# Does there exist quantum algorithms not homotopic to the identity?

Is it possible to operate on a single qubit by a map which has a degree not equal to one?

Let $$c=c_0|0\rangle + c_1|1\rangle$$ represent a qubit state where $$c_0,c_1 \in \mathbb{C}$$ and $$|c_0|^2+|c_1|^2=1$$. Then $$c$$ naturally lives in a 3-sphere $$S^3$$.

There is also the notion that $$c$$ lives in a one dimensional complex projective Hilbert space, $$\mathcal{P}(\mathcal{V}_2) \cong \mathbb{CP}^1 \cong S^2$$. In this case, we have the Hopf fibration $$S^1 \hookrightarrow S^3\rightarrow S^2$$. Generally, $$\mathcal{P}(\mathcal{V}_N)\cong \mathbb{CP}^{N-1} \cong S^{2N-1}/U(1)$$.

Single qubit operations are maps $$f:S^3 \rightarrow S^3$$ or $$g:\mathbb{CP}^1\rightarrow \mathbb{CP}^1$$. It is natural to wonder if $$f$$ or $$g$$ can be not homotopic to the identity. That is $$\deg(f) \ne 1$$ or $$\deg(g)\ne 1$$, given that $$\pi_3(S^3) \cong \mathbb{Z}$$ and $$\pi_2(S^2) \cong \mathbb{Z}$$.

Usually, a single qubit gate is either $$\bf{X}$$, $$\bf{Y}$$, $$\bf{Z}$$, $$\bf{H}$$, $$\bf{S}/\bf{P}$$, $$\bf{T}$$ which are order 2, 4, and 8, respectively and rotate (Pauli), mix (Hadamard), or phase shift states. These must be homotopic to the identity because the unitary group is path connected.

One can imagine an invertible map $$f_n:S^3\rightarrow S^3$$ which maps each point $$e^{i\theta}$$ in the $$S^1$$ fiber of the Hopf bundle to $$f_n(e^{i\theta})=e^{in\theta}$$.

Is $$f_n$$ a valid qubit operation? One could generalize to $$N$$-qubit maps (algorithms).

• It would help if you could include a definition of “qubit operation” in this context. If you are requiring your function $f:S_3\rightarrow S_3$ to be multiplication by some 2×2 unitary matrix A, then the fact that the unitary group is path connected means that there is a path $A_t$ from A to the identity matrix, and then f is homotopic to the identity through the homotopy $f_t(x)=A_t x$, so the degree of f is 1. Sep 18 at 15:41
• @DanRamras Right, if we only allow $f \in U(2)$ or $U(N)$, then nothing interesting happens. This gets at the heart of what I'm asking, which is more of a physics question (this is cross-posted on QC stackexchange) - are we only allowed to act by unitary matrices, or is that just convenient for building circuits? I guess solutions to the Schrödinger equation are of the form $exp(i\hbar\mathcal{H}t)|init\rangle$, so perhaps. I'll propose: a "qubit operation" is a continuous invertible self-map to/from the physical state space of $N$ qubits. Sep 18 at 15:42
• Thanks for clarifying. Note that when you’re writing “homotopically trivial” in the question you mean “homotpic to the identity,” whereas usually “homotopically trivial” means nullhomotopic. This makes the statement “ These must be homotopically trivial because repeated iteration of a nontrivial map raises the degree” a bit confusing. The degree is multiplicative under composition, so if $f$ has degree 0 or 1 then its degree doesn’t grow under iteration (so maybe you’re considering both the identity and the constant map to be homotopically trivial). Sep 18 at 15:54
• @DanRamras Thanks for pointing that out, my topology is a bit rusty. That's exactly what I meant, homotopic to the identity. I had a vague recollection of diagonalizing matrices in $U(N)$ with eigenvalues in $U(1)$ and a parameter $t$ that you then pull to the identity, but I was incorrectly assuming degree was additive so I just quickly wrote that spoof. It's additive under multiplication for $S^1$ I guess. Sep 18 at 16:03

Comment on $$f_n$$: There is no way of continuously identifying the fibers of the Hopf bundle with $$S^1$$, so it is not possible to define a map $$S^3 \rightarrow S^3$$ like this. If $$f_n$$ satisfies this equation in single fiber, then $$f_n(e^{2\pi i/n})= 1 = f(1)$$ so $$f_n$$ is not injective and is not given by a unitary matrix.
A quantum gate acting on $$n$$ qubits is represented by a $$2^n x 2^n$$ unitary matrix (see Wikipedia). Dan Ramras's comment about U(2) works for U(2^n), too, so the maps given by quantum gates are not homotopic to a constant map, but they are all homotopic to the identity map and hence to each other and have degree 1.
• @DanielBruefmann 1) Okay, so what is a map $f$ such that $0\ne\left[f\right]\in \pi_3(S^3)\cong \mathbb{Z}$? That's all I was trying to construct. 2) That's the point of this question - why must a quantum gate be in $\textbf{U}(N)$? Is there no way to manipulate a qubit with a map $f:S^3 \rightarrow S^3$ which is not homotopic to the identity? Sep 18 at 17:38
• 1) Technically ... yes. I meant f not homotopic to the constant map or identity, i.e. $deg(f) \ne 0,1$. For example, the double suspension of $f_k:z\mapsto z^k$ on $S^1$. Then $deg(f)=k$, and it's "nonlinearity" is measured by degree (like a polynomial). 2) Right, I mean this is by definition outside of standard QM I'm just surprised that we're limited to unitary operations. Right, not a single qubit but the "input space" for a quantum computer. Sep 19 at 2:46