A qubit $\vert \psi \rangle = \alpha \vert 0 \rangle + \beta \vert 1 \rangle \in \mathcal{H}^2$. A more general form of $n$-qubits is an element in $$\mathcal{H}^{\otimes 2^{n}} = \underbrace{\mathcal{H}^2\otimes\mathcal{H}^2\otimes\ldots\otimes\mathcal{H}^2}_{n \text{ times}}.$$ Is it possible to have a qubit to be in $\mathcal{H}^{\otimes 2^{n}+1}$ i.e. in an odd dimension?

  • $\begingroup$ I think you mean $\mathbb{C}$ instead of $\mathcal{H}$. $\endgroup$ – tparker Apr 23 at 11:37
  • $\begingroup$ I always thought they lived in $\mathcal{H}$, why do you reckon $\mathbb{C}$ instead? $\endgroup$ – M. Al Jumaily Apr 23 at 11:54
  • $\begingroup$ $\mathcal{H}$ denotes the Hilbert space of a physical system, $\mathbb{C}$ denotes the field of values that the components take on (in this case the complex numbers). So for a qubit, $\mathcal{H} = \mathbb{C}^2$. $\endgroup$ – tparker Apr 23 at 14:14
  • $\begingroup$ Oh, so you are saying that I don't have $\mathcal{H}^2$ and replace it with either $\mathbb{C}^2$ or $\mathcal{H}$? $\endgroup$ – M. Al Jumaily Apr 25 at 6:25
  • $\begingroup$ Yes, although since the point of your question is whether you can have odd-dimensional composite systems, I think that in this case $\mathbb{C}^2$ would make more sense. $\endgroup$ – tparker Apr 26 at 15:09

Sure. This is actually what physicists do when they make a qubit - often they don't have a system of just two levels but many levels. It's just that they choose to only use some of those levels. The simplest case, for example, is the atomic $\Lambda$ system. This has 3 levels. The two lowest energy levels are used as the qubit basis. There's a third level which isn't part of the qubit, but is made use of for making the gates.

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  • $\begingroup$ Excellent comment! Can you please give us some information about the math used in odd dimensions? $\endgroup$ – M. Al Jumaily Apr 26 at 7:14
  • $\begingroup$ There's nothing special about odd dimensions specifically. Generally, you're looking for a Hamiltonian/unitary that has a block-diagonal structure, $H=H_1\oplus H_2$, where $H_1$ is acting just on the two-dimensional space that you want to use as a qubit, and $H_2$ acts on everything else. $\endgroup$ – DaftWullie Apr 27 at 8:32

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