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

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?

## 1 Answer

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 $$L(\mathcal H)$$.

• nice approach, thanks. I wonder if there is a more general way to understand this: we are essentially saying that we have two (or more) orthonormal bases $\{v_i\}_{i=1}^N,\{ w_i\}_{i=1}^N\subset\mathbb R^{N^2}$ whose elements are also mutually orthogonal wrt the different origin $\vec O\equiv \sum_i v_i/N=\sum_i w_i/N$. It's an interesting structure – glS Aug 20 '20 at 7:59