In the linear algebra section of the Qiskit textbook appears the following claim regarding the Bloch sphere:

The surface of this sphere, along with the inner product between qubit state vectors, is a valid Hilbert space.

It is pretty clear that by scaling any quantum statevector we can easily get a vector that points outside or inside the surface of the sphere.. I.e the surface of the sphere isn't a vector space.

From where is this contradiction coming from?


1 Answer 1


The surface of a Bloch sphere is not a Hilbert space.

Maybe they meant to write that it's a valid projective Hilbert space (in particular it's isomorphic to $\mathbb{CP}^1$)? It's not a vector space, so it cannot be a Hilbert space (note that a "projective Hilbert space" is, perhaps somewhat confusingly, not a Hilbert space).

  • $\begingroup$ Really interesting! So what is the right way to describe the Hilbert space that single qubit states live in? $\endgroup$ Sep 28, 2022 at 20:24
  • 1
    $\begingroup$ @RajivKrishnakumar the smallest Hilbert space containing all qubits is $\mathbb{C}^2$. The point is if you want a space where the elements are in bijection (1-to-1) with qubits, then that space is not the Hilbert space $\mathbb{C}^2$, but rather the projective space $\mathbb{CP}^1$, which as glS points out, is not a Hilbert space. I don't think one is more right than the other. $\endgroup$
    – Condo
    Sep 28, 2022 at 22:02
  • $\begingroup$ Oh I see, so if I look at it in the opposite direction, is it correct to say that in addition to qubits, $\mathbb{C}^2$ also includes vectors that are not normalized and/or contain global phases? $\endgroup$ Sep 29, 2022 at 10:04
  • 2
    $\begingroup$ @RajivKrishnakumar the projective space $\mathbb{CP}^1$ is what you get from $\mathbb{C}^2$ modulo an equivalence relation. In other words, if you take $\mathbb{C}^2$, and identify all points which only differ by a global phase and a normalisation factor (and define a suitable topology), you get $\mathbb{CP}^1$. So in this sense, yes, $\mathbb C^2$ includes vectors that are not normalizes and "contain global phases" (the latter does'nt actually mean much; you should say that it contains different vectors differering by a global phase, which all correspond to the same physical state) $\endgroup$
    – glS
    Sep 29, 2022 at 10:19
  • 1
    $\begingroup$ Got it, makes sense, thank you for taking the time everyone for the clear explanations! I gave quick shout to the qiskit folks pointing them to this thread. They may already know this point and decided to "simplify" things when teaching it, but either way, I though it wouldn't hurt to let them know :) $\endgroup$ Sep 29, 2022 at 10:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.