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Direct state measurements under state-preparation-and-measurement errors


Direct state measurement (DSM) is a tomography method that allows for retrieving quantum states’ wave functions directly. However, a shortcoming of current studies on the DSM is that it does not provide access to noisy quantum systems. Here, we attempt to fill the gap by investigating the DSM measurement precision that undergoes the state-preparation-and-measurement (SPAM) errors. We manipulate a quantum controlled measurement framework with various configurations and compare the efficiency between them. Under such SPAM errors, the state to be measured lightly deviates from the true state, and the measurement error in the postselection process results in less accurate in the tomography. Our study could provide a reliable tool for SPAM errors tomography and contribute to understanding and resolving an urgent demand for current quantum technologies.

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This work was supported by JSPS KAKENHI Grant Number 20F20021 and the Vietnam National University under Grant Number QG.20.17. LBH would like to thank Shikano for pointing out Ref. [30].

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An example of noisy quantum state preparation

We provide an example for preparing the quantum state GHZ\(_3\) that contains noise. Consider a quantum circuit as shown in Fig. 7 (inset). Therein, three qubits \(q_0, q_1\), and \(q_2\) are prepared in the ground state, i.e., \(|000\rangle \). Applying a sequence of Hadamard (H) gate onto \(q_0\), control-NOT (CNOT) gate onto \(q_0, q_1\), and control-NOT gate onto \(q_0, q_2\), as shown in the inset figure, respectively, we obtain the output state as

$$\begin{aligned} |000\rangle \xrightarrow []{H} \dfrac{1}{\sqrt{2}}(|000\rangle +|100\rangle ) \xrightarrow []{CNOT} \dfrac{1}{\sqrt{2}}(|000\rangle +|110\rangle ) \xrightarrow []{CNOT} \dfrac{1}{\sqrt{2}}(|000\rangle +|111\rangle ), \end{aligned}$$

which is the GHZ\(_3\) state. We simulate this state in Fig. 7 (left) by using the IBM Qiskis package. It can be seen that the amplitudes of \(|000\rangle \) and \(|111\rangle \) are the same and equal to \(1/\sqrt{2}\).

Fig. 7
figure 7

Inset: Quantum circuit for generating GHZ\(_3\) state. Initially, three qubits \(q_0, q_1\), and \(q_2\) are prepared in the ground state, i.e., \(|000\rangle \). Quantum Hadamard gate and Control-NOT gates are applied to transform the initial state to the desired state. See detailed in Appendix A. Left: the amplitudes of the components in the GHZ\(_3\) state after applying H and CNOT gates onto the initial state \(|000\rangle \). Right: the amplitudes of the components in the GHZ\(_3\) state under the imperfection of the Hadamard gate. These amplitudes are lightly deviated from the true values in the left figure

Now, let us assume the imperfection in the Hadamard gate as follows. We first decompose the Hadamard gate into the two rotations: \(\pi /2\) about the Y-axis, and \(\pi \) about the Z-axis, such that

$$\begin{aligned} -iH = \begin{pmatrix} \cos (\pi /4) &{} -\sin (\pi /4) \\ \sin (\pi /4)&{}\cos (\pi /4) \end{pmatrix} \begin{pmatrix} e^{-i\pi /2} &{} 0 \\ 0 &{} e^{i\pi /2} \end{pmatrix}, \end{aligned}$$

where the rotation matrices are given as

$$\begin{aligned} R_y(\theta ) = \begin{pmatrix} \cos (\theta /2) &{} -\sin (\theta /2) \\ \sin (\theta /2)&{}\cos (\theta /2) \end{pmatrix} \;, R_z(\theta ) = \begin{pmatrix} e^{-i\theta /2} &{} 0 \\ 0 &{} e^{i\theta /2} \end{pmatrix}. \end{aligned}$$

