Why does full state reconstruction require at least $N+1$ MUBs?

Quantum Computing Asked by glS on February 28, 2021

Consider an $N$-dimensional space $mathcal H$. Two orthonormal bases $newcommand{ket}[1]{lvert #1rangle}{ket{u_j}}_{j=1}^N,{ket{v_j}}_{j=1}^Nsubsetmathcal H$ are said to be Mutually Unbiased Bases (MUBs) if $lvertlangle u_ilvert v_jranglervert =1/sqrt N$ for all $i,j$.

Suppose we want to fully reconstruct a state $rho$ by means of projective measurements. A single basis provides us with $N-1$ real parameters (the $N$ outcome probabilities associated with the measurement, minus one for the normalisation constraint).

Intuitively, if two bases are mutually unbiased, they provide fully uncorrelated information (finding a state in some $ket{u_j}$ says nothing about which $ket{v_k}$ would have been found), and thus measuring the probabilities in two different MUBs should characterise $2(N-1)$ real parameters.
If we can measure in $N+1$ different MUBs (assuming they exist), it thus stands to reason that we characterised $(N-1)(N+1)=N^2-1$ independent real parameters of the state, and thus obtained tomographically complete information.
This is also mentioned in passing in this paper (page 2, second column, arXiv:0808.0944).

What is a more rigorous way to see why this is the case?

One Answer

Denote the projections onto basis elements by $P_j^{(k)}=|u_j^{(k)}ranglelangle u_j^{(k)}|$, where superscript indexes different bases. Tomography of a density matrix $rho$ gives us probabilities $text{Tr}(rho P_j^{(k)})$. This is actually a value of the Hilbert-Schmidt inner product between $rho$ and $P_j^{(k)}$ in the space $L(mathcal{H})$ $-$ the complex space of all $Ntimes N$ matrices. Such values can be used to reconstruct a projection of $rho$ onto the $text{span}{P_j^{(k)}}$ in the space $L(mathcal{H})$. For a full reconstruction of $rho$ we must have $text{span}{P_j^{(k)}}_{j,k} = L(mathcal{H})$.

Since $sum_{j=1}^N P_j^{(k)} = I$ we can write $$ text{span}{P_j^{(k)}}_{j=1}^N = text{span}{P_j^{(k)}-I/N}_{j=1}^{N-1} oplus langle Irangle = mathcal S_k oplus langle Irangle, $$ where $mathcal S_k$ is a subspace of dimension $N-1$ in $L(mathcal{H})$.

The element $I$ is special since we a priory know the length of projection on it $text{Tr}(rho I) = 1$ (so we could consider the space $L(mathcal{H}) ominus langle Irangle$ of dimension $N^2-1$, but it's easier for me to work in the full space).

Now note that $$ text{Tr}big((P_i^{(k)}-I/N)(P_j^{(l)}-I/N)big) = 0 $$ whenever $kneq l$. This means that $mathcal S_k perp mathcal S_l$. Hence the dimension of the span of $P_j^{(k)}$ of $m$ MUBs is exactly $m(N-1)+1$ in $L(mathcal H)$.

Correct answer by Danylo Y on February 28, 2021

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