A minimal set of measurements for qudit-state tomography based on unambiguous discrimination

Abstract

Quantum-state tomography (QST) is a fundamental task for reconstructing unknown quantum state from statistics of measurements. We propose a qudit-state tomography based on unambiguous discrimination (UD) of d linearly independent pure states. We then prove that our proposal for QST provides a minimal set of measurements. Our proposal can be used in any finite dimension, and our strategy can be realized by a projective measurement on a system combined with a d-dimensional auxiliary system. In addition, we present another method to improve previously known UD QST of pure quantum state.

Appendices

A Proof of Lemma 1

Proof

Let r be an integer in $\mathbb {Z}_{g}$. Then we have $|\mathbb {Z}_{d}^{m,r}|\le n$ because $nm+r\equiv r(\mathrm{mod}\ d)$. If two integers $k_{1},k_{2}\in \mathbb {Z}_{n}$ satisfy $k_{1}<k_{2}$ and $k_{1}m+r\equiv k_{2}m+r(\mathrm{mod}\ d)$, $k_{2}-k_{1}$ should be a multiple of n because $(k_{2}-k_{1})m \equiv 0(\mathrm{mod}\ d)$. However, this is a contradiction since $0<k_{2}-k_{1}<n$. Therefore, we have $|\mathbb {Z}_{d}^{m,r}|=n$ and $\mathbb {Z}_{d}^{m,r}=\{z_{d,k}^{m,r}\}_{k\in \mathbb {Z}_{n}}$.

Let $r_{1},r_{2}\in \mathbb {Z}_{g}$ be two integers satisfying $r_{1}<r_{2}$. Then we get $\mathbb {Z}_{d}^{m,r_{1}}\cap \mathbb {Z}_{d}^{m,r_{2}}=\emptyset$ because if it is a non-empty set, we have $k_{1}m+r_{1}\equiv k_{2}m+r_{2}(\mathrm{mod}\ d)$ for some different integers $k_{1}$,$k_{2}$. This implies $(k_{1}-k_{2})m\equiv r_{2}-r_{1}(\mathrm{mod}\ d)$, which induces the following contradiction

$$\begin{aligned} g\le \mathrm{gcd}((k_{1}-k_{2})m,d)=\mathrm{gcd}(r_{2}-r_{1},d)< g. \end{aligned}$$

This means $|\bigcup _{r\in \mathbb {Z}_{g}}\mathbb {Z}_{d}^{m,r}|=g\times n=d$ by $|\mathbb {Z}_{d}^{m,r}|=n$. Therefore, we have $\bigcup _{r\in \mathbb {Z}_{g}}\mathbb {Z}_{d}^{m,r}=\mathbb {Z}_{d}$ by $|\mathbb {Z}_{d}|=d$.$\square$

B Derivation of Eq. (14)

Proof

Since $d-n$ pure states $|\psi _{i}\rangle$ except n pure states $|\psi _{x_{k}}\rangle$ are comprised by $d-n$ pure states $|e_{i}\rangle$ without n pure states $|e_{x_{k}}\rangle$, the following relation holds.

$$\begin{aligned} \sum _{k,k'\in \mathbb {Z}_{n}}\tilde{\rho }_{{x}_{k}{x}_{k'}}|\psi _{x_{k}}\rangle \!\langle \psi _{x_{k'}}| ={\Pi }\rho {\Pi }=\sum _{k,k'\in \mathbb {Z}_{n}}{\rho }_{{x}_{k}{x}_{k'}}|e_{x_{k}}\rangle \!\langle e_{x_{k'}}|, \end{aligned}$$

(36)

where

$$\begin{aligned} {\Pi }=\sum _{k\in \mathbb {Z}_{n}}|e_{x_{k}}\rangle \!\langle e_{x_{k}}|. \end{aligned}$$

(37)

By $|e_{x_{k}}\rangle$ of (13), the right-hand side of the above equation can be expressed as follows.

$$\begin{aligned} {\Pi }\rho {\Pi }= & {} \frac{1}{1-\cos n\phi } \sum _{k,k'\in \mathbb {Z}_{n}}\frac{{\rho }_{{x}_{k}{x}_{k'}}}{2\alpha _{x_{k}}\alpha _{x_{k'}}} \left( |\psi _{x_{k}}\rangle -e^{i\phi }|\psi _{x_{k+1}}\rangle \right) \left( \langle \psi _{x_{k'}}|-e^{-i\phi }\langle \psi _{x_{k'+1}}|\right) \nonumber \\= & {} \frac{1}{1-\cos n\phi }\sum _{k,k'\in \mathbb {Z}_{n}}(a_{kk'}-b_{kk'})|\psi _{x_{k+1}}\rangle \!\langle \psi _{x_{k'+1}}|, \end{aligned}$$

