3
$\begingroup$

It is well known that the state of a (pure) qubit can be described as a point on a two-dimensional sphere, the so-called Bloch sphere.

The mapping $\lvert\psi\rangle\mapsto \boldsymbol r_\psi$ that sends each state into its representative in the Bloch sphere has the peculiarity of sending pairs of states that are orthogonal in the original Hilbert space (e.g. $\lvert0\rangle$ and $\lvert1\rangle$) into collinear points in the Bloch sphere. For example, $\lvert0\rangle$ and $\lvert1\rangle$ are sent to the nadir and zenith of the Bloch sphere.

More generally, $d$-dimensional systems can also be represented geometrically, using a generalised Bloch representation (see e.g. this question for how this works). However, on such systems, systems of orthogonal states come in more than pairs, so the property outlined above that held in the Bloch sphere doesn't work anymore.

How (if at all) does this property generalise for $D$-dimensional systems? More precisely, considering a set of orthogonal states (e.g. $\{\lvert k\rangle\}_{k=1}^D$), are the corresponding points in the generalized Bloch representation arranged into some kind of geometrical structure?

$\endgroup$
  • $\begingroup$ In that question you are talking about general (not necessarily pure) states, but here only pure. Also a manifold has dimension that is intrinsic not depending on embedding. So you should say two-dimensional sphere $\endgroup$ – AHusain Jun 4 at 23:35
  • $\begingroup$ @AHusain fair points. For the purity though, if the answer generalises nicely to non-pure states that would be great as well! $\endgroup$ – glS Jun 4 at 23:54
3
$\begingroup$

One natural generalization of that property is that Bloch vectors for a basis set must sum to 0 vector. Though, this property is not a criterion for basis sets in dimensions higher than 2.

If $\rho_i$, $i=1 .. d,$ are corresponding density matrices for a basis set, then $$ \sum_{i=1}^d \rho_i = I $$ Now if $\sigma_i,$ $i=1 .. d^2-1,$ are generalized Pauli matrices then $$ \rho_i = \frac{1}{d}\left(I + \sum_{j=1}^{d^2-1} r_{ij}\sigma_j\right), r_{ij}=\text{Tr}(\rho_i\sigma_j^\dagger) $$ If we add we'll have $$ \frac{1}{d}\sum_{j=1}^{d^2-1}\left(\sum_{i=1}^d r_{ij}\right)\sigma_i = 0 $$ Hence $$ \sum_{i=1}^d \boldsymbol r_i = 0 $$

Update

Actually, we can also calculate scalar products between vectors $\boldsymbol r_i$
(here I assume that $\text{Tr}(\sigma_i\sigma_i^\dagger)=d$, i.e. $\sigma_i$ are normalized in a sense. Also they are orthogonal, so $\text{Tr}(\sigma_i\sigma_j^\dagger)=0, i\neq j$, and traceless $\text{Tr}(\sigma_i)=0$). $$ \delta_{kl} = \text{Tr}(\rho_k\rho_l^\dagger) = \frac{1}{d^2}\text{Tr}\left(\left(I + \sum_{j=1}^{d^2-1} r_{kj}\sigma_j\right)\left(I + \sum_{i=1}^{d^2-1} \bar{r}_{li}\sigma_i^\dagger\right)\right) = $$ $$ = \frac{1}{d^2}\left(\text{Tr}I + \sum_{j=1}^{d^2-1} r_{kj}\bar{r}_{lj}\text{Tr}(\sigma_j\sigma_j^\dagger)\right) = \frac{1}{d}\left(1 + \sum_{j=1}^{d^2-1} r_{kj}\bar{r}_{lj}\right) $$ Hence $$ (\boldsymbol r_k, \boldsymbol r_l) = d\delta_{kl} - 1 $$

Now if we add another condition that all $\boldsymbol r_i$ correspond to pure states (this condition can be extracted from equations $Tr(\rho_i)=1$ and $\rho_i^2=\rho_i=\rho_i^\dagger$) then all those properties of $\boldsymbol r_i$ will form a criterion for a generalized Bloch vectors to be corresponding to some basis set of states.