Under the imperfection, assume that the rotation angles will deviate from their true values, such as \(\pi /2 + \alpha \), and \(\pi + \beta \), where \(\alpha \) and \(\beta \) are small angles. Without loss of generality, we can choose \(\beta = 0\) since the operation of \(R_z(\theta )\) does not affect the amplitudes of the quantum state. As a result, the Hadamard gate becomes

$$\begin{aligned} -iH' = \dfrac{1}{\sqrt{2}} \begin{pmatrix} \cos (\frac{\alpha }{2})-\sin (\frac{\alpha }{2}) &{} \cos (\frac{\alpha }{2})+\sin (\frac{\alpha }{2})\\ \cos (\frac{\alpha }{2})+\sin (\frac{\alpha }{2})&{} -[\cos (\frac{\alpha }{2})-\sin (\frac{\alpha }{2})] \end{pmatrix} = \dfrac{1}{\sqrt{2}} \begin{pmatrix} a &{} b \\ b &{} -a \end{pmatrix}. \end{aligned}$$

Therefore, the final state becomes

$$\begin{aligned} \dfrac{1}{\sqrt{2}}(a|000\rangle +b|111\rangle ), \end{aligned}$$

which slightly different from the true GHZ\(_3\) state. Here, \(\frac{1}{2}(|a|^2 + |b|^2) = 1\). This is an example of a noisy state-preparation process. In Fig. 7 (right), we simulate the GHZ\(_3\) state assuming that the Hadamard gate is under the imperfection at \(\alpha = 0.2\) as an example. Under this noisy state-preparation, the two components \(|000\rangle \) and \(|111\rangle \) are deviated from \(1/\sqrt{2}\). In general, all the components will be deviated from their values as we have modeled from Eq. 3 in the main text.

Quantum controlled measurements for pure states

In this appendix, we closely follow the quantum controlled measurement framework introduced by Ogawa et al. [29] for pure states, and we derive it under the noise using our denotation to make the work self-consistency.

We consider the initial joint state of the target system and the control qubit probe as \(|\varPsi \rangle = |\psi '\rangle \otimes |\xi \rangle \), where

$$\begin{aligned}&|\psi '\rangle = \dfrac{1}{{\mathcal {N}}}\sum _{n=0}^{d-1} \Bigl (\psi _n + \delta _n\Bigr )|n\rangle = \sum _{n=0}^{d-1} \psi '_n |n\rangle , \text { and } \end{aligned}$$
$$\begin{aligned}&|\xi \rangle \equiv |+\rangle = \dfrac{1}{\sqrt{2}} \Bigl (|0\rangle + |1\rangle \Bigr ), \end{aligned}$$

where \(\mathcal {N}\) is the normalization factor, \(\psi '_n = \frac{\psi _n+\delta _n}{\mathcal {N}}\), and \(\delta _n = x_1 + ix_2\) is a complex random number, where \(x_1\) and \(x_2\) are random numbers that follow the normal distribution \(f(x) = \frac{1}{\sigma \sqrt{2\pi }} \exp [-\frac{1}{2}(\frac{x}{\sigma })^2]\).

For C1

For C1, following Ogawa et al. [29], we consider the interaction as

$$\begin{aligned} \varvec{U}_n = \bigl (\varvec{I}_\mathrm{s}-|n\rangle \langle n|\bigr ) \otimes |0\rangle \langle 0| + |n\rangle \langle n| \otimes |1\rangle \langle 1|. \end{aligned}$$

After the interaction, the joint state becomes

$$\begin{aligned} \varvec{U}_n|\varPsi \rangle = \dfrac{1}{\sqrt{2}} \Bigl [ \sum _{m=0}^{d-1}\psi '_{m}|m\rangle - \psi '_n|n\rangle \Bigr ]\otimes |0\rangle +\dfrac{1}{\sqrt{2}} \psi '_n|n\rangle \otimes |1\rangle . \end{aligned}$$

We postselect the target system onto the conjugate basis

$$\begin{aligned} |\mathfrak {c}'_0\rangle = \frac{1}{\mathcal {M}}\sum _{m=0}^{d-1}(1+\kappa _m)|m\rangle = \sum _{m=0}^{d-1} \mathfrak {c}_m |m\rangle , \end{aligned}$$

where \(\mathcal {M}\) is the normalization factor, and \(\mathfrak {c}_m = \frac{1+\kappa _m}{\mathcal {M}}\). Here, \(\kappa _m\) is a real random number and distributes according to the normal distribution. The final control qubit state is given by

$$\begin{aligned} |\eta \rangle&= \bigl (\langle \mathfrak {c}'_0| \otimes \varvec{I}_\mathrm{p}\bigr ) \varvec{U}_n|\varPsi \rangle \nonumber \\&=\dfrac{1}{\sqrt{2}}\Bigl [ \big (\varGamma - \mathfrak {c}_n\psi '_n\big )|0\rangle +\mathfrak {c}_n\psi '_n|1\rangle \Bigr ], \end{aligned}$$

where \(\varGamma = \sum _{m=0}^{d-1} \mathfrak {c}_{m}\psi '_{m}\).

