Perfect NOT and conjugate transformations

Abstract

We obtain the perfect NOT transformation criteria, two necessary and sufficient conditions for realizing a perfect NOT transformation on a quantum state set S of a qubit. Furthermore, we discuss a probabilistic perfect NOT transformation when there is no perfect NOT transformation on a state set S. We also generalize the perfect NOT transformation to the conjugate transformation in the multi-level quantum system and a lower bound of the best possible efficiencies attained by a probabilistic perfect conjugate transformation are obtained.

1 Introduction

The basic building block of any classical information processor is the single bit, which is prepared in one of two possible states, denoted 0 or 1. However, quantum information consists of qubits, each of which has the luxury of being in a superposition of the 0 and 1 states. Since there are an infinite number of superposition states, quantum systems have a much richer and more interesting existence than their classical counterparts. The superposition of states also makes the properties of quantum information quite different from that of its classical counterpart. Whereas the copying of classical information presents no difficulties, owing to the linearity of quantum mechanics, there is a quantum no-cloning theorem [1, 2] which asserts that it is impossible to construct a device that will perfectly copy an arbitrary (unknown) state of a two-level particle. However, the quantum no-cloning theorem does not rule out the possibility of either imperfect cloning [3, 4] or probabilistic cloning [5]. Some applications of cloning have been presented [6–8]. With the progress of a quantum information theory, quantum cloning has become a quite interesting field.

There is another difference between classical and quantum information systems. It is very easy to complement a classical bit, i.e., to change the value of a bit, a 0 to a 1 and vice versa. Usually, this operation can be accomplished by a NOT transformation (gate). However, in quantum information systems, changing an unknown state |Ψ〉=α|0〉+β|1〉 of a qubit to its orthogonal complement |Ψ^⊥〉=α^∗|1〉−β^∗|0〉 that is orthogonal to |Ψ〉 (i.e., inverting the state of a two-level quantum system) is impossible [9, 10]. The result is that one can not design a device that will take an arbitrary qubit and transform it into its orthogonal qubit. This is because complex conjugation of the coefficients in the NOT transformation of a qubit must be accomplished by an antiunitary transformation and cannot be performed by a unitary one. In other words, it is impossible to achieve the perfect NOT gate in quantum information systems.

However, the NOT transformation can be achieved on some states while leaving other states unchanged. Alternatively, there can be a transformation operation that approximates, at best, the NOT gate on all states, called the universal NOT gate [9, 10]. In fact, the output of a quantum cloning machine, the ancilla, carries the optimal anticlone of the input state so the universal NOT gate can be accomplished as a by-product of cloning [4].

A combination of unitary evolution together with measurements is an important method in quantum information processing and often achieves very interesting results. It has been used in quantum programming [11], the purification of entanglement [12], quantum teleportation [13], and the preparation of quantum states [14]. Recently, by using this method, Duan and Guo designed a probabilistic quantum cloning machine [5]. With a postselection of the measurement results, the machine outputs perfect copies of the input states.

In this paper, we present two necessary and sufficient conditions for realizing a perfect NOT transformation on the state set S. We also discuss the probabilistic perfect NOT transformations when there is no perfect NOT transformation on the state set S. We generalize the probabilistic NOT transformation to the conjugate transformation in the multi-level quantum system. Furthermore, the lower bound of the best possible efficiencies attained by a probabilistic perfect conjugate transformation is obtained.

2 Conditions required for perfect NOT transformations

In order to aid the analysis, we first state a Lemma of Duan and Guo [5]:

Lemma 1

If two sets of states |ϕ₁〉,|ϕ₂〉,⋯,|ϕ_n〉 and $|\tilde {\phi _{1}}\rangle, |\tilde {\phi _{2}}\rangle, \cdots, |\tilde {\phi _{n}}\rangle$ satisfy the condition

$$\begin{array}{cc} \langle\phi_{i}|\phi_{j}\rangle=\langle\tilde{\phi_{i}}|\tilde{\phi_{j}}\rangle,& (i,j=1,2,\cdots, n), \end{array} $$

(1)

then there exists a unitary operator U such that $U|\phi _{i}\rangle =|\tilde {\phi }_{i}\rangle, (i=1,2,\cdots,n)$.

