# What theorem is behind writing an operator in matrix form as outer products?

Consider a Hamiltonian of a two state system that follows, for the two eigenstates:

$$H|\phi_1\rangle = E_1|\phi_1\rangle \ ; \ H|\phi_2\rangle = E_2|\phi_2\rangle$$

Its matrix can be represented as:

$$H = \begin{pmatrix}E_1 &0 \\0 &E_2 \end{pmatrix}$$

But in terms of the outer product, it can also be written as:

$$H = E_1|\phi_1\rangle\langle\phi_1| +E_2|\phi_2\rangle\langle\phi_2| = H|\phi_1\rangle\langle\phi_1|+ H|\phi_2\rangle\langle\phi_2|$$

But from what theorem does this result come from?

• It comes from the fact that your eigenstates are linearly independent, and that the outer product is a linear transformation on the Hilbert space. There is not much difference between ur two expressions, one book to read in order to get used to this is Sakurai Modern QM Jun 10, 2023 at 18:18
• You should understand what expressions like $|\phi_1\rangle\langle \phi_2|$ mean (these are linear operators). You can then find the matrix representations of these operators. Jun 10, 2023 at 18:32
• It's a special case of the Riesz Representation theorem, or maybe an application of such. Jun 10, 2023 at 18:47
• The spectral theorem states that any normal (this includes Hermitian) operator can be diagonalised - in the matrix representation this means that there exists a vector basis in which the matrix representing the operator can be written as a diagonal matrix (this is the eigenbasis of the operator). Your last expression is called the spectral decomposition of H and the outer products are called the projectors of H. Jun 10, 2023 at 18:53

The two expressions you have given are exactly the same, just different notation. Just as you can choose different ways to notationally define your basis vectors on a Hilbert space, $$|\phi_{1}\rangle=\begin{pmatrix}1\\0\end{pmatrix},$$ $$|\phi_{2}\rangle=\begin{pmatrix}0\\1\end{pmatrix}$$ you can also choose different ways to notationally define the basis elements of the space of operators acting on the Hilbert space. This space of operators is itself a Hilbert space, known as the Liouville Space. In the case of the Hilbert space written in your question, Liouville space can be associated to basis elements, $$|\phi_{1}\rangle\langle\phi_{1}|=\begin{pmatrix}1&0\\0&0\end{pmatrix},|\phi_{1}\rangle\langle\phi_{2}|=\begin{pmatrix}0&1\\0&0\end{pmatrix},$$ $$|\phi_{2}\rangle\langle\phi_{1}|=\begin{pmatrix}0&0\\1&0\end{pmatrix},|\phi_{2}\rangle\langle\phi_{2}|=\begin{pmatrix}0&0\\0&1\end{pmatrix},$$ such that any operator acting on a general vector on the Hilbert space, $$|\psi\rangle=c_{1}|\phi_{1}\rangle+c_{2}|\phi_{2}\rangle=\begin{pmatrix}c_{1}\\c_{2}\end{pmatrix}$$ can be expressed as $$\hat{O}=\alpha|\phi_{1}\rangle\langle\phi_{1}|+\beta|\phi_{1}\rangle\langle\phi_{2}|+\gamma|\phi_{2}\rangle\langle\phi_{1}|+\delta|\phi_{2}\rangle\langle\phi_{2}|=\begin{pmatrix}\alpha&\beta\\\gamma&\delta\end{pmatrix}$$ for some $$\alpha$$, $$\beta$$, $$\gamma$$, $$\delta$$.

The key is that in both representations, you can write any general vector on the Hilbert space, and any general operator acting on a vector in the Hilbert space.

As pll04 says in their comment: the spectral theorem for Hermitian operators.

Let $$\mathcal{H}$$ be a Hilbert space. It is perhaps helpful to note that $$\lvert \psi \rangle \in \mathcal{H}$$ is simply a suggestive relabeling of an abstract vector. When represented w.r.t. to some basis $$\{b_n\}$$, $$\lvert \psi \rangle$$ is represented as a column vector:

$$\lvert \psi \rangle \xrightarrow{\text{basis}} \begin{pmatrix} \langle b_1 \lvert \psi\rangle \\ \langle b_2 \lvert \psi\rangle \\ ... \end{pmatrix}.$$

Via the dual correspondence, $$\lvert \psi \rangle^\dagger = \langle\psi\lvert,$$

so a bra is a row vector when represented w.r.t. to a basis: $$\langle \psi \lvert \xrightarrow{\text{basis}} \begin{pmatrix} \langle b_1 \lvert \psi\rangle^* & \langle b_2 \lvert \psi\rangle^* & ... \end{pmatrix}.$$

Fix your basis to be the basis which makes your Hermitian operator diagonal. Then, you can prove to yourself that the spectral decomposition of the operator and the diagonal form of the operator as represented as matrices are equal in this basis. Since representing vectors w.r.t. to a basis is performed via an isomorphism you can then abstract the matrices back into operators.

This result comes from the fact that the eigenvectors $$|\phi_k\rangle$$ of a Hermitian operator $$H$$, when properly normalized, form a complete basis of the underlying Hilbert space. This is called the "completeness relation" (see for example Intermediate Quantum Mechanics, Lecture 3, page 1). $$\sum_k |\phi_k\rangle\langle\phi_k|=1$$

Using this relation (and putting aside any mathematical rigorousness) we can easily derive the expansion of $$H$$ in terms of its matrix elements and outer products: \begin{align} H &= 1\ H\ 1 \\ &= \left(\sum_k |\phi_k\rangle\langle\phi_k|\right) H \left(\sum_l |\phi_l\rangle\langle\phi_l|\right) \\ &= \sum_k \sum_l |\phi_k\rangle \langle\phi_k|H|\phi_l\rangle \langle\phi_l| \\ &= \sum_k \sum_l |\phi_k\rangle H_{kl} \langle\phi_l| \end{align}