# Linear optics and the problem of Bell-like state discrimination

Jatin Ghai1,2<EMAIL_ADDRESS>Sibasish Ghosh1,2<EMAIL_ADDRESS>1Optics
and Quantum Information Group, The Institute of Mathematical Sciences, C. I.
T. Campus, Taramani, Chennai 600113, India  2Homi Bhabha National Institute,
Training School Complex, Anushaktinagar, Mumbai 400094, India

###### Abstract

A linear optics-based scheme to implement various quantum information
processing tasks is of paramount importance due to ease of implementation and
low noise. Many information-theoretic tasks depend on the successful
discrimination of Bell states. A no-go theorem has been proved in literature
which tells that it is not possible to perfectly discriminate among the four
Bell states by restricting measurement apparatus to linear optical elements.
The success probability is only $50\%$. Through using extra resources such as
hyper entanglement, ancillary entanglement, and even a minimum amount of non-
linearity complete Bell-state discrimination can be achieved. The success
probability for Bell-like state discrimination is only $25\%$. We find that
this can be boosted up to $50\%$ using hyperentanglement in polarization,
momentum, or OAM degrees of freedom of the photons which is in contrast to the
Bell-state discrimination scenario where $100\%$ can be achieved. Furthermore,
we find that by using correlation in time of the photons all four Bell states
can be distinguished with $100\%$ success probability while for the Bell-like
state discrimination, it strictly lies between $25\%$ and $50\%$ depending on
the state parameter with only three Bell-like states being distinguishable. We
also observe a similar contrast when we use ancillary entangled photons. While
the success probability for all four Bell-state discrimination increases as
$1-\frac{1}{2^{N}}$ where N is the number of ancillary photons for Bell-like
states it depends again on the state parameters and can be less than $25\%$ in
some cases. Also adding further ancillary photons decreases the success
probability. We then show that using non-linear gadgets namely SFG $100\%$
success probability can be achieved even for Bell-like state discrimination.

##  I Introduction

Linear optics-based schemes are very promising candidates for building devices
for computation[2] and communication[3] purposes. Photons are very robust to
noise. They have large coherence times. It is also known that KLM scheme
offers a universal model of quantum computing just using linear-optical
elements[4, 5, 6]. Any N-port unitary operation can be implemented using an
array of $\frac{N(N-1)}{2}$ beam splitters[7, 8]. This number grows only
quadratically with N, thus giving a realistic hope for implementing various
quantum operations using linear optical elements. But there are fundamental
limitations on what kind of operations are possible to implement
deterministically if we stick to linear optics only[9, 10, 11, 12]. A very
simple but important example of the inadequacy of the linear optical-based
schemes is that it is unable to distinguish among maximally entangled Bell
states if no ancillary photons and resources are not allowed[1]. By relaxing
the constraints of perfect Bell state discrimination by allowing it to fail
with some probability it can be proved that the maximum success probability of
this scheme is $50\%$ given that there are zero auxiliary photons and no
conditional measurements[13]. Thus linear optics-based schemes are restrictive
in that sense. But various information-theoretic tasks namely quantum
teleportation[14], entanglement swapping[15, 16], super-dense coding[17, 18],
etc. depend on the successful discrimination of Bell states. Hence it is
important to look for ways of increasing the probability of successful Bell
state discrimination from $50\%$. As it turns it can be increased by using
extra ancillary photons[19], hyperentanglement[20, 21, 17], or some minimal
amount of non-linearity[22, 23]. All these methods can boost the success
probability to $100\%$.

A more general version of this problem is distinguishing among the Bell-like
states unambiguously. On the same lines as the treatment in [13] it was shown
in [24] that the optimal success probability for discrimination of Bell-like
states without using any ancillary photons and conditional measurements is
$25\%$. In practical scenarios, one might not be provided with Bell states.
Instead, a weaker resource, namely non-maximally entangled states, might be
present. Thus it is important to distinguish among non-maximally entangled
states. Also it is recently shown that in order to prepare the state with the
highest teleportation fidelity through entanglement swapping sometimes non-
maximally entangled measurements are required instead of Bell
measurements[25]. Thus it is important to distinguish among non-maximally
entangled states.

In this work, we try to devise methods to enhance the success probability of
unambiguous Bell-like state discrimination using extra resources on the same
lines as done in the case of Bell-state discrimination. We for the time being
are not concerned about the specific preparation procedure (that is, using
linear optical or otherwise) of these states. We should also mention here that
throughout the work we are using the dual-rail encoding for the photons and
while distinguishing the two-photon entangled states, no restriction is going
to be imposed on the accessibility of both the photons: they can be either in
one lab or in spatially separated labs. The brief outline of the paper is as
follows:

In Sec. (II) ancillary entanglement in the polarization basis of photons is
used as a resource in the form of hyperentanglement while system qubits are
represented by their momentum DOF. In Sec. (III) and (IV) similar thing is
done for ancillary entanglement in momentum DOF and entanglement in OAM
degrees of freedom respectively with system qubits in polarisation basis in
both the cases. In Sec. (V) correlations in time are used to assist in
distinguishing Bell-like states. Sec. (VI) includes the effect of ancillary
entangled physical qubits on the success probability of unambiguous
discrimination of Bell-like states while Sec. (VII) deals with the effect of
using a non-linear resource called SFG in the context of distinguishing bell-
like states. Finally, we conclude and try to give some future directions in
Sec. (VIII). All results are compared with the corresponding scenario of the
Bell-state discrimination.

##  II Ancillary entanglement in polarisation degrees of freedom

We are given the following four mutually orthogonal Bell-like states:

$\displaystyle|\psi_{1}\rangle=(\sin(\theta)\hat{a}^{{\dagger}}_{1}\hat{a}^{{\dagger}}_{2}+\cos(\theta)\hat{b}^{{\dagger}}_{1}\hat{b}^{{\dagger}}_{2})\ket{0}$
(1)
$\displaystyle|\psi_{2}\rangle=(\cos(\theta)\hat{a}^{{\dagger}}_{1}\hat{a}^{{\dagger}}_{2}-\sin(\theta)\hat{b}^{{\dagger}}_{1}\hat{b}^{{\dagger}}_{2})\ket{0}$
(2)
$\displaystyle|\psi_{3}\rangle=(\sin(\theta)\hat{a}^{{\dagger}}_{1}\hat{b}^{{\dagger}}_{2}+\cos(\theta)\hat{b}^{{\dagger}}_{1}\hat{a}^{{\dagger}}_{2})\ket{0}$
(3)
$\displaystyle|\psi_{4}\rangle=(\cos(\theta)\hat{a}^{{\dagger}}_{1}\hat{b}^{{\dagger}}_{2}-\sin(\theta)\hat{b}^{{\dagger}}_{1}\hat{a}^{{\dagger}}_{2})\ket{0}$
(4)

where $\\{\hat{a}^{{\dagger}}_{1(2)},\hat{b}^{{\dagger}}_{1(2)}\\}$ are the
creation operators for the first (second) photon. These states can be produced
experimentally using the process of SPDC[20]. Our job is to distinguish
between them unambiguously using passive linear optical elements like beam
splitters, wave plates etc.. It has been shown in [24] that by confining
ourselves to the regime of linear optics, we can distinguish only between two
Bell-like states $|\psi_{1}\rangle$ and $|\psi_{2}\rangle$ (or
$|\psi_{3}\rangle$ and $|\psi_{4}\rangle$) with a success probability of
$25\%$. This is half compared to the optimal success probability for Bell
states[13] which is $50\%$ under the same settings. For Bell states it is
known that by using ancillary pair of entangled photons this success
probability can be boosted to $75\%$[19]. Besides this, using extra
entanglement in other degrees of freedom of the same photon pair like momentum
[20], time [26, 21], orbital angular momentum[17] etc. it is known that this
can be boosted to $100\%$. A similar kind of approach is used here also.
Momentum degrees of freedom are chosen as system degrees of freedom while
polarisation degrees of freedom are used for ancillary entanglement. We will
follow the procedure described in [20] very closely. The joint state entangled
in both momentum and polarization degrees of freedom is written as

$\displaystyle|\Theta_{1}\rangle=(\sin(\theta)\hat{a}^{{\dagger}}_{1}\hat{a}^{{\dagger}}_{2}+\cos(\theta)\hat{b}^{{\dagger}}_{1}\hat{b}^{{\dagger}}_{2})\otimes\frac{(\hat{h}^{{\dagger}}_{1}\hat{v}^{{\dagger}}_{2}+\hat{v}^{{\dagger}}_{1}\hat{h}^{{\dagger}}_{2})}{\sqrt{2}}\ket{0}$
(5)
$\displaystyle|\Theta_{2}\rangle=(\cos(\theta)\hat{a}^{{\dagger}}_{1}\hat{a}^{{\dagger}}_{2}-\sin(\theta)\hat{b}^{{\dagger}}_{1}\hat{b}^{{\dagger}}_{2})\otimes\frac{(\hat{h}^{{\dagger}}_{1}\hat{v}^{{\dagger}}_{2}+\hat{v}^{{\dagger}}_{1}\hat{h}^{{\dagger}}_{2})}{\sqrt{2}}\ket{0}$
(6)
$\displaystyle|\Theta_{3}\rangle=(\sin(\theta)\hat{a}^{{\dagger}}_{1}\hat{b}^{{\dagger}}_{2}+\cos(\theta)\hat{b}^{{\dagger}}_{1}\hat{a}^{{\dagger}}_{2})\otimes\frac{(\hat{h}^{{\dagger}}_{1}\hat{v}^{{\dagger}}_{2}+\hat{v}^{{\dagger}}_{1}\hat{h}^{{\dagger}}_{2})}{\sqrt{2}}\ket{0}$
(7)
$\displaystyle|\Theta_{4}\rangle=(\cos(\theta)\hat{a}^{{\dagger}}_{1}\hat{b}^{{\dagger}}_{2}-\sin(\theta)\hat{b}^{{\dagger}}_{1}\hat{a}^{{\dagger}}_{2})\otimes\frac{(\hat{h}^{{\dagger}}_{1}\hat{v}^{{\dagger}}_{2}+\hat{v}^{{\dagger}}_{1}\hat{h}^{{\dagger}}_{2})}{\sqrt{2}}\ket{0}$
(8)

Thus, we have here the task of distinguishing – in a linear optical setup –
the four non-maximally entangled states of two photons, given in Eqns. (1) –
(4) (but, this time, entangled in momentum degrees of freedom instead of
polarization), assisted with a fixed maximally entangled (in polarization)
state of the same two photons. We pass them through the circuit in Fig.1. The
circuit closely resembles the one used in [20]. Two half-wave plates mounted
at an incidence angle of $45^{\circ}$ are placed in paths $b_{1}$ and $b_{2}$.
This scheme differs from the one in [20] as instead of using balanced beam
splitters in paths of both $(a_{1},b_{1})$ and $(a_{2},b_{2})$ here we use a
balanced beam splitter while the beam splitter used in the path of
$(a_{2},b_{2})$ is unbalanced having transmission coefficient
$\cos^{2}(\theta)$. Then a polarising beam splitter is used and single photon
detectors are placed at both of its ends.

In Fig. (1) HWP acts as a CNOT gate where momentum DOFs act as control qubits
and polarisation DOFs act as target qubits. The action of this operation can
be summarised mathematically through the following transformations:

$\displaystyle\hat{a}^{{\dagger}}_{1}\hat{h}^{{\dagger}}_{1}\ket{0}\longrightarrow\hat{a}^{{\dagger}}_{1}\hat{h}^{{\dagger}}_{1}\ket{0}$
(9)
$\displaystyle\hat{a}^{{\dagger}}_{1}\hat{v}^{{\dagger}}_{1}\ket{0}\longrightarrow\hat{a}^{{\dagger}}_{1}\hat{v}^{{\dagger}}_{1}\ket{0}$
(10)
$\displaystyle\hat{a}^{{\dagger}}_{2}\hat{h}^{{\dagger}}_{2}\ket{0}\longrightarrow\hat{a}^{{\dagger}}_{2}\hat{h}^{{\dagger}}_{2}\ket{0}$
(11)
$\displaystyle\hat{a}^{{\dagger}}_{2}\hat{v}^{{\dagger}}_{2}\ket{0}\longrightarrow\hat{a}^{{\dagger}}_{2}\hat{v}^{{\dagger}}_{2}\ket{0}$
(12)
$\displaystyle\hat{b}^{{\dagger}}_{1}\hat{h}^{{\dagger}}_{1}\ket{0}\longrightarrow\hat{b}^{{\dagger}}_{1}\hat{v}^{{\dagger}}_{1}\ket{0}$
(13)
$\displaystyle\hat{b}^{{\dagger}}_{1}\hat{v}^{{\dagger}}_{1}\ket{0}\longrightarrow\hat{b}^{{\dagger}}_{1}\hat{h}^{{\dagger}}_{1}\ket{0}$
(14)
$\displaystyle\hat{b}^{{\dagger}}_{2}\hat{h}^{{\dagger}}_{2}\ket{0}\longrightarrow\hat{b}^{{\dagger}}_{2}\hat{v}^{{\dagger}}_{2}\ket{0}$
(15)
$\displaystyle\hat{b}^{{\dagger}}_{2}\hat{v}^{{\dagger}}_{2}\ket{0}\longrightarrow\hat{b}^{{\dagger}}_{2}\hat{h}^{{\dagger}}_{2}\ket{0}$
(16)

Figure 1: Circuit for the hyper-entangled Bell-like state analyzer where
momentum DOF act as system qubits while polarisation DOF act as control
qubits.