Let S={|Ψ₁〉,|Ψ₂〉,⋯,|Ψ_n〉} be a set of states of a qubit. When the quantum states of S satisfy

$$\begin{array}{cc} \langle \Psi_{i}|\Psi_{j}\rangle=\left\langle \Psi_{i}^{\perp}| \Psi_{j}^{\perp}\right\rangle,& (i,j=1,2,\cdots,n),\end{array} $$

(2)

based on Lemma we can find a unitary transformation U such that $\begin {array}{ccc} |\Psi ^{\perp }_{i}\rangle =U|\Psi _{i}\rangle & \text {for} & i=1,2,\cdots, n.\end {array}$

It is easy to see that $\langle \Psi _{i}^{\perp }|\Psi _{j}^{\perp }\rangle =(\alpha _{i}\langle 1| -\beta _{i}\langle 0|) \left (\alpha _{j}^{*}|1\rangle -\beta _{j}^{*}|0\rangle \right)= \alpha _{i}\alpha _{j}^{*}+\beta _{i}\beta _{j}^{*}=\langle \Psi _{i}|\Psi _{j}\rangle ^{*}$, which shows that the condition Eq.(2) is equivalent to 〈Ψ_i|Ψ_j〉=〈Ψ_i|Ψ_j〉^∗ for i,j=1,2,⋯,n. This, in turn, implies that all inner-products of the quantum states in the set S are real. Hence we arrive at the following conclusion:

Theorem 1

Suppose that S={|Ψ₁〉,|Ψ₂〉,⋯,|Ψ_n〉} is the set of quantum states. Then, a perfect NOT transformation (gate) U on the set S can be realized by a unitary transformation (i.e., there is a unitary transformation U such that $U|\Psi _{i}\rangle =|\Psi _{i}^{\perp }\rangle$) if and only if

$$ \langle \Psi_{i}|\Psi_{j}\rangle=\langle \Psi_{i}|\Psi_{j}\rangle^{*}, ~~ i,j=1,2,\cdots, n. $$

(3)

It turns out that if S contains all points of a Bloch sphere [15] of a qubit, one can not realize the perfect NOT transformation on the set S.

Obviously, in Theorem 1, we only consider the case without an ancilla (probe). Now, let us introduce a probe P with the initial state |P⁽⁰⁾〉. By Lemma, if

$$\begin{array}{@{}rcl@{}} \left\langle P^{(0)}|\langle\Psi_{i}|\Psi_{j}\rangle|P^{(0)}\right\rangle &=&\langle\Psi_{i}|\Psi_{j}\rangle\\ &=&\left\langle\Psi_{i}^{\perp}|\Psi_{j}^{\perp}\right\rangle\left\langle P^{(i)}|P^{(j)}\right\rangle\\ &=&\langle\Psi_{i}|\Psi_{j}\rangle^{*}\left\langle P^{(i)}|P^{(j)}\right\rangle, \end{array} $$

(4)

for arbitrary i,j=1,2,⋯,n, then there exists a unitary transformation U, such that

$$ U\left(|\Psi_{i}\rangle|P^{(0)}\rangle\right)=\left|\Psi_{i}^{\perp}\right\rangle\left|P^{(i)}\right\rangle. $$

(5)

It means we can realize the perfect NOT transformation on the quantum state set S={|Ψ₁〉,|Ψ₂〉,⋯,|Ψ_n〉} with the assistance of the ancilla (probe).

Next, we discuss the case in which 〈Ψ_i|Ψ_j〉≠0 for arbitrary i,j=1,2,⋯,n. Let $\phantom {\dot {i}\!}\langle \Psi _{i}|\Psi _{j}\rangle =t_{ij}\texttt {e}^{\texttt {i}\theta _{ij}}$, and $\left \langle P^{(i)}|P^{(j)}\right \rangle =p_{ij}\texttt {e}^{\texttt {i}\varphi _{ij}}$. Here 0<t_ij,p_ij≤1, and 0≤θ_ij,φ_ij<2π. Eq.(4) is then equivalent to

$$ t_{ij}\texttt{e}^{\texttt{i}\theta_{ij}}=t_{ij}\texttt{e}^{-\texttt{i}\theta_{ij}}p_{ij}\texttt{e}^{\texttt{i}\varphi_{ij}}. $$

(6)

It implies that

$$ \begin{array}{cc}p_{ij}=1,& 2\theta_{ij}=\varphi_{ij}+2k_{ij}\pi, \end{array} $$

(7)

where k_ij=0 or 1. Because p_1j=1, there must be

$$ \left|P^{(i)}\right\rangle=\texttt{e}^{\texttt{i}\varphi_{i}}\left|P^{(1)}\right\rangle $$

(8)

for i=2,3,⋯,n, and φ₁=0. It follows that

$$ \varphi_{ij}=\varphi_{j}-\varphi_{i}+2m_{ij}\pi, $$

(9)

where m_ij=0 or 1. By Eq.(7), we obtain

$$ \varphi_{j}-\varphi_{i}=2\theta_{ij}-2(k_{ij}+m_{ij})\pi. $$

(10)

Thus,

$$ \varphi_{j}=2\theta_{1j}-2(k_{1j}+m_{1j})\pi. $$

(11)

