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A physical qubit, for example, an electron with energy level (ground state and excited state) represents a simple quantum system. I was curious whether a physical system is a basis for mathematically representing qubit in 2D space?

Why not a qubit state be in 3D Hilbert space? Is there a mathematical formulation for this even though an equivalent quantum system doesn't exist?

Note: I don't have a formal background in physics or mathematics, this question came to mind when I started reading about quantum computation.

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    $\begingroup$ You'd probably find photon polarization more intuitive than electron spin (I recommend watching that video). There's only two basis states for an electron (up and down spin states), so it's natural that they live in a 2D Hilbert space, rather than a 3D or higher space. The dimension of a vector space (complex Hilbert space, in this context) is determined solely from the number of basis states. $\endgroup$
    – Sanchayan Dutta
    Commented Dec 14, 2019 at 15:19
  • $\begingroup$ @SanchayanDutta Thanks for the link. I will watch. Still, based on response, I feel question is not conveyed clearly or I missed the point in the answer. I mentioned "electron example" just to bring physical and mathematical world. If we ignore physical realization, then extending dimension makes sense (just as a theory)? $\endgroup$
    – Guruprasad Hegde
    Commented Dec 14, 2019 at 15:50
  • $\begingroup$ In theory, you could consider a higher dimensional complex Hilbert space in which the 2-dimensional Hilbert space of a qubit is embedded. But I don't see the use. We generally consider the smallest vector space structure in which the physical description is meaningful. $\endgroup$
    – Sanchayan Dutta
    Commented Dec 14, 2019 at 15:56
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    $\begingroup$ Ground state and excited state are two states which span 2D Hilbert space; if you have 3 independent states (say ground state, first excited state and second excited state) then you get 3D Hilbert space. $\endgroup$
    – kludg
    Commented Dec 14, 2019 at 16:11
  • $\begingroup$ Thanks for the clarification. How to use such a formulation would be interesting I guess! $\endgroup$
    – Guruprasad Hegde
    Commented Dec 14, 2019 at 16:26

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A two-dimensional Hilbert space is very different from two spatial dimensions. In particular, two spatial directions (up/down and left/right) are described by two real parameters. A two-dimensional Hilbert space corresponds to two complex parameters, i.e. 4 real parameters. That said, quantum states have a constraint (normalisation) which means that there are effectively only three relevant parameters. This is why, if we want to visualise a qubit, we use something called the Bloch sphere in three spatial dimensions.

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  • $\begingroup$ Could you explain this: That said, quantum states have a constraint (normalisation) which means that there are effectively only three relevant parameters. $\endgroup$
    – Kittu A
    Commented Nov 26, 2021 at 17:49
  • $\begingroup$ @KittuA When you measure a qubit, you have different possible outcomes, and corresponding probabilities of getting them. The probabilities are derived from the parameters describing the state, but there is the constraint that the probabilities always sum to 1. So that constraints the parameters, so that lets you rewrite the thing using fewer parameters. $\endgroup$
    – DaftWullie
    Commented Nov 29, 2021 at 10:29
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    $\begingroup$ @KittuA For example, if you want to describe the density matrix of a single qubit, that is a Hermitian, non-negative matrix with trace 1. (It's the trace that ensures the probabilities sum to 1). So you can write it as $\left(\begin{array}{cc} z & x+iy \\ x-iy & 1-z\end{array}\right)$i.e. just 3 real parameters $x,y,z$. $\endgroup$
    – DaftWullie
    Commented Nov 29, 2021 at 10:29

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