The action of the balanced beam splitter is given as:

$\displaystyle\hat{a}^{{\dagger}}_{1}\ket{0}\longrightarrow\frac{(\hat{a}^{{\dagger}}_{1}+\hat{b}^{{\dagger}}_{1})}{\sqrt{2}}\ket{0}$
(17)
$\displaystyle\hat{b}^{{\dagger}}_{1}\ket{0}\longrightarrow\frac{(\hat{a}^{{\dagger}}_{1}-\hat{b}^{{\dagger}}_{1})}{\sqrt{2}}\ket{0}$
(18)

Similarly, the action of an unbalanced beam splitter can be written as:

$\displaystyle\hat{a}^{{\dagger}}_{2}\ket{0}\longrightarrow(\cos(\theta)\hat{a}^{{\dagger}}_{2}+\sin(\theta)\hat{b}^{{\dagger}}_{2})\ket{0}$
(19)
$\displaystyle\hat{b}^{{\dagger}}_{2}\ket{0}\longrightarrow(\sin(\theta)\hat{a}^{{\dagger}}_{2}-\cos(\theta)\hat{b}^{{\dagger}}_{2})\ket{0}$
(20)

Finally, the action of PBS is mathematically represented as:

$\displaystyle\hat{a}^{{\dagger}}_{1}\hat{h}^{{\dagger}}_{1}\ket{0}\longrightarrow\hat{a}^{{\dagger}}_{1H}\hat{h}^{{\dagger}}_{1}\ket{0}$
(21)
$\displaystyle\hat{a}^{{\dagger}}_{1}\hat{v}^{{\dagger}}_{1}\ket{0}\longrightarrow\hat{a}^{{\dagger}}_{1V}\hat{v}^{{\dagger}}_{1}\ket{0}$
(22)
$\displaystyle\hat{a}^{{\dagger}}_{2}\hat{h}^{{\dagger}}_{2}\ket{0}\longrightarrow\hat{a}^{{\dagger}}_{2H}\hat{h}^{{\dagger}}_{2}\ket{0}$
(23)
$\displaystyle\hat{a}^{{\dagger}}_{2}\hat{v}^{{\dagger}}_{2}\ket{0}\longrightarrow\hat{a}^{{\dagger}}_{2V}\hat{v}^{{\dagger}}_{2}\ket{0}$
(24)
$\displaystyle\hat{b}^{{\dagger}}_{1}\hat{h}^{{\dagger}}_{1}\ket{0}\longrightarrow\hat{b}^{{\dagger}}_{1H}\hat{h}^{{\dagger}}_{1}\ket{0}$
(25)
$\displaystyle\hat{b}^{{\dagger}}_{1}\hat{v}^{{\dagger}}_{1}\ket{0}\longrightarrow\hat{b}^{{\dagger}}_{1V}\hat{v}^{{\dagger}}_{1}\ket{0}$
(26)
$\displaystyle\hat{b}^{{\dagger}}_{2}\hat{h}^{{\dagger}}_{2}\ket{0}\longrightarrow\hat{b}^{{\dagger}}_{2H}\hat{h}^{{\dagger}}_{2}\ket{0}$
(27)
$\displaystyle\hat{b}^{{\dagger}}_{2}\hat{v}^{{\dagger}}_{2}\ket{0}\longrightarrow\hat{b}^{{\dagger}}_{2V}\hat{v}^{{\dagger}}_{2}\ket{0}$
(28)

Being equipped with all the necessary transformations we can now analyze what
will happen to the state in Eqn.(LABEL:Belllikehyperm) when they are passed
through the circuit in Fig.1. We can easily verify that after passing through
the circuit before entering the PBS for detection

$\displaystyle|\Theta_{1}\rangle\longrightarrow\Big{(}$
$\displaystyle\frac{1}{2}\hat{a}^{{\dagger}}_{1}\hat{a}^{{\dagger}}_{2}\hat{h}^{{\dagger}}_{1}\hat{v}^{{\dagger}}_{2}\sin\left(2\theta\right)+\frac{1}{2}\hat{a}^{{\dagger}}_{1}\hat{a}^{{\dagger}}_{2}\hat{h}^{{\dagger}}_{2}\hat{v}^{{\dagger}}_{1}\sin\left(2\theta\right)$
$\displaystyle-\frac{1}{2}\hat{a}^{{\dagger}}_{1}\hat{b}^{{\dagger}}_{2}\hat{h}^{{\dagger}}_{1}\hat{v}^{{\dagger}}_{2}\cos\left(2\theta\right)-\frac{1}{2}\hat{a}^{{\dagger}}_{1}\hat{b}^{{\dagger}}_{2}\hat{h}^{{\dagger}}_{2}\hat{v}^{{\dagger}}_{1}\cos\left(2\theta\right)$
$\displaystyle+\frac{1}{2}\hat{b}^{{\dagger}}_{1}\hat{b}^{{\dagger}}_{2}\hat{h}^{{\dagger}}_{1}\hat{v}^{{\dagger}}_{2}+\frac{1}{2}\hat{b}^{{\dagger}}_{1}\hat{b}^{{\dagger}}_{2}\hat{h}^{{\dagger}}_{2}\hat{v}^{{\dagger}}_{1}\Big{)}\ket{0}$
(29) $\displaystyle|\Theta_{2}\rangle\longrightarrow\Big{(}$
$\displaystyle\frac{1}{2}\hat{a}^{{\dagger}}_{1}\hat{a}^{{\dagger}}_{2}\hat{h}^{{\dagger}}_{1}\hat{v}^{{\dagger}}_{2}\cos\left(2\theta\right)+\frac{1}{2}\hat{a}^{{\dagger}}_{1}\hat{a}^{{\dagger}}_{2}\hat{h}^{{\dagger}}_{2}\hat{v}^{{\dagger}}_{1}\cos\left(2\theta\right)$
$\displaystyle+\frac{1}{2}\hat{a}^{{\dagger}}_{1}\hat{b}^{{\dagger}}_{2}\hat{h}^{{\dagger}}_{1}\hat{v}^{{\dagger}}_{2}\sin\left(2\theta\right)+\frac{1}{2}\hat{a}^{{\dagger}}_{1}\hat{b}^{{\dagger}}_{2}\hat{h}^{{\dagger}}_{2}\hat{v}^{{\dagger}}_{1}\sin\left(2\theta\right)$
$\displaystyle+\frac{1}{2}\hat{a}^{{\dagger}}_{2}\hat{b}^{{\dagger}}_{1}\hat{h}^{{\dagger}}_{1}\hat{v}^{{\dagger}}_{2}+\frac{1}{2}\hat{a}^{{\dagger}}_{2}\hat{b}^{{\dagger}}_{1}\hat{h}^{{\dagger}}_{2}\hat{v}^{{\dagger}}_{1}\Big{)}\ket{0}$
(30) $\displaystyle|\Theta_{3}\rangle\longrightarrow\Big{(}$
$\displaystyle\frac{1}{2}\hat{a}^{{\dagger}}_{1}\hat{a}^{{\dagger}}_{2}\hat{h}^{{\dagger}}_{1}\hat{h}^{{\dagger}}_{2}+\frac{1}{2}\hat{a}^{{\dagger}}_{1}\hat{a}^{{\dagger}}_{2}\hat{v}^{{\dagger}}_{1}\hat{v}^{{\dagger}}_{2}$
$\displaystyle-\frac{1}{2}\hat{a}^{{\dagger}}_{2}\hat{b}^{{\dagger}}_{1}\hat{h}^{{\dagger}}_{1}\hat{h}^{{\dagger}}_{2}\cos\left(2\theta\right)-\frac{1}{2}\hat{a}^{{\dagger}}_{2}\hat{b}^{{\dagger}}_{1}\hat{v}^{{\dagger}}_{1}\hat{v}^{{\dagger}}_{2}\cos\left(2\theta\right)$
$\displaystyle-\frac{1}{2}\hat{b}^{{\dagger}}_{1}\hat{b}^{{\dagger}}_{2}\hat{h}^{{\dagger}}_{1}\hat{h}^{{\dagger}}_{2}\sin\left(2\theta\right)-\frac{1}{2}\hat{b}^{{\dagger}}_{1}\hat{b}^{{\dagger}}_{2}\hat{v}^{{\dagger}}_{1}\hat{v}^{{\dagger}}_{2}\sin\left(2\theta\right)\Big{)}\ket{0}$
(31) $\displaystyle|\Theta_{4}\rangle\longrightarrow\Big{(}$
$\displaystyle-\frac{1}{2}\hat{a}^{{\dagger}}_{1}\hat{b}^{{\dagger}}_{2}\hat{h}^{{\dagger}}_{1}\hat{h}^{{\dagger}}_{2}-\frac{1}{2}\hat{a}^{{\dagger}}_{1}\hat{b}^{{\dagger}}_{2}\hat{v}^{{\dagger}}_{1}\hat{v}^{{\dagger}}_{2}$
$\displaystyle+\frac{1}{2}\hat{a}^{{\dagger}}_{2}\hat{b}^{{\dagger}}_{1}\hat{h}^{{\dagger}}_{1}\hat{h}^{{\dagger}}_{2}\sin\left(2\theta\right)+\frac{1}{2}\hat{a}^{{\dagger}}_{2}\hat{b}^{{\dagger}}_{1}\hat{v}^{{\dagger}}_{1}\hat{v}^{{\dagger}}_{2}\sin\left(2\theta\right)$
$\displaystyle-\frac{1}{2}\hat{b}^{{\dagger}}_{1}\hat{b}^{{\dagger}}_{2}\hat{h}^{{\dagger}}_{1}\hat{h}^{{\dagger}}_{2}\cos\left(2\theta\right)-\frac{1}{2}\hat{b}^{{\dagger}}_{1}\hat{b}^{{\dagger}}_{2}\hat{v}^{{\dagger}}_{1}\hat{v}^{{\dagger}}_{2}\cos\left(2\theta\right)\Big{)}\ket{0}$
(32)

Photons in these states then pass through PBS and are detected by single
photon detectors. We here make two observations:

  * •

For states $|\Theta_{1}\rangle$ and $|\Theta_{2}\rangle$ both the photons are
always detected in detectors oriented in directions perpendicular to each
other. Furthermore,in this case, we have the following observation:

    1. 1.

The terms
$\\{\hat{b}^{{\dagger}}_{1}\hat{b}^{{\dagger}}_{2}\hat{h}^{{\dagger}}_{2}\hat{v}^{{\dagger}}_{1},\hat{b}^{{\dagger}}_{1}\hat{b}^{{\dagger}}_{2}\hat{h}^{{\dagger}}_{1}\hat{v}^{{\dagger}}_{2}\\}$
are uniquely present in $|\Theta_{1}\rangle$. Thus clicking of detectors
$\hat{b}^{{\dagger}}_{1V}$ and $\hat{b}^{{\dagger}}_{2H}$ or
$\hat{b}^{{\dagger}}_{1H}$ and $\hat{b}^{{\dagger}}_{2V}$ indicates
unambiguous detection of this state.

    2. 2.

The erms
$\\{\hat{b}^{{\dagger}}_{1}\hat{a}^{{\dagger}}_{2}\hat{h}^{{\dagger}}_{2}\hat{v}^{{\dagger}}_{1},\hat{b}^{{\dagger}}_{1}\hat{a}^{{\dagger}}_{2}\hat{h}^{{\dagger}}_{1}\hat{v}^{{\dagger}}_{2}\\}$
are uniquely present in $|\Theta_{2}\rangle$. Thus clicking of detectors
$\hat{b}^{{\dagger}}_{1V}$ and $\hat{a}^{{\dagger}}_{2H}$ or
$\hat{b}^{{\dagger}}_{1H}$ and $\hat{a}^{{\dagger}}_{2V}$ indicates
unambiguous detection of this state.

  * •

For states $|\Theta_{3}\rangle$ and $|\Theta_{4}\rangle$ both the photons are
always detected in detectors oriented in directions parallel to each other.
Furthermore, we have, in this case, the following observation:

    1. 1.

Terms
$\\{\hat{a}^{{\dagger}}_{1}\hat{a}^{{\dagger}}_{2}\hat{h}^{{\dagger}}_{2}\hat{h}^{{\dagger}}_{1},\hat{a}^{{\dagger}}_{1}\hat{a}^{{\dagger}}_{2}\hat{v}^{{\dagger}}_{1}\hat{v}^{{\dagger}}_{2}\\}$
are uniquely present in $|\Theta_{3}\rangle$. Thus clicking of detectors
$\hat{a}^{{\dagger}}_{1H}$ and $\hat{a}^{{\dagger}}_{2H}$ or
$\hat{a}^{{\dagger}}_{1V}$ and $\hat{a}^{{\dagger}}_{2V}$ indicates
unambiguous detection of this state.

    2. 2.

