# Transformation of Bell states using linear optics

Sarika Mishra<EMAIL_ADDRESS>Quantum Science and Technology Laboratory,
Physical Research Laboratory, Ahmedabad, India 380009.  Indian Institute of
Technology, Gandhinagar, India 382355.  R. P. Singh<EMAIL_ADDRESS>Quantum Science and Technology Laboratory, Physical Research Laboratory,
Ahmedabad, India 380009.

###### Abstract

Bell states form a complete set of four maximally polarization entangled two-
qubit quantum state. Being a key ingredient of many quantum applications such
as entanglement based quantum key distribution protocols, superdense coding,
quantum teleportation, entanglement swapping etc, Bell states have to be
prepared and measured. Spontaneous parametric down conversion is the easiest
way of preparing Bell states and a desired Bell state can be prepared from any
entangled photon pair through single-qubit logic gates. In this paper, we
present the generation of complete set of Bell states, only by using unitary
transformations of half-wave plate (HWP). The initial entangled state is
prepared using a combination of a nonlinear crystal and a beam-splitter (BS)
and the rest of the Bell states are created by applying single-qubit logic
gates on the entangled photon pairs using HWPs. Our results can be useful in
many quantum applications, especially in superdense coding where control over
basis of maximally entangled state is required.

Quantum entanglement, Bell states, Quantum logic gate, Unitary transformation,
Quantum tomography, Quantum information processing

##  I Introduction

Quantum entanglement [1, 2, 3] is one of the most interesting phenomena in
quantum world which laid the foundation of many quantum applications such as
quantum superdense coding [4, 5], quantum teleportation [6, 7], entanglement
swapping [8, 9], and quantum key distribution (QKD) protocols [10, 11, 12, 13,
14, 15], quantum secret sharing (QSS)[16, 17], quantum secure direct
communication (QSDC)[18, 19, 20, 21] etc. The simplest way to create an
entangled photon pair is spontaneous parametric down-conversion (SPDC) process
[22, 23, 24]. In this process, a nonlinear $\chi^{(2)}$ crystal is used to
produce two correlated photons. These photon paires can be correlated in any
degree of freedom such as, polarization, orbital angular momentum (OAM),
energy, time, frequency etc. Among these, polarization entangled states are
easy to prepare and measure. Hence it is the most widely used resource for all
the quantum communication protocols.

A set of all four maximally entangled polarization states in two dimensional
Hilbert space is known as Bell states [3]. In many quantum information schemes
such as quantum superdense coding, teleportation, and entanglement swapping,
the generation and discrimination of all the four maximally entangled states
is required. For example, in superdense coding control over basis of maximally
entangled state is required. A desired Bell state can be prepared through
unitary transformations on an entangled pair of photons generated in SPDC. In
this case, transformation of one Bell state into another by using single-qubit
gate (Pauli gate) can be used. For example, $\ket{\Psi}^{+}$ can be
transformed into $\ket{\Psi}^{-}$ and $\ket{\Phi}^{+}$ with the help of
polarization-dependent phase shift (phase-flip) and polarization exchange
(bit-flip) respectively. Conventionally, A quarter-wave plate (QWP) (which
performs a Pauli-Z transformation) and half-wave plate (HWP) (which performs a
Pauli-X transformation) is used for phase-flip and bit-flip respectively [25,
26, 27]. In this work, we show that only a single half-wave plate is enough to
perform Pauli-X and Pauli-Z transformation. A pair of HWP would be required to
perform Pauli-Y transformation.

This paper is organized as follows: In section II, we explain the theoretical
model of our proposed scheme where we first explain the single qubit gate
(Pauli gate) and how these gates are analogous to the rotation of HWP. In
section III we present the experimental setup for both generation and
transformation of Bell states. In section IV, we discuss the experimental
results. We end the paper with concluding remarks in last section V.

##  II Theory

Consider two photons, signal and idler, are entangled in polarization and
propagating along two different directions. This entangled state can be
expressed as in four different ways,

$\ket{\Psi}^{\pm}=\frac{1}{\sqrt{2}}(\ket{H}_{s}\ket{V}_{i}\pm\ket{V}_{s}\ket{H}_{i}),$
(1a)
$\ket{\Phi}^{\pm}=\frac{1}{\sqrt{2}}(\ket{H}_{s}\ket{H}_{i}\pm\ket{V}_{s}\ket{V}_{i}).$
(1b)

where $\ket{\Psi}^{-}$ is anti-symmetric entangled state and rest are
symmetric states.These polarization entangled states are popularly known as
Bell states. $\ket{H}$ and $\ket{V}$ are the horizontal and vertical
polarization state of photons, respectively. Using Jones vector notation, the
polarization states can be represented by column vector,

$\ket{H}=\left(\begin{array}[]{cc}1\\\ 0\\\
\end{array}\right),\,\,\ket{V}=\left(\begin{array}[]{cc}0\\\ 1\\\
\end{array}\right)$ (2a) $\ket{D}=\left(\begin{array}[]{cc}1\\\ 1\\\
\end{array}\right),\,\,\ket{A}=\left(\begin{array}[]{cc}1\\\ -1\\\
\end{array}\right)$ (2b)

Using single-qubit quantum logic gate in Eqn. 1b, one Bell state can be easily
manipulated and transformed into another Bell state. Such a gate operation
transforms input state (say $\ket{\psi}$) to an output state (say
$\ket{\phi}$) ( Figure 1).

