# What is meant by the term "completeness relation"

A Hilbert space $\cal H$ is *complete* which means that every *Cauchy sequence* of vectors admits a limit in the space itself.

Under this hypothesis there exist Hilbert bases also known as *complete orthonormal systems* of vectors in $\cal H$. A set of vectors $\{\psi_i\}_{i\in I}\subset \cal H$ is called an **orthonormal system** if $\langle \psi_i |\psi_j \rangle = \delta_{ij}$. It is also said to be **complete** if a certain set of equivalent conditions hold. One of them is
$$\langle \psi | \phi \rangle = \sum_{i\in I}\langle \psi| \psi_i\rangle \langle \psi_i| \phi \rangle\quad \forall \psi, \phi \in \cal H\tag{1}\:.$$
(This sum is absolutely convergent and must be interpreted if $I$ is not countable, but I will not enter into these details here.)
Since $\psi,\phi$ are arbitrary, (1) is often written
$$I = \sum_{i\in I}| \psi_i\rangle \langle \psi_i|\tag{2}\:.$$

This completeness relation of the basis means that you can reach all possible directions in the Hilbert space. It means that *any* $|\psi \rangle$ can be made up from these basis vectors.

If the sum of the projectors (the ket-bras) would not be the unit matrix, the vector $|\psi\rangle$ could have components which cannot be represented within your basis.

Take a three dimensional example. Taking the three canonical basis vectors as your $|\phi_n\rangle$, like $|\phi_1\rangle = (1, 0, 0)^\mathrm T$ and so on, you can see the completeness relation. If one of them is missing, your basis would not span the entire $\mathbb R^3$ space.