Terms
$\\{\hat{a}^{{\dagger}}_{1}\hat{b}^{{\dagger}}_{2}\hat{h}^{{\dagger}}_{2}\hat{h}^{{\dagger}}_{1},\hat{a}^{{\dagger}}_{1}\hat{b}^{{\dagger}}_{2}\hat{v}^{{\dagger}}_{1}\hat{v}^{{\dagger}}_{2}\\}$
are uniquely present in $|\Theta_{4}\rangle$. Thus clicking of detectors
$\hat{a}^{{\dagger}}_{1H}$ and $\hat{b}^{{\dagger}}_{2H}$ or
$\hat{a}^{{\dagger}}_{1V}$ and $\hat{b}^{{\dagger}}_{2V}$ indicates
unambiguous detection of this state.

There will be detection events where some ambiguity will be there about which
state is present. Thus success probability is necessarily not $100\%$. If we
consider that all four states are provided to us with equal probability then
the success probability is:

$\displaystyle P_{succ}$
$\displaystyle=\frac{1}{4}\Big{(}\frac{1}{4}+\frac{1}{4}\Big{)}+\frac{1}{4}\Big{(}\frac{1}{4}+\frac{1}{4}\Big{)}+\frac{1}{4}\Big{(}\frac{1}{4}+\frac{1}{4}\Big{)}+\frac{1}{4}\Big{(}\frac{1}{4}+\frac{1}{4}\Big{)}$
$\displaystyle=\frac{1}{2}$ (33)

Thus success probability of unambiguous state discrimination is $50\%$.
Comparing it with [24] we note that not only success probability is boosted
from $25\%$, but also all four Bell-like states can be distinguished as there,
only two of them at a time ($|\psi_{3}\rangle$ and $|\psi_{4}\rangle)$ or
$(|\psi_{1}\rangle$ and $|\psi_{2}\rangle$) can be distinguished. Thus
hyperentanglement acts as a resource in order to discriminate among them just
like it happens in Bell state analysis(BSA) although success probability
doesn’t reach $100\%$ as it reaches for Bell states. It is, of course,
important to mention here that the optimal success probability in the case of
unambiguous discrimination of the four Bell-like states (supplied with equal
apriori probabilities) of two photons in their momenta degrees of freedom
while the two photons are also maximally entangled in their polarization
degrees of freedom – as given in Eqns. (5) - (8) – may turn out to be more
than $50\%$.

##  III Ancillary entanglement in momentum degrees of freedom

We can also consider polarisation degrees of freedom as system qubits while
momentum degrees of freedom serve as a source of ancillary entanglement.
Similar to the previous section joint state of four Bell-like states in
polarisation basis and ancillary qubits in the momentum basis can be written
as

$\displaystyle|\Xi_{1}\rangle=(\sin(\theta)\hat{h}^{{\dagger}}_{1}\hat{h}^{{\dagger}}_{2}+\cos(\theta)\hat{v}^{{\dagger}}_{1}\hat{v}^{{\dagger}}_{2})\otimes\frac{(\hat{a}^{{\dagger}}_{1}\hat{b}^{{\dagger}}_{2}+\hat{b}^{{\dagger}}_{1}\hat{a}^{{\dagger}}_{2})}{\sqrt{2}}\ket{0}$
(34)
$\displaystyle|\Xi_{2}\rangle=(\cos(\theta)\hat{h}^{{\dagger}}_{1}\hat{h}^{{\dagger}}_{2}-\sin(\theta)\hat{v}^{{\dagger}}_{1}\hat{v}^{{\dagger}}_{2})\otimes\frac{(\hat{a}^{{\dagger}}_{1}\hat{b}^{{\dagger}}_{2}+\hat{b}^{{\dagger}}_{1}\hat{a}^{{\dagger}}_{2})}{\sqrt{2}}\ket{0}$
(35)
$\displaystyle|\Xi_{3}\rangle=(\sin(\theta)\hat{h}^{{\dagger}}_{1}\hat{v}^{{\dagger}}_{2}+\cos(\theta)\hat{v}^{{\dagger}}_{1}\hat{h}^{{\dagger}}_{2})\otimes\frac{(\hat{a}^{{\dagger}}_{1}\hat{b}^{{\dagger}}_{2}+\hat{b}^{{\dagger}}_{1}\hat{a}^{{\dagger}}_{2})}{\sqrt{2}}\ket{0}$
(36)
$\displaystyle|\Xi_{4}\rangle=(\cos(\theta)\hat{h}^{{\dagger}}_{1}\hat{v}^{{\dagger}}_{2}-\sin(\theta)\hat{v}^{{\dagger}}_{1}\hat{h}^{{\dagger}}_{2})\otimes\frac{(\hat{a}^{{\dagger}}_{1}\hat{b}^{{\dagger}}_{2}+\hat{b}^{{\dagger}}_{1}\hat{a}^{{\dagger}}_{2})}{\sqrt{2}}\ket{0}$
(37)

The circuit used for state discrimination is shown in Fig.2. Here first,
photons pass through PBS which acts as a CNOT operation with polarisation DOF
as the system qubit while momentum DOF as the control qubit. The
transformation relations are given as:

$\displaystyle\hat{h}^{{\dagger}}_{1}\hat{a}^{{\dagger}}_{1}\ket{0}\longrightarrow\hat{h}^{{\dagger}}_{1}\hat{a}^{{\dagger}}_{1}\ket{0}$
(38)
$\displaystyle\hat{h}^{{\dagger}}_{1}\hat{b}^{{\dagger}}_{1}\ket{0}\longrightarrow\hat{h}^{{\dagger}}_{1}\hat{b}^{{\dagger}}_{1}\ket{0}$
(39)
$\displaystyle\hat{h}^{{\dagger}}_{2}\hat{a}^{{\dagger}}_{2}\ket{0}\longrightarrow\hat{h}^{{\dagger}}_{2}\hat{a}^{{\dagger}}_{2}\ket{0}$
(40)
$\displaystyle\hat{h}^{{\dagger}}_{2}\hat{b}^{{\dagger}}_{2}\ket{0}\longrightarrow\hat{h}^{{\dagger}}_{2}\hat{b}^{{\dagger}}_{2}\ket{0}$
(41)
$\displaystyle\hat{v}^{{\dagger}}_{1}\hat{a}^{{\dagger}}_{1}\ket{0}\longrightarrow\hat{v}^{{\dagger}}_{1}\hat{b}^{{\dagger}}_{1}\ket{0}$
(42)
$\displaystyle\hat{v}^{{\dagger}}_{1}\hat{b}^{{\dagger}}_{1}\ket{0}\longrightarrow\hat{v}^{{\dagger}}_{1}\hat{a}^{{\dagger}}_{1}\ket{0}$
(43)
$\displaystyle\hat{v}^{{\dagger}}_{2}\hat{a}^{{\dagger}}_{2}\ket{0}\longrightarrow\hat{v}^{{\dagger}}_{2}\hat{b}^{{\dagger}}_{2}\ket{0}$
(44)
$\displaystyle\hat{v}^{{\dagger}}_{2}\hat{b}^{{\dagger}}_{2}\ket{0}\longrightarrow\hat{v}^{{\dagger}}_{2}\hat{a}^{{\dagger}}_{2}\ket{0}$
(45)

Figure 2: Circuit for the hyper-entangled Bell-like state analyzer where
polarisation DOF act as system qubits while momentum DOF act as control
qubits.

After that, another PBS is used in the circuit involving modes
$\hat{a}^{{\dagger}}_{1}$ and $\hat{b}^{{\dagger}}_{1}$. It is used so that
HWP mounted at angle $\frac{\theta}{2}$ acts on mode $\hat{h}^{{\dagger}}_{1}$
while HWP mounted at angle $45^{\circ}-\frac{\theta}{2}$ acts on mode
$\hat{h}^{{\dagger}}_{2}$. This kind of asymmetry is necessary for certain
cancellations which render Bell-like states unambiguously distinguishable. In
the circuit involving modes $\hat{a}^{{\dagger}}_{2}$ and
$\hat{b}^{{\dagger}}_{2}$ both are passed through HWP mounted at an angle
$\frac{\theta}{2}$. The transformation rules for a modes
$\hat{h}^{{\dagger}}_{2}$ and $\hat{v}^{{\dagger}}_{2}$are

$\displaystyle\hat{h}^{{\dagger}}_{2}\ket{0}\longrightarrow(\cos(\theta)\hat{h}^{{\dagger}}_{2}+\sin(\theta)\hat{v}^{{\dagger}}_{2})\ket{0}$
(46)
$\displaystyle\hat{v}^{{\dagger}}_{2}\ket{0}\longrightarrow(\sin(\theta)\hat{h}^{{\dagger}}_{2}-\cos(\theta)\hat{v}^{{\dagger}}_{2})\ket{0}$
(47)

While modes $\hat{h}^{{\dagger}}_{1}$ and $\hat{v}^{{\dagger}}_{1}$ transform
as

$\displaystyle\hat{h}^{{\dagger}}_{1}\ket{0}\longrightarrow(\cos(\theta)\hat{h}^{{\dagger}}_{1}+\sin(\theta)\hat{v}^{{\dagger}}_{1})\ket{0}$
(48)
$\displaystyle\hat{v}^{{\dagger}}_{1}\ket{0}\longrightarrow(\cos(\theta)\hat{h}^{{\dagger}}_{1}-\sin(\theta)\hat{v}^{{\dagger}}_{1})\ket{0}$
(49)

A beam splitter is used to then merge the splitted modes
$\hat{h}^{{\dagger}}_{1}$ and $\hat{v}^{{\dagger}}_{1}$ after HWPs act on
them. The output is recorded from only one arm of the beam splitter resulting
in the $50\%$ loss. It is worth mentioning that collecting output from either
arm of the beam splitter is equivalent from the state discrimination point of
view. After passing through the circuit the states from Eqn. (34)— (37) just
before entering PBS are given as:

$\displaystyle|\Xi_{1}\rangle\longrightarrow\Big{(}$
$\displaystyle\frac{\hat{a}^{{\dagger}}_{1}\hat{b}^{{\dagger}}_{2}\hat{h}^{{\dagger}}_{1}\hat{h}^{{\dagger}}_{2}\sin\left(\theta\right)}{\sqrt{2}}+\frac{\hat{a}^{{\dagger}}_{1}\hat{b}^{{\dagger}}_{2}\hat{h}^{{\dagger}}_{1}\hat{h}^{{\dagger}}_{2}\sin\left(\theta\right)\cos\left(2\theta\right)}{\sqrt{2}}$
$\displaystyle-\frac{\hat{a}^{{\dagger}}_{1}\hat{b}^{{\dagger}}_{2}\hat{h}^{{\dagger}}_{1}\hat{v}^{{\dagger}}_{2}\cos\left(\theta\right)}{2\sqrt{2}}-\frac{\hat{a}^{{\dagger}}_{1}\hat{b}^{{\dagger}}_{2}\hat{h}^{{\dagger}}_{1}\hat{v}^{{\dagger}}_{2}\cos\left(3\theta\right)}{2\sqrt{2}}$
$\displaystyle+\frac{\hat{a}^{{\dagger}}_{1}\hat{b}^{{\dagger}}_{2}\hat{v}^{{\dagger}}_{1}\hat{v}^{{\dagger}}_{2}\sin\left(\theta\right)}{\sqrt{2}}+\frac{\hat{a}^{{\dagger}}_{2}\hat{b}^{{\dagger}}_{1}\hat{h}^{{\dagger}}_{1}\hat{h}^{{\dagger}}_{2}\sin\left(\theta\right)}{\sqrt{2}}$
$\displaystyle+\frac{\hat{a}^{{\dagger}}_{2}\hat{b}^{{\dagger}}_{1}\hat{h}^{{\dagger}}_{1}\hat{h}^{{\dagger}}_{2}\sin\left(\theta\right)\cos\left(2\theta\right)}{\sqrt{2}}-\frac{\hat{a}^{{\dagger}}_{2}\hat{b}^{{\dagger}}_{1}\hat{h}^{{\dagger}}_{1}\hat{v}^{{\dagger}}_{2}\cos\left(\theta\right)}{2\sqrt{2}}$
$\displaystyle-\frac{\hat{a}^{{\dagger}}_{2}\hat{b}^{{\dagger}}_{1}\hat{h}^{{\dagger}}_{1}\hat{v}^{{\dagger}}_{2}\cos\left(3\theta\right)}{2\sqrt{2}}+\frac{\hat{a}^{{\dagger}}_{2}\hat{b}^{{\dagger}}_{1}\hat{v}^{{\dagger}}_{1}\hat{v}^{{\dagger}}_{2}\sin\left(\theta\right)}{\sqrt{2}}\Big{)}\ket{0}$
(50) $\displaystyle|\Xi_{2}\rangle\longrightarrow\Big{(}$
$\displaystyle\frac{\hat{a}^{{\dagger}}_{1}\hat{b}^{{\dagger}}_{2}\hat{h}^{{\dagger}}_{1}\hat{h}^{{\dagger}}_{2}\cos\left(\theta\right)}{2\sqrt{2}}+\frac{\hat{a}^{{\dagger}}_{1}\hat{b}^{{\dagger}}_{2}\hat{h}^{{\dagger}}_{1}\hat{h}^{{\dagger}}_{2}\cos\left(3\theta\right)}{2\sqrt{2}}$
$\displaystyle+\frac{\hat{a}^{{\dagger}}_{1}\hat{b}^{{\dagger}}_{2}\hat{h}^{{\dagger}}_{1}\hat{v}^{{\dagger}}_{2}\sin\left(\theta\right)}{\sqrt{2}}+\frac{\hat{a}^{{\dagger}}_{1}\hat{b}^{{\dagger}}_{2}\hat{h}^{{\dagger}}_{1}\hat{v}^{{\dagger}}_{2}\sin\left(\theta\right)\cos\left(2\theta\right)}{\sqrt{2}}$
$\displaystyle+\frac{\hat{a}^{{\dagger}}_{1}\hat{b}^{{\dagger}}_{2}\hat{h}^{{\dagger}}_{2}\hat{v}^{{\dagger}}_{1}\sin\left(\theta\right)}{\sqrt{2}}+\frac{\hat{a}^{{\dagger}}_{2}\hat{b}^{{\dagger}}_{1}\hat{h}^{{\dagger}}_{1}\hat{h}^{{\dagger}}_{2}\cos\left(\theta\right)}{2\sqrt{2}}$
$\displaystyle+\frac{\hat{a}^{{\dagger}}_{2}\hat{b}^{{\dagger}}_{1}\hat{h}^{{\dagger}}_{1}\hat{h}^{{\dagger}}_{2}\cos\left(3\theta\right)}{2\sqrt{2}}+\frac{\hat{a}^{{\dagger}}_{2}\hat{b}^{{\dagger}}_{1}\hat{h}^{{\dagger}}_{1}\hat{v}^{{\dagger}}_{2}\sin\left(\theta\right)}{\sqrt{2}}$
$\displaystyle+\frac{\hat{a}^{{\dagger}}_{2}\hat{b}^{{\dagger}}_{1}\hat{h}^{{\dagger}}_{1}\hat{v}^{{\dagger}}_{2}\sin\left(\theta\right)\cos\left(2\theta\right)}{\sqrt{2}}+\frac{\hat{a}^{{\dagger}}_{2}\hat{b}^{{\dagger}}_{1}\hat{h}^{{\dagger}}_{2}\hat{v}^{{\dagger}}_{1}\sin\left(\theta\right)}{\sqrt{2}}\Big{)}\ket{0}$
(51) $\displaystyle|\Xi_{3}\rangle\longrightarrow\Big{(}$
$\displaystyle\frac{\hat{a}^{{\dagger}}_{1}\hat{a}^{{\dagger}}_{2}\hat{h}^{{\dagger}}_{1}\hat{h}^{{\dagger}}_{2}\cos\left(\theta\right)}{\sqrt{2}}+\frac{\hat{a}^{{\dagger}}_{1}\hat{a}^{{\dagger}}_{2}\hat{h}^{{\dagger}}_{2}\hat{v}^{{\dagger}}_{1}\sin\left(\theta\right)}{2\sqrt{2}}$
$\displaystyle-\frac{\hat{a}^{{\dagger}}_{1}\hat{a}^{{\dagger}}_{2}\hat{h}^{{\dagger}}_{2}\hat{v}^{{\dagger}}_{1}\sin\left(3\theta\right)}{2\sqrt{2}}-\frac{\hat{a}^{{\dagger}}_{1}\hat{a}^{{\dagger}}_{2}\hat{v}^{{\dagger}}_{1}\hat{v}^{{\dagger}}_{2}\cos\left(\theta_{1}\right)}{2\sqrt{2}}$
$\displaystyle+\frac{\hat{a}^{{\dagger}}_{1}\hat{a}^{{\dagger}}_{2}\hat{v}^{{\dagger}}_{1}\hat{v}^{{\dagger}}_{2}\cos\left(3\theta\right)}{2\sqrt{2}}+\frac{\hat{b}^{{\dagger}}_{1}\hat{b}^{{\dagger}}_{2}\hat{h}^{{\dagger}}_{1}\hat{h}^{{\dagger}}_{2}\cos\left(\theta\right)}{\sqrt{2}}$
$\displaystyle+\frac{\hat{b}^{{\dagger}}_{1}\hat{b}^{{\dagger}}_{2}\hat{h}^{{\dagger}}_{2}\hat{v}^{{\dagger}}_{1}\sin\left(\theta\right)}{2\sqrt{2}}-\frac{\hat{b}^{{\dagger}}_{1}\hat{b}^{{\dagger}}_{2}\hat{h}^{{\dagger}}_{2}\hat{v}^{{\dagger}}_{1}\sin\left(3\theta\right)}{2\sqrt{2}}$
$\displaystyle-\frac{\hat{b}^{{\dagger}}_{1}\hat{b}^{{\dagger}}_{2}\hat{v}^{{\dagger}}_{1}\hat{v}^{{\dagger}}_{2}\cos\left(\theta\right)}{2\sqrt{2}}+\frac{\hat{b}^{{\dagger}}_{1}\hat{b}^{{\dagger}}_{2}\hat{v}^{{\dagger}}_{1}\hat{v}^{{\dagger}}_{2}\cos\left(3\theta\right)}{2\sqrt{2}}\Big{)}\ket{0}$
(52) $\displaystyle|\Xi_{4}\rangle\longrightarrow\Big{(}$
$\displaystyle-\frac{\hat{a}^{{\dagger}}_{1}\hat{a}^{{\dagger}}_{2}\hat{h}^{{\dagger}}_{1}\hat{v}^{{\dagger}}_{2}\cos\left(\theta\right)}{\sqrt{2}}+\frac{\hat{a}^{{\dagger}}_{1}\hat{a}^{{\dagger}}_{2}\hat{h}^{{\dagger}}_{2}\hat{v}^{{\dagger}}_{1}\cos\left(\theta\right)}{2\sqrt{2}}$
$\displaystyle-\frac{\hat{a}^{{\dagger}}_{1}\hat{a}^{{\dagger}}_{2}\hat{h}^{{\dagger}}_{2}\hat{v}^{{\dagger}}_{1}\cos\left(3\theta\right)}{2\sqrt{2}}+\frac{\hat{a}^{{\dagger}}_{1}\hat{a}^{{\dagger}}_{2}\hat{v}^{{\dagger}}_{1}\hat{v}^{{\dagger}}_{2}\sin\left(\theta_{1}\right)}{2\sqrt{2}}$
$\displaystyle-\frac{\hat{a}^{{\dagger}}_{1}\hat{a}^{{\dagger}}_{2}\hat{v}^{{\dagger}}_{1}\hat{v}^{{\dagger}}_{2}\sin\left(3\theta\right)}{2\sqrt{2}}-\frac{\hat{b}^{{\dagger}}_{1}\hat{b}^{{\dagger}}_{2}\hat{h}^{{\dagger}}_{1}\hat{v}^{{\dagger}}_{2}\cos\left(\theta\right)}{\sqrt{2}}$
$\displaystyle+\frac{\hat{b}^{{\dagger}}_{1}\hat{b}^{{\dagger}}_{2}\hat{h}^{{\dagger}}_{2}\hat{v}^{{\dagger}}_{1}\cos\left(\theta\right)}{2\sqrt{2}}-\frac{\hat{b}^{{\dagger}}_{1}\hat{b}^{{\dagger}}_{2}\hat{h}^{{\dagger}}_{2}\hat{v}^{{\dagger}}_{1}\cos\left(3\theta\right)}{2\sqrt{2}}$
$\displaystyle+\frac{\hat{b}^{{\dagger}}_{1}\hat{b}^{{\dagger}}_{2}\hat{v}^{{\dagger}}_{1}\hat{v}^{{\dagger}}_{2}\sin\left(\theta\right)}{2\sqrt{2}}-\frac{\hat{b}^{{\dagger}}_{1}\hat{b}^{{\dagger}}_{2}\hat{v}^{{\dagger}}_{1}\hat{v}^{{\dagger}}_{2}\sin\left(3\theta\right)}{2\sqrt{2}}\Big{)}\ket{0}$
(54)

We make the following observations

  * •

For states $|\Xi_{1}\rangle$ and $|\Xi_{2}\rangle$ both the photons are always
detected as coincidence detections in modes
$\\{\hat{a}^{{\dagger}}_{1},\hat{b}^{{\dagger}}_{2}\\}$ or
$\\{\hat{b}^{{\dagger}}_{1},\hat{a}^{{\dagger}}_{2}\\}$. Furthermore between
them

    1. 1.

Terms
$\\{\hat{b}^{{\dagger}}_{1}\hat{a}^{{\dagger}}_{2}\hat{v}^{{\dagger}}_{2}\hat{v}^{{\dagger}}_{1},\hat{a}^{{\dagger}}_{1}\hat{b}^{{\dagger}}_{2}\hat{v}^{{\dagger}}_{1}\hat{v}^{{\dagger}}_{2}\\}$
are uniquely present in $|\Xi_{1}\rangle$. Thus clicking of detectors
$\hat{b}^{{\dagger}}_{1V}$ and $\hat{a}^{{\dagger}}_{2V}$ or
$\hat{a}^{{\dagger}}_{1V}$ and $\hat{b}^{{\dagger}}_{2V}$ indicates
unambiguous detection of this state.

    2. 2.

Terms
$\\{\hat{b}^{{\dagger}}_{1}\hat{a}^{{\dagger}}_{2}\hat{h}^{{\dagger}}_{2}\hat{v}^{{\dagger}}_{1},\hat{a}^{{\dagger}}_{1}\hat{b}^{{\dagger}}_{2}\hat{h}^{{\dagger}}_{1}\hat{v}^{{\dagger}}_{2}\\}$
are uniquely present in $|\Xi_{2}\rangle$. Thus clicking of detectors
$\hat{a}^{{\dagger}}_{1V}$ and $\hat{b}^{{\dagger}}_{2H}$ or
$\hat{b}^{{\dagger}}_{1H}$ and $\hat{a}^{{\dagger}}_{2V}$ indicates
unambiguous detection of this state.

  * •

For states $|\Xi_{3}\rangle$ and $|\Xi_{4}\rangle$ both the photons are always
detected as coincidence detections in modes
$\\{\hat{a}^{{\dagger}}_{1},\hat{a}^{{\dagger}}_{2}\\}$ or
$\\{\hat{b}^{{\dagger}}_{1},\hat{b}^{{\dagger}}_{2}\\}$.

    1. 1.

Terms
$\\{\hat{a}^{{\dagger}}_{1}\hat{a}^{{\dagger}}_{2}\hat{h}^{{\dagger}}_{2}\hat{h}^{{\dagger}}_{1},\hat{b}^{{\dagger}}_{1}\hat{b}^{{\dagger}}_{2}\hat{h}^{{\dagger}}_{1}\hat{h}^{{\dagger}}_{2}\\}$
are uniquely present in $|\Xi_{3}\rangle$. Thus clicking of detectors
$\hat{a}^{{\dagger}}_{1H}$ and $\hat{a}^{{\dagger}}_{2H}$ or
$\hat{b}^{{\dagger}}_{1V}$ and $\hat{b}^{{\dagger}}_{2V}$ indicates
unambiguous detection of this state.

    2. 2.

Terms
$\\{\hat{a}^{{\dagger}}_{1}\hat{a}^{{\dagger}}_{2}\hat{v}^{{\dagger}}_{2}\hat{h}^{{\dagger}}_{1},\hat{b}^{{\dagger}}_{1}\hat{b}^{{\dagger}}_{2}\hat{h}^{{\dagger}}_{1}\hat{v}^{{\dagger}}_{2}\\}$
are uniquely present in $|\Xi_{4}\rangle$. Thus clicking of detectors
$\hat{a}^{{\dagger}}_{1H}$ and $\hat{a}^{{\dagger}}_{2V}$ or
$\hat{b}^{{\dagger}}_{1H}$ and $\hat{b}^{{\dagger}}_{2V}$ indicates
unambiguous detection of this state.

Again there will be some detection that will render the discrimination of
these states ambiguous. Thus the success probability again won’t be $100\%$.
The success probability can be calculated in a similar way as the previous
section. It comes out to be

$\displaystyle P_{succ}$
$\displaystyle=\frac{1}{4}(\frac{\sin^{2}(\theta)}{2}+\frac{\sin^{2}(\theta)}{2})+\frac{1}{4}(\frac{\sin^{2}(\theta)}{2}+\frac{\sin^{2}(\theta)}{2})$
$\displaystyle+\frac{1}{4}(\frac{\cos^{2}(\theta)}{2}+\frac{\cos^{2}(\theta)}{2})+\frac{1}{4}(\frac{\cos^{2}(\theta)}{2}+\frac{\cos^{2}(\theta)}{2})$
$\displaystyle=\frac{1}{2}$ (55)

Thus, in this case, we see that the success probability of unambiguous state
discrimination is again $50\%$.

##  IV Ancillary entanglement in OAM degrees of freedom

Till now we have discussed how momentum DOF act as system qubit while
polarisation DOF acts as ancillary qubits to assist in the unambiguous
discrimination of Bell-like states. But photons have orbital angular momentum
(OAM) DOF also. They can act as maximally entangled ancillary qubits also
while polarisation DOF bein the system qubits. Such a scheme has been adopted
by Kwiat et.al. in [17] to implement super-dense coding. The circuit presented
here is on the same lines. It is shown in Fig. (3)

Figure 3: Circuit for the hyper-entangled Bell-like state analyzer where
polarisation DOF act as system qubits while OAM DOF act as ancillary qubits.