Therefore,

$$ \begin{array}{c} \theta_{ij}=\theta_{1j}-\theta_{1i}+(k_{ij}+m_{ij}-k_{1j}-m_{1j}+k_{1i}+m_{1i})\pi,\\ i,j=1,2,\cdots,n, \end{array} $$

(12)

which implies that

$$ \begin{array}{c} \theta_{lj}-\theta_{li}=\theta_{ij}+(k_{lj}+m_{lj}-k_{li}-m_{li}-k_{ij}-m_{ij})\pi, \\ i,j,l=1,2,\cdots,n;~k_{ij},m_{ij}=0 ~ \text{or} ~ 1. \end{array} $$

(13)

Furthermore, starting from Eq.(13), we can reverse the process. This means that if the quantum state set S={|Ψ₁〉,|Ψ₂〉,⋯,|Ψ_n〉} satisfies 〈Ψ_i|Ψ_j〉≠0 and Eq.(13), one can find a unitary transformation U such that Eq.(5) holds and a perfect NOT transformation can be realized.

Based on the above argument we obtain following conclusion:

Theorem 2

Suppose that the quantum state set S={|Ψ₁〉,|Ψ₂〉,⋯,|Ψ_n〉} satisfies 〈Ψ_i|Ψ_j〉≠0. Then, a perfect NOT transformation (gate) on the state set S can be realized by a unitary transformation acting on the system and a probe if and only if Eq.(13) holds.

Note that, when the quantum state set S contains only two quantum states |Ψ₁〉,|Ψ₂〉, Eq.(13) can always hold. Therefore, the perfect NOT transformation (gate) on the state set S of two arbitrary quantum states |Ψ₁〉,|Ψ₂〉 can always be realized.

Clearly, if there are no quantum states |P⁽¹⁾〉,|P⁽²⁾〉,⋯,|P⁽ⁿ⁾〉 satisfying Eq.(4) for the quantum state set S, then one can not design a perfect NOT gate for this state set S. In this case, one can only consider the universal-NOT or the probabilistic perfect NOT gate. As the universal-NOT gate has been well studied [9, 10], in the next section, we will only discuss the probabilistic perfect NOT gate.

3 Probabilistic perfect NOT transformation

The definition of a probabilistic perfect NOT gate is that for a quantum state set S={|Ψ₁〉,|Ψ₂〉,⋯,|Ψ_n〉}, there is a unitary transformation together with a measurement, which when combined with a postselection of measurement results, makes an arbitrarily unknown input quantum state |Ψ_i〉 transform into its orthogonal complement $|\Psi _{i}^{\perp }\rangle$ with certain nonzero probability of success. That is, for a quantum state set S={|Ψ₁〉,|Ψ₂〉,⋯,|Ψ_n〉}, if there exists a unitary operation U and a measurement M, which together yield the following evolution:

$$ |\Psi_{i}\rangle\begin{array}{c} U+M\\\longrightarrow \end{array} \left|\Psi_{i}^{\bot}\right\rangle, $$

(14)

then a probabilistic NOT gate is said to have been built.

The combination of a unitary evolution operation and a measurement is very general and can be used to describe any operation in quantum mechanics [16]. Unitary operation with measurement is also an important tool in quantum information sciences, for instance in quantum programming. A specific model of quantum programming of such kind is the measurement-based quantum computation [17]. Another example of such kind is the quantum convolutional neural network based on linear combination of unitaries [18]. As for entanglement purification, in addition to the pioneering work of Bennett et al [12], the recent experimental advancement [19] is worth noting. The use of parity-check measurement is used in concentration of W-state in ref. [20]. Of course, unitary operation and measurement are extensively used in quantum key distribution [21]. It is also used in quantum steering [22]. As for preparation of quantum state, two examples are given in refs. [23,24]. It even plays a vital role in deciding if a theory of entanglement-entropy is renormalizable [25].

Obviously, we can not build a probabilistic NOT gate for any arbitrary quantum state set S={|Ψ₁〉,|Ψ₂〉,⋯,|Ψ_n〉}, so it is very important to find the conditions that the quantum state set S should be satisfied in order to construct a probabilistic perfect NOT gate.

