Can we rotate Bloch vectors for qudits like we do with qubits in the Bloch sphere?

Question

I have been looking into the Bloch vectors for qudits and have been wondering if we can do rotations that are similar to the rotations in the qubit Bloch sphere.

Like, once we create a Bloch vector for a qutrit, it exists in 8-dimensional space. In this 8 dimensional space, we could do its rotation about a 6-dimensional hyperspace (like look here). One possible method to come up with this is to exponentiate a flip matrix, they should flip the sign of two of the basis states and should let the others stay the same.

But I can not find a matrix that does that. Using the qutrit Z gate we can flip the signs of 4 of the 6 antisymmetric and symmetric matrices but not 2 at a time, which is the problem. Furthermore, a matrix that flips diagonal Gell–Mann matrices changes the other basis states.

Answer 1

A few remarks:

1. You can define a "Bloch representation" for higher-dimensional states. This amounts to representing density matrices by their coefficients in some basis of Hermitian (generally traceless) operators. It is however worth stressing that in more than two dimensions, you don't get a hypersphere but a much more involved kind of surface. See e.g. Why is the boundary of the set of states in the generalised Bloch representation comprised of singular matrices?, Geometry of qutrit and Gell-Mann matrices, and links therein.

2. In a Euclidean space $\mathbb R^N$, "rotations" are (special) orthogonal matrices, for reasons discussed e.g. in Why are orthogonal matrices generalizations of rotations and reflections?. Unitary operations on states are represented as orthogonal operations in the associated Bloch representation. This happens because if $U\in\mathbf{U}(N)$ is unitary and $\{\sigma_i\}_i\subset\mathfrak{su}(N)\simeq\mathfrak{so}(N)$ is a basis of traceless and skew-Hermitian matrices, then $U\sigma_i U^\dagger=\sum_j B_{ji}\sigma_j$ with $B\equiv\operatorname{Ad}(U)\in\mathbf{SO}(N^2-1)$. More formally, this is the observation that the Lie group $\mathbf{U}(N)$ acts on its Lie algebra $\mathfrak u(N)$, and also on the Lie algebra $\mathfrak{su}(N)$, via the adjoint representation, as special orthogonal matrices. More concretely, if $\rho\in\mathrm{Herm}(\mathbb C^N)$ has representative $\vec r_\rho\in\mathbb R^{N^2-1}$ in the Bloch representation, then $U\rho U^\dagger$ has representative $\vec r_{U\rho U^\dagger}={\rm Ad}(U)\vec r_\rho$.

   It is also worth noting that while this process, for $N=2$, gives the full $\mathbf{SO}(3)$ rotation group, more generally the dimensions don't quite match, as $\mathbf U(N)$ is being represented in $\mathfrak{su}(N)$, which is a $(N^2-1)$-dimensional vector space, whereas $\mathbf{SO}(N^2-1)$ (as a real manifold) has dimension $(N^2-1)(N^2-2)/2$. Thus in general $\mathbf U(N)$ is represented as a subgroup of $\mathbf{SO}(N^2-1)$ via its adjoint.

3. Rotations in more than three dimensions don't generally look as what you might be picturing in 3D. You don't always have a single rotation axis and a periodic motion around said axis. See also the relevant discussion on math.SE: Can rotations in 4D be given an explicit matrix form?. For example, in $\mathbf{SO}(8)$ you might consider a path of the form $S(\theta)=R(\theta)\oplus R(\theta/\pi)\oplus I_4$ with $R(\theta)$ a two-dimensional $\mathbf{SO}(2)$ rotation by an angle $\theta$, thus such that $R(2\pi n)=I$ for any $n\in\mathbb Z$. You might observe that there is no $\theta\neq0$ such that $S(\theta)=I_8$, as you would expect from a rotation in 2D or 3D.

Answer 2

This is an addendum to glS's answer and concern the issue of an "axis" in higher dimensions.

