule belongs to the point group \(C_{2v}\), and we want to know whether a transition dipole integral such as

\[ \langle \psi_i \,|\, z \,|\, \psi_f \rangle \;=\; \int \psi_i \, z \, \psi_f \, d\tau \]

can be nonzero. The symmetry rule says: the integrand must contain the totally symmetric representation (\(A_1\) in \(C_{2v}\)).

Step 1: Identify the symmetries of each factor

In \(C_{2v}\), the coordinate \(z\) transforms as \(A_1\). Therefore the operator \(\hat{\mu}_z \propto z\) also has symmetry \(A_1\).

Assume:

Step 2: Multiply the symmetries (direct products)

\[ \Gamma(\text{integrand}) = \Gamma(\psi_i)\times\Gamma(z)\times\Gamma(\psi_f) = B_1 \times A_1 \times A_2. \]

Because \(A_1\) is totally symmetric, multiplying by \(A_1\) does not change the symmetry: \(B_1\times A_1 = B_1\). So:

\[ \Gamma(\text{integrand}) = B_1 \times A_2. \]

Using the \(C_{2v}\) multiplication rules (or character table), \(B_1\times A_2 = B_2\).

Step 3: Check for the totally symmetric representation

The integrand transforms as \(B_2\), not \(A_1\). Since the integrand does not contain an \(A_1\) component, the integral is forced to be zero by symmetry:

\[ \langle \psi_i \,|\, z \,|\, \psi_f \rangle = 0. \]

Conclusion: for this pair of states, the \(z\)-polarized transition is symmetry-forbidden.

Big idea: you can often determine whether an integral is exactly zero before doing any math—just by checking whether the direct product contains the totally symmetric irreducible representation.

Your turn

Problem 1
Using the character table for \(C_{2v}\), evaluate the direct product \(B_1 \times B_2\).

\[ \begin{array}{c|cccc} C_{2v} & E & C_2 & \sigma_v & \sigma_v' \\ \hline A_1 & 1 & 1 & 1 & 1 \\ A_2 & 1 & 1 & -1 & -1 \\ B_1 & 1 & -1 & 1 & -1 \\ B_2 & 1 & -1 & -1 & 1 \end{array} \]
\(A_1\) \(A_2\) \(B_1\) \(B_2\)
Problem 2
In \(C_{4v}\), decompose the direct product \(E \times E\).

\[ \begin{array}{c|ccccc} C_{4v} & E & 2C_4 & C_2 & 2\sigma_v & 2\sigma_d \\ \hline A_1 & 1 & 1 & 1 & 1 & 1 \\ A_2 & 1 & 1 & 1 & -1 & -1 \\ B_1 & 1 & -1 & 1 & 1 & -1 \\ B_2 & 1 & -1 & 1 & -1 & 1 \\ E & 2 & 0 & -2 & 0 & 0 \end{array} \]
\(A_1 + A_2\) \(A_1 + B_1 + B_2\) \(A_1 + A_2 + B_1 + B_2\) \(A_1 + E\)
Problem 3
Using the character table for \(D_{2d}\), what is \(E \times E\)?

\[ \begin{array}{c|ccccc} D_{2d} & E & 2S_4 & C_2 & 2C_2' & 2\sigma_d \\ \hline A_1 & 1 & 1 & 1 & 1 & 1 \\ A_2 & 1 & 1 & 1 & -1 & -1 \\ B_1 & 1 & -1 & 1 & 1 & -1 \\ B_2 & 1 & -1 & 1 & -1 & 1 \\ E & 2 & 0 & -2 & 0 & 0 \end{array} \]
\(A_1 + A_2\) \(A_1 + B_1 + B_2\) \(A_1 + A_2 + B_1 + B_2\) \(E\)
Problem 4
In \(C_{4h}\), what is the result of \(A_u \times B_g\)?

\[ \begin{array}{c|cccccc|cc} C_{4h} & E & 2C_4 & C_2 & i & 2S_4 & \sigma_h & \text{Linear} & \text{Rotations} \\ \hline A_g & 1 & 1 & 1 & 1 & 1 & 1 & & \\ B_g & 1 & -1 & 1 & 1 & -1 & 1 & & \\ E_g & 2 & 0 & -2 & 2 & 0 & -2 & (x,y) & (R_x,R_y) \\ A_u & 1 & 1 & 1 & -1 & -1 & -1 & z & \\ B_u & 1 & -1 & 1 & -1 & 1 & -1 & & R_z \\ E_u & 2 & 0 & -2 & -2 & 0 & 2 & & \\ \end{array} \]
\(A_g\) \(B_g\) \(A_u\) \(B_u\)
Problem 5
Why are direct products useful in molecular symmetry?
They determine bond lengths They reveal which combinations of functions are symmetry-allowed They list all symmetry elements They replace character tables
Problem 6
In an abelian point group, why do all irreducible representations have dimension 1?
Because all characters are zero Because all operations commute Because the group order is small Because no mirror planes exist
Problem 7
What does it mean if a direct product contains the totally symmetric representation?
The interaction is forbidden The interaction is symmetry-allowed The representation is reducible The molecule is nonpolar
Problem 8
Why must a reducible representation be decomposed before it can be used for selection rules?
Selection rules apply only to irreducible representations Reducible representations have incorrect characters Reducible representations violate orthogonality Reducible representations are always two-dimensional

Key points (one glance)

Big picture: symmetry allows you to determine whether a quantity is exactly zero without performing any explicit integration.