Figure 1: Schematic diagram of transformation of state

These single-qubit logic gates are Paulie $X$, $Y$, and $Z$ gate and are
analogous to Pauli spin matrices $\sigma_{x}$, $\sigma_{y}$ and $\sigma_{z}$
respectively. The matrix representation for these gates is given as,

$\displaystyle I$ $\displaystyle=\sigma_{0}=\left(\begin{array}[]{cc}1&0\\\
0&1\\\ \end{array}\right),\,\,\hskip
5.0ptX=\sigma_{x}=\left(\begin{array}[]{cc}0&1\\\ 1&0\\\ \end{array}\right)$
(3) $\displaystyle Y$
$\displaystyle=\sigma_{y}=\left(\begin{array}[]{cc}0&-i\\\ i&0\\\
\end{array}\right),\,\,Z=\sigma_{z}=\left(\begin{array}[]{cc}1&0\\\ 0&-1\\\
\end{array}\right)$

Here, $I$ is the identity gate and it does not change the input state. $X$
gate is known as bit-flip because it flips the state $\ket{H}$ to $\ket{V}$
and vice-versa, whereas $Z$ gate is known as phase-flip because it flips the
sign of $\ket{V}$, while leaving $\ket{H}$ unchanged. $iY$ gate is used to
flip the bit and phase simultaneously and it can be realized using the
property $[\sigma_{x},\,\sigma_{z}]=2i\sigma_{y}$. Therefore,

$\displaystyle\sigma_{x}\sigma_{z}$ $\displaystyle=i\sigma_{y}$ (4)
$\displaystyle XZ$ $\displaystyle=iY=\left(\begin{array}[]{cc}0&1\\\ -1&0\\\
\end{array}\right)$

Now consider our initial state is $\ket{\Psi}^{+}$ (Eqn. 1b). Applying $Z$ and
$X$ gate operation on signal photon of state $\ket{\Psi}^{+}$ results
$\ket{\Psi}^{-}$ and $\ket{\Phi}^{+}$ respectively,

$\displaystyle Z_{s}\ket{\Psi}^{+}$ $\displaystyle=\ket{\Psi}^{-}$ (5a)
$\displaystyle X_{s}\ket{\Psi}^{+}$ $\displaystyle=\ket{\Phi}^{+}$ (5b) Again,
applying $Y$ gate operation on signal photon results $\ket{\Phi}^{+}$ and
$\ket{\Phi}^{-}$ respectively, $\displaystyle iY\ket{\Psi}^{+}$
$\displaystyle=X_{s}Z_{s}\ket{\Psi}^{+}=\ket{\Phi}^{-}$ (5c) Or one can apply
$X$ gate in signal photon and $Z$ gate on idler photon and vice-versa. Hence
$Y$ gate operation can also be written as,
$iY\ket{\Psi}^{+}=X_{s}Z_{i}\ket{\Psi}^{+}=\ket{\Phi}^{-}$ (5d)

These states transformation are illustrated in Fig. 2.

Figure 2: Transformation of Bell states using single-qubit logic gates

These gates can be experimentally realized in laboratory using a half-wave
plate (HWP). HWP is used to rotate the polarization of photons [28, 29]. It
retards the polarization state of photons by adding a phase difference between
the two orthogonal polarization components $\ket{H}$ and $\ket{V}$ and the
phase difference depends on wavelength $\lambda$ of incident photons,
birefringence properties ($\Delta n$) of the crystal, and thickness $d$ of the
HWP,

$\Delta\phi=\frac{2\pi d\Delta n}{\lambda}$ (6)

where $\Delta n$ is the difference of refractive index along slow axis and
fast axis. $\lambda$, $\Delta n$, and $d$ are chosen in such a way that the
phase difference between polarization components is $\pi$. The action of HWP
can be represented by Jones matrix notation,

$\hat{U}_{\text{hwp}}(\theta)=\left(\begin{array}[]{cc}\text{cos}\,2\theta&\text{sin}\,2\theta\\\
\text{sin}\,2\theta&-\text{cos}\,2\theta\\\ \end{array}\right)$ (7)

where $\theta$ is the angle between fast axis and horizontal axis. When the
photons of particular polarization transmit through the HWP, the polarization
changes according to HWP angle $\theta$. $Z$ and $X$ gate can be realized by
aligning the fast axis of HWP at an angle 0∘ and 45∘ with respect to
horizontal axis and $i$Y gate can be realized by using two HWPs at two
different angles 0∘ and 45∘.