Finally, we measure the control qubit probe in the Pauli basis \(|j\rangle \in \{ |0\rangle , |1\rangle , |+\rangle , |-\rangle , |L\rangle , |R\rangle \}\), where \(|\pm \rangle = \frac{1}{\sqrt{2}}\bigl (|0\rangle \pm |1\rangle \bigr ), |L\rangle = \frac{1}{\sqrt{2}}\bigl (|0\rangle + i|1\rangle \bigr ), |R\rangle = \frac{1}{\sqrt{2}}\bigl (|0\rangle - i|1\rangle \bigr )\). The probability for measuring \(|j\rangle \langle j|\) is \(P_j = |\langle j|\eta \rangle |^2\) explicitly give

$$\begin{aligned}&P_0=\frac{1}{2} \Bigl [\varGamma ^2 - 2\mathfrak {c}_n\varGamma \, \mathrm{Re}\psi '_n +\mathfrak {c}_n^2|\psi '_n|^2\Bigr ], \end{aligned}$$
$$\begin{aligned}&P_1=\frac{1}{2}\mathfrak {c}_n^2|\psi '_n|^2, \end{aligned}$$
$$\begin{aligned}&P_+=\frac{1}{4}\varGamma ^2, \end{aligned}$$
$$\begin{aligned}&P_-=\frac{1}{4} \Bigl [\varGamma ^2-4\varGamma \mathfrak {c}_n \, \mathrm{Re}\psi '_n +4 \mathfrak {c}_n^2|\psi '_n|^2 \Bigr ], \end{aligned}$$
$$\begin{aligned}&P_L=\frac{1}{4} \Bigl [\varGamma ^2-2\varGamma \mathfrak {c}_n \, \mathrm{Re}\psi '_n +2\varGamma \mathfrak {c}_n \, \mathrm{Im}\psi '_n + 2\mathfrak {c}_n^2|\psi '_n|^2 \Bigr ], \end{aligned}$$
$$\begin{aligned}&P_R=\frac{1}{4} \Bigl [\varGamma ^2-2\varGamma \mathfrak {c}_n \, \mathrm{Re}\psi '_n -2\varGamma \mathfrak {c}_n \, \mathrm{Im}\psi '_n + 2 \mathfrak {c}_n^2 |\psi '_n|^2 \Bigr ]. \end{aligned}$$

As a result, the real and imaginary parts of the amplitude \(\psi '_n\) are reproduced as

$$\begin{aligned} \mathrm{Re}\psi '_n=\frac{P_+-P_-+2P_1}{\mathfrak {c}_n\varGamma }, \ \text{ and } \mathrm{Im}\psi '_n=\frac{P_L-P_R}{\mathfrak {c}_n\varGamma }. \end{aligned}$$

For C2

For C2, the interaction is given by [29]

$$\begin{aligned} \varvec{U}=(\varvec{I}_s-|\mathfrak {c}'_0\rangle \langle \mathfrak {c}'_0|)\otimes |0\rangle \langle 0| +|\mathfrak {c}'_0\rangle \langle \mathfrak {c}'_0|\otimes |1\rangle \langle 1|. \end{aligned}$$

The joint state after the interaction is given by

$$\begin{aligned} \varvec{U}|\varPsi \rangle = \frac{1}{\sqrt{2}}\Bigl [\sum _{m=0}^{d-1}\psi '_{m}|m\rangle -\sum _{m=0}^{d-1}\psi '_{m} \mathfrak {c}_{m}|\mathfrak {c}'_0\rangle \Bigr ] \otimes |0\rangle +\frac{1}{\sqrt{2}}\Bigl [\sum _{m=0}^{d-1}\psi '_{m} \mathfrak {c}_{m}|\mathfrak {c}'_0\rangle \Bigr ] \otimes |1\rangle . \end{aligned}$$

