6
$\begingroup$

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?

$\endgroup$
0
4
$\begingroup$

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 $L(\mathcal H)$.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.