The joint states of the system and ancillary qubits are given as:

$\displaystyle|\Pi_{1}\rangle=(\sin(\theta)\hat{h}^{{\dagger}}_{1}\hat{h}^{{\dagger}}_{2}+\cos(\theta)\hat{v}^{{\dagger}}_{1}\hat{v}^{{\dagger}}_{2})\otimes\frac{(\hat{p}^{{\dagger}}_{1}\hat{m}^{{\dagger}}_{2}+\hat{m}^{{\dagger}}_{1}\hat{p}^{{\dagger}}_{2})}{\sqrt{2}}\ket{0}$
(56)
$\displaystyle|\Pi_{2}\rangle=(\cos(\theta)\hat{h}^{{\dagger}}_{1}\hat{h}^{{\dagger}}_{2}-\sin(\theta)\hat{v}^{{\dagger}}_{1}\hat{v}^{{\dagger}}_{2})\otimes\frac{(\hat{p}^{{\dagger}}_{1}\hat{m}^{{\dagger}}_{2}+\hat{m}^{{\dagger}}_{1}\hat{p}^{{\dagger}}_{2})}{\sqrt{2}}\ket{0}$
(57)
$\displaystyle|\Pi_{3}\rangle=(\sin(\theta)\hat{h}^{{\dagger}}_{1}\hat{v}^{{\dagger}}_{2}+\cos(\theta)\hat{v}^{{\dagger}}_{1}\hat{h}^{{\dagger}}_{2})\otimes\frac{(\hat{p}^{{\dagger}}_{1}\hat{m}^{{\dagger}}_{2}+\hat{m}^{{\dagger}}_{1}\hat{p}^{{\dagger}}_{2})}{\sqrt{2}}\ket{0}$
(58)
$\displaystyle|\Pi_{4}\rangle=(\cos(\theta)\hat{h}^{{\dagger}}_{1}\hat{v}^{{\dagger}}_{2}-\sin(\theta)\hat{v}^{{\dagger}}_{1}\hat{h}^{{\dagger}}_{2})\otimes\frac{(\hat{p}^{{\dagger}}_{1}\hat{m}^{{\dagger}}_{2}+\hat{m}^{{\dagger}}_{1}\hat{p}^{{\dagger}}_{2})}{\sqrt{2}}\ket{0}$
(59)

where $\hat{p}^{{\dagger}}$ represent creation operator for OAM mode $+1$
while $\hat{m}^{{\dagger}}$ represent creation operator for OAM mode $-1$.
Photons fall on a forked binary hologram and according to the OAM values $+1$
or $-1$ they will split into two beams marked by $(u_{1},d_{1})$ and
$(u_{2},d_{2})$ for the first and second photon respectively. Then they pass
through an assembly of PBSs and HWPs and are finally detected. To discriminate
among Bell-like states it is necessary that HWPs act differently on both
horizontally and vertically polarised first photon. Hence a PBS is used to
split these two and then merge again by only recording output from one arm of
a balanced beam splitter. This costs us a $50\%$ loss in the beam. The action
of the forked binary hologram is given as:

$\displaystyle\hat{h}^{{\dagger}}_{i}\hat{p}^{{\dagger}}_{i}\longrightarrow\hat{h}^{{\dagger}u_{i}}_{i}\hat{g}^{{\dagger}}_{i}$
$\displaystyle\hat{h}^{{\dagger}}_{i}\hat{m}^{{\dagger}}_{i}\longrightarrow\hat{h}^{{\dagger}d_{i}}_{i}\hat{g}^{{\dagger}}_{i}$
$\displaystyle\hat{v}^{{\dagger}}_{i}\hat{p}^{{\dagger}}_{i}\longrightarrow\hat{v}^{{\dagger}u_{i}}_{i}\hat{g}^{{\dagger}}_{i}$
$\displaystyle\hat{v}^{{\dagger}}_{i}\hat{m}^{{\dagger}}_{i}\longrightarrow\hat{v}^{{\dagger}d_{i}}_{i}\hat{g}^{{\dagger}}_{i}$
(60)

where $\hat{g}^{{\dagger}}_{i}$ represent creation operator for $0$ OAM mode
of ith particle. Thus the final states just before the final detection at PBS
becomes

$\displaystyle|\Pi_{1}\rangle\longrightarrow\Big{(}$
$\displaystyle\frac{\hat{h}^{{\dagger}d_{1}}_{1}\hat{h}^{{\dagger}d_{2}}_{2}\cos\left(\theta\right)}{\sqrt{2}}+\frac{\hat{h}^{{\dagger}u_{1}}_{1}\hat{h}^{{\dagger}u_{2}}_{2}\cos\left(\theta\right)}{\sqrt{2}}$
$\displaystyle+\frac{\hat{h}^{{\dagger}d_{2}}_{2}\hat{v}^{{\dagger}d_{1}}_{1}\sin\left(\theta\right)}{2\sqrt{2}}-\frac{\hat{h}^{{\dagger}d_{2}}_{2}\hat{v}^{{\dagger}d_{1}}_{1}\sin\left(3\theta\right)}{2\sqrt{2}}$
$\displaystyle+\frac{\hat{h}^{{\dagger}u_{2}}_{2}\hat{v}^{{\dagger}u_{1}}_{1}\sin\left(\theta\right)}{2\sqrt{2}}-\frac{\hat{h}^{{\dagger}u_{2}}_{2}\hat{v}^{{\dagger}u_{1}}_{1}\sin\left(3\theta\right)}{2\sqrt{2}}$
$\displaystyle-\frac{\hat{v}^{{\dagger}d_{1}}_{1}\hat{v}^{{\dagger}d_{2}}_{2}\cos\left(\theta\right)}{2\sqrt{2}}+\frac{\hat{v}^{{\dagger}d_{1}}_{1}\hat{v}^{{\dagger}d_{2}}_{2}\cos\left(3\theta\right)}{2\sqrt{2}}$
$\displaystyle-\frac{\hat{v}^{{\dagger}u_{1}}_{1}\hat{v}^{{\dagger}u_{2}}_{2}\cos\left(\theta\right)}{2\sqrt{2}}+\frac{\hat{v}^{{\dagger}u_{1}}_{1}\hat{v}^{{\dagger}u_{2}}_{2}\cos\left(3\theta\right)}{2\sqrt{2}}\Big{)}\otimes\hat{g}^{{\dagger}}_{1}\hat{g}^{{\dagger}}_{2}\ket{0}$
(61) $\displaystyle|\Pi_{2}\rangle\longrightarrow\Big{(}$
$\displaystyle-\frac{\hat{h}^{{\dagger}d_{1}}_{1}\hat{v}^{{\dagger}d_{2}}_{2}\cos\left(\theta\right)}{\sqrt{2}}-\frac{\hat{h}^{{\dagger}u_{1}}_{1}\hat{v}^{{\dagger}u_{2}}_{2}\cos\left(\theta\right)}{\sqrt{2}}$
$\displaystyle+\frac{\hat{h}^{{\dagger}d_{2}}_{2}\hat{v}^{{\dagger}d_{1}}_{1}\cos\left(\theta\right)}{2\sqrt{2}}-\frac{\hat{h}^{{\dagger}d_{2}}_{2}\hat{v}^{{\dagger}d_{1}}_{1}\cos\left(3\theta\right)}{2\sqrt{2}}$
$\displaystyle+\frac{\hat{h}^{{\dagger}u_{2}}_{2}\hat{v}^{{\dagger}u_{1}}_{1}\cos\left(\theta\right)}{2\sqrt{2}}-\frac{\hat{h}^{{\dagger}u_{2}}_{2}\hat{v}^{{\dagger}u_{1}}_{1}\cos\left(3\theta\right)}{2\sqrt{2}}$
$\displaystyle+\frac{\hat{v}^{{\dagger}d_{1}}_{1}\hat{v}^{{\dagger}d_{2}}_{2}\sin\left(\theta\right)}{2\sqrt{2}}-\frac{\hat{v}^{{\dagger}d_{1}}_{1}\hat{v}^{{\dagger}d_{2}}_{2}\sin\left(3\theta\right)}{2\sqrt{2}}$
$\displaystyle+\frac{\hat{v}^{{\dagger}u_{1}}_{1}\hat{v}^{{\dagger}u_{2}}_{2}\sin\left(\theta\right)}{2\sqrt{2}}-\frac{\hat{v}^{{\dagger}u_{1}}_{1}\hat{v}^{{\dagger}u_{2}}_{2}\sin\left(3\theta\right)}{2\sqrt{2}}\Big{)}\otimes\hat{g}^{{\dagger}}_{1}\hat{g}^{{\dagger}}_{2}\ket{0}$
(62) $\displaystyle|\Pi_{3}\rangle\longrightarrow\Big{(}$
$\displaystyle\frac{\hat{h}^{{\dagger}d_{1}}_{1}\hat{h}^{{\dagger}u_{2}}_{2}\sin\left(\theta\right)}{\sqrt{2}}+\frac{\hat{h}^{{\dagger}d_{1}}_{1}\hat{h}^{{\dagger}u_{2}}_{2}\sin\left(\theta\right)\cos\left(2\theta_{1}\right)}{\sqrt{2}}$
$\displaystyle-\frac{\hat{h}^{{\dagger}d_{1}}_{1}\hat{v}^{{\dagger}u_{2}}_{2}\cos\left(\theta\right)}{2\sqrt{2}}-\frac{\hat{h}^{{\dagger}d_{1}}_{1}\hat{v}^{{\dagger}u_{2}}_{2}\cos\left(3\theta\right)}{2\sqrt{2}}$
$\displaystyle+\frac{\hat{h}^{{\dagger}u_{1}}_{1}\hat{h}^{{\dagger}d_{2}}_{2}\sin\left(\theta\right)}{\sqrt{2}}+\frac{\hat{h}^{{\dagger}u_{1}}_{1}\hat{h}^{{\dagger}d_{2}}_{2}\sin\left(\theta\right)\cos\left(2\theta_{1}\right)}{\sqrt{2}}$
$\displaystyle-\frac{\hat{h}^{{\dagger}u_{1}}_{1}\hat{v}^{{\dagger}d_{2}}_{2}\cos\left(\theta\right)}{2\sqrt{2}}-\frac{\hat{h}^{{\dagger}u_{1}}_{1}\hat{v}^{{\dagger}d_{2}}_{2}\cos\left(3\theta\right)}{2\sqrt{2}}$
$\displaystyle+\frac{\hat{v}^{{\dagger}d_{1}}_{1}\hat{v}^{{\dagger}u_{2}}_{2}\sin\left(\theta\right)}{\sqrt{2}}+\frac{\hat{v}^{{\dagger}u_{1}}_{1}\hat{v}^{{\dagger}d_{2}}_{2}\sin\left(\theta\right)}{\sqrt{2}}\Big{)}\otimes\hat{g}^{{\dagger}}_{1}\hat{g}^{{\dagger}}_{2}\ket{0}$
(64) $\displaystyle|\Pi_{4}\rangle\longrightarrow\Big{(}$
$\displaystyle\frac{\hat{h}^{{\dagger}d_{1}}_{1}\hat{h}^{{\dagger}u_{2}}_{2}\cos\left(\theta\right)}{2\sqrt{2}}+\frac{\hat{h}^{{\dagger}d_{1}}_{1}\hat{h}^{{\dagger}u_{2}}_{2}\cos\left(3\theta\right)}{2\sqrt{2}}$
$\displaystyle+\frac{\hat{h}^{{\dagger}d_{1}}_{1}\hat{v}^{{\dagger}u_{2}}_{2}\sin\left(\theta\right)}{2\sqrt{2}}+\frac{\hat{h}^{{\dagger}d_{1}}_{1}\hat{v}^{{\dagger}u_{2}}_{2}\sin\left(3\theta\right)}{2\sqrt{2}}$
$\displaystyle+\frac{\hat{h}^{{\dagger}u_{1}}_{1}\hat{h}^{{\dagger}d_{2}}_{2}\cos\left(\theta_{1}\right)}{2\sqrt{2}}+\frac{\hat{h}^{{\dagger}u_{1}}_{1}\hat{h}^{{\dagger}d_{2}}_{2}\cos\left(3\theta\right)}{2\sqrt{2}}$
$\displaystyle+\frac{\hat{h}^{{\dagger}u_{1}}_{1}\hat{v}^{{\dagger}d_{2}}_{2}\sin\left(\theta\right)}{2\sqrt{2}}+\frac{\hat{h}^{{\dagger}u_{1}}_{1}\hat{v}^{{\dagger}d_{2}}_{2}\sin\left(3\theta\right)}{2\sqrt{2}}$
$\displaystyle+\frac{\hat{h}^{{\dagger}d_{2}}_{2}\hat{v}^{{\dagger}u_{1}}_{1}\sin\left(\theta\right)}{\sqrt{2}}+\frac{\hat{h}^{{\dagger}u_{2}}_{2}\hat{v}^{{\dagger}d_{1}}_{1}\sin\left(\theta\right)}{\sqrt{2}}\Big{)}\otimes\hat{g}^{{\dagger}}_{1}\hat{g}^{{\dagger}}_{2}\ket{0}$
(65)

We make the following observations

  * •

For states $|\Pi_{1}\rangle$ and $|\Pi_{2}\rangle$ both the photons are always
detected as coincidence detections in either path $(u_{1},u_{2})$ or paths
$(d_{1},d_{2})$. Furthermore between them

    1. 1.