$\displaystyle\hat{U}_{\text{hwp}}(0^{\circ})$
$\displaystyle=\left(\begin{array}[]{cc}1&0\\\ 0&-1\\\
\end{array}\right)=\sigma_{z}$ (8)
$\displaystyle\hat{U}_{\text{hwp}}(45^{\circ})$
$\displaystyle=\left(\begin{array}[]{cc}0&1\\\ 1&0\\\
\end{array}\right)=\sigma_{x}$
$\displaystyle\hat{U}_{\text{hwp}}(0^{\circ})\hat{U}_{\text{hwp}}(45^{\circ})$
$\displaystyle=\left(\begin{array}[]{cc}0&1\\\ -1&0\\\
\end{array}\right)=i\sigma_{y}$

Using Eqn. 2b and 8, we will show that how one can use only HWP to perform all
quantum gates,

I. Verification of Z gate:

$\displaystyle\hat{U}_{\text{hwp}}(0^{\circ})\ket{H}$
$\displaystyle=\;\ket{H},\hskip
10.0pt\hat{U}_{\text{hwp}}(0^{\circ})\ket{V}=-\ket{V},$ (9a)
$\displaystyle\hat{U}_{\text{hwp}}(0^{\circ})\ket{D}$
$\displaystyle=\;\ket{A},\hskip
10.0pt\hat{U}_{\text{hwp}}(0^{\circ})\ket{A}=\;\ket{D}.$ (9b)

II. Verification of X gate:

$\displaystyle\hat{U}_{\text{hwp}}(45^{\circ})\ket{H}$
$\displaystyle=\ket{V},\hskip
10.0pt\hat{U}_{\text{hwp}}(45^{\circ})\ket{V}=\ket{H}$ (10a)
$\displaystyle\hat{U}_{\text{hwp}}(45^{\circ})\ket{D}$
$\displaystyle=\ket{D},\hskip
10.0pt\hat{U}_{\text{hwp}}(45^{\circ})\ket{A}=-\ket{A}$ (10b)

Eqn. 9 shows that apart from quarter-wave plate (QWP), HWP can also be used
for phase-flip. Using Eqn. 9 and 10, one can transform bell states using HWPs
only. Table 1 shows the transformation of Bell state using HWP.

Input state |  Quantum gate |  HWPa |  HWPb |  Output state
---|---|---|---|---
$\ket{\Psi}^{+}$ |  I |  |  |  $\ket{\Psi}^{+}$
$\ket{\Psi}^{+}$ |  Z |  0∘ |  |  $\ket{\Psi}^{-}$
$\ket{\Psi}^{+}$ |  X |  45∘ |  |  $\ket{\Phi}^{+}$
$\ket{\Psi}^{+}$ |  XZ |  0∘ |  45∘ |  $\ket{\Phi}^{-}$
Table 1: Operation of quantum gate and corresponding HWP angles. The gate
transforms an input state to an output state.

##  III Experimental setup

###  III.1 Entangled photon pair generation

Fig. 3 shows the experimental setup for the generation of polarization
entangled Bell state $\ket{\Psi}^{+}$. Horizontally polarized continuous wave
laser (Toptica iBeam smart) of wavelength 405 nm and pump power 2.5 mW is used
to pump the nonlinear type-I $\beta$-barium borate ($\beta$-BBO) crystal of
thickness 2 mm and transverse dimensions of 6 mm × 6 mm with an optic axis
oriented at $29.97^{\circ}$ to the normal incidence. This will produce two
vertically polarized photon pair of wavelength 810 nm each. A band pas filter
(BPF) is used to block 405 nm photons while transmitting the 810 nm photons.
Both the photons are separated by prism mirror (PM) in two different arms.
Both the photons are again combined at the input port of the 50:50 beam
splitter (BS). A half-wave plate (HWP1) is used in one of the input arm of
beam-splitter (BS) to flip the polarization state of one photon so that both
the photons become orthogonally polarized to each other. A translation delay
stage is used to compensate the path difference between signal and idler.

Figure 3: Experimental setup for the generation of entangled state
$\ket{\Psi}^{+}$. L1, L2 \- Lenses, BBO - $\beta$-Barium borate crystal, BPF -
Band pass filter, PM - Prism mirror, M - Mirror, BS - 50:50 Beam splitter, PBS
- Polarizing beam splitter, HWP - Half-wave plate, FC - Fiber coupler, SMF -
Single mode fiber, SPCM - single photon counting module, CC - Coincidence
counter

For the detection part, we have used combination of HWP and PBS at each output
port of BS. After combining at the BS both the photons will exit from two
different output ports and get coupled into single mode fiber (SMF)
(P1-780A-FC-2, Thorlabs) with the help of fiber couplers (FC) (CFC-5X-B,
Thorlabs) . These SMFs are connected to the single photon counting module
(SPCM) detectors (SPCM-AQRH-16-FC, Excelitas) having a time resolution 350 ps.
To measure the number of correlated photon pairs, Both the SPCMs are then
connected to a coincidence counter (CC), IDQuantique-ID800, having a time
resolution of 81 ps. The photons pairs that exit from same output port will
not contribute to coincidence counts. This setup can directly produce
$\ket{\Psi}^{\pm}$ state at the output ports of BS by simply adjusting the
phase difference between both the photons.