The unitary evolution of the qubit A and probe P can be described by the following equation

$$ {}\begin{array}{c} U(|\Psi_{i}\rangle|P_{0}\rangle)=\sqrt {\gamma_{i}}\left|\Psi_{i}^{\perp}\right\rangle\left|P^{(i)}\right\rangle+\sqrt {1-\gamma_{i}}\left|\Phi^{(i)}_{AP}\right\rangle,\\ (i=1,2,...,n), \end{array} $$

(15)

where |P₀〉 and |P⁽ⁱ⁾〉 are normalized states of the probe P (not generally orthogonal) and $\left |\Phi ^{(1)}_{AP}\right \rangle, \left |\Phi ^{(2)}_{AP}\right \rangle, \cdots$, and $\left |\Phi ^{(n)}_{AP}\right \rangle$ are n normalized states of the composite system AP (not generally orthogonal). We assume that in Eq.(15) the coefficients before the states $\left |\Psi _{i}^{\perp }\right \rangle \left |P^{(i)}\right \rangle$, and $\left |\Phi ^{(i)}_{AP}\right \rangle$ are positive real numbers. Let S₀ be the subspace spanned by the states |P⁽¹⁾〉,|P⁽²⁾〉,⋯,|P⁽ⁿ⁾〉. In order to realize the probabilistic perfect NOT transformation, we must require that after the unitary evolution a measurement of the probe with a postselection of the measurement results should project its state into the subspace S₀. After this projection, the state of the system A should be $\left |\Psi _{i}^{\perp }\right \rangle$. Therefore, all of the states $\left |\Phi ^{(i)}_{AP}\right \rangle$, lie in a space orthogonal to S₀ and can be represented by the following equation

$$\begin{array}{c} \left|P^{(i)}\right\rangle\left\langle P^{(i)}\left|\Phi_{AP}^{(j)}\right.\right\rangle=0,\\ (i,j=1,2,...,n). \end{array} $$

(16)

With above restriction, inter-inner-products of Eq.(15) yield the following matrix equation

$$ X^{(1)}=\sqrt \Gamma X^{(\perp)}_{P}\sqrt {\Gamma^{+}}+\sqrt {E_{n}-\Gamma}Y\sqrt {E_{n}-\Gamma^{+}}, $$

(17)

where $X^{(1)}=[\langle \Psi _{i}|\Psi _{j}\rangle ], Y=\left [\left \langle \Phi ^{(i)}_{AP}\left |\Phi ^{(j)}_{AP}\right.\right \rangle \right ], X^{(\perp)}_{P}=\left [\left \langle \Psi _{i}^{\perp }|\Psi _{j}^{\perp }\right \rangle \left \langle P^{(i)}\left |P^{(j)}\right.\right \rangle \right ]$ are n×n matrices and E_n is the n×n identity matrix. The diagonal efficiency matrix Γ is defined by Γ=diag(γ₁,γ₂,...,γ_n); therefore, ${\sqrt \Gamma }={\sqrt {\Gamma ^{+}}}= \text {diag}\left ({\sqrt \gamma _{1}}, {\sqrt \gamma _{2}},...,{\sqrt \gamma _{n}}\right)$. According to result of Duan and Guo [5], Y is a positive-semidefinite matrix. Thus, $\sqrt {E_{n}-\Gamma }Y\sqrt {E_{n}-\Gamma ^{+}}$ is also a positive-semidefinite matrix. Based on Eq.(17), $X^{(1)}-\sqrt \Gamma X^{(\perp)}_{P}\sqrt {\Gamma ^{+}}$ is a positive-semidefinite matrix. Conversely, if $X^{(1)}-\sqrt \Gamma X^{(\perp)}_{P}\sqrt {\Gamma ^{+}}$ is a positive-semidefinite matrix, one can choose $|\Phi ^{(i)}_{AP}\rangle$ such that Eq.(17) holds. By Lemma the states |Ψ₁〉,|Ψ₂〉,..., and |Ψ_n〉 are able to be probabilistically transformed to their respective orthogonal complement states. Thus we have the following theorem:

Theorem 3

The states |Ψ₁〉,|Ψ₂〉,..., and |Ψ_n〉 can be probabilistically perfectly transformed to their respective orthogonal complement states if and only if there exist a diagonal positive-definite matrix Γ and |P⁽ⁱ⁾〉 (i=1,2,...,n) such that the matrix $X^{(1)}-\sqrt \Gamma X^{(\perp)}_{P}\sqrt {\Gamma ^{+}}$ is positive-semidefinite. Here X⁽¹⁾=[〈Ψ_i|Ψ_j〉] and $X^{(\perp)}_{P}=\left [\left \langle \Psi _{i}^{\perp }\left |\Psi _{j}^{\perp }\right.\right \rangle \left \langle P^{(i)}\left |P^{(j)}\right.\right \rangle \right ]$ are n×n matrices, and |P⁽ⁱ⁾〉(i=1,2,⋯,n) are quantum states of a probe.

Theorem 3 is very general, and for the linearly independent quantum state set S, Qiu showed the following conclusion [26].

Theorem 4

The states secretly chosen from the set S={|Ψ₁〉,|Ψ₂〉,⋯,|Ψ_n〉} can be probabilistically transformed into their respective orthogonal complements by a general unitary-reduction operation, if |Ψ₁〉,|Ψ₂〉,⋯, and |Ψ_n〉 are linearly independent.

