# Can a qubit live in $\mathcal{H}^{\otimes 2^{n}+1}$?

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?

• I think you mean $\mathbb{C}$ instead of $\mathcal{H}$. – tparker Apr 23 at 11:37
• I always thought they lived in $\mathcal{H}$, why do you reckon $\mathbb{C}$ instead? – M. Al Jumaily Apr 23 at 11:54
• $\mathcal{H}$ 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, $\mathcal{H} = \mathbb{C}^2$. – tparker Apr 23 at 14:14
• Oh, so you are saying that I don't have $\mathcal{H}^2$ and replace it with either $\mathbb{C}^2$ or $\mathcal{H}$? – M. Al Jumaily Apr 25 at 6:25
• 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. – 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.
• 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. – DaftWullie Apr 27 at 8:32