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

Question

Consider an $N$-dimensional space $\mathcal H$. Two orthonormal bases $\newcommand{\ket}[1]{\lvert #1\rangle}\{\ket{u_j}\}_{j=1}^N,\{\ket{v_j}\}_{j=1}^N\subset\mathcal H$ are said to be Mutually Unbiased Bases (MUBs) if $\lvert\langle u_i\lvert v_j\rangle\rvert =1/\sqrt N$ 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?

Answer 1

Denote the projections onto basis elements by $P_j^{(k)}=|u_j^{(k)}\rangle\langle u_j^{(k)}|$, 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 $N\times N$ 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 $$ \text{span}\{P_j^{(k)}\}_{j=1}^N = \text{span}\{P_j^{(k)}-I/N\}_{j=1}^{N-1} \oplus \langle I\rangle = \mathcal S_k \oplus \langle I\rangle, $$ 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 $L(\mathcal{H}) \ominus \langle I\rangle$ 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 $k\neq l$. 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 $L(\mathcal H)$.