Suppose the Hilbert space of the probe P is an n_p-dimensional space, where n_p≥n+1. We use |P₀〉,|P₁〉,..., and |P_n〉 to denote n+1 orthonormal states of a probe P. By extended the above Theorem, we know that if |Ψ₁〉,|Ψ₂〉,⋯, and |Ψ_n〉 are linearly independent, there exists a unitary operator U that satisfies

$$\begin{array}{c} {}U(|\Psi_{i}\rangle|P_{0}\rangle)=\sqrt \gamma_{i} \left|\Psi_{i}^{\perp}\right\rangle \texttt{e}^{\texttt{i}\varphi_{i}}|P_{0}\rangle +\sum_{j=1}^{n} c_{ij}\left|\Phi^{(j)}\right\rangle |P_{j}\rangle, \\ (i=1,2,\cdots, n), \end{array} $$

(18)

where |Φ^(j)〉(j=1,2,⋯,n) stand for n normalized states of the system (not generally orthogonal) and φ_i are real numbers. Eq.(18) is a special case of Eq.(15). The NOT transformation is successful, and the output state of the system is $\left |\Psi _{i}^{\perp }\right \rangle$, if and only if the measurement outcome of the probe is |P₀〉. Evidently, the probability of success (obtaining |P₀〉) is γ_i.

Next, we consider probabilistic perfect NOT transformation of the quantum state set {|Ψ₁〉,|Ψ₂〉,|Ψ₃〉}. Suppose that |Ψ₁〉 and |Ψ₂〉 are linearly independent and that

$$ |\Psi_{3}\rangle=\alpha|\Psi_{1}\rangle+\beta|\Psi_{2}\rangle. $$

(19)

Here, α and β satisfy the normalizing condition

$$ \alpha\alpha^{*}+\beta\beta^{*}+\alpha^{*}\beta\langle\Psi_{1}|\Psi_{2}\rangle+\alpha\beta^{*}\langle\Psi_{2}|\Psi_{1}\rangle=1. $$

(20)

There exists a unitary transformation U such that

$$\begin{array}{@{}rcl@{}} && {}U(|\Psi_{1}\rangle|P_{0}\rangle)=\sqrt {\gamma_{1}}\left|\Psi_{1}^{\perp}\right\rangle|P_{0}\rangle+\sqrt {1-\gamma_{1}}\left|\Phi^{(1)}_{AP}\right\rangle,\\ &&{}U(|\Psi_{2}\rangle|P_{0}\rangle)=\sqrt {\gamma_{2}}\left|\Psi_{2}^{\perp}\right\rangle \texttt{e}^{\texttt{i} \varphi}|P_{0}\rangle+\sqrt {1-\gamma_{2}}\left|\Phi^{(2)}_{AP}\right\rangle.\\ \end{array} $$

(21)

The linearity of U implies that

$$ {}\begin{aligned} U(|\Psi_{3}\rangle|P_{0}\rangle) &=\left(\alpha\sqrt{\gamma_{1}}\left|\Psi_{1}^{\perp}\right\rangle + \beta\sqrt{\gamma_{2}}\left|\Psi_{2}^{\perp}\right\rangle \texttt{e}^{\texttt{i}\varphi}\right)|P_{0}\rangle\\ &\quad+\alpha\sqrt{1-\gamma_{1}}\left|\Phi^{(1)}_{AP}\right\rangle+\beta\sqrt{1-\gamma_{2}}\left|\Phi^{(2)}_{AP}\right\rangle. \end{aligned} $$

(22)

Hence, if

$$ \alpha\sqrt {\gamma_{1}}\left|\Psi_{1}^{\perp}\right\rangle+\beta\sqrt {\gamma_{2}}\left|\Psi_{2}^{\perp}\right\rangle \texttt{e}^{\texttt{i} \varphi}=\sqrt {\gamma_{3}}\left|\Psi_{3}^{\perp}\right\rangle \texttt{e}^{\texttt{i}\chi}, $$

(23)

one obtains

$$\begin{array}{@{}rcl@{}} U(|\Psi_{3}\rangle|P_{0}\rangle)=\sqrt{\gamma_{3}}\left|\Psi_{3}^{\perp}\right\rangle|P_{0}\rangle\texttt{e}^{\texttt{i}\chi}+\sqrt{1-\gamma_{3}}\left|\Phi^{(3)}_{AP}\right\rangle.\\ \end{array} $$

(24)

Here, χ is a real number. Therefore, we can realize the probabilistic perfect NOT transformation on the set {|Ψ₁〉,|Ψ₂〉,|Ψ₃〉} in case of Eq.(23) being satisfied.

