# 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?

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

Get help from others!