Figure 4: Experimental setup for the transformation of entangled state
$\ket{\Psi^{+}}$ into $\ket{\Psi}^{-}\,\text{and}\,\ket{\Phi}^{+}$ by
performing $Z$ and $X$ gate operation respectively using HWPa only. Additional
HWPb along with HWPa is used to transform the state $\ket{\Psi^{+}}$ into
$\ket{\Phi}^{-}$.

###  III.2 Transformation of Bell state

Fig. 4 shows the experimental setup of the transformation of Bell state. First
we generated Bell state $\ket{\Psi^{+}}$ using Fig. 3 and then in the same
setup, we kept HWPa and HWPb in the output port of BS as shown in Fig. 4. We
transform state $\ket{\Psi^{+}}$ into states $\ket{\Psi}^{-}$ and
$\ket{\Phi}^{+}$ by performing $Z$ and $X$ gate respectively in output port
$a$ using HWPa. One extra HWPb along with HWPa is used to produce
$\ket{\Phi}^{-}$ state by performing $Y$ gate operation. In this case both the
HWPs, a and b, are rotated at different orientations 0∘ and 45∘ respectively.
In our case, both HWP is placed at output port $a$ and $b$, but one can put
both HWPs at same output port. Both the cases will give the same result.

##  IV Results and discussion

In this section, we present experimental results obtained for all the four
polarization-entangled Bell states. In Fig. 4a, when HWPa is aligned to 0∘
then it will act as Z gate and hence it will flip the sign of $\ket{V}_{a}$,
while leaving $\ket{H}_{a}$ unchanged. In D/A basis, $\ket{D}_{a}$ will
convert to $\ket{A}_{a}$ and vice-versa. This will transferred the state from
$\ket{\Psi}^{+}$ to $\ket{\Psi}^{-}$. The initial state is given by,

$\ket{\Psi}^{+}=\left\\{\begin{array}[]{lr}\frac{1}{\sqrt{2}}(\ket{H}_{a}\ket{V}_{b}+\ket{V}_{a}\ket{H}_{b})&\,H/V\,\text{basis}\\\
\frac{1}{\sqrt{2}}(\ket{D}_{a}\ket{D}_{b}-\ket{A}_{a}\ket{A}_{b})&\,D/A\,\text{basis}\end{array}\right.$
(11)

For $\theta_{a}=0^{\circ}$, output state will be,

$U_{\text{hwp}}(\theta_{a}=0^{\circ})\ket{\Psi}^{+}=\ket{\Psi}^{-}$
$\ket{\Psi}^{-}=\left\\{\begin{array}[]{lr}\frac{1}{\sqrt{2}}(\ket{H}_{a}\ket{V}_{b}-\ket{V}_{a}\ket{H}_{b})&\,H/V\,\text{basis}\\\
\frac{1}{\sqrt{2}}(\ket{A}_{a}\ket{D}_{b}-\ket{D}_{a}\ket{A}_{b})&\,D/A\,\text{basis}\end{array}\right.$
(12)

Now if we again rotate the HWPa to 45∘ then it will act as a X gate and hence
it will convert $\ket{H}_{a}$ to $\ket{V}_{a}$ and vice-versa. Also it will
flip the sign of $\ket{A}_{a}$, while leaving $\ket{D}_{a}$
unchanged.Therefore, for $\theta_{a}=45^{\circ}$, the state will become, For
$\theta_{a}=45^{\circ}$,

$U_{\text{hwp}}(\theta_{a}=45^{\circ})\ket{\Psi}^{+}=\ket{\Phi}^{+}$
$\ket{\Phi}^{+}=\left\\{\begin{array}[]{lr}\frac{1}{\sqrt{2}}(\ket{H}_{a}\ket{H}_{b}+\ket{V}_{a}\ket{V}_{b})&\,H/V\,\text{basis}\\\
\frac{1}{\sqrt{2}}(\ket{D}_{a}\ket{D}_{b}+\ket{A}_{a}\ket{A}_{b})&\,D/A\,\text{basis}\end{array}\right.$
(13)

Now to transfer $\ket{\Psi}^{+}$ state to $\ket{\Phi}^{-}$ state we used one
more HWPb in output port b of BS (Fig. 4b). we aligned HWPa to 0∘ and HWPb to
45∘ to get $\ket{\Phi}^{-}$ state. Therefore, for $\theta_{a}=0^{\circ}$ and
$\theta_{b}=45^{\circ}$, the output state will be,

$U_{\text{hwp}}(\theta_{a}=0^{\circ})U_{\text{hwp}}(\theta_{b}=45^{\circ})\ket{\Psi}^{+}=\ket{\Phi}^{-}$
$\ket{\Phi}^{-}=\left\\{\begin{array}[]{lr}\frac{1}{\sqrt{2}}(\ket{H}_{a}\ket{H}_{b}-\ket{V}_{a}\ket{V}_{b})&\,H/V\,\text{basis}\\\
\frac{1}{\sqrt{2}}(\ket{A}_{a}\ket{D}_{b}+\ket{D}_{a}\ket{A}_{b})&\,D/A\,\text{basis}\end{array}\right.$
(14)

(a)  (b)

Figure 5: Experimental observation of polarization correlations for (a)
$\ket{\Psi}^{+}$ and (b) $\ket{\Psi}^{-}$ state. Error bars indicate
statistical uncertainty of one standard deviation.