After the interaction, the target system is postselected onto \(|n\rangle \) while the remaining state of the control qubit probe is given as

$$\begin{aligned} |\eta \rangle =\dfrac{1}{\sqrt{2}}\Bigl [\big ( \psi '_n-\mathfrak {c}_n\varGamma \big )|0\rangle + \mathfrak {c}_n\varGamma |1\rangle \Big ]. \end{aligned}$$

Measuring the control qubit probe in the Pauli basis as above, we obtain

$$\begin{aligned} P_0&= \dfrac{1}{2}\Bigl [|\psi '_n|^2- 2\mathfrak {c}_n\varGamma \mathrm{Re}\psi '_n +\mathfrak {c}_n^2\varGamma ^2\Bigr ], \end{aligned}$$
$$\begin{aligned} P_1&= \dfrac{1}{2}\mathfrak {c}_n^2\varGamma ^2, \end{aligned}$$
$$\begin{aligned} P_+&=\frac{1}{4}|\psi '_n|^2, \end{aligned}$$
$$\begin{aligned} P_-&= \dfrac{1}{4}\Bigl [|\psi '_n|^2-4 \mathfrak {c}_n\varGamma \mathrm{Re}\psi '_n +4 \mathfrak {c}_n^2\varGamma ^2\Bigr ], \end{aligned}$$
$$\begin{aligned} P_L&= \dfrac{1}{4}\Bigl [|\psi '_n|^2-2 \mathfrak {c}_n\varGamma \mathrm{Re}\psi '_n +2\mathfrak {c}_n\varGamma \mathrm{Im}\psi '_n +2\mathfrak {c}_n^2\varGamma ^2\Bigr ], \end{aligned}$$
$$\begin{aligned} P_R&= \dfrac{1}{4}\Bigl [|\psi '_n|^2- 2\mathfrak {c}_n\varGamma \mathrm{Re}\psi '_n -2\mathfrak {c}_n\varGamma \mathrm{Im}\psi '_n +2\mathfrak {c}_n^2\varGamma ^2\Bigr ]. \end{aligned}$$

Then, we have

$$\begin{aligned} \mathrm{Re}\psi '_n=\frac{P_+-P_-+2P_1}{\mathfrak {c}_n\varGamma },\ \text {and } \mathrm{Im}\psi '_n=\frac{P_L-P_R}{\mathfrak {c}_n\varGamma }. \end{aligned}$$

Quantum controlled measurements for mixed states

We consider the joint state \(\varLambda \) as following

$$\begin{aligned} \varLambda = \rho '_0 \otimes |+\rangle \langle +|, \quad \text {with}\; \rho '_0 = \sum _{n,m=0}^{d-1}\rho '_{nm}|n\rangle \langle m|. \end{aligned}$$

For C1

The interaction operator is given the same as above:

$$\begin{aligned} \varvec{U}_n = \big (\varvec{I}_\mathrm{s} - |n\rangle \langle n|\big ) \otimes |0\rangle \langle 0 |+ |n\rangle \langle n|\otimes |1\rangle \langle 1|. \end{aligned}$$

After the interaction, the joint state evolves to

$$\begin{aligned} \varLambda ' = \varvec{U}_n\varLambda \varvec{U}_n^{\dag }\, , \end{aligned}$$

which is explicitly written as

$$\begin{aligned} \varLambda ' =&\Big [\rho '_0 -\Bigl (\sum _{m=0}^{d-1}\rho '_{nm}|n\rangle \langle m|+c.c\Bigr )+\rho '_{nn}|n\rangle \langle n|\Big ]\otimes \frac{1}{2}|0\rangle \langle 0|\nonumber \\&+ \Big [\sum _{m=0}^{d-1}\rho '_{mn}|m\rangle \langle n|-\rho '_{nn}|n\rangle \langle n|\Big ]\otimes \frac{1}{2}|0\rangle \langle 1|\nonumber \\&+ \Big [\sum _{m=0}^{d-1}\rho '_{nm}|n\rangle \langle m|-\rho '_{nn}|n\rangle \langle n|\Big ]\otimes \frac{1}{2}|1\rangle \langle 0|\nonumber \\&+ \Big [\rho '_{nn}|n\rangle \langle n|\Big ]\otimes \frac{1}{2}|1\rangle \langle 1|. \end{aligned}$$

