# Meaning of "Metrically defined" matrix

I got stuck on understanding stuff from Methods of molecular quantum mechanics written by R. McWeeny.

The content is:

It must be stressed that the representation property embodied in $\mathcal{A} \mathcal{B} = \mathcal{C}$ and $\mathbf{A} \mathbf{B}= \mathbf{C}$, with the matrix elements defined as in $H_{ji}=\langle\phi_{j}|\mathcal{H}|\phi_{i}\rangle$, depends on the use of an orthonormal basis $\left\{ \phi_{i} \right\}$. If the basis is non-orthonormal, with a metric $\mathbf{S}$ defined as $S_{ij}=\langle\phi_{i}|\phi_{j}\rangle$, then the "metrically defined" matrices ($\mathbf{A}^{S}, \: \mathbf{B}^{S} , \cdots$, say) do not reflect the properties of the operators; when $\mathcal{A} \mathcal{B}= \mathcal{C}$ it is not generally true that $\mathbf{A}^{S} \mathbf{B}^{S} = \mathbf{C}^{S}$.

[Matrices are written in boldface characters that represent the corresponding operators written in calligraphic style.]

I already studied some undergraduate-level Linear Algebra, but I'm not familiar with "metric" and the related notation superscipted $S$. From what material can I learn about this?

• Could you elaborate on the meaning of your symbols? Jul 26, 2016 at 21:01
• Matrices are written in boldface characters that represent the corresponding operators written in calligraphic style. @Sanya Jul 26, 2016 at 21:05
• then we'll have to wait for someone smarter than me - in general, operator equations should be independent of the basis chosen, as the linear maps represented by the operators do not depend on the choice of a basis. Jul 26, 2016 at 21:09

Suppose $V$ is a real or complex vector space with inner product, $\{\left|\phi_i\right\rangle\}_i$ is a basis, not necessarily orthogonal, $A$ is an operator on this space, so $A:V\to V$.

Since $\{\left|\phi_i\right\rangle\}_i$ is a basis we can certainly expand as follows

$A\left|\phi_i\right\rangle =\sum_j a_{ji}\big|\phi_{j}\rangle$

The numbers $\{a_{ji}\}_{i,j}$ are simply the matrix of $A$ in this basis. We can label the following for convenience $S_{ij}=\langle\phi_i\big|\phi_j\rangle$ as you do.

But, notice the $S_{ij}$ are just some real or complex numbers, not necessarily just $1$'s and $0$'s like $\delta_{ij}$ so when we consider the following

$$\langle\phi_k\big|\,A\big|\phi_i\rangle=\langle\phi_k\big|\,\sum_j a_{ji}\big|\phi_{j}\rangle=\sum_j a_{ji}\langle\phi_k\big|\phi_{j}\rangle=\sum_j a_{ji}S_{kj}\neq a_{ki}$$

The last part is not true in general hence writing $\neq$, and this I believe is the point.

If you took the matrix $A^S$ to be the collection of numbers $\langle\phi_k\big|\,A\big|\phi_i\rangle$, which do not actually agree with the matrix of the operator in this basis, then theres no reason to expect matrix multiplication to work either.

As in suppose we take $A,B,C$ as operators with matrices $(A)_{ij}=a_{ij}$, $(B)_{ij}=b_{ij}$, $(C)_{ij}=c_{ij}$, and the following matrices $(A^S)_{ij}=\langle\phi_i\big|\,A\big|\phi_j\rangle$, $(B^S)_{ij}=\langle\phi_i\big|\,B\big|\phi_j\rangle$ and $(C^S)_{ij}=\langle\phi_i\big|\,C\big|\phi_j\rangle$, and suppose that $AB=C$.

This means that $c_{ij}=\sum_k a_{ik}b_{kj}$, but now consider the $i,j$'th entry of $A^SB^S$ denoted $(A^SB^S)_{ij}$.

It is a matrix of numbers so matrix multiplying we get that:

$$(A^SB^S)_{ij}=\sum_k (A^S)_{ik}(B^S)_{kj}=\sum_k \left(\sum_m a_{mk}S_{im}\right)\left(\sum_n b_{nj}S_{kn}\right)$$

While

$$(C^S)_{ij}=\sum_k c_{kj}S_{ik}$$

Notice though when $S_{ij}=\delta_{ij}$ we do reproduce the correct matrix multiplication formula.

\begin{align} (A^SB^S)_{ij}=\sum_k (A^S)_{ik}(B^S)_{kj}&=\sum_k \left(\sum_m a_{mk}S_{im}\right)\left(\sum_n b_{nj}S_{kn}\right)\\ &=\sum_k \left(\sum_m a_{mk}\delta_{im}\right)\left(\sum_n b_{nj}\delta_{kn}\right)\\ &=\sum_k a_{ik}b_{kj} \end{align}

I'm going to adjust notation so that the whole thing becomes much clearer for you. Let's work with linear operators $A$, $B$ and $C$, which act on our Hilbert space, and use no notation for matrices.

So instead of "$\mathcal{A} \mathcal{B} = \mathcal{C}$", I write $AB=C$. Instead of "$\mathbf{A} \mathbf{B}= \mathbf{C}$", I will write that for all $i$, $k$: $$\sum_j\langle \psi_i | A | \psi_j\rangle\langle \psi_j | B | \psi_k\rangle=\langle \psi_i | C | \psi_k\rangle$$

This is what the author means by "$\mathbf{A} \mathbf{B} = \mathbf{C}$". This only holds for an orthonormal basis, because it is only for orthonormal bases that $$\mathbf{1}=\sum_j | \psi_j\rangle\langle \psi_j |$$ You can prove this statement* by requiring that $\sum_j | \psi_j\rangle\langle \psi_j | \psi_k \rangle=|\psi_k\rangle$ for all $k$. For any fixed/given $k$, you can solve this equation with linear algebra. The homogeneous part of the solution $\sum_j | \psi_j\rangle c_{jk}=0$ has only the zero solution, by the linear independence of the kets. The inhomogeneous part has the solution $c_{jk}=\delta_{jk}$, plus the homogeneous part. That is to say, the only solution is $c_{jk}=\delta_{jk}$. Therefore, $\langle \psi_j | \psi_k \rangle=\delta_{jk}$.

*the statement being proved is that: "Given a basis of linearly independent kets $|\psi_k\rangle$, $\mathbf{1}=\sum_j | \psi_j\rangle\langle \psi_j |$ implies that $|\psi_k\rangle$ is an orthonormal basis."