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Let $\mathbb{C}^{d}$ be a complex Euclidean space.

Let $\mathsf{H}(\mathbb{C}^{d})$ be the set of all Hermitian operators, mapping vectors from $\mathbb{C}^{d}$ to $\mathbb{C}^{d}$. I had some questions about the notation introduced in this paper (page 2.)

A quasiprobability representation of a qudit over $\Lambda$ is defined by a frame $\{F(λ) : λ \in \Lambda\}$ and a dual frame $\{G(λ) : λ ∈ Λ\}$, which are (generally over-complete) bases for $\mathsf{H}(\mathbb{C}^{d})$, satisfying \begin{equation} A = \sum_{\lambda \in \Lambda} G(\lambda) ~\text{Tr}[A F(\lambda)], \end{equation} for all $A$.

Here are my questions.

  1. What is the cardinality of the set $\Lambda$? Each $A \in \mathsf{H} (\mathbb{C}^{d})$ has $d^{2}$ elements. If the frames are to be overcomplete bases of $\mathsf{H} (\mathbb{C}^{d})$, does it mean that \begin{equation} |\Lambda| \geq d^{2}? \end{equation}

  2. What is the relation between $F(\lambda)$ and $G(\lambda)$?

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  1. The authors are certainly thinking about finite frames. In this case, your statement is correct, since the number of elements in every spanning set is at least the vector space dimension. As glS already pointed out, the frames could be infinite:
  • for countable frames, the summation is then meant as convergence in the $L^2$-sense (i.e. w.r.t. Hilbert-Schmidt norm)
  • for uncountable frames, we would need a measure on the set $\Lambda$ and the sum is replaced with integration w.r.t. that measure
  1. $G(\lambda)$ is a dual frame to the frame $F(\lambda)$. The defining condition is exactly the equation $$ A = \sum_{\lambda\in\Lambda} \mathrm{tr}(AF(\lambda))\, G(\lambda) = \sum_{\lambda\in\Lambda} \mathrm{tr}(AG(\lambda))\, F(\lambda). $$ If we use "operator braket notation", this is $(B|A):= \mathrm{tr}(AB)$, we can write this as $$ \sum_{\lambda\in\Lambda} |F(\lambda))(G(\lambda)| = \sum_{\lambda\in\Lambda} |G(\lambda))(F(\lambda)| = \mathrm{id}_{\mathsf{H}(\mathbb{C}^d)}. $$ Note that the dual frame is not unique! However, there is always the canonical dual frame which is constructed as follows. Consider the frame operator (it's a superoperator) $$ S = \sum_{\lambda\in\Lambda} |F(\lambda))(F(\lambda)|. $$ It is always a positive operator in the sense that it is self-adjoint and has positive eigenvalues. Thus, it is invertible and we can define $$ G^{\mathrm{can}}(\lambda) := S^{-1}\left( F(\lambda) \right). $$ It is not hard to check that this is a frame dual to $F(\lambda)$. In practice, tight frames are very important. These are frames for which $S = a\, \mathrm{id}_{\mathsf{H}(\mathbb{C}^d)}$ for $a\in\mathbb R$. In this case, the canonical dual frame is $a^{-1}F(\lambda)$. Tight frames are nice since they behave similar to an orthogonal basis (which is also self-dual up to a constant in this sense).

If you want to know more (almost everything) about frame theory, I recommend

Shayne F. D. Waldron, "An Introduction to Finite Tight Frames". Springer (2017)

It also connects a bit with the quantum info literature in the end (MUBs etc).

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  • $\begingroup$ Thanks for an excellent answer! A quick question, since you brought up a basis while discussing tight frames: qualitatively, what is the difference between a frame and a basis? Are all basis sets --- be they orthogonal or non-orthogonal --- frames, but the reverse isn't true? $\endgroup$ – BlackHat18 Jun 16 at 18:43
  • $\begingroup$ Additionally, the "operator bra-ket notation" was a bit hard for me to follow. When you write $|F(\lambda)(G(\lambda)|$, do you mean $|F(\lambda))(G(\lambda)|$ instead, and you mean that the $|\cdot)$ operator is a "ket" and the $(\cdot|$ operator is a bra? $\endgroup$ – BlackHat18 Jun 16 at 18:45
  • $\begingroup$ @BlackHat18 yes, every basis is a frame but not vice versa. Synonymous words for 'frame' are 'generating set' or 'overcomplete basis'. In the sense, frames are more general. $\endgroup$ – Markus Heinrich Jun 17 at 4:59
  • $\begingroup$ @BlackHat18 sorry I messed up the brackets, it should be $|A)(B|$ (I corrected my answer). This is basically the same notation as for ordinary bra-kets, but instead of taking vectors in the Hilbert space $\mathbb{C}^d$, we take vectors in the Hilbert space $L(\mathbb C^d)$ (or $\mathsf{H}(\mathbb C^d)$ in this case). The superoperator $|A)(B|$ acts as $C\mapsto |A)(B|C) \equiv \mathrm{tr}(B^\dagger C) A$ (where can leave out the dagger for Hermitian operators). $\endgroup$ – Markus Heinrich Jun 17 at 5:03

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