# What is meant by the term “completeness relation”

From my humble (physicist) mathematics training, I have a vague notion of what a Hilbert space actually is mathematically, i.e. an inner product space that is complete, with completeness in this sense heuristically meaning that all possible sequences of elements within this space have a well-defined limit that is itself an element of this space (I think this is right?!). This is a useful property as it enables one to do calculus in this space.

Now, in quantum mechanics Hilbert spaces play an important role in that they are the spaces in which the (pure) states of quantum mechanical systems "live". Given a set of orthonormal basis vectors, $\lbrace\lvert\phi_{n}\rangle\rbrace$ for such a Hilbert space, one can express a given state vector, $\lvert\psi\rangle$ as a linear combination of these basis states, $$\lvert\psi\rangle=\sum_{n}c_{n}\lvert\phi_{n}\rangle$$ since the basis states are orthonormal, i.e. $\langle\phi_{n}\lvert\phi_{m}\rangle =\delta_{nm}$ we find that $c_{n}=\langle\phi_{n}\lvert\psi\rangle$, and hence $$\lvert\psi\rangle=\sum_{n}c_{n}\lvert\phi_{n}\rangle =\sum_{n}\langle\phi_{n}\lvert\psi\rangle\lvert\phi_{n}\rangle =\left(\sum_{n}\lvert\phi_{n}\rangle\langle\phi_{n}\lvert\right)\lvert\psi\rangle$$ which implies that $$\sum_{n}\lvert\phi_{n}\rangle\langle\phi_{n}\lvert =\mathbf{1}$$ This is referred to as a completeness relation, but I'm unsure what this is referring to? I've also read that the basis must be complete. Is this referring to the notion of completeness associated with limits of sequences, or is there something else I'm missing?

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 $$I = \sum_{i\in I}| \psi_i\rangle \langle \psi_i|\tag{2}\:.$$

• The condition (1) is in fact equivalent to $\psi = \sum_{i\in I} \langle \psi_i|\psi\rangle \psi_i$ for all $\psi \in \cal H$. In this situation the relevant Cauchy sequence is made of all vectors $\Psi_N := \sum_{|i|<N}\langle \psi_i|\psi\rangle \psi_i$. So, yes, the completeness relation is equivalent to the fact that the basis spans the whole space when considering all infinite sequences in the Hilbert space topology. – Valter Moretti Apr 24 '16 at 17:09
• @user35305 there are different definitions of "basis" and "span" at play here. The one you might have first learned about in a Linear Algebra class is a "Hamel basis", where only finite sums are allowed. In this sort of context, we're dealing with a Schauder/Hilbert basis where infinite sums are allowed. – Mark S. Apr 24 '16 at 21:29
• @MarkS. How do they differ in this case? Is a Schauder/Hilbert basis not necessarily complete then, i.e. it doesn't span the entire vector space? – user35305 Apr 24 '16 at 21:41
• @user35305 A Schauder/Hilbert basis does not "span" in the sense that not every vector is a finite linear combination of things in the basis. It does "span" in the sense that every vector is a limit of finite linear combinations (an "infinite linear combination"). – Mark S. Apr 24 '16 at 21:45
• @user35305 The completeness relation is presumably what Valter Moretti said it was. Intuitively, it's a relation that says "this orthonormal system you have? It's indeed a Schauder/Hilbert basis. You don't need any more to write everything as an infinite linear combination of stuff in the system." – Mark S. Apr 24 '16 at 21:51

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.

• I kind of got the example case you gave before, but I've been unsure as to the general reasoning behind why a basis is complete if it's outer product sums to the identity matrix? I mean, doesn't a basis by definition span the entire vector space?! – user35305 Apr 24 '16 at 18:11
• A set of vectors have to fulfill the completeness relation in order to be a basis. – Martin Ueding Apr 24 '16 at 18:22
• Ah ok, so is that the point then, that one first considers a set of vectors, expresses an arbitrary vector in terms of this set and then finds that in order for this set of vectors to be a basis set and (in this sense) complete, they must necessarily satisfy the completeness relation?! Does this have anything to do with Cauchy sequences and the sense of completeness defining a Hilbert space? – user35305 Apr 24 '16 at 18:26
• To your first question, I think that the completeness relation is equivalent to the requirement that a basis spans the whole space. The property with the Cauchy sequences is probably rather something regarding the underlying field ($\mathbb C$); I am not very confident about that part. – Martin Ueding Apr 24 '16 at 20:09

This is just a mathematical trick to decompose a vector into components of the space. consider like $i,j,k$ of the Cartesian space $\sum_{c=i,j,k} |c\rangle \langle c|$. A vector can be decomposed in Cartesian space. $$\psi=\sum_{c=i,j,k}|c\rangle \langle c|\psi$$, when we will apply $|i\rangle\langle i|$ on $ψ$. i.e $|i\rangle\langle i| ψ\rangle$, the $\langle i|ψ\rangle$ will give us value of the vector $“a”$ and $|i\rangle$ gives direction. this is just for one component. similarly for $j$ and $k$, so $ψ=ai+bj+ck$. But for infinite dimensional problem, we need infinite dimensional space, i.e. Hilbert space. there will be infinite components, $i,j,k,l,m,n,o……..$