Here, c.c stands for ‘complex conjugate.’ After postselecting this state onto

$$\begin{aligned} |\mathfrak {c}'_k\rangle \langle \mathfrak {c}'_k|&= \frac{1}{\mathcal {M}^2}\sum _{n,m=0}^{d-1} e^{i2\pi (m-n)k/d} (1+\kappa _{m})(1+\kappa _{n})|m\rangle \langle n|\nonumber \\&=\sum _{n,m = 0}^{d-1} e^{i2\pi (m-n)k/d} \mathfrak {c}_m\mathfrak {c}_n |m\rangle \langle n|\;, \end{aligned}$$

the final state of the control qubit probe becomes

$$\begin{aligned} \varLambda '' = \langle \mathfrak {c}'_k| \varLambda '|\mathfrak {c}'_k\rangle = \begin{pmatrix} \varLambda _{00}''(n,k) &{} \varLambda _{01}''(n,k) \\ \varLambda _{10}''(n,k) &{} \varLambda _{11}''(n,k) \end{pmatrix}. \end{aligned}$$


$$\begin{aligned} \varLambda _{00}''(n,k)&= \frac{1}{2}\Big [\sum _{n,m=0}^{d-1} \rho '_{nm}e^{i2\pi (m-n)k/d} \mathfrak {c}_{m} \mathfrak {c}_{n} -\Big (\sum _{m=0}^{d-1} \rho '_{nm}e^{i2\pi (m-n)k/d}\mathfrak {c}_{n} \mathfrak {c}_{m}+c.c\Big ) +\rho '_{nn} \mathfrak {c}_n^2 \Big ] \end{aligned}$$
$$\begin{aligned} \varLambda _{01}''(n,k)&= \frac{1}{2}\Big [ \sum _{m=0}^{d-1}\rho '_{mn}e^{i2\pi (n-m)k/d} \mathfrak {c}_n \mathfrak {c}_{m}-\rho '_{nn} \mathfrak {c}_n^2 \Big ] \end{aligned}$$
$$\begin{aligned} \varLambda _{10}''(n,k)&= [\varLambda _{01}''(n,k)]^* \end{aligned}$$
$$\begin{aligned} \varLambda _{11}''(n,k)&= \frac{1}{2}\rho '_{nn} \mathfrak {c}_n^2. \end{aligned}$$

Using Fourier transformation on \(\varLambda ''_{10}(n,k)\), we obtain

$$\begin{aligned} \rho '_{nm} \varpropto \frac{1}{\mathfrak {c}_{n}\mathfrak {c}_{m}} \Big [d\delta _{n,m} \varLambda ''_{11}(n,k)+\sum _{k=0}^{d-1}e^{i2\pi (n-m)k/d} \varLambda _{10}''(n,k) \Big ]. \end{aligned}$$

To get \(\varLambda _{10}''(n,k)\) and \(\varLambda ''_{11}(n,k)\), the control qubit is measured as follows:

$$\begin{aligned}&\varLambda ''_{10}(n,k)= & {} \frac{1}{2}\Big [(P_+ - P_-) + i(P_L-P_R) \Big ], \, \text {and} \end{aligned}$$
$$\begin{aligned}&\varLambda ''_{11}(n,k)= & {} P_1, \end{aligned}$$

where \(P_j = \mathrm{Tr}[|j\rangle \langle j|\varLambda '']\) is the probability when measuring the control qubit probe in the element j of the Pauli basis.