Terms
$\\{\hat{h}^{{\dagger}u_{1}}_{1}\hat{h}^{{\dagger}u_{2}}_{2},\hat{h}^{{\dagger}d_{1}}_{1}\hat{h}^{{\dagger}d_{2}}_{2}\\}$
are uniquely present in $|\Pi_{1}\rangle$. Thus both the photons being
horizontally polarised in paths $u_{1}(d_{1})$ and $u_{2}(d_{2})$ indicates
unambiguous detection of this state.

    2. 2.

Terms
$\\{\hat{h}^{{\dagger}u_{1}}_{1}\hat{v}^{{\dagger}u_{2}}_{2},\hat{h}^{{\dagger}d_{1}}_{1}\hat{v}^{{\dagger}d_{2}}_{2}\\}$
are uniquely present in $|\Pi_{2}\rangle$. Thus horizontally polarised first
photon in paths $u_{1}(d_{1})$ and vertically polarised second photon in paths
$u_{2}(d_{2})$ indicates unambiguous detection of this state.

  * •

For states $|\Pi_{3}\rangle$ and $|\Pi_{4}\rangle$ both the photons are always
simultaneously detected in both the paths u and d. Furthermore

    1. 1.

Terms
$\\{\hat{v}^{{\dagger}u_{1}}_{1}\hat{v}^{{\dagger}d_{2}}_{2},\hat{v}^{{\dagger}d_{1}}_{1}\hat{v}^{{\dagger}u_{2}}_{2}\\}$
are uniquely present in $|\Pi_{3}\rangle$. Thus both the photons being
vertically polarised in paths $u_{1}(d_{1})$ and $d_{2}(u_{2})$ indicates
unambiguous detection of this state.

    2. 2.

Terms
$\\{\hat{v}^{{\dagger}u_{1}}_{1}\hat{h}^{{\dagger}d_{2}}_{2},\hat{v}^{{\dagger}d_{1}}_{1}\hat{h}^{{\dagger}u_{2}}_{2}\\}$
are uniquely present in $|\Pi_{4}\rangle$. Thus vertically polarised first
photon in paths $u_{1}(d_{1})$ and horizontally polarised second photon in
paths $d_{2}(u_{2})$ indicates unambiguous detection of this state.

There will also be some detection events that will render the discrimination
of the Bell-like states ambiguous. Thus success probability, unlike Bell state
discrimination, will not be $100\%$. If all Bell-like states are provided with
equal probability then success probability can be calculated as

$\displaystyle P_{succ}$
$\displaystyle=\frac{1}{4}(\frac{\sin^{2}(\theta)}{2}+\frac{\sin^{2}(\theta)}{2})+\frac{1}{4}(\frac{\sin^{2}(\theta)}{2}+\frac{\sin^{2}(\theta)}{2})$
$\displaystyle+\frac{1}{4}(\frac{\cos^{2}(\theta)}{2}+\frac{\cos^{2}(\theta)}{2})+\frac{1}{4}(\frac{\cos^{2}(\theta)}{2}+\frac{\cos^{2}(\theta)}{2})$
$\displaystyle=\frac{1}{2}$ (66)

Thus success probability of unambiguous discrimination is essentially $50\%$.
We would again like to stress the point that the aforesaid protocol may not be
the optimal one.

##  V Using correlation in time DOF for Bell-like state analysis

It is well known that the photon pair coming out of a spontaneous down
conversion source is strongly correlated in time. This correlation can be used
to distinguish Bell-like states. It was first proposed by Kwiat and Weinfurter
[26] for Bell state analysis. The circuit is shown in Fig.(5)

Figure 4: This is the circuit for a Bell-like state analyzer where correlation
in time DOF of photon pair is used.

An entangled photon pair is generated by passing a pump photon beam through
non-linear crystals where it undergoes the SPDC process. Horizontally
polarised photons are passed through a balanced beam splitter while vertically
polarised photons are passed through an unbalanced beam splitter. The output
of both beam splitters is then passed through birefringent crystals. These
crystals introduce the delay in the passing photon depending on their
polarization. Horizontally polarised photon picks up a delay of, say, time
$t_{h}$ while the vertically polarised photon is delayed by time $t_{v}$.
Mathematically it can be written as:

$\displaystyle\hat{h}^{{\dagger}}_{1}\longrightarrow\hat{h}^{{\dagger}}_{1}(t_{h})$
$\displaystyle\hat{h}^{{\dagger}}_{2}\longrightarrow\hat{h}^{{\dagger}}_{2}(t_{h})$
$\displaystyle\hat{v}^{{\dagger}}_{1}\longrightarrow\hat{v}^{{\dagger}}_{1}(t_{v})$
$\displaystyle\hat{v}^{{\dagger}}_{2}\longrightarrow\hat{v}^{{\dagger}}_{2}(t_{v})$
(67)

The amount of delay depends on the length of the crystal. We have to bear two
things in mind while introducing this delay:

  * •

Both the delay times $t_{h}$ and $t_{v}$ should be greater than the coherence
time of the entangled photon pair. This ensures that both the photons
(horizontally and vertically polarised) can be distinguished by their arrival
times $t_{h}$ and $t_{v}$ at the detectors.

  * •

Both of these delay times should be less than the coherence time of the pump
photon. this ensures that the photon pair as a whole cannot be distinguished
by their times of creation i.e. terms like
$\hat{x}^{{\dagger}}_{1}(t_{h})\hat{x}^{{\dagger}}_{2}(t_{h})$ and
$\hat{x}^{{\dagger}}_{1}(t_{v})\hat{x}^{{\dagger}}_{2}(t_{v})$ are
indistinguishable from each other hence will be replaced by
$\hat{x}^{{\dagger}}_{1}\hat{x}^{{\dagger}}_{2}$ where
$\hat{x}^{{\dagger}}\in\\{\hat{h}^{{\dagger}},\hat{v}^{{\dagger}}\\}$.

The horizontally polarised photons then pass through an HWP aligned at an
angle $\frac{\theta}{2}$ while vertically polarised photons are passed through
an HWP aligned at the angle of $22.5^{\circ}$. After merging two photons at
the beam splitter the Bell-like states transform as

$\displaystyle|\psi_{1}\rangle=$
$\displaystyle(\sin(\theta)\hat{h}^{{\dagger}}_{1}\hat{v}^{{\dagger}}_{2}+\cos(\theta)\hat{v}^{{\dagger}}_{1}\hat{h}^{{\dagger}}_{2})\ket{0}$
$\displaystyle\longrightarrow$
$\displaystyle\Big{(}\frac{\cos(\theta)}{2}\hat{h}^{{\dagger}}_{1}(t_{h})\hat{h}^{{\dagger}}_{1}(t_{v})-\frac{\cos(\theta)}{2}\hat{h}^{{\dagger}}_{1}(t_{h})\hat{v}^{{\dagger}}_{1}(t_{v})$
$\displaystyle+\frac{\sin(\theta)}{2}\hat{v}^{{\dagger}}_{1}(t_{h})\hat{h}^{{\dagger}}_{1}(t_{v})-\frac{\sin(\theta)}{2}\hat{v}^{{\dagger}}_{1}(t_{h})\hat{v}^{{\dagger}}_{1}(t_{v})$
$\displaystyle-\frac{\cos(\theta)\cos(2\theta)}{2}\hat{h}^{{\dagger}}_{2}(t_{h})\hat{h}^{{\dagger}}_{1}(t_{v})+\frac{\cos(\theta)\cos(2\theta)}{2}\hat{h}^{{\dagger}}_{2}(t_{h})\hat{v}^{{\dagger}}_{1}(t_{v})$
$\displaystyle-\frac{\sin(\theta)\cos(2\theta)}{2}\hat{v}^{{\dagger}}_{2}(t_{h})\hat{h}^{{\dagger}}_{1}(t_{v})+\frac{\sin(\theta)\cos(2\theta)}{2}\hat{v}^{{\dagger}}_{2}(t_{h})\hat{v}^{{\dagger}}_{1}(t_{v})$
$\displaystyle-\cos^{2}(\theta)\sin(\theta)\hat{h}^{{\dagger}}_{2}(t_{h})\hat{h}^{{\dagger}}_{2}(t_{v})+\cos^{2}(\theta)\sin(\theta)\hat{h}^{{\dagger}}_{2}(t_{h})\hat{v}^{{\dagger}}_{2}(t_{v})$
$\displaystyle-\sin^{2}(\theta)\cos(\theta)\hat{v}^{{\dagger}}_{2}(t_{h})\hat{h}^{{\dagger}}_{2}(t_{v})+\sin^{2}(\theta)\cos(\theta)\hat{v}^{{\dagger}}_{2}(t_{h})\hat{v}^{{\dagger}}_{2}(t_{v})\Big{)}\ket{0}$
(68) $\displaystyle|\psi_{2}\rangle=$
$\displaystyle(\cos(\theta)\hat{h}^{{\dagger}}_{1}\hat{v}^{{\dagger}}_{2}-\sin(\theta)\hat{v}^{{\dagger}}_{1}\hat{h}^{{\dagger}}_{2})\ket{0}$
$\displaystyle\longrightarrow$
$\displaystyle-\Big{(}\frac{\cos(\theta)}{2}\hat{h}^{{\dagger}}_{1}(t_{h})\hat{h}^{{\dagger}}_{2}(t_{v})+\frac{\cos(\theta)}{2}\hat{h}^{{\dagger}}_{1}(t_{h})\hat{v}^{{\dagger}}_{2}(t_{v})$
$\displaystyle-\frac{\sin(\theta)}{2}\hat{v}^{{\dagger}}_{1}(t_{h})\hat{h}^{{\dagger}}_{2}(t_{v})+\frac{\sin(\theta)}{2}\hat{v}^{{\dagger}}_{1}(t_{h})\hat{v}^{{\dagger}}_{2}(t_{v})$
$\displaystyle-\frac{\cos(\theta)\cos(2\theta)}{2}\hat{h}^{{\dagger}}_{2}(t_{h})\hat{h}^{{\dagger}}_{2}(t_{v})+\frac{\cos(\theta)\cos(2\theta)}{2}\hat{h}^{{\dagger}}_{2}(t_{h})\hat{v}^{{\dagger}}_{2}(t_{v})$
$\displaystyle-\frac{\sin(\theta)\cos(2\theta)}{2}\hat{v}^{{\dagger}}_{2}(t_{h})\hat{h}^{{\dagger}}_{2}(t_{v})+\frac{\sin(\theta)\cos(2\theta)}{2}\hat{v}^{{\dagger}}_{2}(t_{h})\hat{v}^{{\dagger}}_{2}(t_{v})$
$\displaystyle+\cos^{2}(\theta)\sin(\theta)\hat{h}^{{\dagger}}_{2}(t_{h})\hat{h}^{{\dagger}}_{1}(t_{v})-\cos^{2}(\theta)\sin(\theta)\hat{h}^{{\dagger}}_{2}(t_{h})\hat{v}^{{\dagger}}_{1}(t_{v})$
$\displaystyle+\sin^{2}(\theta)\cos(\theta)\hat{v}^{{\dagger}}_{2}(t_{h})\hat{h}^{{\dagger}}_{1}(t_{v})-\sin^{2}(\theta)\cos(\theta)\hat{v}^{{\dagger}}_{2}(t_{h})\hat{v}^{{\dagger}}_{1}(t_{v})\Big{)}\ket{0}$
(69) $\displaystyle|\psi_{3}\rangle=$
$\displaystyle(\sin(\theta)\hat{h}^{{\dagger}}_{1}\hat{h}^{{\dagger}}_{2}+\cos(\theta)\hat{v}^{{\dagger}}_{1}\hat{v}^{{\dagger}}_{2})\ket{0}$
$\displaystyle\longrightarrow$
$\displaystyle\Big{(}\sin(\theta)\cos^{2}(\theta))\hat{h}^{{\dagger}2}_{1}-\sin(\theta)\cos^{2}(\theta))\hat{h}^{{\dagger}2}_{2}+\frac{\sin(\theta)}{2}\hat{v}^{{\dagger}2}_{1}$
$\displaystyle-\frac{\sin(\theta)}{2}\hat{v}^{{\dagger}2}_{2}+\big{(}\sin^{2}(\theta)\cos(\theta)-\cos^{2}(\theta)\sin(\theta)\big{)}\hat{h}^{{\dagger}}_{1}\hat{v}^{{\dagger}}_{1}$
$\displaystyle-\big{(}\sin^{2}(\theta)\cos(\theta)-\cos^{2}(\theta)\sin(\theta)\big{)}\hat{h}^{{\dagger}}_{2}\hat{v}^{{\dagger}}_{2}$
$\displaystyle-\frac{\cos(\theta)\cos(2\theta)}{2}\hat{h}^{{\dagger}}_{1}\hat{h}^{{\dagger}}_{2}+\frac{\cos(\theta)\cos(2\theta)}{2}\hat{h}^{{\dagger}}_{1}\hat{v}^{{\dagger}}_{2}$
$\displaystyle+\frac{\cos(\theta)\cos(2\theta)}{2}\hat{v}^{{\dagger}}_{1}\hat{h}^{{\dagger}}_{2}-\frac{\cos(\theta)\cos(2\theta)}{2}\hat{v}^{{\dagger}}_{1}\hat{v}^{{\dagger}}_{2}\Big{)}\ket{0}$
(71) $\displaystyle|\psi_{4}\rangle=$
$\displaystyle(\cos(\theta)\hat{h}^{{\dagger}}_{1}\hat{h}^{{\dagger}}_{2}-\sin(\theta)\hat{v}^{{\dagger}}_{1}\hat{v}^{{\dagger}}_{2})\ket{0}$
$\displaystyle\longrightarrow$
$\displaystyle\Big{(}\frac{\cos(\theta)\cos(2\theta)}{2}\hat{h}^{{\dagger}2}_{1}-\frac{\cos(\theta)\cos(2\theta)}{2}\hat{h}^{{\dagger}2}_{2}$
$\displaystyle+\big{(}\sin^{2}(\theta)\cos(\theta)+\cos^{2}(\theta)\sin(\theta)\big{)}\hat{h}^{{\dagger}}_{1}\hat{v}^{{\dagger}}_{1}$
$\displaystyle-\big{(}\sin^{2}(\theta)\cos(\theta)+\cos^{2}(\theta)\sin(\theta)\big{)}\hat{h}^{{\dagger}}_{2}\hat{v}^{{\dagger}}_{2}$
$\displaystyle+\frac{\sin(\theta)\cos(2\theta)}{2}\hat{h}^{{\dagger}}_{1}\hat{h}^{{\dagger}}_{2}-\frac{\sin(\theta)\cos(2\theta)}{2}\hat{h}^{{\dagger}}_{1}\hat{v}^{{\dagger}}_{2}$
$\displaystyle-\frac{\sin(\theta)\cos(2\theta)}{2}\hat{v}^{{\dagger}}_{1}\hat{h}^{{\dagger}}_{2}+\frac{\sin(\theta)\cos(2\theta)}{2}\hat{v}^{{\dagger}}_{1}\hat{v}^{{\dagger}}_{2}\Big{)}\ket{0}$
(72)