We have recorded the measurement in coincidence basis. Polarization
measurements are carried out using a combination of HWP and PBS in each arm.
This combination will act as a projection operator. The measurement is
performed by fixing the angle of HWP2, $\alpha$ in one output port, while
changing the angle of HWP3, $\beta$ in another port. The transmitted part of
PBS is coupled into a single mode fiber. For example, if we want to measure
the diagonal polarization (D) of photon then first we will rotate the diagonal
polarization (D) to horizontal (H) with the help of HWP so that photons get
transmitted through the PBS and get coupled into single mode fiber.

Coincidence detection |  $\ket{\Psi}^{+}$ |  $\ket{\Psi}^{-}$ |  $\ket{\Phi}^{+}$ |  $\ket{\Phi}^{-}$
---|---|---|---|---
HH/VV |  - |  - |  ✓ |  ✓
HV/VH |  ✓ |  ✓ |  - |  -
DD/AA |  ✓ |  - |  ✓ |  -
DA/AD |  - |  ✓ |  - |  ✓
Table 2: Theoretical prediction of Detecting entangled photon pair in H/V/D/A basis Coincidence detection |  $\ket{\Psi}^{+}$ |  $\ket{\Psi}^{-}$ |  $\ket{\Phi}^{+}$ |  $\ket{\Phi}^{-}$
---|---|---|---|---
HH/VV |  13/14 |  18/15 |  698/751 |  731/784
HV/VH |  660/771 |  763/724 |  21/14 |  18/15
DD/AA |  798/675 |  17/14 |  680/738 |  17/15
DA/AD |  13/16 |  653/768 |  14/15 |  825/735
Table 3: Experimental observation of detecting counts of entangled photon pair
in H/V/D/A basis. The coincidence counts are recorded with a coincidence
window of 1.62 ns and integration time of 1 s.

(a)  (b)

Figure 6: Experimental observation of polarization correlations for (a)
$\ket{\Phi}^{+}$ and (b) $\ket{\Phi}^{-}$ state. Error bars indicate
statistical uncertainty of one standard deviation.

Table 2 shows the theoretical discrimination between all the four Bell state.
To verify the theoretical prediction, we have recorded coincidence counts per
second in H/V and D/A basis for each Bell state. Coincidence detection is
experimentally measured number of photon pairs for certain polarization state.
Table 3 shows the experimental data. Our experimental result is in excellent
agreement with the theoretical prediction and by comparing table 2 and 3, one
can predict which bell state is measured. The experimentally observed
polarization correlations are given as visibility curves in Fig. 5 and 6. The
total coincidence counts per second are plotted in y axis as a function of
HWP3 angle $2\beta$, while HWP2 angle $2\alpha$ is fixed.

To check the quality of entanglement, we calculated the Bell-CHSH parameter S.
The S value greater than 2 ensures the existence of quantum entanglement
between two photons. In our experiment, the average value of visibilities and
the estimated Bell-CHSH parameter S for $\ket{\Psi}^{+}$,
$\ket{\Psi}^{-}$,$\ket{\Phi}^{+}$, and $\ket{\Phi}^{-}$ are shown in table 4.
Because of the unitary transformation the S value for all the four Bell state
is remains unchanged.

Figure 7: Density matrices are calculated for entangled states (a)$\ket{\Psi}^{+}$ and (b) $\ket{\Psi}^{-}$ (c)$\ket{\Phi}^{+}$ and (d) $\ket{\Phi}^{-}$. F is the state fidelity which is the measure of state overlap  Bell state | Avg. Vis.($\%$) | S Value | Fidelity (F)
---|---|---|---
$\ket{\Psi}^{+}$ | 96.18 $\pm$ 5.03 | 2.72 $\pm$ 0.07 | 0.9625
$\ket{\Psi}^{-}$ | 95.66$\pm$ 4.28 | 2.71 $\pm$ 0.06 | 0.9626
$\ket{\Phi}^{+}$ | 95.61 $\pm$ 3.75 | 2.70 $\pm$ 0.05 | 0.9647
$\ket{\Phi}^{-}$ | 95.84 $\pm$ 3.17 | 2.71 $\pm$ 0.05 | 0.9608
Table 4: Calculated average visibility,Bell-CHSH parameter S, and fidelity F.

Quantum state tomography is performed to verify the theoretical prediction by
calculating the state fidelity. The density matrix for all the four Bell
states are shown in Fig.7. The state fidelity is around $\sim 96\%$ for each
Bell state which shows the successful transformation of Bell states using
HWPs.