$\endgroup$
4
$\begingroup$

"Generalized Bloch" manifolds are synonyms to coherent state manifolds. The points of these manifolds do not correspond, in general, to orthonormal vectors, as there are much more points than the dimension of the system's Hilbert space. Points on the manifold correspond rather to generalized coherent states.

These states are actually classical, they satisfy properties analogous to the ordinary coherent states of the harmonic oscillator such as (appropriately defined) minimal uncertainty (please see for example, Ali, Antoine, Gazeau and Mueller). More, importantly, they satisfy Bell's inequality upon splitting, please see Brief Mann and Revzen. However, they do span the full Hilbert space: Uncountable number of orthonormal bases can be constructed from linear combinations of these states.

The question is, given a coherent state manifold; does it contain a set of special points whose corresponding coherent states form an orthonormal basis of the corresponding Hilbert space. In the case of an $N+1$-level system (qudit), there is a coherent state manifold for which the answer is positive: The complex projective space: $\mathbb{C}P^N$ (as well as other coherent state manifolds where the answer is negative).

For an $N+1$-level system, a coherent state vector parametrized by $\mathbb{C}P^N$ has the following explicit expression: $$|\psi(z, \bar{z}) \rangle= \frac{1}{\sqrt{1 + \bar{z}_1z_1+ … +\bar{z}_N z_N}} \begin{bmatrix} 1\\ z_1\\ .\\ .\\ .\\ z_N\\ \end{bmatrix}$$ Where the coordinates, $z_i$ are global coordinates of $\mathbb{C}P^N$ : $z_i\in \mathbb{C}\cup \{\infty\}$.

Now please observe that the coherent state vectors corresponding to the $N+1$ points:

\begin{matrix} [z_1=0,...,z_N=0]\\ [z_1=0,z_2=\infty,...,z_N=0]\\ [z_1=0,z_2=0, z_3=\infty,...,z_N=0]\\ .\\ .\\ .\\ [z_1=0,...,z_N=\infty]\\ \end{matrix} form the standard orthonormal vector spanning the $N+1$ dimensional Hilbert space.

These special points enjoy unique geometric properties on $\mathbb{C}P^N$, please see Berceanu: Taking the first point as the origin, then the rest of the points

1) Form a cut locus, whose points are the first on the geodesic joining them to the origin beyond which the geodesic ceases to minimize the arc length (i.e., it maximizes it).

2) They form also the conjugate locus which is the set of points which can be connected to the origin with an infinite number of geodesics.

3) In addition, since $\mathbb{C}P^N$, is an $SU(N+1)$, homogeneous space. The transformations which exchange a pair of points are just the matrices which exchange a pair of rows: These transformations generate a discrete group called the Weyl group. The order of the Weyl group for $SU(N+1)$ is $(N+1)!$, and its action on the orthonormal basis is just by permutations.

4) In addition, the $\mathbb{C}P^N$ has a cell structure, i.e., it can be decomposed into a disjoint set of affine spaces: $$\mathbb{C}P^N = \mathbb{C}^N \cup \mathbb{C}^{N-1}…\cup \mathbb{C} \cup \{0\}$$ Such that each special point resides on a different cell.

There are other choices of coherent state manifolds which do not allow an orthonormal basis of coherent states. For example, we can span an $N+1$ Hilbert space by the Bloch sphere $\mathbb{C}^1=S^2$ as follows: $$|\psi(z, \bar{z}) \rangle= \frac{1}{\sqrt{(1 + \bar{z}z)^N}} \begin{bmatrix} 1\\ \sqrt{N} z\\ .\\ .\\ .\\ \sqrt{\begin{pmatrix} N\\ i \end{pmatrix}}z^i\\ .\\ .\\ .\\ z^N\\ \end{bmatrix}$$ Where: $z\in \mathbb{C}\cup \{\infty\}$ isthe stereographic projection coordinate of the Bloch sphere.