For C2

In this case, the interaction is

$$\begin{aligned} \varvec{U} = \big ({\varvec{I}}_\mathrm{s} - |\mathfrak {c}'_k\rangle \langle \mathfrak {c}'_k|\big ) \otimes |0\rangle \langle 0 |+ |\mathfrak {c}'_k\rangle \langle \mathfrak {c}'_k|\otimes |1\rangle \langle 1|. \end{aligned}$$

After applying this interaction U, the initial joint state becomes

$$\begin{aligned} \varLambda ' = \varvec{U}\varLambda \varvec{U}^{\dag }, \end{aligned}$$

which is explicitly given as

$$\begin{aligned} \varLambda '&= \Big [\rho '_{0} - \sum _{n,m=0}^{d-1} \rho '_{nm}\mathfrak {c}_{m} e^{i2\pi km/d}|n\rangle \langle \mathfrak {c}'_k | - |\mathfrak {c}'_k\rangle \sum _{n,m=0}^{d-1} \rho '_{nm}\mathfrak {c}_{n} e^{-i2\pi kn/d}\langle m| \nonumber \\&\quad + \sum _{n,m=0}^{d-1} \rho '_{nm}\mathfrak {c}_{n}\mathfrak {c}_{m} e^{i2\pi k(m-n)/d} |\mathfrak {c}'_k\rangle \langle \mathfrak {c}'_k| \Big ]\otimes \frac{1}{2}|0\rangle \langle 0|\nonumber \\&\quad + \Big [ \sum _{n,m=0}^{d-1} \rho '_{nm}\mathfrak {c}_{m} e^{i2\pi km/d}|n\rangle \langle \mathfrak {c}'_k| - \sum _{n,m=0}^{d-1} \rho '_{nm}\mathfrak {c}_{n}\mathfrak {c}_{m} e^{i2\pi k(m-n)/d} |\mathfrak {c}'_k\rangle \langle \mathfrak {c}'_k| \Big ] \otimes \frac{1}{2}|0\rangle \langle 1|\nonumber \\&\quad + \Big [|\mathfrak {c}'_k\rangle \sum _{n,m=0}^{d-1} \rho '_{nm}\mathfrak {c}_{n} e^{-i2\pi kn/d}\langle m| - \sum _{n,m=0}^{d-1} \rho '_{nm}\mathfrak {c}_{n}\mathfrak {c}_{m} e^{i2\pi k(m-n)/d} |\mathfrak {c}'_k\rangle \langle \mathfrak {c}'_k| \Big ]\otimes \frac{1}{2}|1\rangle \langle 0|\nonumber \\&\quad + \Big [\sum _{n,m=0}^{d-1} \rho '_{nm}\mathfrak {c}_{n}\mathfrak {c}_{m} e^{i2\pi k(m-n)/d} |\mathfrak {c}'_k\rangle \langle \mathfrak {c}'_k| \Big ]\otimes \frac{1}{2}|1\rangle \langle 1|. \end{aligned}$$

Next, we postselect this state onto \(|n\rangle \langle n|\) and get

$$\begin{aligned} \varLambda '' = \langle n|\varLambda '|n\rangle = \begin{pmatrix} \varLambda _{00}''(n,k) &{} \varLambda _{01}''(n,k) \\ \varLambda _{10}''(n,k)&{} \varLambda _{11}''(n,k) \end{pmatrix}. \end{aligned}$$


$$\begin{aligned} \varLambda _{00}''(n,k)&= \frac{1}{2}\Big [ \rho '_{nn} - \Big (\sum _{m=0}^{d-1} \rho '_{nm}\mathfrak {c}_{m}\mathfrak {c}_{n} e^{i2\pi k(m-n)/d} + c.c.\Big ) + \Big (\sum _{n,m=0}^{d-1} \rho '_{nm}\mathfrak {c}_{n}\mathfrak {c}_{m} e^{i2\pi k(m-n)/d} \Big )\mathfrak {c}_{n}^{2}\Big ], \end{aligned}$$
$$\begin{aligned} \varLambda _{01}''(n,k)&=\frac{1}{2}\Big [ \Big (\sum _{m=0}^{d-1} \rho '_{nm}\mathfrak {c}_{m}\mathfrak {c}_{n} e^{i2\pi k(m-n)/d} \Big ) - \Big (\sum _{n,m=0}^{d-1} \rho '_{nm}\mathfrak {c}_{n}\mathfrak {c}_{m}e^{i2\pi k(m-n)/d} \Big )\mathfrak {c}_{n}^{2}\Big ], \end{aligned}$$
$$\begin{aligned} \varLambda _{10}''(n,k)&= [\varLambda _{01}''(n,k)]^*, \end{aligned}$$
$$\begin{aligned} \varLambda _{11}''(n,k)&= \frac{1}{2} \Big [\sum _{n,m=0}^{d-1} \rho '_{nm}\mathfrak {c}_{n}\mathfrak {c}_{m}e^{i2\pi k(m-n)/d} \Big ]\mathfrak {c}_{n}^{2}. \end{aligned}$$

