# Do pure qudit states lie on a hypersphere in the Bloch representation?

It is known that every state $$\rho$$ of a $$d$$-level system (or if you prefer, qudits living in a $$d$$-dimensional Hilbert space) can be mapped into elements of $$\mathbb R^{d^2-1}$$ through the mapping provided by the Bloch representation, by writing it as $$\rho=\frac{1}{d}\left(I+\sum_{k=1}^{d^2-1} c_k \sigma_k\right),$$ with $$\{\sigma_k\}$$ traceless Hermitian matrices satisfying $$\operatorname{Tr}(\sigma_j\sigma_k)=d\delta_{jk}$$ and $$c_k\in\mathbb R$$. We can then consider the mapping $$f:\rho\mapsto f(\rho)\equiv\boldsymbol c\in\mathbb R^{d^2-1}$$. One can also show that the set of all states maps into a compact set in $$\mathbb R^{d^2-1}$$, and furthermore characterise the boundary of this set in "spherical coordinates", as shown for example in this answer.

Moreover, as shown for example in the answers to this question, the purity of a state $$\rho$$ translates into the norm of $$f(\rho)$$ (i.e. its distance from the origin): $$\operatorname{Tr}(\rho^2)=(1+\|f(\rho)\|^2)/d$$. It follows that if $$\rho$$ is pure, that is $$\operatorname{Tr}(\rho^2)=1$$, then $$\|f(\rho)\|=\sqrt{d-1}$$. This means that the set of pure states is a subset of the hypersphere $$S^{d^2-2}\subseteq\mathbb R^{d^2-1}$$ with radius $$\sqrt{d-1}$$.

However, the pure states do not cover this hypersphere: the boundary of the set of states is not comprised of pure states (see e.g. this question), and many (most) elements on $$S^{d^2-2}$$ do not correspond to physical states. This tells us that the set of pure states does not lie on a hypersphere in the Bloch representation (which must obviously be true, as we know that the set of pures has a much smaller dimension).

But then again, this does not (I think?) rule out the possibility that the set of pure states lies on some lower-dimensional hypersphere $$S^{2d-3}$$ embedded in $$\mathbb R^{d^2-1}$$. Is this the case?

Upon some reflection, the answer is that no, they most definitely do not.

The easiest way to see this is to observe that there are $$d^2-1$$ orthogonal directions in the Bloch representation (i.e. orthogonal Hermitian traceless operators) containing pure states. This means that that the pure states are not contained in any linear subspace of dimension less than $$d^2-1$$, and thus in particular cannot be contained in any lower-dimensional hypersphere.

More specifically, given any orthogonal basis of Hermitian traceless operators $$\{\boldsymbol\sigma_j\}_{j=1}^{d^2-1}$$, and any versor $$\hat{\mathbf n}\in\mathbb R^{d^2-1}$$ with $$\|\hat{\mathbf n}\|=1$$, there are always pure states in the direction $$\boldsymbol\sigma_{\hat{\mathbf n}}\equiv\hat{\mathbf n}\cdot\boldsymbol\sigma$$ (this follows from the fact that the operator norm satisfies $$\|\boldsymbol\sigma_{\hat{\mathbf n}}\|>0$$ and the characterisation in spherical coordinates explained e.g. here). It follows that there cannot be less than $$d^2-1$$ elements of $$\mathbb R^{d^2-1}$$ whose span contains the set of pure states.

# A concrete example: three-level systems

Consider a generic pure state of a three-level system: $$|\psi\rangle=\cos\alpha|0\rangle+e^{i\phi}\sin\alpha \cos\beta |1\rangle+e^{i\theta}\sin\alpha\sin\beta|2\rangle,$$ for all $$\alpha,\beta,\phi,\theta\in\mathbb R$$. Let us also use the standard operatorial basis for this space (the matrices used at the bottom of this answer):

$$Z^{(1)}=\sqrt{\frac{3}{2}}\begin{pmatrix}1 & 0&0 \\ 0 & -1&0\\0&0&0\end{pmatrix}, \quad Z^{(2)}=\sqrt{\frac{3}{6}}\begin{pmatrix}1 & 0&0 \\ 0 & 1&0\\0&0&-2\end{pmatrix},$$ \begin{align} X^{(12)}=\sqrt{\frac{3}{2}}\begin{pmatrix}0 & 1&0 \\ 1 & 0&0\\0&0&0\end{pmatrix}, \quad X^{(13)}=\sqrt{\frac{3}{2}}\begin{pmatrix}0 & 0&1 \\ 0 & 0&0\\1&0&0\end{pmatrix}, \quad X^{(23)}=\sqrt{\frac{3}{2}}\begin{pmatrix}0 & 0&0 \\ 0 & 0&1\\0&1&0\end{pmatrix} \end{align} \begin{align} Y^{(12)}=\sqrt{\frac{3}{2}}\begin{pmatrix}0 & -i&0 \\ i & 0&0\\0&0&0\end{pmatrix}, \quad Y^{(13)}=\sqrt{\frac{3}{2}}\begin{pmatrix}0 & 0&-i \\ 0 & 0&0\\i&0&0\end{pmatrix}, \quad Y^{(23)}=\sqrt{\frac{3}{2}}\begin{pmatrix}0 & 0&0 \\ 0 & 0&-i\\0&i&0\end{pmatrix}. \end{align} Then, the surface covered by the pure states in $$\mathbb R^8$$ has the following parametrisation: $$\begin{cases} \langle Z^{(1)}\rangle&=\sqrt{3/2} (\cos^2\alpha-\sin^2\alpha\cos^2\beta),\\ \langle Z^{(2)}\rangle&=\sqrt{3/6} [\cos^2\alpha+\sin^2\alpha(\cos^2\beta-2\sin^2\beta)],\\\hline \langle X^{(12)}\rangle&=\sqrt{3/2}\sin(2\alpha)\cos\beta \cos\phi,\\ \langle X^{(13)}\rangle&=\sqrt{3/2}\sin(2\alpha)\sin\beta \cos\theta,\\ \langle X^{(23)}\rangle&=\sqrt{3/2}\sin^2(\alpha)\sin(2\beta) \cos(\phi-\theta),\\\hline \langle Y^{(12)}\rangle&=\sqrt{3/2}\sin(2\alpha)\cos\beta \sin\phi,\\ \langle Y^{(13)}\rangle&=\sqrt{3/2}\sin(2\alpha)\sin\beta \sin\theta,\\ \langle Y^{(23)}\rangle&=\sqrt{3/2}\sin^2(\alpha)\sin(2\beta) \sin(\phi-\theta).\tag A \end{cases}$$

To easily check that these points do indeed lie on a hypersphere (which we also know from this answer must have a radius of $$\sqrt2$$), just run the following snippet in Mathematica: