# Why do we need Hilbert spaces when talking about qubits and quantum computation?

I was just curious to know why do we need Hilbert Spaces when talking about the qubits and quantum computation in general. I mean why can't we just work with inner product spaces, rather than going for complete inner product spaces. I am guessing it has to do with some convergence of an infinite series. Can somebody explain?

• It's worth mentioning that any finite-dimensional inner product space is trivially Hilbert. “Only” in the infinite-dimensional case is completeness relevant. Now, in physics you're usually working in infinite-dimensional Hilbert spaces and the finite-dimensional case is special, but I suppose in quantum computing it's the other way around. Jan 1 at 15:07
• @leftaroundabout This is not strictly speaking true. For example, $\mathbb{A}^n$ with standard dot product $\langle u,v\rangle=\sum_k \overline{u_k}v_k$ where $\mathbb{A}$ denotes the field of algebraic numbers is a finite dimensional inner product space which is not complete. Generally, a finite-dimensional normed vector space is complete if and only if its underlying field of scalars is complete. $\mathbb{A}$ is a subfield of complex numbers that are roots of polynomials with integer coefficients and which excludes the transcendental numbers like $\pi$ and $e$ and therefore is not complete. Jan 1 at 16:31

Completeness ensures that Cauchy sequences have limits. This guarantees the existence of objects defined via limits of such sequences. These include derivatives, exponentials, trigonometric functions etc. Perhaps the most spectacular consequence of dropping the completeness assumption is the inability to describe state evolution.

## State evolution

By the Schrödinger equation, the evolution of a quantum system initially in state $$|\psi(0)\rangle$$ under the constant Hamiltonian $$H$$ is given$$^1$$ by

$$|\psi(t)\rangle = e^{-iHt}|\psi(0)\rangle = \sum_{k=0}^\infty\frac{(-it)^k}{k!}H^k|\psi(0)\rangle=\lim_{n\to\infty}\sum_{k=0}^n\frac{(-it)^k}{k!}H^k|\psi(0)\rangle$$

and we need completeness to guarantee that the limit, and hence $$|\psi(t)\rangle$$, exists.

$$^1$$ In units with $$\hbar=1$$.

• That could have been possible just by having a Banach space. Jan 1 at 21:43
• Banach space is generally not an inner product space. Jan 1 at 21:53
• Another interesting aspect is that probing states with measurements uses the fact that the best one perform a particular measurement process, closer and closer we expect to be from the desired description. This is I think a physically interesting way to see converging arguments even if perfect time evolution may not be possible to probe.
– R.W
Jan 2 at 12:44
• Adam Zalcman I know that not all Banach spacee are Hilbert. That is what I was asking why need hilbert, if we can do with Banach spaces. Is it beacuse we need bra-ket $\langle x|y\rangle$ to be scalar. And in Banach we cannot define product Jan 2 at 15:59
• @Upstart From an axiomatic point of view, the proper generalization to arbitrary, including infinite-dimensional systems, would be via $C^*$-algebras. This comes with only a few axioms that allow you to recover e.g. the Born rule. But then, there is the Gelfand-Naimark theorem which basically says that any $C^*$-algebra is isomorphic to a subalgebra of $\mathcal B(H)$, the $C^*$-algebra of bounded operators on a suitable Hilbert space. Jan 3 at 9:14

I'll consider vector spaces over $$\mathbb R$$ or $$\mathbb C$$.

In finite dimensions, an inner product space is automatically complete, and thus a Hilbert space, so no issues there.

More generally, one important feature of a Hilbert space $$V$$ is the isomorphism between $$V$$ and its (continuous) dual. This duality is what allows to work seamlessly in bra-ket notation, switching between and manipulating kets and bras without a care in the world. But to have duality, completeness is required. In fact, for a generic Hausdorff topological vector space $$V$$, its continuous dual $$V^*$$ is equal to the continuous dual of the completion of $$V$$.

More concretely, suppose $$V$$ is an inner product space, over a field $$\mathbb F$$, that is not complete. For example, we could take $$V=\text{span}(\{e_1,e_2,...\})$$, the linear span of the basis of a Hilbert space (remember that a linear span contains all finite linear combinations, whereas the associated Hilbert space is the closure of such set). In such a space, there are Cauchy sequences that do not converge, e.g. $$(x_n)_{n\in\mathbb N}$$ with $$x_n\equiv\sum_{i=1}^n\frac{1}{i}e_i$$ (this example is taken from this post on math.SE).

But the continuous dual $$V^*$$ contains by definition all continuous linear maps $$V\to \mathbb F$$. In particular, it contains the functional $$\alpha(e_i)=1/i$$ for all $$i\in\mathbb N$$. This is clearly a continuous functional, as $$\|\alpha(x)\|\le\|x\|$$ for all $$x\in V$$. But such $$\alpha\in V^*$$ does not correspond to an element of $$V$$: that element would have been the limit of the sequence $$(x_n)$$, which is not in $$V$$.

Aside from the, in principle, merely practical hassle of not being able to switch between the space and its dual, this also makes it hard to interpret a number of things. For example, if $$\psi\in V$$ represents a quantum (pure) state, we would generally represent with $$\psi^*\in V^*$$ the map sending any other state $$\phi\in V$$ to the probability of finding $$\psi$$ when measuring $$\phi$$ in a suitable measurement basis. But if $$\psi\notin V$$, what would this even mean? I suppose you could simply say that $$\psi^*$$ does not correspond to a physical operation, but then there would be sequences of functionals converging to $$\psi^*$$, which all correspond to actual states, but whose limit doesn't, which would make for a rather odd description of things I think.

As a further example, observe that things such as coherent states, which have the form $$|\beta\rangle=\sum_{n=0}^\infty \frac{\beta^n}{\sqrt{n!}}|n\rangle$$, are limits of Cauchy sequences, which wouldn't necessarily be contained in a non-complete space spanned by $$\{|n\rangle\}_{n\ge0}$$.

• I agree that duality is indeed key in QM. It captures the physically reasonable situation where for every state $|\psi\rangle$, we have a corresponding effect $\langle\psi|$ which provides the means to (non-deterministically) test for $|\psi\rangle$. That said, there is another interesting argument for where duality comes from in this paper where it is shown that from among all $Lp$ norms, $L2$ is the only one that is preserved by non-trivial linear maps (and $L2$ is the only such norm that gives rise to an inner product). Jan 2 at 20:33

The completeness requirement is for mathematical convenience. On infinite-dimensional pre-Hilbert spaces you can't meaningfully do calculus. You also don't have a reasonable notion of the spectral decomposition theorem for self-adjoint operators. You need a spectral theory of Hilbert spaces to make sense of self-adjoint operators as observables in quantum mechanics.