##  V Conclusion

In this article, we have shown that only half-wave plate (HWP) is enough to
perform Pauli Z gate and X gate and it become the easiest way to transform
entangled state. We first prepared the polarization-entangled Bell state
$\ket{\Psi}^{+}$ using type-I $\beta$-BBO crystal and 50:50 beam-splitter (BS)
and then manipulated the initial state in order to achieve rest of the
entangled state. We calculated the Bell-CHSH parameter S for each Bell state
and since the transformation is unitary, we found out that S value,
coincidence counts, and visibility for all the Bell state remains unchanged.
Quantum state tomography is also performed to check the state fidelity with
the desired state. The presented result may find applications in quantum
communication and quantum information protocols, especially where the control
over basis of maximally entangled state is required.

## Disclosures

The authors declare no conflicts of interest related to this article

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\land (\sigma_2\in R(F_2) \land \sigma_3 \in R(F_3) \land
\sigma_2 \approx \sigma_3) \land \sigma_1 \approx (\sigma_2\cup\sigma_3)\} \\
& = \{(\sigma_1\cup\sigma_2)\cup\sigma_3\mid (\sigma_1\in R(F_1)
\land \sigma_2\in R(F_2)) \land \sigma_3 \in R(F_3) \land
\sigma_2 \approx \sigma_3 \land \sigma_1 \approx
\tag{by \Cref{frm:proof:lem:homo1}} \\
& = \{(\sigma_1\cup\sigma_2)\cup\sigma_3\mid (\sigma_1\in R(F_1)
\land \sigma_2\in R(F_2)) \land \sigma_3 \in R(F_3) \land
\sigma_2 \approx \sigma_3 \land \sigma_1\approx\sigma_2 \land
\sigma_1 \approx \sigma_3\}\\
& = \{(\sigma_1\cup\sigma_2)\cup\sigma_3\mid (\sigma_1\in R(F_1)
\land \sigma_2\in R(F_2) \land \sigma_1 \approx \sigma_2) \land \sigma_3 \in R(F_3) \land
(\sigma_1 \cup \sigma_2)\approx\sigma_3\} \\
& = \{(\sigma_1\cup\sigma_2)\cup\sigma_3\mid
(\sigma_1\cup\sigma_2)\in R(F_1) \otimes R(F_2) \land \sigma_3 \in R(F_3) \land
(\sigma_1 \cup \sigma_2)\approx\sigma_3\} \\
= (R(F_1) \otimes R(F_2)) \otimes R(F_3)
\end{align*}

Hence, $(\mathcal{R},\otimes)$ is a semi-group, too.
<Ref> then completes the proof.

For this we only need to show that any two risk states
$\sigma,\sigma'\in R$ are

* comparable and

* antisymmetric:
$[\sigma]_{\sim_{\mathsf{s}}} \leq_{\mathsf{m}} [\sigma']_{\sim_{\mathsf{s}}} \land
[\sigma']_{\sim_{\mathsf{s}}} \leq_{\mathsf{m}} [\sigma]_{\sim_{\mathsf{s}}} \Rightarrow [\sigma]_{\sim_{\mathsf{s}}} =_{\mathsf{m}}

We show (i) by showing that the conditions (a) and (b) guarantee the
comparability of any two risk states in $R$ based on the interval
order $\leq$ as defined above and use the fact that comparability
can be dropped from
$R/\sim_{\mathsf{s}}$ to $R$ using
$[\sigma]_{\sim_{\mathsf{s}}} \leq_{\mathsf{m}} [\sigma']_{\sim_{\mathsf{s}}} \iff
\forall
\sigma\in[\sigma]_{\sim_{\mathsf{s}}},\sigma''\in[\sigma']_{\sim_{\mathsf{s}}} \colon
\sigma \leq_{\mathsf{m}} \sigma''$ (<Ref>).  We have
to consider the following three
cases to complete the sub-proof of (i):

Case “no risk factors activated”:
$\varnothing \geq \varnothing \lor \varnothing \subset \varnothing$
validates (b) and lack of comparability invalidates (a),
yet we have $[\sigma]_{\sim_{\mathsf{s}}} \leq_{\mathsf{m}} [\sigma']_{\sim_{\mathsf{s}}}$.

Case “at least one risk factor activated only in $\sigma$”:
Let $(a,b) = S(\sigma)$.
$(a,b) \geq \varnothing \lor (a,b) \subset \varnothing$ invalidates
(a) and (b).  However, the observation
$\varnothing \geq (a,b) \lor \varnothing \subset (a,b)$ yields
$[\sigma']_{\sim_{\mathsf{s}}} \leq_{\mathsf{m}} [\sigma]_{\sim_{\mathsf{s}}}$.  The dual of this
case works analogously.

Case “at least one risk factor activated in both $\sigma,\sigma'$”:
Let $(a,b) = S(\sigma)$ and $(c,d) = S(\sigma')$.  We need to show
$(a,b) \geq (c,d) \lor (a,b) \subset (c,d)$ for all
$a,b,c,d \in\mathbb{R}$: We can assume $a\leq b \land c\leq d$ by
definition of intervals. Then, we face the following cases:

* $c \leq a \land d \leq b$ validates (a) by definition of $\leq$
over intervals.

* $c > a \land d \leq b$ validates (b).