Let us set $D:=N^2-1$ such that we are interested in elements $R\in SO(D)$. $R$ is in general only diagonalizable over the complex numbers and since it is a special orthogonal matrix, its eigenvalues generally come in conjugate pairs $\lambda_i = \omega_i$ and $\bar\lambda_i = \bar\omega_i$, where $\omega_i$ is a root of unity. More precisely, the spectrum depends on $D$ being even or odd.

$D$ even: Then, the spectrum consists entirely of complex conjugate pairs. In particular, the fixed point subspace of $R$ has to be even-dimensional and can be $\{0\}$. There is no axis of rotation in the sense of a one-dimensional fixed point space.

$D$ odd: In this case, $\lambda=1$ is always an eigenvalue. This is because $\det R = \det R^{-1} = 1$ and thus $$ \det(R-I) = \det(R^{-1}R - R^{-1}) = \det(I - R)^\top = \det(I-R) = (-1)^D \det(R-I). $$ Hence, if $D$ is odd, then $\lambda=1$ is an eigenvalue of each $R\in SO(D)$. The remaining spectrum comes in conjugate pairs. In particular, the fixed point space of $R$ is always odd-dimensional and an axis might or might not exist.

What does that imply for unitary rotations? The qudit dimension $N$ is even/odd if and only if $D=N^2-1$ is odd/even. Hence, for qubits, $N=2$, we have $D=3$ and the possible fixed point spaces are one-dimensional or full-dimensional (only for $R=I$ of course). In contrast, for odd-dimensional qudits, like $N=3$, we have $D=8$, so there are 5 possible dimensions for the fixed point spaces.

But as already pointed out, not all elements $R\in SO(N^2-1)$ come from unitaries $U\in U(N)$. Here, we can say a bit more. As any unitary is normal, we have an orthonormal eigenbasis $\psi_i$ for $i=1,\dots,N$. Note that $U|\psi_i\rangle\langle \psi_i|U^\dagger = |\psi_i\rangle\langle \psi_i| =: A_i$ since any eigenvalue has modulus 1. We will now project these pure states onto the subspace of traceless, Hermitian matrices. To this end, it is not necessary to introduce an orthogonal basis like the Gell-Mann matrices so I will avoid this. The projection is $$ a_i := A_i - \frac{I}{N}. $$ Since the operators $A_i$ are pure states, they have unit 2-norm. The decomposition into traceless subspace and identity subspace is orthogonal hence $$ 1 = \|A_i\|_2^2 = \| a_i \|_2^2 + \frac{1}{N} \quad \Rightarrow \quad \| a_i \|_2^2 = 1 - \frac{1}{N} = \frac{N-1}{N}. $$ Furthermore, since the eigenstates are orthogonal, $$ \langle a_i, a_j \rangle = \mathrm{tr}(a_i a_j) = \mathrm{tr}(A_i A_j) - \frac{2}{N} + \frac{1}{N} = - \frac{1}{N}. $$ Defining a normalized version of the $a_i$'s as $\hat a_i := \sqrt{\frac{N}{N-1}} a_i$, we have $$ \langle \hat a_i, \hat a_j \rangle = \begin{cases} - \frac{1}{N-1} & i \neq j, \\ 1 & \text{ else}. \end{cases} $$ Note that for $N=2$, this is the fact that orthogonal states are mapped to antipodal points on the Bloch sphere!

In general, this is enough to conclude that the equiangular vectors $\hat a_i$ are isomorphic to the vertices of a $(N-1)$-simplex, in particular, the span of the $\hat a_i$ is $(N-1)$-dimensional.

Thus, we have shown that the fixed point space of the special orthogonal matrix $R=\mathrm{Ad}(U)$ is (at least) $(N-1)$-dimensional. But again, for $U=I$, this subspace is $(N^2-1)$-dimensional, so I suspect that it could also be larger. It might be possible to find a $(N-1)$-simplex of pure states in every $(N-1)$-dimensional subspace and thus every rotation fixing such a subspace could be induced by a unitary. But that's only speculation.