Now, suppose that

$$\begin{array}{@{}rcl@{}} |\Psi_{1}\rangle=\cos{\frac {\theta_{1}}{2}}|0\rangle+\sin{\frac {\theta_{1}}{2}}\texttt{e}^{\texttt{i}\phi_{1}}|1\rangle,\\ |\Psi_{2}\rangle=\cos{\frac {\theta_{2}}{2}}|0\rangle+\sin{\frac {\theta_{2}}{2}}\texttt{e}^{\texttt{i}\phi_{2}}|1\rangle. \end{array} $$

(25)

Eq.(20) then becomes

$$ {}\begin{aligned} & |\alpha|^{2}\,+\,|\beta|^{2}\,+\,\alpha^{*}\beta\left(\cos{\frac{\theta_{1}}{2}}\cos{\frac {\theta_{2}}{2}} \,+\, \sin{\frac{\theta_{1}}{2}}\sin{\frac{\theta_{2}}{2}}\texttt{e}^{\texttt{i}(\phi_{2}-\phi_{1})}\right)\\ &+\alpha\beta^{*}\left(\cos{\frac{\theta_{1}}{2}}\cos{\frac {\theta_{2}}{2}}+\sin{\frac{\theta_{1}}{2}}\sin{\frac{\theta_{2}}{2}}\texttt{e}^{\texttt{-i}(\phi_{2}-\phi_{1})}\right)=1, \end{aligned} $$

(26)

and Eq.(23) changes to

$$ \begin{aligned} &\alpha\sqrt {\gamma_{1}}\cos{\frac{\theta_{1}}{2}}+\beta\texttt{e}^{\texttt{i}\varphi}\sqrt{\gamma_{2}}\cos{\frac {\theta_{2}}{2}}\\ &=\sqrt{\gamma_{3}}\texttt{e}^{\texttt{i}\chi}\left(\alpha^{*}\cos{\frac{\theta_{1}}{2}}+\beta^{*}\cos{\frac{\theta_{2}}{2}}\right), \\ &\alpha\sqrt {\gamma_{1}}\sin{\frac{\theta_{1}}{2}}\texttt{e}^{\texttt{-i}\phi_{1}}+\beta\texttt{e}^{\texttt{i}\varphi}\sqrt{\gamma_{2}}\sin{\frac{\theta_{2}}{2}}\texttt{e}^{\texttt{-i}\phi_{2}}\\ &=\sqrt{\gamma_{3}}\texttt{e}^{\texttt{i}\chi}\left(\alpha^{*}\sin{\frac{\theta_{1}}{2}}\texttt{e}^{\texttt{-i}\phi_{1}}+\beta^{*}\sin{\frac{\theta_{2}}{2}}\texttt{e}^{\texttt{-i}\phi_{2}}\right). \end{aligned} $$

(27)

Therefore, if Eqs.(26) and (27) are satisfied by {|Ψ₁〉,|Ψ₂〉,|Ψ₃〉}, then probabilistic perfect NOT transformation on this quantum state set can be realized.

For example, we can realize a probabilistic perfect NOT transformation with probability γ on the quantum states

$$\begin{array}{@{}rcl@{}} &&|\Psi_{1}\rangle=\frac {1}{\sqrt 2}(|0\rangle+|1\rangle),\\ &&|\Psi_{2}\rangle=\frac {1}{\sqrt 2}(|0\rangle+\texttt{i}|1\rangle),\\ &&|\Psi_{3}\rangle=\frac {1}{\sqrt 2}\left[\left(q+r \texttt{e}^{-\texttt{i}\frac {\varphi}{2}}\right)|0\rangle +\left(q+\texttt{i} r \texttt{e}^{-\texttt{i}\frac {\varphi}{2}}\right)|1\rangle)\right],\\\end{array} $$

(28)

where q,r,γ,φ are real and satisfy

$$ \begin{array}{l} q^{2}+r^{2}+\sqrt 2qr\cos\left(\frac {\pi}{4}-\frac {\varphi}{2}\right)=1,\\ \frac {1}{2}-2\gamma+\frac {\gamma^{2}}{2}+\gamma\sin\varphi\geq 0,\\ 1-\gamma\geq 0.\\ \end{array} $$

(29)

Obviously, $\varphi =\frac {\pi }{2}$ corresponds to the perfect NOT transformation case.