We make the following observations:

  1. 1.

Terms
$\\{\hat{h}^{{\dagger}}_{1}(t_{h})\hat{h}^{{\dagger}}_{1}(t_{v}),\hat{h}^{{\dagger}}_{1}(t_{h})\hat{v}^{{\dagger}}_{1}(t_{v}),\hat{v}^{{\dagger}}_{1}(t_{h})\hat{h}^{{\dagger}}_{1}(t_{v}),\hat{v}^{{\dagger}}_{1}(t_{h})\hat{v}^{{\dagger}}_{1}(t_{v})\\}$
are uniquely present in $|\psi_{1}\rangle$. Thus detection of both the photon
in detector $A(Ah_{1},Av_{1})$ with the delay of $|t_{h}-t_{v}|$ between them
indicates the presence of this state.

  2. 2.

Terms
$\\{\hat{h}^{{\dagger}}_{1}(t_{h})\hat{h}^{{\dagger}}_{2}(t_{v}),\hat{h}^{{\dagger}}_{1}(t_{h})\hat{v}^{{\dagger}}_{2}(t_{v}),\hat{v}^{{\dagger}}_{1}(t_{h})\hat{h}^{{\dagger}}_{2}(t_{v}),\hat{v}^{{\dagger}}_{1}(t_{h})\hat{v}^{{\dagger}}_{2}(t_{v})\\}$
are uniquely present in $|\psi_{2}\rangle$.Thus detection of the first photon
in detector $A(Ah_{1},Av_{1})$ and the second photon in detector
$B(Bh_{2},Bv_{2})$ with the delay of $|t_{h}-t_{v}|$ between them indicates
the presence of this state.

  3. 3.

Terms $\hat{v}^{{\dagger}2}_{1},\hat{v}^{{\dagger}2}_{2}$ are uniquely present
in $|\psi_{3}\rangle$. Thus if both photons are detected simultaneously
without any delay in any of the detectors $Av_{1},Bv_{2}$ the presence of this
state is confirmed.

Evidently, the present scheme does not provide an unambiguous discrimination
of all the four Bell-Like states. In particular, if $\hat{h}^{{\dagger}2}_{1}$
or $\hat{h}^{{\dagger}2}_{2}$ clicks, it can happen for both $|{\psi}3\rangle$
as well as $|\psi_{4}\rangle$.

The success probability is necessarily not $100\%$. If we consider that all
four states are provided to us with equal probability then the success
probability is:

$\displaystyle P_{succ}$
$\displaystyle=\frac{1}{4}\Big{(}\frac{\sin^{2}(\theta)}{4}+\frac{\sin^{2}(\theta)}{4}+\frac{\cos^{2}(\theta)}{4}+\frac{\cos^{2}(\theta)}{4}\Big{)}$
$\displaystyle+\frac{1}{4}\Big{(}\frac{\sin^{2}(\theta)}{4}+\frac{\sin^{2}(\theta)}{4}+\frac{\cos^{2}(\theta)}{4}+\frac{\cos^{2}(\theta)}{4}\Big{)}$
$\displaystyle+\frac{1}{4}\Big{(}\frac{\sin^{2}(\theta)}{2}+\frac{\sin^{2}(\theta)}{2}\Big{)}+\frac{1}{4}(0)$
$\displaystyle=\frac{(1+\sin^{2}(\theta))}{4}$ (73)

Figure 5: Variation of the success probability of state discrimination w.r.t.
the state parameter $\theta$.

Interestingly success probability depends on the state parameter $\theta$ of
the given Bell-like states which we want to discriminate. The expression
success probability in Eqn. (V) is valid for $\theta\neq\frac{\pi}{4}$ and
$0<\theta<\frac{\pi}{2}$.

Thus we observe that in case correlations in time are used to assist in the
discrimination of four bell-like states the success probability although
higher than $25\%$ unlike previous cases never reaches $50\%$ let alone
$100\%$. Another interesting observation is that here, even ambiguously, we
can only discriminate among three Bell-like states $\psi_{1},\psi_{2}$ and
$\psi_{3}$. The state $\psi_{4}$ doesn’t have a unique detection signature.
This is in stark contrast to Bell-state discrimination where we can
discriminate among all 4 Bell states with $100\%$ success probability.

##  VI Bell-like state analysis using Ancillary qubits

Aside from using hyperentanglement extra ancillary photons can come to our aid
for Bell-like state analysis. Grice found out that by using two ancillary
maximally entangled photons [19] the success probability for Bell state
discrimination can be enhanced to $75\%$ from $50\%$. Only two bell states
($\psi^{+}$ and $\psi^{-}$) are distinguishable without using any resource in
a linear optical setting. The introduction of ancillary entangled qubits
renders the remaining two ($\phi^{+}$ and $\phi^{-}$) also distinguishable
(not unambiguously as there will be some detection signatures that will be
common for both states). Thus success probability of Bell state discrimination
increases from $50\%$ (two states are distinguishable without any ambiguity)
to $75\%$ (all four can be distinguished with some ambiguity in ($\phi^{+}$
and $\phi^{-}$) ). We try to do a similar kind of thing here. The Bell-like
states with ancillary photons are given as

$\displaystyle|\Gamma_{1}\rangle=$
$\displaystyle\left(\hat{a}^{{\dagger}}_{1}\hat{a}^{{\dagger}}_{4}\sin\left(\theta_{2}\right)+\hat{a}^{{\dagger}}_{2}\hat{a}^{{\dagger}}_{3}\cos\left(\theta_{2}\right)\right)$
$\displaystyle\qquad\qquad\qquad\otimes\left(\hat{a}^{{\dagger}}_{5}\hat{a}^{{\dagger}}_{7}\sin\left(\theta_{1}\right)+\hat{a}^{{\dagger}}_{6}\hat{a}^{{\dagger}}_{8}\cos\left(\theta_{1}\right)\right)\ket{0}$
(74) $\displaystyle|\Gamma_{2}\rangle=$
$\displaystyle\left(\hat{a}^{{\dagger}}_{1}\hat{a}^{{\dagger}}_{4}\cos\left(\theta_{2}\right)-\hat{a}^{{\dagger}}_{2}\hat{a}^{{\dagger}}_{3}\sin\left(\theta_{2}\right)\right)$
$\displaystyle\qquad\qquad\qquad\otimes\left(\hat{a}^{{\dagger}}_{5}\hat{a}^{{\dagger}}_{7}\sin\left(\theta_{1}\right)+\hat{a}^{{\dagger}}_{6}\hat{a}^{{\dagger}}_{8}\cos\left(\theta_{1}\right)\right)\ket{0}$
(75) $\displaystyle|\Gamma_{3}\rangle=$
$\displaystyle\left(\hat{a}^{{\dagger}}_{1}\hat{a}^{{\dagger}}_{3}\sin\left(\theta_{1}\right)+\hat{a}^{{\dagger}}_{2}\hat{a}^{{\dagger}}_{4}\cos\left(\theta_{1}\right)\right)$
$\displaystyle\qquad\qquad\qquad\otimes\left(\hat{a}^{{\dagger}}_{5}\hat{a}^{{\dagger}}_{7}\sin\left(\theta_{1}\right)+\hat{a}^{{\dagger}}_{6}\hat{a}^{{\dagger}}_{8}\cos\left(\theta_{1}\right)\right)\ket{0}$
(76) $\displaystyle|\Gamma_{4}\rangle=$
$\displaystyle\left(\hat{a}^{{\dagger}}_{1}\hat{a}^{{\dagger}}_{3}\cos\left(\theta_{1}\right)-\hat{a}^{{\dagger}}_{2}\hat{a}^{{\dagger}}_{4}\sin\left(\theta_{1}\right)\right)$
$\displaystyle\qquad\qquad\qquad\otimes\left(\hat{a}^{{\dagger}}_{5}\hat{a}^{{\dagger}}_{7}\sin\left(\theta_{1}\right)+\hat{a}^{{\dagger}}_{6}\hat{a}^{{\dagger}}_{8}\cos\left(\theta_{1}\right)\right)\ket{0}$
(77)

$\theta_{1}$ and $\theta_{2}$ are the state parameters. The circuit for Bell-
like state analysis is given in Fig.(6).

Figure 6: Circuit for the Bell-like state analyzer using ancillary qubits for
assistance.

Here $\phi_{1}^{o}=\phi_{2}^{o}=\phi_{3}^{o}=\phi_{4}^{o}=\frac{\pi}{4}$ and
$\phi_{1}^{e}=\phi_{2}^{e}=\phi_{3}^{e}=\phi_{4}^{e}=\theta_{2}$. It turns out
that we can discriminate among $|\Gamma_{1}\rangle,|\Gamma_{2}\rangle$ and
$|\Gamma_{4}\rangle$ with the success probability of

$P_{succ}=\frac{1}{32}\left(-2\cos\left(4\theta_{2}\right)-\cos\left(2\theta_{1}\right)\left(\cos\left(4\theta_{2}\right)+3\right)+6\right)$
(78)

Detection signatures can be summarised as:

  * •

If output modes
$\\{\hat{a}^{{\dagger}}_{1}\hat{a}^{{\dagger}}_{3}\hat{a}^{{\dagger}}_{5}\hat{a}^{{\dagger}}_{7}\\}$
and
$\\{\hat{a}^{{\dagger}}_{2}\hat{a}^{{\dagger}}_{4}\hat{a}^{{\dagger}}_{6}\hat{a}^{{\dagger}}_{8}\\}$
have odd number of photons in total, states $|\Gamma_{1}\rangle$ and
$|\Gamma_{2}\rangle$ are present.

    1. 1.

If
$\\{\hat{a}^{{\dagger}}_{1}\hat{a}^{{\dagger}}_{2}\hat{a}^{{\dagger}}_{5}\hat{a}^{{\dagger}}_{6}\\}$
have even no. of photons, we can unambiguously detect $|\Gamma_{1}\rangle$.

    2. 2.

If
$\\{\hat{a}^{{\dagger}}_{1}\hat{a}^{{\dagger}}_{2}\hat{a}^{{\dagger}}_{5}\hat{a}^{{\dagger}}_{6}\\}$
have odd no. of photons we can unambiguously detect $|\Gamma_{2}\rangle$.

  * •

If
$\\{\hat{a}^{{\dagger}}_{1}\hat{a}^{{\dagger}}_{3}\hat{a}^{{\dagger}}_{5}\hat{a}^{{\dagger}}_{7}\\}$
and
$\\{\hat{a}^{{\dagger}}_{2}\hat{a}^{{\dagger}}_{4}\hat{a}^{{\dagger}}_{6}\hat{a}^{{\dagger}}_{8}\\}$
have even number of photons, states $|\Gamma_{3}\rangle$ and
$|\Gamma_{4}\rangle$ are present.