* $c \leq a \land d > b$ implies (b) for the dual case
$[\sigma']_{\sim_{\mathsf{s}}} \leq_{\mathsf{m}} [\sigma]_{\sim_{\mathsf{s}}}$.

* $c > a \land d > b$ implies (a) for the dual case.

Hence, from each of these four cases, either $[\sigma]_{\sim_{\mathsf{s}}}
\leq_{\mathsf{m}} [\sigma']_{\sim_{\mathsf{s}}}$
or $[\sigma']_{\sim_{\mathsf{s}}} \leq_{\mathsf{m}} [\sigma]_{\sim_{\mathsf{s}}}$ follows.

Then, we show (ii) by contradiction: Assuming
$[\sigma]_{\sim_{\mathsf{s}}} \leq_{\mathsf{m}} [\sigma']_{\sim_{\mathsf{s}}} \land
[\sigma']_{\sim_{\mathsf{s}}} \leq_{\mathsf{m}} [\sigma]_{\sim_{\mathsf{s}}}$, we have
$\forall \sigma\in[\sigma]_{\sim_{\mathsf{s}}},\sigma''\in[\sigma']_{\sim_{\mathsf{s}}}
\colon \sigma \leq_{\mathsf{m}} \sigma'' \land \sigma \geq_{\mathsf{m}} \sigma''$ by
dropping from $R/\sim_{\mathsf{s}}$.  Now, we claim that
$[\sigma]_{\sim_{\mathsf{s}}} \not=_{\mathsf{m}} [\sigma']_{\sim_{\mathsf{s}}}$.  Hence,
$\exists \sigma\in[\sigma]_{\sim_{\mathsf{s}}},
\sigma''\in[\sigma']_{\sim_{\mathsf{s}}}\colon \sigma \not=_{\mathsf{m}} \sigma''$ and,
$\exists \sigma\in[\sigma]_{\sim_{\mathsf{s}}},
\sigma''\in[\sigma']_{\sim_{\mathsf{s}}}\colon \sigma \not\leq_{\mathsf{m}} \sigma'' \lor \sigma
\not\geq_{\mathsf{m}} \sigma''$.  The latter contradicts our assumption.

Fix a finite $F \subseteq \mathbb{F}$ and a pair
$\sigma, \sigma' \in R(F)$.  The proof is by induction over $F$ and
relies on the assumption that, for any $\rfct\in F$, the phase $\activ$ is the
unique maximal element of $\preceq_{\mathsf{\rfct}}$ and that
$\mathit{active}$ only returns such elements:

Induction start $F_0 = \varnothing$:
$\sigma|_{\varnothing} \preceq_{\mathsf{m}} \sigma'|_{\varnothing}$ holds
trivially and so does
$active(\sigma'|_{\varnothing}) \subseteq

Induction step (IS) $F_{n+1} = F_n \cup \{\rfct\}$ where $n \geq 0$ and
$\rfct \in F\setminus F_n$: For the induction hypothesis (IH), assume
$\sigma|_{F_n} \preceq_{\mathsf{m}} \sigma'|_{F_n} \Rightarrow
active(\sigma'|_{F_n}) \subseteq active(\sigma|_{F_n})$.  Then, we
show that
$\sigma|_{F_{n+1}} \preceq_{\mathsf{m}} \sigma'|_{F_{n+1}} \Rightarrow
active(\sigma'|_{F_{n+1}}) \subseteq
active(\sigma|_{F_{n+1}})$ (i.e.,the IS).

For this, prove

* the case of incomparable phases
$(\sigma(\rfct),\sigma'(\rfct)) \not\in\; \preceq_{\mathsf{\rfct}}$: In this case, the
state pair gets incomparable and the implication is trivially fulfilled,

* the inverse case $\sigma(\rfct) \succ_{\rfct} \sigma'(\rfct)$: In
this case, the state pair gets incomparable and the implication is again
trivially fulfilled, and

* the aligned case $\sigma(\rfct) \preceq_{\mathsf{\rfct}} \sigma'(\rfct)$: In
this case, the state pair stays comparable.  However, $\sigma'(\rfct)$ can
either be $\activ$ (hence, $\sigma(\rfct) = \activ$ and maintaining
$\subseteq$), $\mitig$ (hence,
$\sigma(\rfct) \in \{\activ,\mitig,\inact\}$ and maintaining
$\subseteq$), or $\inact$ (see former sub-case).

Having proved these cases completes the induction step by
establishing IS.

For $\precsim_{\mathsf{m}}$, we only have to substitute case 1 of the IS: For
$\rfct$, the smallest $\preceq_{\mathsf{\rfct}}$ after <Ref>
implies incomparability at most for
$(\inact,\mitig)$ and $(\mitig,\inact)$.  These two phase
pairs of $\rfct$ would not alter $\mathit{active}$ of both states and hence
maintain $\subseteq$ as well.