4 Conjugate transformation of a multi-level quantum system

In this section, we discuss conjugate transformation of a multi-level quantum system (qudit). Suppose that the dimension of a Hilbert space for the quantum system is d. An arbitrary quantum state of the system can be written as

$$ |\Psi\rangle=\sum_{i=0}^{d-1}\alpha_{i}|i\rangle, $$

(30)

where α_i are complex numbers and {|i〉} is an orthonormal basis. Let us define a conjugate transformation T as

$$ T|\Psi\rangle=T\left(\sum_{i=0}^{d-1} \alpha_{i}|i\rangle\right)=\sum_{i=0}^{d-1} \alpha^{*}_{i}|i\rangle\equiv \left|\Psi^{T}\right\rangle. $$

(31)

Obviously, a perfect NOT transformation equals UT for a qubit, where $U=\left (\begin {array}{cc}0&-1\\ 1&0\end {array}\right)$is a unitary transformation. We call |Ψ^T〉 the conjugate state of quantum state |Ψ〉. Evidently, one can not design a machine that will take an arbitrary quantum state |Ψ〉 and transform it into its conjugate state |Ψ^T〉 because of the need for complex conjugation of the coefficients in the transformation, which must be accomplished by an antiunitary transformation and cannot be performed by a unitary one. By Lemma 1, we can also assert that this kind transformation is impossible on a general quantum state set S={|Ψ₁〉,|Ψ₂〉,⋯,|Ψ_n〉} of a qudit, since $\langle \Psi _{i}|\Psi _{j}\rangle \neq \langle \Psi _{i}^{T}|\Psi _{j}^{T}\rangle =\langle \Psi _{i}|\Psi _{j}\rangle ^{*}$ for two arbitrary quantum states in the set S.

However, by the argument similar to the qubit case, we do have the following conclusions:

Theorem 5

A perfect conjugate transformation on the state set S={|Ψ₁〉,|Ψ₂〉,⋯,|Ψ_n〉} of a qudit can be realized by a unitary transformation if and only if all inner-products of the quantum states in the set S are real.

Theorem 6

Suppose that the quantum state set S={|Ψ₁〉,|Ψ₂〉,⋯,|Ψ_n〉} of a qudit satisfies 〈Ψ_i|Ψ_j〉≠0. Then a conjugate transformation on the state set S can be realized by a unitary transformation acting on the system and a probe if and only if Eq.(13) holds.

Theorem 7

The states |Ψ₁〉,|Ψ₂〉,..., and |Ψ_n〉 can be probabilistically perfectly transformed to their respective conjugate states if and only if there exist a diagonal positive-definite matrix Γ and |P⁽ⁱ⁾〉 (i=1,2,...,n) such that the matrix $X^{(1)}-\sqrt \Gamma X^{(T)}_{P}\sqrt {\Gamma ^{+}}$ is positive-semidefinite. Here, X⁽¹⁾=[〈Ψ_i|Ψ_j〉] and $X^{(T)}_{P}=\left [\left \langle \Psi _{i}^{\perp }\left |\Psi _{j}^{\perp }\right.\right \rangle \left \langle P^{(i)}\left |P^{(j)}\right.\right \rangle \right ]$ are n×n matrices, and |P⁽ⁱ⁾〉(i=1,2,⋯,n) are quantum states of a probe.

Next, we investigate the best possible efficiencies γ_i attained by a probabilistic conjugate transformation.

For the sake of simplicity, we only discuss the special case S={|Ψ₁〉,|Ψ₂〉,|Ψ₃〉}. In this case, $X^{(1)}-\sqrt \Gamma X^{(T)}_{P}\sqrt {\Gamma ^{+}}$ becomes

$$\begin{array}{*{20}l} {} \left(\begin{array}{ccc} 1-\gamma_{1} &\langle\Psi_{1}|\Psi_{2}\rangle-\sqrt {\gamma_{1}\gamma_{2}}\langle\Psi_{1}|\Psi_{2}\rangle^{*}\left\langle P^{(1)}|P^{(2)}\right\rangle & \langle\Psi_{1}|\Psi_{3}\rangle-\sqrt {\gamma_{1} \gamma_{3}}\langle\Psi_{1}|\Psi_{3}\rangle^{*}\langle P^{(1)}|P^{(3)}\rangle\\ \langle\Psi_{2}|\Psi_{1}\rangle-\sqrt {\gamma_{1}\gamma_{2}}\langle\Psi_{2}|\Psi_{1}\rangle^{*}\langle P^{(2)}|P^{(1)}\rangle & 1- \gamma_{2} &\langle\Psi_{2}|\Psi_{3}\rangle-\sqrt {\gamma_{2}\gamma_{3}}\langle\Psi_{2}|\Psi_{3}\rangle^{*}\langle P^{(2)}|P^{(3)}\rangle\\ \langle\Psi_{3}|\Psi_{1}\rangle-\sqrt {\gamma_{1}\gamma_{3}}\langle\Psi_{3}|\Psi_{1}\rangle^{*}\langle P^{(3)}|P^{(1)}\rangle & \langle\Psi_{3}|\Psi_{2}\rangle-\sqrt { \gamma_{2}\gamma_{3}}\langle\Psi_{3}|\Psi_{2}\rangle^{*}\left\langle P^{(3)}|P^{(2)}\right\rangle & 1-\gamma_{3}\end{array}\right). \end{array} $$