    1. 1.

Furthermore if two creation operators are from the set
$\\{\hat{a}^{{\dagger}}_{1},\hat{a}^{{\dagger}}_{3},\hat{a}^{{\dagger}}_{5},\hat{a}^{{\dagger}}_{7}\\}$
and the other two are from the set
$\\{\hat{a}^{{\dagger}}_{2},\hat{a}^{{\dagger}}_{4},\hat{a}^{{\dagger}}_{6},\hat{a}^{{\dagger}}_{8}\\}$
only then we can uniquely detect $|\Gamma_{4}\rangle$

    2. 2.

$\\{\hat{a}^{{\dagger}}_{1}\hat{a}^{{\dagger}}_{2}\hat{a}^{{\dagger}}_{3}\hat{a}^{{\dagger}}_{4}\\}$
has odd photons and presence of exactly two photons in any one of the
$\\{\hat{a}^{{\dagger}}_{1},\hat{a}^{{\dagger}}_{3},\hat{a}^{{\dagger}}_{5},\hat{a}^{{\dagger}}_{7}\\}$
is the signature for unambiguous detection for this state.

A similar thing can be done if we add another 4 extra ancillary entangled
photons of the form
$\big{(}\hat{a}^{{\dagger}}_{9}\hat{a}^{{\dagger}}_{11}\hat{a}^{{\dagger}}_{13}\hat{a}^{{\dagger}}_{15}\sin\left(\theta_{1}\right)+\hat{a}^{{\dagger}}_{10}\hat{a}^{{\dagger}}_{12}\hat{a}^{{\dagger}}_{14}\hat{a}^{{\dagger}}_{16}\cos\left(\theta_{1}\right)\Big{)}\ket{0}$.
The variation of success probability with state parameters $\theta_{1}$ and
$\theta_{2}$ is given in Fig.(7)

Figure 7: Variation of success probability of Bell-like state discrimination
against state parameters – in the context of using ancilliary photon pairs in
entangled states.

We observe two things:

  * •

The success probability goes beyond $25\%$ for certain values of state
parameters $\theta_{1}$ and $\theta_{2}$.

  * •

Counterintuitively, as we increase the number of ancillary photons the success
probability decreases for given values of state parameters $\theta_{1}$ and
$\theta_{2}$. It still goes beyond $25\%$ in certain state parameter regimes.

If we take $\theta_{1}=\theta_{2}$ then we can unambiguously discriminate all
four Bell-like states. The detection signatures remain exactly the same with
extra addition being

  * •

$\\{\hat{a}^{{\dagger}}_{1}\hat{a}^{{\dagger}}_{2}\hat{a}^{{\dagger}}_{3}\hat{a}^{{\dagger}}_{4}\\}$
has even photons and presence of exactly two photons in any one of the modes
$\\{\hat{a}^{{\dagger}}_{1},\hat{a}^{{\dagger}}_{3},\hat{a}^{{\dagger}}_{5},\hat{a}^{{\dagger}}_{7}\\}$
is the signature for unambiguous detection for $|\Gamma_{3}\rangle$.

The success probability of the Bell-like state discrimination using 2
ancillary entangled photons can be calculated as

$P_{succ}=\frac{1}{16}\sin^{2}\left(\theta_{2}\right)\left(7\cos\left(2\theta_{2}\right)+\cos\left(4\theta_{2}\right)+10\right)$
(79)

Just like we did previously, we can add extra ancillary entangled photons of
the same form for assistance and calculate the success probability. these are
plotted in the Fig.(8)

Figure 8: Variation of success probability of Bell-like state discrimination
against state parameter

We again observe that for a certain value of state parameter success
probability can go beyond $25\%$. Also as we increase the number of ancillary
photons the success probability for a given value of the state parameter
decreases. The only difference being that in latter case we can discriminate
all four Bell-like states.

##  VII Bell-like state analysis using Non-linear elements

Besides using hyperentanglement or extra ancillary photons we can take the aid
of non-linear gadgets in an aim to completely discriminate among Bell states.
Four Bell-like states in a polarization basis are given as

$\displaystyle|\psi_{1}\rangle=(\sin(\theta_{1})\hat{h}^{{\dagger}}_{1}\hat{h}^{{\dagger}}_{2}+\cos(\theta_{1})\hat{v}^{{\dagger}}_{1}\hat{v}^{{\dagger}}_{2})\ket{0}$
(80)
$\displaystyle|\psi_{2}\rangle=(\cos(\theta_{1})\hat{h}^{{\dagger}}_{1}\hat{h}^{{\dagger}}_{2}-\sin(\theta_{1})\hat{v}^{{\dagger}}_{1}\hat{v}^{{\dagger}}_{2})\ket{0}$
(81)
$\displaystyle|\psi_{3}\rangle=(\sin(\theta_{2})\hat{h}^{{\dagger}}_{1}\hat{v}^{{\dagger}}_{2}+\cos(\theta_{2})\hat{v}^{{\dagger}}_{1}\hat{h}^{{\dagger}}_{2})\ket{0}$
(82)
$\displaystyle|\psi_{4}\rangle=(\cos(\theta_{2})\hat{h}^{{\dagger}}_{1}\hat{v}^{{\dagger}}_{2}-\sin(\theta_{2})\hat{v}^{{\dagger}}_{1}\hat{h}^{{\dagger}}_{2})\ket{0}$
(83)

### Using Sum Frequency Generation

A non-linear process known as sum-frequency generation(SFG) is used in [22] to
implement complete Bell-state analysis and then used to implement quantum
teleportation with $100\%$ efficiency. A similar scheme is used here. The
circuit looks like Fig.(9)

Figure 9: Circuit for the Bell-like state analyzer where non-linear processes
like SFG are used to assist.

The photon pair is first passed through a non-linear crystal where it
undergoes type-I SFG. Basically, pair of photons is converted into a single
photon with double the frequency. Mathematically it enables the following
transformation

$\displaystyle\hat{h}^{{\dagger}}_{1}\hat{h}^{{\dagger}}_{2}\longrightarrow\hat{v}^{{\dagger}}_{3}$
(84)
$\displaystyle\hat{v}^{{\dagger}}_{1}\hat{v}^{{\dagger}}_{2}\longrightarrow\hat{h}^{{\dagger}}_{3}$
(85)

This means it will only act on $|\psi_{1}\rangle$ and $|\psi_{2}\rangle$ and
leave the remaining two states unchanged. The transformed states are

$\displaystyle|\psi_{1}\rangle\longrightarrow(\sin(\theta_{1})\hat{v}^{{\dagger}}_{3}+\cos(\theta_{1})\hat{h}^{{\dagger}}_{3})\ket{0}$
(86)
$\displaystyle|\psi_{2}\rangle\longrightarrow(\cos(\theta_{1})\hat{v}^{{\dagger}}_{3}-\sin(\theta_{1})\hat{h}^{{\dagger}}_{3})\ket{0}$
(87)

An important thing to note is that these states have double the frequency of
the original photons. Then a dichoric mirror reflects this frequency in
direction ”a” while all other frequencies are transmitted on path ”b”. On path
”a” the photon is passed through an HWP at an angle $\frac{\theta_{1}}{2}$.
The final transformation looks like

$\displaystyle|\psi_{1}\rangle\longrightarrow\hat{h}^{{\dagger}a}_{3}\ket{0}$
(88)
$\displaystyle|\psi_{2}\rangle\longrightarrow\hat{v}^{{\dagger}a}_{3}\ket{0}$
(89)

Similarly in path ”b” $|\psi_{3}\rangle$ and $|\psi_{4}\rangle$ undergo type-
II SFG which can be mathematically written as

$\displaystyle\hat{v}^{{\dagger}}_{1}\hat{h}^{{\dagger}}_{2}\longrightarrow\hat{v}^{{\dagger}}_{3}$
(90)
$\displaystyle\hat{h}^{{\dagger}}_{1}\hat{v}^{{\dagger}}_{2}\longrightarrow\hat{h}^{{\dagger}}_{3}$
(91)

Then the beam is allowed to fall on an HWP at an angle
$45^{\circ}-\frac{\theta_{3}}{2}$. The states thus transform as

$\displaystyle|\psi_{3}\rangle\longrightarrow\hat{h}^{{\dagger}b}_{3}\ket{0}$
(92)
$\displaystyle|\psi_{4}\rangle\longrightarrow\hat{v}^{{\dagger}b}_{3}\ket{0}$
(93)

Thus we have complete Bell-like state discrimination. The signature is:

  * •

The state is $|\psi_{1}\rangle$ if detector $h_{1}^{a}$ clicks.

  * •

The state is $|\psi_{2}\rangle$ if detector $v_{1}^{a}$ clicks.

  * •

The state is $|\psi_{3}\rangle$ if detector $h_{1}^{b}$ clicks.

  * •

The state is $|\psi_{4}\rangle$ if detector $v_{1}^{b}$ clicks.

Thus we have $100\%$ success probability as compared to a scheme involving
hyperentanglement where success probability is only $50\%$. Here we want to
point out that the circuit used in [22] and above only differ in the
orientations of the half-wave plates. Thus without the use of any extra
resource in comparison to the Bell state discrimination, we can distinguish
Bell-like states.

All these results can be summarised in the form of the following table:

Discrimination prtocol | Success Probability
---|---
Using Hyperentanglement |  |
System qubits | Ancillary qubits | Bell states | Bell-like states
Polarisation DOF | Spatial DOF | $100\%$ | $50\%$
Spatial DOF | Polarisation DOF | $100\%$ | $50\%$
Polarisation DOF | OAM DOF | $100\%$ | $50\%$
Polarisation DOF | Time DOF | $100\%$ |  $>25\%$ but $<50\%$
|  |  | (depends on $\theta$)
Using extra ancillary | $75\%$ | $>25\%$
photon pair |  | (depends on $\theta_{1},\theta_{2}$)
Using SFG | $100\%$ | $100\%$

##  VIII Conclusion and Outlook:

In this work, we explored various methods of increasing the success
probability of discrimination of Bell-like states using extra resources such
as hyperentanglement (entanglement in other DOF of photons), correlation in
time, and ancillary entangled qubits in the backdrop of linear optics. Besides
that, we have also considered the minimum amount of non-linear resources
required to increase this success probability (particularly SFG).

In the case where momentum (or spatial) DOF of photons act as system qubits
while polarisation DOF acts as ancillary qubits, we have seen that the success
probability of Bell-like state discrimination increases from $25\%$ to $50\%$.
Also, all four Bell-like states can be distinguished from each other.
Comparing it to the case where we have Bell states instead we observe there
also all four Bell states can be distinguished from one another. The only
difference is that it is a perfect discrimination i.e. the success probability
is $100\%$. Instead of the above scenario if we have polarisation DOF as
system qubits and momentum DOF as system qubits we obtain the similar result.
The success probability is $50\%$. Using OAM degrees of freedom as ancillary
qubits while keeping polarisation DOF fixed as system qubits leads to the same
observation. In all three cases, the success probability of unambiguous Bell-
like state discrimination becomes $50\%$. Here, it is important to reiterate
the point that the results obtained above are yet to be shown to be optimal.
The optimal fidelity may be higher than $50\%$ but we guess that it is not the
case. The results obtained in this paper are indeed optimal.

We have also tried to tackle the problem of Bell-like state discrimination by
introducing the correlations in time DOF of two photons through the use of
birefringent materials. here we obtain an interesting result. In the case of
Bell state discrimination, $100\%$ success probability can be achieved through
this method which automatically implies that all four Bell states can be
perfectly distinguished. What we have observed is that in case of Bell-like
states only three of them can be distinguished from one another. Furthermore,
the success probability, although greater than $25\%$, is less than even
$50\%$. Another interesting observation arises when we use ancillary qubits to
distinguish between Bell-like states. In the Bell state scenario by adding two
ancillary photons the optimal success probability reaches $75\%$ from $50\%$
and further increases with the number of ancillary photons. However, this is
not the case with Bell-like states. firstly just by using two ancillary
photons the success probability can be higher or lower than $25\%$ depending
upon the state parameters $\theta_{1}$ and $\theta_{2}$. Secondly, by adding
more number of ancillary photons, instead of increasing, in stark contrast to
Bell states scenario, it decreases.

From the above discussion, one can conclude that discrimination of Bell-like
states requires more resources as compared to discrimination of Bell states in
the linear optical setting. However, we have found that when we use a non-
linear element (namely SFG) for Bell-like state discrimination it can be
accomplished in the same amount of resources as required for Bell state
discrimination. The only difference indeed, between both setups is the
orientation of the half-wave plates.

It has been shown that in a generalized three-party entanglement-swapping
scenario between Alice, Bob, and Charlie where the aim is to create the state
having the highest teleportation fidelity between Alice and Charlie, in some
cases, it is preferable for Bob to do non-maximally entangled measurements
(namely Bell-like measurements) instead of Bell measurements or any maximally
entangled measurement[25]. The experimental realization of such measurements
can be done with higher efficiency using the results mentioned in this paper.
Also as mentioned above the optimality of the given results has still not been
analyzed rigorously which can be another worthwhile direction to explore in
order to understand the limitations posed by linear optical schemes in the
implementation of various information-theoretical tasks.

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