A Framework For Estimating Amplitudes of Quantum State With Single-Qubit Measurement

We begin by discussing two examples: $\ket{\psi}_{1}\in\mathbb{C}^{4}$ being the quantum state of 2-qubits and $\ket{\psi}_{2}\in\mathbb{C}^{8}$ being the quantum state of 3-qubits.

Let $\ket{\psi}_{1}=a_{00}\ket{00}+a_{01}\ket{01}+a_{10}\ket{10}+a_{11}\ket{11}$ be the decomposition of $\ket{\psi}_{1}$ in the computational basis state. If we measure two qubits, then it is fundamental, that is, Born’s rule, that the probability of measuring string $(i,j)$ ($i,j=0,1$) is $p_{ij}=\operatorname{Tr}(\ket{ij}\bra{ij}\cdot\ket{\psi}_{1}\bra{\psi}_{1})=|a_{ij}|^{2}$. If we want to estimate all of this probability $p_{ij}$, then we need to repeat the measurement multiple times and use the results to statistically estimate all probabilities $p_{ij}$.

To see how the outcomes can be used for estimating all probabilities $p_{ij}$ from measuring $\ket{\psi}_{1}$ in computational basis, we can first consider the estimation of $p_{00}$. Then the measurement outcomes form a Bernoulli distribution for $00$, as we simply treat all the outcomes $\{01,10,11\}$ as a discrete variable with corresponding probability $1-p_{00}$. It is known (via Chernoff bound) that the samples required to estimate $p_{00}$ to accurary $\epsilon$ with probability at least $1-\delta$ is $\mathcal{O}(\ln(1/\delta)/\epsilon^{2})$. The same strategy can be applied to estimate $p_{01},p_{10},p_{11}$ by treating each of them and the remaining probabilities as Bernoulli distribution. In the end, the whole procedure takes $\mathcal{O}(4\ln(1/\delta)/\epsilon^{2})$ repetitions to estimate all probabilities to additive accuracy $\epsilon$, with high probability.

The same result is applied for 3-qubits case. Let $\ket{\psi}_{2}=a_{000}\ket{000}+a_{001}\ket{001}+a_{010}\ket{010}+a_{011}\ket{011}+a_{100}\ket{100}+a_{101}\ket{101}+a_{110}\ket{110}+a_{111}\ket{111}$. Then repeating the measurements in computational basis state allows us to estimate each probability $p_{ijk}=|a_{ijk}|^{2}$ (for $i,j,k=0,1$) to error at most $\epsilon$ with success probability at least $1-\delta$, with $\mathcal{O}(8\ln(1/\delta)/\epsilon^{2})$ repetitions according to the above lemma. It is straightforward to see that even in $n$-qubits case, the procedure is the same for estimating all amplitudes, hence the sample complexity is $\mathcal{O}(2^{n}\ln(1/\delta)/\epsilon^{2})$ to achieve the desired precision.

Now for $\ket{\psi}_{1}$, instead of measuring all qubits in computational basis state (with four possible outcomes $00,01,10,11$), we measure the first qubit in computational basis state. Then there will be two possible outcomes, which are 0 and 1 (or more like $\ket{0}$ or $\ket{1}$ in the first qubit), with corresponding probability $p_{0}$ and $p_{1}$:

	$\displaystyle p_{0}=\operatorname{Tr}(\ket{0}\bra{0}\otimes\mathbb{I}_{2}\cdot\ket{\psi}_{1}\bra{\psi}_{1})$		(1)
	$\displaystyle p_{1}=\operatorname{Tr}(\ket{1}\bra{1}\otimes\mathbb{I}_{2}\cdot\ket{\psi}_{1}\bra{\psi}_{1})$		(2)

where $\mathbb{I}_{2}$ refers to the identity matrix on the second qubit system. To avoid the confusion, as subsequently we also have $p_{0}$ and $p_{1}$ but those probabilities are for the measurement’s outcome on the second qubit, we use the superscript to denote the qubit explicitly: $p_{0}^{1(or2)}$ and $p_{1}^{1(or2)}$ refers to the probability of obtaining $0$ and $1$ on the first, or second qubit respectively. We remark the following crucial information, which is essentially a marginal probability property:

	$\displaystyle p_{00}+p_{01}=p_{0}^{1}$		(3)
	$\displaystyle p_{10}+p_{11}=p_{1}^{1}$		(4)

where $p_{ij}=|a_{ij}|^{2}$ is the usual squared amplitude of state $\ket{\psi}_{1}$. Likewise, if we measure the second qubit, there would also be two possible outcomes 0 and 1, with probability $p_{0}^{2}$ and $p_{1}^{2}$. Then we also have that, for second qubit:

	$\displaystyle p_{00}+p_{10}=p_{0}^{2}$		(5)
	$\displaystyle p_{01}+p_{11}=p_{1}^{2}$		(6)

It is straightforward to see that the above equation form a linear system:

$\displaystyle\begin{pmatrix}1&1&0&0\\ 0&0&1&1\\ 1&0&1&0\\ 0&1&0&1\end{pmatrix}\begin{pmatrix}p_{00}\\ p_{01}\\ p_{10}\\ p_{11}\end{pmatrix}=\begin{pmatrix}p_{0}^{1}\\ p_{1}^{1}\\ p_{0}^{2}\\ p_{1}^{2}\end{pmatrix}$

(7)

As we have discussed, $\mathcal{O}(\ln(1/\delta)/\epsilon^{2})$ repetition is required to estimate $p_{0}^{1},p_{0}^{2}$ and $p_{1}^{1},p_{1}^{2}$ to the precision $\epsilon$. This means that the right-hand side of the above linear system is erroneous. Therefore, if we wish to solve the linear system above to obtain the probability $p_{00},p_{01},p_{10},p_{11}$, then the solution appears to be erroneous, which we will discuss in more detail later.

Now we examine the same procedure as above for 3-qubits cases. We measure each qubit of the state $\ket{\psi}_{2}$ $\mathcal{O}(1/\epsilon^{2})$ times, and obtain the outcomes $p_{0}^{1},p_{1}^{1},p_{0}^{2},p_{1}^{2},p_{0}^{3},p_{1}^{3}$. Recall that

$\displaystyle\ket{\psi}_{2}$

$\displaystyle=a_{000}\ket{000}+a_{001}\ket{001}+a_{010}\ket{010}+a_{011}\ket{011}+a_{100}\ket{100}+a_{101}\ket{101}+a_{110}\ket{110}+a_{111}\ket{111}$

(8)

where each amplitude squared $|a_{ijk}|^{2}=p_{ijk}$ is the probability of obtaining corresponding outcome $(i,j,k)$. Then due to marginal probability property, we have that:

	$\displaystyle p_{000}+p_{001}+p_{010}+p_{011}=p_{0}^{1}$		(9)
	$\displaystyle p_{100}+p_{101}+p_{110}+p_{111}=p_{1}^{1}$		(10)
	$\displaystyle p_{000}+p_{001}+p_{100}+p_{101}=p_{0}^{2}$		(11)
	$\displaystyle p_{010}+p_{011}+p_{110}+p_{111}=p_{1}^{2}$		(12)
	$\displaystyle p_{000}+p_{010}+p_{100}+p_{110}=p_{0}^{3}$		(13)
	$\displaystyle p_{001}+p_{011}+p_{101}+p_{111}=p_{1}^{3}$		(14)

The above equations can be written as a linear system:

$\displaystyle\begin{pmatrix}1&1&1&1&0&0&0&0\\ 0&0&0&0&1&1&1&1\\ 1&1&0&0&1&1&0&0\\ 0&0&1&1&0&0&1&1\\ 1&0&1&0&1&0&1&0\\ 0&1&0&1&0&1&0&1\end{pmatrix}\begin{pmatrix}p_{000}\\ p_{001}\\ p_{010}\\ p_{011}\\ p_{100}\\ p_{101}\\ p_{110}\\ p_{111}\end{pmatrix}=\begin{pmatrix}p_{0}^{1}\\ p_{1}^{1}\\ p_{0}^{2}\\ p_{1}^{2}\\ p_{0}^{3}\\ p_{1}^{3}\end{pmatrix}$

(15)

The above system is rectangular, rather than square system comparing to Eqn. 7, which makes it a bit trickier to solve, as there are more variables than the number of equations. In the context of linear algebra, it is referred as underdetermined systems, and unique solution might not exists, rather, there can be infinitely many solutions. To make it square, we need to have more equations. However, the used procedure employs a single qubit measurement only, and we already measure all individual qubit. In order to collect more information as to obtain more equations, we need to have more measurement outcomes.