Using Fourier transformation on \(\varLambda _{01}''(n,k)\), we obtain:

$$\begin{aligned} \rho '_{nm} \varpropto \frac{1}{\mathfrak {c}_{m}\mathfrak {c}_{n}} \Big [ \sum _{k=0}^{d-1} e^{i2\pi k(n-m)/d} \Big (\varLambda ''_{01}(n,k) + \varLambda ''_{11}(n,k) \Big ) \Big ], \end{aligned}$$

where \(\varLambda _{01}''(n,k)\) is obtained by measuring the control qubit probe as follows:

$$\begin{aligned} \varLambda ''_{01}(n,k) = \frac{1}{2}\Big [(P_+ - P_-) - i(P_L-P_R) \Big ]. \end{aligned}$$

Quantum fisher information

In this section, we show how to calculate the total quantum Fisher information (QFI) for \(|\psi '\rangle \) state:

$$\begin{aligned} |\psi '\rangle = \dfrac{1}{\mathcal {N}} \sum _{n=0}^{d-1} \bigl (\psi _n + \delta _n\bigr ) |n\rangle , \end{aligned}$$

where \(\psi _n\) is unknown. The normalization constant is

$$\begin{aligned} \mathcal {N}^2 = \sum _{n=0}^{d-1} \Bigl (\psi _n + \delta _n\Bigr )^2. \end{aligned}$$

Here, note that we consider both \(\psi _n\) and \(\delta _n\) are real for simplicity. First, we calculate \(\partial _{\psi _n}|\psi '\rangle \), where we are using \(\partial _{\psi _n}\) as shorthand for \(\partial /\partial {\psi _n}\). We have

$$\begin{aligned} \partial _{\psi _n}|\psi '\rangle&= \dfrac{\partial }{\partial {\psi _n}} \Bigl (\dfrac{1}{{\mathcal {N}}}\Bigr ) \sum _{n=0}^{d-1} \bigl (\psi _n + \delta _n\bigr ) |n\rangle + \dfrac{1}{{\mathcal {N}}} \dfrac{\partial }{\partial {\psi _n}} \Bigl (\sum _{n=0}^{d-1} \bigl (\psi _n + \delta _n\bigr ) |n\rangle \Bigr )\nonumber \\&= -\dfrac{\psi _n+\delta _n}{{\mathcal {N}}^2} |\psi '\rangle + \dfrac{1}{{\mathcal {N}}} |n\rangle . \end{aligned}$$

The Quantum Fisher Information (QFI) is given by

$$\begin{aligned} Q'_n&= 4 \Bigl [\dfrac{\langle \partial \psi '|}{\partial _{\psi _n}} \dfrac{|\partial \psi '\rangle }{\partial _{\psi _n}} -\Bigl |\dfrac{\langle \partial \psi '|}{\partial _{\psi _n}}|\psi '\rangle \Bigr |^2\Bigr ],\nonumber \\&= \dfrac{4}{{\mathcal {N}}^2} \Bigl [ 1 - \dfrac{(\psi _n+\delta _n)^2}{{\mathcal {N}}^2} \Bigr ]. \end{aligned}$$

Then, the total QFI is

$$\begin{aligned} Q' = \sum _{n=0}^{d-1}Q'_n = \dfrac{4}{{\mathcal {N}}^2} (d-1). \end{aligned}$$

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Tuan, K.Q., Nguyen, H.Q. & Ho, L.B. Direct state measurements under state-preparation-and-measurement errors. Quantum Inf Process 20, 197 (2021).

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  • Quantum state tomography
  • Direct state measurement
  • Quantum controlled measurement