(32)

Let $\phantom {\dot {i}\!}\langle \Psi _{1}|\Psi _{2}\rangle =t_{12}\texttt {e}^{\texttt {i}\theta _{12}}, \langle \Psi _{1}|\Psi _{3}\rangle =t_{13}\texttt {e}^{\texttt {i}\theta _{13}}$ and $\phantom {\dot {i}\!}\langle \Psi _{2}|\Psi _{3}\rangle =t_{23}\texttt {e}^{\texttt {i}\theta _{23}}$. We choose $\gamma _{1}=\gamma _{2}=\gamma _{3}=\gamma, \left \langle P^{(1)}|P^{(2)}\right \rangle =\texttt {e}^{2\texttt {i}\theta _{12}}$ and $\left \langle P^{(1)}|P^{(3)}\right \rangle =\texttt {e}^{2\texttt {i}\theta _{13}}$. So Eq. (32) becomes

$$ (1-\gamma)\left(\begin{array}{ccc} 1 &\langle\Psi_{1}|\Psi_{2}\rangle & \langle\Psi_{1}|\Psi_{3}\rangle\\ \langle\Psi_{2}|\Psi_{1}\rangle & 1 & A \\ \langle\Psi_{3}|\Psi_{1}\rangle & A^{*} & 1\end{array}\right), $$

(33)

where $A=\frac {\langle \Psi _{2}|\Psi _{3}\rangle -\gamma \langle \Psi _{2}|\Psi _{3}\rangle ^{*}\texttt {e}^{2\texttt {i}(\theta _{13}-\theta _{12})}}{1-\gamma }$. The positive-semidefinite condition of Eq.(33) requires

$$ \texttt{Det}\left(\begin{array}{ccc} 1 &\langle\Psi_{1}|\Psi_{2}\rangle & \langle\Psi_{1}|\Psi_{3}\rangle\\ \langle\Psi_{2}|\Psi_{1}\rangle & 1 & A \\ \langle\Psi_{3}|\Psi_{1}\rangle & A^{*} & 1\end{array}\right)\geq 0, $$

(34)

and

$$ \texttt{Det}\left(\begin{array}{cc} 1 & A \\ A^{*} & 1\end{array}\right)\geq 0. $$

(35)

Let $\delta =\theta _{12}-\theta _{13}+\theta _{23}, a=-\texttt {Det} X^{(1)}=-1+t_{12}^{2}+t_{13}^{2}+t_{23}^{2}-2t_{12}t_{13}t_{23}\cos \delta, b=t_{23}^{2}-1$. Eq.(34) means

$$ a\gamma^{2}+2\gamma(2t_{23}^{2}\sin^{2}\delta-a)+a\leq 0. $$

(36)

So, we have

$$ 0<\gamma\leq 1+\frac {2\sqrt{t_{23}^{4}\sin^{4}\delta-at_{23}^{2}\sin^{2}\delta }-2t_{23}^{2}\sin^{2}\delta }{a}. $$

(37)

By Eq.(35) we obtain

$$ b\gamma^{2}+2\gamma(2t_{23}^{2}\sin^{2}\delta-b)+b\leq 0. $$

(38)

Therefore, γ satisfies

$$ 0<\gamma\leq 1+ \frac {2\sqrt{t_{23}^{4}\sin^{4}\delta-bt_{23}^{2}\sin^{2}\delta }-2t_{23}^{2}\sin^{2}\delta}{b}. $$

(39)

Since a>b, and $1+2\frac {\sqrt {1-x}-1}{x}$ is a monotone function, the maximum of γ in this special case is

$$ \gamma_{\text{max}}=1+\frac {2\sqrt{t_{23}^{4}\sin^{4}\delta-at_{23}^{2}\sin^{2}\delta }-2t_{23}^{2}\sin^{2}\delta }{a}. $$

(40)

Hence a lower bound of $\text {Max}\left (\frac {\gamma _{1}+\gamma _{2}+\gamma _{3}}{3}\right)$ is γ_max.

5 Summary

In conclusion, we have investigated a perfect NOT transformation on a quantum state set S of a qubit and derived two necessary and sufficient conditions for realizing a perfect NOT transformation on S. A probabilistic perfect NOT transformation (gate) was discussed. Furthermore, we generalize the probabilistic NOT transformation to the conjugate transformation in a multi-level quantum system. The lower bound of the best possible efficiencies attained by a probabilistic perfect conjugate transformation was obtained.