There are two possible ways to solve the aforementioned challenge. The first one is, we can perform joint measurement on $2$ qubits. Suppose we perform measurement on the first two qubits of $\ket{\psi}_{2}$, then there are four possible outcomes: 00,01,10,11. Let $p_{ij}^{1,2}$ with $i,j=0,1$ denote the probability of measuring $(i,j)$ on the qubit 1 and 2 of $\ket{\psi}_{2}$. Then because of the decomposition of $\ket{\psi}_{2}=a_{000}\ket{000}+a_{001}\ket{001}+a_{010}\ket{010}+a_{011}\ket{011}+a_{100}\ket{100}+a_{101}\ket{101}+a_{110}\ket{110}+a_{111}\ket{111}$, and $|a_{ijk}|^{2}=p_{ijk}$ is the corresponding probability of measuring outcome $(i,j,k)$. Using marginal probability property as in previous 2-qubit case, we have that:

	$\displaystyle p_{000}+p_{001}=p_{00}^{1,2}$		(16)
	$\displaystyle p_{010}+p_{011}=p_{01}^{1,2}$		(17)

Note that there are two other outcomes $(1,0)$ and $(1,1)$, however, we don’t need to use them, as the above equations, once combined with equations Eqn.14, yields:

	$\displaystyle p_{000}+p_{001}+p_{010}+p_{011}=p_{0}^{1}$		(18)
	$\displaystyle p_{100}+p_{101}+p_{110}+p_{111}=p_{1}^{1}$		(19)
	$\displaystyle p_{000}+p_{001}+p_{100}+p_{101}=p_{0}^{2}$		(20)
	$\displaystyle p_{010}+p_{011}+p_{110}+p_{111}=p_{1}^{2}$		(21)
	$\displaystyle p_{000}+p_{010}+p_{100}+p_{110}=p_{0}^{3}$		(22)
	$\displaystyle p_{001}+p_{011}+p_{101}+p_{111}=p_{1}^{3}$		(23)
	$\displaystyle p_{000}+p_{001}=p_{00}^{1,2}$		(24)
	$\displaystyle p_{010}+p_{011}=p_{01}^{1,2}$		(25)

which is a square linear system:

$\displaystyle\begin{pmatrix}1&1&1&1&0&0&0&0\\ 0&0&0&0&1&1&1&1\\ 1&1&0&0&1&1&0&0\\ 0&0&1&1&0&0&1&1\\ 1&0&1&0&1&0&1&0\\ 0&1&0&1&0&1&0&1\\ 1&1&0&0&0&0&0&0\\ 0&0&1&1&0&0&0&0\end{pmatrix}\begin{pmatrix}p_{000}\\ p_{001}\\ p_{010}\\ p_{011}\\ p_{100}\\ p_{101}\\ p_{110}\\ p_{111}\end{pmatrix}=\begin{pmatrix}p_{0}^{1}\\ p_{1}^{1}\\ p_{0}^{2}\\ p_{1}^{2}\\ p_{0}^{3}\\ p_{1}^{3}\\ p_{00}^{1,2}\\ p_{01}^{1,2}\end{pmatrix}$

(26)

In principle, it is sufficient from the above square system to solve for desired probability from the two qubits measurement outcomes $(0,0)$ and $(0,1)$, in addition to single qubit measurements.

While the above described joint qubit measurement seems efficient enough to do, however, the whole procedure gives us a linear system with probabilities, or amplitude square, as variables. We remark that, all amplitudes are complex number, so for this 3-qubit case, we have $2^{3}$ amplitudes, corresponding with $2^{3+1}$ complex variables. Therefore, we can only find the amplitude square, meanwhile, amplitude could be complex and the above linear system isn’t giving us enough information to find the amplitude, including its real and complex part. Performing more joint 2-qubit measurements, in principle, can give us more equations. However, to our surprise, solution based on only single qubit measurement exists. Remind that, in order to obtain the equation Eqn. 15, we performed single qubit measurement on $Z$ basis. In fact, we can also perform measurement on arbitrary basis, and it should be able to give us more information. To proceed, let $\theta$ denotes the angle for another basis measurement, and $\ket{\theta}_{0/1}$ be the eigenstate of corresponding measurement operator. Then the probability of obtaining $\ket{\theta}_{0}$ or $\ket{\theta}_{1}$ is:

$\displaystyle p(\theta)_{0/1}=\operatorname{Tr}(\ket{\theta}_{0/1}\bra{\theta}_{0/1}\otimes\mathbb{I}_{n-1}\cdot\ket{\psi}_{2}\bra{\psi}_{2})$

(27)

where $\mathbb{I}_{4}$ refers to identity matrix in the remaining $2$ qubits system. Let

	$\displaystyle\ket{\theta}_{0}=\cos(\theta)\ket{0}+\sin(\theta)\ket{1}$		(28)
	$\displaystyle\ket{\theta}_{1}=\sin(\theta)\ket{0}-\cos(\theta)\ket{1}$		(29)

be the decomposition of $\ket{\theta}_{0/1}$ in regular computational basis. From the above equation, it is easy to rewrite:

	$\displaystyle\ket{0}=(\cos(\theta)\ket{\theta}_{0}+\sin(\theta)\ket{\theta}_{1})$		(30)
	$\displaystyle\ket{1}=(\sin(\theta)\ket{\theta}_{0}-\cos(\theta)\ket{\theta}_{1})$		(31)

Recall that

	$\displaystyle\ket{\psi}_{2}$	$\displaystyle=a_{000}\ket{000}+a_{001}\ket{001}+a_{010}\ket{010}+a_{011}\ket{011}+a_{100}\ket{100}+a_{101}\ket{101}+a_{110}\ket{110}+a_{111}\ket{111}$		(32)
		$\displaystyle=\ket{0}\big{(}a_{000}\ket{00}+a_{001}\ket{01}+a_{010}\ket{10}+a_{011}\ket{11}\big{)}+\ket{1}\big{(}a_{100}\ket{00}+a_{101}\ket{01}+a_{110}\ket{10}+a_{111}\ket{11}\big{)}$		(33)
		$\displaystyle=(\cos(\theta)\ket{\theta}_{0}+\sin(\theta)\ket{\theta}_{1})\big{(}a_{000}\ket{00}+a_{001}\ket{01}+a_{010}\ket{10}+a_{11}\ket{11}\big{)}+$		(34)
		$\displaystyle(\sin(\theta)\ket{\theta}_{0}-\cos(\theta)\ket{\theta}_{1})\big{(}a_{100}\ket{00}+a_{101}\ket{01}+a_{110}\ket{10}+a_{111}\ket{11}\big{)}$		(35)

Let $\ket{\phi}_{1}=a_{000}\ket{00}+a_{001}\ket{01}+a_{010}\ket{10}+a_{011}\ket{11}$ and $\ket{\phi}_{2}=a_{100}\ket{00}+a_{101}\ket{01}+a_{110}\ket{10}+a_{111}\ket{11}$. Then we have:

$\displaystyle\ket{\psi}_{2}$

$\displaystyle=\ket{\theta}_{0}\big{(}\cos(\theta)\ket{\phi}_{1}+\sin(\theta)\ket{\phi}_{2}\big{)}+\ket{\theta}_{1}\big{(}\cos(\theta)\ket{\phi}_{1}-\sin(\theta)\ket{\phi}_{2}\big{)}$

(36)

If we measure the first qubit, then the probability of measuring $\ket{\theta}_{0},\ket{\theta}_{1}$ is:

	$\displaystyle p(\theta)_{0}=|\cos(\theta)\ket{\phi}_{1}+\sin(\theta)\ket{\phi}_{2}|^{2}$		(37)
	$\displaystyle p(\theta)_{1}=|\cos(\theta)\ket{\phi}_{1}-\sin(\theta)\ket{\phi}_{2}|^{2}$		(38)

We have that:

	$\displaystyle\cos(\theta)\ket{\phi}_{1}+\sin(\theta)\ket{\phi}_{2}=$	$\displaystyle(\cos(\theta)a_{000}+\sin(\theta)a_{100})\ket{00}+$		(39)
		$\displaystyle(\cos(\theta)a_{001}+\sin(\theta)a_{101})\ket{01}+$		(40)
		$\displaystyle(\cos(\theta)a_{010}+\sin(\theta)a_{110})\ket{10}+$		(41)
		$\displaystyle(\cos(\theta)a_{011}+\sin(\theta)a_{111})\ket{11}$		(42)

Then:

	$\displaystyle|\cos(\theta)\ket{\phi}_{1}+\sin(\theta)\ket{\phi}_{2}|^{2}$	$\displaystyle=|\cos(\theta)a_{000}+\sin(\theta)a_{100}|^{2}+$		(43)
		$\displaystyle|\cos(\theta)a_{001}+\sin(\theta)a_{101}|^{2}+$		(44)
		$\displaystyle\cos(\theta)a_{010}+\sin(\theta)a_{110}|^{2}+$		(45)
		$\displaystyle|\cos(\theta)a_{011}+\sin(\theta)a_{111}|^{2}$		(46)

Likewise:

	$\displaystyle|\cos(\theta)\ket{\phi}_{1}-\sin(\theta)\ket{\phi}_{2}|^{2}$	$\displaystyle=|\cos(\theta)a_{000}-\sin(\theta)a_{100}|^{2}+$		(48)
		$\displaystyle|\cos(\theta)a_{001}-\sin(\theta)a_{101}|^{2}+$		(49)
		$\displaystyle|\cos(\theta)a_{010}-\sin(\theta)a_{110}|^{2}+$		(50)
		$\displaystyle|\cos(\theta)a_{011}-\sin(\theta)a_{111}|^{2}$		(51)

Recall that from equation Eqn. 14:

	$\displaystyle p_{000}+p_{001}+p_{010}+p_{011}=p_{0}^{1}$		(53)
	$\displaystyle p_{100}+p_{101}+p_{110}+p_{111}=p_{1}^{1}$		(54)
	$\displaystyle p_{000}+p_{001}+p_{100}+p_{101}=p_{0}^{2}$		(55)
	$\displaystyle p_{010}+p_{011}+p_{110}+p_{111}=p_{1}^{2}$		(56)
	$\displaystyle p_{000}+p_{010}+p_{100}+p_{110}=p_{0}^{3}$		(57)
	$\displaystyle p_{001}+p_{011}+p_{101}+p_{111}=p_{1}^{3}$		(58)

if we replace $p_{ijk}=|a_{ijk}|^{2}$ for $i,j,k=0,1$, we equivalently have:

	$\displaystyle|a_{000}|^{2}+|a_{001}|^{2}+|a_{010}|^{2}+|a_{011}|^{2}=p_{0}^{1}$		(59)
	$\displaystyle|a_{100}|^{2}+|a_{101}|^{2}+|a_{110}|^{2}+|a_{111}|^{2}=p_{1}^{1}$		(60)
	$\displaystyle|a_{000}|^{2}+|a_{001}|^{2}+|a_{100}|^{2}+|a_{101}|^{2}=p_{0}^{2}$		(61)
	$\displaystyle|a_{010}|^{2}+|a_{011}|^{2}+|a_{110}|^{2}+|a_{111}|^{2}=p_{1}^{2}$		(62)
	$\displaystyle|a_{000}|^{2}+|a_{010}|^{2}+|a_{100}|^{2}+|a_{110}|^{2}=p_{0}^{3}$		(63)
	$\displaystyle|a_{001}|^{2}+|a_{011}|^{2}+|a_{101}|^{2}+|a_{111}|^{2}=p_{1}^{3}$		(64)

From the above measurement on $\theta$-angle basis, we have:

	$\displaystyle|\cos(\theta)a_{000}+\sin(\theta)a_{100}|^{2}+|\cos(\theta)a_{001}+\sin(\theta)a_{101}|^{2}+$		(65)
	$\displaystyle|\cos(\theta)a_{010}+\sin(\theta)a_{110}|^{2}+|\cos(\theta)a_{011}+\sin(\theta)a_{111}|^{2}=p(\theta)_{0}$		(66)
	$\displaystyle|\cos(\theta)a_{000}-\sin(\theta)a_{100}|^{2}+|\cos(\theta)a_{001}-\sin(\theta)a_{101}|^{2}+$		(67)
	$\displaystyle|\cos(\theta)a_{010}-\sin(\theta)a_{110}|^{2}+|\cos(\theta)a_{011}-\sin(\theta)a_{111}|^{2}=p(\theta)_{1}$		(68)

Putting everything together, we have the following set of equations:

