4
$\begingroup$

Is it possible to operate on a single qubit by a map which has a nonzero degree?

Let $|c\rangle=c_0|0\rangle + c_1|1\rangle$ 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 $|c\rangle$ 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 homotopic to something other than the identity or a constant map, that is $\deg(f) \ne 0,1$ or $\deg(g)\ne 0,1$, 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 $U(N)$ is path connected.

Let map $f_n:S^3\rightarrow S^3$ such that $\left[f_n\right]=n\ne 0,1$. Is $f_n$ a valid "qubit operation"?

One could generalize to $N$-qubit maps (algorithms).

$\endgroup$
6
  • $\begingroup$ shouldn't you have $f:S^2\to S^2$ here, coming from $S^3/U(1)\simeq S^2$? Unless I'm misunderstanding what you mean. Essentially the mapping on qubits when represented on the Bloch sphere $\endgroup$
    – glS
    Sep 18, 2023 at 15:21
  • $\begingroup$ @glS well, that's what $g:\mathbb{CP}^1 \rightarrow \mathbb{CP}^1$ is. Good catch because I'm confused. As a mathematician, "Bloch sphere" seems overloaded. I can't tell if we're talking about "pure states", which do like on an $S^2$ and are parameterized by $\theta, \phi$. However, just having two complex amplitudes puts you on a 3-sphere. $\endgroup$ Sep 18, 2023 at 15:23
  • $\begingroup$ well, yes, I thought you meant $f$ and $g$ as different representations of the same thing, but maybe I misunderstood. Two complex amplitudes plus normalization puts you on $S^3$, sure, but then you take into account the phase "gauge freedom", and thus get the sphere thanks to the "magical" homeomorphism $S^3/U(1)\simeq S^2$. So if you mean with $f$ a map on $S^3$ before global phases are gauged out, I guess you could have whatever you want on the $S^1$ fibers there, for any gate; but it won't matter because each fiber collapses to a point when removing the global phase $\endgroup$
    – glS
    Sep 18, 2023 at 15:27
  • $\begingroup$ $f$ and $g$ aren't necessarily different representations of the same map. I'm just separating them because I don't know, for instance, if I should view $\textbf{H} \in U(2)$ as a map from $S^3 \rightarrow S^3$ or $S^2 \rightarrow S^2$. For example, $\textbf{H}(\lambda v)=\lambda \textbf{H}(v) \ne \textbf{H}(v)$ because $\textbf(H)$ is linear. Is the "gauge freedom" the same as the one in QED? We're saying a different $U(1)$ phase represents the same state? Even if the phase on the $S^1$ fiber is gauged out, one could still have a degree for $g:\mathbb{CP}^1 \rightarrow \mathbb{CP}^1$, no? $\endgroup$ Sep 18, 2023 at 15:33
  • $\begingroup$ also, I don't know if it helps here, but you can "easily" represent single-qubit gates projectively via stereographic projections as mobius transformations, see eg quantumcomputing.stackexchange.com/a/25834/55. See also quantumcomputing.stackexchange.com/a/17763/55 about ${\bf U}(2)$ vs ${\bf SU}(2)$ for gates (which seems like what the last comment is hinting at: "gates" are also elements of the projective linear group if you think them as acting on states represented as rays in projective space). Gauge freedom here just means identifying kets differing by a global phase $\endgroup$
    – glS
    Sep 18, 2023 at 15:33

1 Answer 1

4
$\begingroup$

TL;DR: The state space of an $n$-level quantum system is $\mathbb{CP}^{n-1}$. In particular, the state space of a qubit is $\mathbb{CP}^1$, not $S^3$. Time evolution of a closed quantum system is described by unitary operations. Consequently, if $g:\mathbb{CP}^{n-1}\to\mathbb{CP}^{n-1}$ is an operation that can be realized as the time evolution of a closed quantum system, then $\deg f=1$. In fact, the required homotopy can easily be constructed by interpreting the time parameter as the homotopy parameter and exploiting the physically obvious boundary condition that the zero-duration evolution from $t_0$ to $t_0$ is the identity map.

Quantum state space

Quantum mechanics can be summarized in three to four postulates that identify mathematical objects used to model the states, dynamics and observables of quantum systems. The first postulate describes the state space of an $n$-level quantum system as the complex projective Hilbert space $\mathbb{CP}^{n-1}$.