In this case (for $N>2$), we can only generate an orthonormal basis by forming linear combinations of the coherent vectors for different values of $z$.

$\endgroup$
  • $\begingroup$ wow, ok so this looks very interesting, but I'm a bit confused. I might have unknowingly been using the wrong terminology here. When you say "Generalized Bloch manifolds are synonyms to coherent state manifolds.", I'm not sure what you are referring to. In my mind, I simply meant the decomposition of density matrices in orthogonal bases of Hermitians, that is, the mapping $\rho\mapsto r$ if $\rho=1/d(I+r\cdot \boldsymbol\sigma)$. Are these "CSMs" a generalisation of this? Your mentioning $\mathbb{CP}^n$ being a possible CSM, which makes me think this is the case $\endgroup$ – glS Jun 5 at 20:06
  • $\begingroup$ Sorry, I was under the impression that you were interested in manifolds of pure states, so the answer treated only pure states. The complex projective space is the manifold of pure states of an $N+1$ level system; It is a $SU(N+1)$ orbit of a single density matrix of rank $1$. One possible parametrization of the full space of density matrices is a union of $SU(N+1)$ orbits through the set of diagonal density matrices. In this case, we need to complement the complex projective space with all the other orbits. In the full space, every density matrix will correspond to a single point... $\endgroup$ – David Bar Moshe Jun 6 at 11:55
  • $\begingroup$ This space is not a manifold, as it will have singularities whenever the density matrices change ranks, and it will not be a coherent state manifold. One place to read about this parametrization is in the book by Bengtsson and Życzkowski: Geometry of Quantum States: pdfs.semanticscholar.org/3f28/… section 8.5. They do not give an explicit parametrization, although. $\endgroup$ – David Bar Moshe Jun 6 at 11:59
  • $\begingroup$ In the qutrit case, the orbit type of the mixed (but not maximally mixed) states will be the flag manifold $FL_2$, an explicit parametrization for it is given for example in equation (23) in the work by Daoud and Jellal: arxiv.org/abs/hep-th/0610157v2 . In any case, the sets corresponding to the orthonormal bases will appear only in the pure space subset, i.e., in the complex projective space. I can complement the answer with the construction of the full parametrization for the qutrit case, but it will require some writing to do. $\endgroup$ – David Bar Moshe Jun 6 at 11:59
3
$\begingroup$

This other answer already gave a nice proof that orthogonal bases are mapped into vectors $r_i$ such that $\sum_i r_i=0$. Here I'll work out explicitly the coordinates in a few cases, to show what kind of geometrical figure exactly comes out.

I'll consider in particular the computational basis in $d$ dimensions: $\{\lvert 1\rangle,...,\lvert d\rangle\}$. Any other orthonormal basis is just a rotation away from the computational basis anyway, so the geometrical features should be the same.

I'll consider the following standard basis of traceless Hermitians operators: $$Z^{(j)}\equiv\sqrt{\frac{d}{j(j+1)}}\left(\sum_{k=1}^j\lvert j\rangle\!\langle j\rvert-j\lvert j+1\rangle\!\langle j+1\rvert\right), \,\, j=1,...,d-1,$$ $$X^{(jk)}\equiv\sqrt{d/2}[\lvert j\rangle\!\langle k\rvert + \lvert k\rangle\!\langle j\rvert], \quad Y^{(jk)}\equiv i\sqrt{d/2}[\lvert j\rangle\!\langle k\rvert - \lvert k\rangle\!\langle j\rvert],\,\, j>k.$$

