I have an extremely ridiculous doubt that has been bothering me, since I started learning quantum mechanics.

If we consider the finite dimensional vector space for the spin$\frac{1}{2}$ particles, I guess it is nothing but $\mathbb{C}^2$. Each vector has two components (which is why it is two dimensional right ?), each of which can be any complex number.

Now coming to the case of position space (say one-dimension). I was taught this LVS is infinite dimensional (also continuously infinite, unlike the number operator basis). I am not able to understand this subtle thing that it is infinte dimensional (is it something like $\mathbb{R}^\infty$?). It is quite confusing every time I encounter this kind of space. Also in this each component (of the infinite no. of them) can take any real value (infinite number of them)? I learnt that the way to represent these can be in term of complex-valued functions, I would like have it elucidated.


1 Answer 1


Your doubt is not ridiculous, it is probably simply due to the confused way often mathematics is taught in physics. (I am a physicist too and, during my career, I had to bear ridiculous misconceptions, wasting lot of time in tackling non-existent pseudo-mathematical problems instead of focusing on genuine physical issues). There are sensible mathematical definitions, but there is also a practical use of math in physics. Disasters arise, in my view, when the two levels are confused, especially while teaching students.

The Hilbert space of a particle in QM is not continuous: it is a separable Hilbert space, $L^2(\mathbb R)$ which, just in view of being separable, admits discrete countable orthogonal bases.

Moreover, a well known theorem proves that if a Hilbert space admits a countable orthonormal basis, then every other basis is countable (more generally, all Hilbert bases have same cardinality).

In $L^2(\mathbb R)$, a countable basis with physical meaning is, for instance, that made of the eigenvectors $\psi_n$ of the harmonic oscillator Hamiltonian operator.

However, it is convenient for practical computations also speaking of formal eigenvectors of, for example, the position operator: $|x\rangle$. In this case, $x \in \mathbb R$ so it could seem that $L^2(\mathbb R)$ admits also uncountable bases. It is false! $\{|x\rangle\}_{x\in \mathbb R}$ is not an orthonormal basis. It is just a formal object, (very) useful in computations.

If you want to make rigorous these objects, you should picture the space of the states as a direct integral over $\mathbb R$ of finite dimensional spaces $\mathbb C$, or as a rigged Hilbert space. In both cases however $\{|x\rangle\}_{x\in \mathbb R}$ is not an orthonormal Hilbertian basis. And $|x\rangle$ does not belong to $L^2(\mathbb R)$.

As a final remark, I would like to stress that the vectors of $L^2(\mathbb R)$ are equivalence classes of functions: $\psi$ is equivalent to $\phi$ iff $\int| \psi(x)−\phi(x)|^2 dx=0$, so if $\psi(x)\neq \phi(x)$ on a set whose measure vanishes, they define however the same vector of $L^2$. Consequently the value an element of the space assumes at a given $x$ does not make any sense, since each set $\{x\}$ has zero measure.

  • $\begingroup$ What is the theorem you alluded to concerning countable orthonormal basis for Hilbert spaces? @V.Moretti $\endgroup$
    – sunspots
    Mar 25, 2014 at 19:42
  • $\begingroup$ It is just the statement I wrote: All orthonormal bases (Hilbert bases) of a given Hilbert space have the same cardinality. So if one of them is countable each of them is countable. $\endgroup$ Mar 25, 2014 at 20:48
  • $\begingroup$ I should have been clearer in what I was asking, does this theorem have a name? @V.Moretti $\endgroup$
    – sunspots
    Mar 25, 2014 at 21:24
  • $\begingroup$ I do not think so, it is a quite elementary proposition of Hilbert space theory. See for instance Rudin's book, "Real and complex analysis". $\endgroup$ Mar 25, 2014 at 21:35
  • $\begingroup$ I found it. See page 86 of 3rd edition of Rudin's book. $\endgroup$ Mar 25, 2014 at 21:41

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