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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?

Qmechanic
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user35305
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3 Answers3

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

Valter Moretti
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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.

Martin Ueding
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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, $i,j,k,l,m,n,o……..$

DanielC
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arif
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