(38)

where

$$\begin{aligned} a_{kk'}=\frac{\rho _{x_{k}x_{k'}}}{2\alpha _{x_{k}}\alpha _{x_{k'}}} +\frac{\rho _{x_{k+1}x_{k'+1}}}{2\alpha _{x_{k+1}}\alpha _{x_{k'+1}}},\quad b_{kk'}=\frac{e^{i\phi }\rho _{x_{k}x_{k'+1}}}{2\alpha _{x_{k}}\alpha _{x_{k'+1}}} +\frac{e^{-i\phi }\rho _{x_{k+1}x_{k'}}}{2\alpha _{x_{k+1}}\alpha _{x_{k'}}}. \end{aligned}$$

(39)

Therefore, $\tilde{\rho }_{x_{k+1}x_{k+1}}$ becomes $(a_{kk}-b_{kk})/(1-\cos n\phi )$. $\square$

C Derivation of Eq. (23)

Proof

Since $d-2$ pure states $|\psi _{i}\rangle$ excluding two pure states $|\psi _{x}\rangle$,$|\psi _{y}\rangle$ are comprised by $d-2$ pure states $|e_{i}\rangle$ except two pure states $|e_{x}\rangle$,$|e_{y}\rangle$, the following relation holds:

$$\begin{aligned} \sum _{i,j\in \{x,y\}}\tilde{\rho }_{ij}|\psi _{i}\rangle \!\langle \psi _{j}| ={\Pi }\rho {\Pi }=\sum _{i,j\in \{x,y\}}{\rho }_{ij}|e_{i}\rangle \!\langle e_{j}|, \end{aligned}$$

(40)

where

$$\begin{aligned} {\Pi }=|e_{x}\rangle \!\langle e_{x}|+|e_{y}\rangle \!\langle e_{y}|. \end{aligned}$$

(41)

Using (22), $|e_{x}\rangle$ and $|e_{y}\rangle$ are expressed by $|\psi _{x}\rangle ,|\psi _{y}\rangle$ as follows:

$$\begin{aligned} |e_{x}\rangle =\frac{1}{(1+i)\beta _{x}}(|\psi _{x}\rangle +i|\psi _{y}\rangle ),\quad |e_{y}\rangle =\frac{1}{(1+i)\beta _{y}}(|\psi _{x}\rangle -|\psi _{y}\rangle ). \end{aligned}$$

(42)

Then, by (42) the right-hand side of (40) is expressed as follows.

$$\begin{aligned} {\Pi }\rho {\Pi }= & {} \frac{\rho _{xx}}{2\beta _{x}^{2}}(|\psi _{x}\rangle +i|\psi _{y}\rangle )(\langle \psi _{x}|-i\langle \psi _{y}|) +\frac{\rho _{yy}}{2\beta _{y}^{2}}(|\psi _{x}\rangle -|\psi _{y}\rangle )(\langle \psi _{x}|-\langle \psi _{y}|)\\&+\,\frac{\rho _{xy}}{2\beta _{x}\beta _{y}}(|\psi _{x}\rangle +i|\psi _{y}\rangle )(\langle \psi _{x}|-\langle \psi _{y}|)\\&+\frac{\rho _{yx}}{2\beta _{x}\beta _{y}}(|\psi _{x}\rangle -|\psi _{y}\rangle )(\langle \psi _{x}|-i\langle \psi _{y}|)\\= & {} \left[ \frac{\rho _{xx}}{2\beta _{x}^{2}}+\frac{\rho _{yy}}{2\beta _{y}^{2}}+\frac{\rho _{xy}+\rho _{yx}}{2\beta _{x}\beta _{y}}\right] |\psi _{x}\rangle \!\langle \psi _{x}|\\&+\left[ \frac{\rho _{xx}}{2\beta _{x}^{2}}+\frac{\rho _{yy}}{2\beta _{y}^{2}}-\frac{i\rho _{xy}-i\rho _{yx}}{2\beta _{x}\beta _{y}}\right] |\psi _{y}\rangle \!\langle \psi _{y}|\\&-\,\left[ \frac{i\rho _{xx}}{2\beta _{x}^{2}}+\frac{\rho _{yy}}{2\beta _{y}^{2}}+\frac{\rho _{xy}+i\rho _{yx}}{2\beta _{x}\beta _{y}}\right] |\psi _{x}\rangle \!\langle \psi _{y}|\\&+\left[ \frac{i\rho _{xx}}{2\beta _{x}^{2}}-\frac{\rho _{yy}}{2\beta _{y}^{2}}+\frac{i\rho _{xy}-\rho _{yx}}{2\beta _{x}\beta _{y}}\right] |\psi _{y}\rangle \!\langle \psi _{x}|. \end{aligned}$$

The above equation provides (23). $\square$