From associativity of set intersection in
<Ref>, we know that the order in which
constraints are applied to $R \times R$ does not matter.
Consequently, we also have
$[\hspace{-.15em}[c]\hspace{-.15em}]_c \supseteq [\hspace{-.15em}[\{c,c'\}]\hspace{-.15em}]_c$ for any
$c,c' \in \mathcal{C}$, that is,constraints only prune
$R \times R$.
First, we prove the following lemma:

\begin{equation}
\label{eq:risk:constraints:exchange:helper}
[[\mathfrak{R}]_{\{c\}}]_{C} = [\mathfrak{R}]_{\{c\} \cup C}
\end{equation}

Fix $\tran[]{\sigma}{e}{\sigma'}$.

* If $\tran[]{\sigma}{e}{\sigma'} \in [\hspace{-.15em}[c]\hspace{-.15em}]_c$ then
$[\cdot]_C$-step is applicable to $[\mathfrak{R}]_{\{c\}}$, and if
$\tran[]{\sigma}{e}{\sigma'} \in [\hspace{-.15em}[C]\hspace{-.15em}]_c$ then to
$[[\mathfrak{R}]_{\{c\}}]_C$.  Then, $\tran[]{\sigma}{e}{\sigma'}$ must be
in $[\hspace{-.15em}[c]\hspace{-.15em}]_c \cap [\hspace{-.15em}[C]\hspace{-.15em}]_c$ which by
<Ref> is $[\hspace{-.15em}[\{c\} \cup C]\hspace{-.15em}]_c$.
Because of
$\tran[]{\sigma}{e}{\sigma'} \in [\hspace{-.15em}[\{c\} \cup C]\hspace{-.15em}]_c$,
$[\cdot]_C$-step applies to $[\mathfrak{R}]_{\{c\} \cup C}$.

* If $\tran[]{\sigma}{e}{\sigma'} \not\in [\hspace{-.15em}[c]\hspace{-.15em}]_c$ or
$\tran[]{\sigma}{e}{\sigma'} \not\in [\hspace{-.15em}[C]\hspace{-.15em}]_c$ then the
$[\cdot]_C$-step is not applicable to $[[\mathfrak{R}]_{\{c\}}]_C$.
Because of
$[\hspace{-.15em}[\{c\} \cup C]\hspace{-.15em}]_c = [\hspace{-.15em}[c]\hspace{-.15em}]_c \cap [\hspace{-.15em}[C]\hspace{-.15em}]_c$, this also
holds of $[\mathfrak{R}]_{\{c\} \cup C}$.

* If $\tran[]{\sigma}{e}{\sigma'} \in [\hspace{-.15em}[\{c\} \cup C]\hspace{-.15em}]_c$
then $[\cdot]_C$-step is applicable to $[\mathfrak{R}]_{\{c\} \cup C}$ and,
because of $[\hspace{-.15em}[c]\hspace{-.15em}]_c \cap [\hspace{-.15em}[C]\hspace{-.15em}]_c$, twice to

* If
$\tran[]{\sigma}{e}{\sigma'} \not\in [\hspace{-.15em}[\{c\} \cup C]\hspace{-.15em}]_c$ then
$[\cdot]_C$-step is not applicable to $[\mathfrak{R}]_{\{c\} \cup C}$ and,
because of
$\tran[]{\sigma}{e}{\sigma'} \not\in [\hspace{-.15em}[c]\hspace{-.15em}]_c \cap
[\hspace{-.15em}[C]\hspace{-.15em}]_c$, at most once to $[[\mathfrak{R}]_{\{c\}}]_C$.

Second, we show by induction over $C_1$ that the $[\cdot]_C$-step
rule applies in the same way to both sides of
Induction start with $C_1^0 = \varnothing$: In this case, we have
$[[\mathfrak{R}]_{\varnothing}]_{C_2} = [\mathfrak{R}]_{\varnothing \cup C_2}$ and
because $[\cdot]_C$-step is always applicable,
$[\mathfrak{R}]_{C_2} = [\mathfrak{R}]_{C_2}$.

Induction step (IS) with $C_1^{n+1} = C_1^n \cup \{c\}$ where
$n \geq 0$ and $c \in C_1 \setminus C_1^{n}$:
By assuming $[[\mathfrak{R}]_{C_1^n}]_{C_2} = [\mathfrak{R}]_{C_1^n \cup C_2}$ (IH), we show

\begin{align*}
& [[\mathfrak{R}]_{C_1^{n+1}}]_{C_2}
\tag{by def of IS}
\\
&= [[\mathfrak{R}]_{C_1^n \cup \{c\}}]_{C_2}
\tag{by $\cup$-comm and \Cref{eq:risk:constraints:exchange:helper}}
\\
&= [[[\mathfrak{R}_{\{c\}}]_{C_1^n}]_{C_2}
\tag{by IH}
\\
&= [[\mathfrak{R}_{\{c\}}]_{C_1^n \cup C_2}
\tag{by \Cref{eq:risk:constraints:exchange:helper}}
\\
&= [\mathfrak{R}]_{\{c\} \cup C_1^n \cup C_2}
\tag{by def}
\\
&= [\mathfrak{R}]_{C_1^{n+1} \cup C_2}
\end{align*}