It is instructive to consider the number of real degrees of freedom in $|c\rangle=c_0|0\rangle + c_1|1\rangle$. The real and imaginary parts of $c_0$ and $c_1$ provide four. Accounting for normalization $|c_0|^2+|c_1|^2=1$ brings the count down to three. At this stage it seems like the qubit's state space might look like $S^3$. It does not. We also need to recognize that multiplying $c_0$ and $c_1$ by any complex number $e^{i\theta}$ with unit magnitude leaves all predictions of the theory unchanged. This is what people mean when they say that the "global phase is unobservable". Indeed, in the density matrix formulation of quantum mechanics the global phase does not make any appearance at all suggesting that it is a feature of the language used to describe phenomena and not a feature of the phenomena themselves. In any case, eliminating the global phase factor, e.g. by choosing $c_0$ to be a non-negative real number, brings the total down to two.

This two-stage procedure gives rise to the equivalence relation $$ (c_0,c_1)\sim(\lambda c_0,\lambda c_1)\tag1 $$ for any non-zero $\lambda\in\mathbb{C}$. Thus, the numbers $c_0$ and $c_1$, called amplitudes in quantum physics, are in fact just normalized homogeneous coordinates in the complex projective space $\mathbb{CP}^1$. The magnitude of $\lambda$ gives us normalization and the phase angle of $\lambda$ gives us global phase invariance.

The upshot is that the state space of a qubit is $\mathbb{CP}^1\cong S^2$, not $S^3$.

Homotopy between identity and quantum evolution

Another postulate of quantum mechanics says that a closed$^1$ quantum system evolves according to the Schrödinger equation $$ i\hbar\frac{d}{dt}|\phi(t)\rangle=H(t)|\phi(t)\rangle\tag2 $$ where $H(t)$ is a Hermitian operator corresponding to the system's designated observable called the Hamiltonian. The solution to $(2)$ defines the time evolution operator $U(t_1, t_0)$ which sends a state at time $t_0$ to a state at time $t_1$ $$ |\psi(t_1)\rangle=U(t_1,t_0)|\psi(t_0)\rangle.\tag3 $$ It is physically obvious that we must have $$ U(t,t)=I\tag4 $$ for every $t\in\mathbb{R}$. Moreover, it can be shown that $$ U(t_1,t_0)=\mathcal{T}\exp\left(-\frac{i}{\hbar}\int_{t_0}^{t_1}H(t)dt\right)\tag5 $$ where $\mathcal{T}\exp$ is the time-ordered operator exponential.

The boundary condition $(4)$ suggests a way of constructing a homotopy between the identity and any map that describes a physically realizable time evolution. Indeed, consider a map $g:\mathbb{CP}^{n-1}\to\mathbb{CP}^{n-1}$. By the above postulate, if $g$ describes a physically realizable time evolution of an $n$-level quantum system, then there exists a Hamiltonian $H(t)$ and $t_0,t_1\in\mathbb{R}$ such that the time evolution operator corresponding to the solution of Schrödinger equation satisfies $$ U(t_1, t_0)=g.\tag6 $$ Define $G:\mathbb{CP}^{n-1}\times[0,1]\to\mathbb{CP}^{n-1}$ as $$ G(|\psi\rangle, t):=U(t_0+(t_1-t_0)\cdot t, t_0)|\psi\rangle\tag7 $$ for every $|\psi\rangle\in\mathbb{CP}^{n-1}$ and every $t\in[0,1]$. By $(4)$ we have $G(.,0)=I$ and by $(6)$ we have $G(.,1)=g$. Moreover, $(5)$ implies that $G$ is continuous. Therefore, $G$ is a homotopy between the identity and $g$. Finally, since the degree is homotopy invariant, we have $$ \deg g=\deg I=1.\tag8 $$ Thus, every physically realizable function on $\mathbb{CP}^{n-1}$ has degree $1$.


$^1$ Here, "closed" means that the system is isolated from its environment.

$\endgroup$
3
  • $\begingroup$ Thanks so much for this thorough answer! While that clears up two of the main points (global phase and our model of nature), I'm left wondering why something like the "sphere wrapping" map (the suspension of $z \mapsto z^k$ to $S^2$) is disallowed by nature. It is continuous, invertible, preserves probabilities (as it's a map $\mathbb{CP}^1 \rightarrow \mathbb{CP}^1$), but nonlinear of degree $k$. Why is nature forcing us to use linear, unitary maps? I suppose it boils down to the Schrödinger or Dirac equations being used as the model. $\endgroup$ Sep 19, 2023 at 15:51
  • $\begingroup$ Non-linear maps have some rather staggering implications, such as faster-than-light (FTL) communication and computers able to solve #P-hard problems in polynomial time, see e.g. this post. $\endgroup$ Sep 20, 2023 at 2:07
  • $\begingroup$ Thank you, that is exactly what I've been looking for. $\endgroup$ Sep 20, 2023 at 21:47

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.