It is easy to check that for all the elements of this basis, we have $$\langle X^{(jk)}\rangle=\langle Y^{(jk)}\rangle=0,$$ while for the $z$ coordinates we have \begin{align} z_j(\lvert\ell\rangle) &=\sqrt{d/[j(j+1)]}, \text{ for all } j\ge\ell,\\ z_{\ell-1}(\lvert\ell\rangle) &=-\sqrt{d(\ell-1)/\ell}, \text{ for all }, \\ z_{j}(\lvert \ell\rangle) &=0, \text{ for all }j<\ell-1. \end{align}

Explicit $d=3$ example

To get a better understanding of what this means, let us work out explicitly the $d=3$ case. We have two nonzero coordinates, corresponding to the operators $Z^{(1)}$ and $Z^{(2)}$. Remember that these matrices are $$Z^{(1)}=\sqrt{3/2}\begin{pmatrix}1&0&0\\0& -1&0\\0&0&0\end{pmatrix},\qquad Z^{(2)}=\sqrt{1/2}\begin{pmatrix}1&0&0\\0& 1&0\\0&0&-2\end{pmatrix}.$$ so that the corresponding coordinates are: \begin{array}{cc} & z_1 & z_2 \\ \hline \lvert1\rangle & \sqrt{3/2}& \sqrt{1/2}\\ \lvert2\rangle & -\sqrt{3/2}& \sqrt{1/2}\\ \lvert3\rangle & 0& -\sqrt2 \end{array} It is easier to recognise what these points represent by normalising the vectors, which gives $$\lvert1\rangle\doteq\begin{pmatrix}\sqrt3/2 \\ 1/2\end{pmatrix}=\begin{pmatrix}\sin(\pi/3) \\ \cos(\pi/3)\end{pmatrix},\quad \lvert2\rangle\doteq\begin{pmatrix}-\sqrt3/2 \\ 1/2\end{pmatrix}=\begin{pmatrix}\sin(2\pi/3) \\ \cos(2\pi/3)\end{pmatrix},\quad \lvert3\rangle\doteq\begin{pmatrix}0 \\-1\end{pmatrix}.$$ In other words, the states $\lvert1\rangle,\lvert2\rangle,\lvert3\rangle$ are mapped into an equilateral triangle: enter image description here

Explicit $d=4$ example

In four dimensions, we have

$$ Z^{(1)}=\sqrt{2}\begin{pmatrix}1&0&0&0\\0& -1&0&0\\0&0&0&0\\0&0&0&0\end{pmatrix},\quad Z^{(2)}=\sqrt{2/3}\begin{pmatrix}1&0&0&0\\0& 1&0&0\\0&0&-2&0\\0&0&0&0\end{pmatrix},\quad Z^{(3)}=\sqrt{1/3}\begin{pmatrix}1&0&0&0\\0& 1&0&0\\0&0&1&0\\0&0&0&-3\end{pmatrix},$$ and thus the computational basis is sent to

\begin{array}{ccc} & z_1 & z_2 & z_3 \\ \hline \lvert1\rangle & \sqrt{2}& \sqrt{2/3} & \sqrt{1/3} \\ \lvert2\rangle & -\sqrt{2}& \sqrt{2/3} & \sqrt{1/3} \\ \lvert3\rangle & 0& -2\sqrt{2/3} & \sqrt{1/3} \\ \lvert4\rangle & 0 & 0 & -\sqrt3 \end{array}

Renormalising the lengths and working out the angles we find that again the elements of the basis are mapped into a regular figure, in this case a tetrahedron (as in, the states are mapped into the vertices of a tetrahedron):

enter image description here

General $d$ case

The natural next step is to ask whether this feature (that is, that the computational basis is mapped into the regular $(d-1)$-dimensional $d$-vertices polytope in which the pairwise distance between vertices is constant) still holds increasing the number of dimensions. It seems almost obvious that it does, but I haven't found a nice way to show it yet.

$\endgroup$
  • 1
    $\begingroup$ I updated my answer. Scalar products are indeed constant. $\endgroup$ – Danylo Y Jun 6 at 15:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.