$\displaystyle\begin{cases}&|a_{000}|^{2}+|a_{001}|^{2}+|a_{010}|^{2}+|a_{011}|^{2}=p_{0}^{1}\\ &|a_{100}|^{2}+|a_{101}|^{2}+|a_{110}|^{2}+|a_{111}|^{2}=p_{1}^{1}\\ &|a_{000}|^{2}+|a_{001}|^{2}+|a_{100}|^{2}+|a_{101}|^{2}=p_{0}^{2}\\ &|a_{010}|^{2}+|a_{011}|^{2}+|a_{110}|^{2}+|a_{111}|^{2}=p_{1}^{2}\\ &|a_{000}|^{2}+|a_{010}|^{2}+|a_{100}|^{2}+|a_{110}|^{2}=p_{0}^{3}\\ &|a_{001}|^{2}+|a_{011}|^{2}+|a_{101}|^{2}+|a_{111}|^{2}=p_{1}^{3}\\ &|\cos(\theta)a_{000}+\sin(\theta)a_{100}|^{2}+|\cos(\theta)a_{001}+\sin(\theta)a_{101}|^{2}+\\ &|\cos(\theta)a_{010}+\sin(\theta)a_{110}|^{2}+|\cos(\theta)a_{011}+\sin(\theta)a_{111}|^{2}=p(\theta)_{0}\\ &|\cos(\theta)a_{000}-\sin(\theta)a_{100}|^{2}+|\cos(\theta)a_{001}-\sin(\theta)a_{101}|^{2}+\\ &|\cos(\theta)a_{010}-\sin(\theta)a_{110}|^{2}+|\cos(\theta)a_{011}-\sin(\theta)a_{111}|^{2}=p(\theta)_{1}\\ \end{cases}$

(70)

We remind that all amplitudes $a_{ijk}$ are complex numbers, i.e., $a_{ijk}=\Re(a_{ijk})+i\Im(a_{ijk})$. At this point, we still don’t have enough information to solve because as mentioned, all amplitudes are complex number, so this 3-qubit case, we have $2^{3}$ amplitudes, corresponding with $2^{3+1}$ complex variables. The above equations contain 8 equations, so we need 8 more. In fact, there are many ways to collect more equations. Previously we have showed that performing joint 2-qubit measurements can also work, however, if we restrict ourselves to single qubit, then measuring a single qubit can also work well. If we measure the same first qubit in another basis, for example, different angle $\theta^{\prime}$, then we will have 2 more equations, corresponding with two possible outcomes. By choosing more angles, or we can opt to measure another qubit, then we can obtain more outcomes, thus forming further equations.

The above examples on two-qubits and three-qubits cases have illustrated the key idea behind our proposal, as instead of measuring all qubits, we only need to measure some subsets of qubits, and even a single qubit measurement is sufficient. The outcomes are then being processed classically, e.g., solving a nonlinear system, to obtain the desired amplitudes. Now we proceed to provide a formal description for our method, generalizing the above examples into $n$-qubits state.

From the above examples, it is straightforward to generalize the procedure to the $n$-qubit state. As there are $2^{n}$ amplitudes and $2^{n+1}$ variables, therefore we need $2^{n+1}$ equations. Each qubit measurement yields two possible outcomes, corresponding to the equations $2$ (see the back equation 7, 14), then measuring each qubit (among $n$ qubits) on a given basis yields $2n$ equations. If, as a first step, we choose to measure each individual qubit in Z basis, then we have $2n$ equations. Then, we continue to measure only the first qubit, then if in total $M$ is a different basis to measure, then we require $2M\geq 2^{n+1}-2n$. Given that $M$ is an integer, one can opt for $M=\lceil 2^{n}-n\rceil$, which is the total of different bases, or different angles, which one needs to choose and measure.

The last and very important point we need to discuss is the measurement-induced error and how it affects the solution we solve from such erroneous equations. Recall that the right-hand side of Eqn. 71 are estimated by collecting measurement results from single-qubit measurement. From an earlier discussion, we knew that by obeying the Bernoulli distribution, $\mathcal{O}(1/\epsilon^{2})$ is necessary and sufficient to estimate, for example, $p_{0}^{1},p_{1}^{1}$ to accuracy $\epsilon$ if we measure the first qubit. Likewise, all remaining probabilities are estimated up to $\epsilon$. As the right-hand side is erroneous, the solution to the system of nonlinear algebraic equations is also deviating from the real, ideal solution. The following theorem characterizes the error deviation of the erroneous system from the ideal system: