aa r X i v : . [ qu a n t - ph ] S e p Equal Superposition Transformationsand Quantum Random Walks
Preeti ParasharPhysics and Applied Mathematics UnitIndian Statistical Institute203 B T Road, Kolkata 700 108, Indiae-mail: [email protected]
Abstract
The largest ensemble of qubits which satisfy the general transformation of equalsuperposition is obtained by different methods, namely, linearity, no-superluminalsignalling and non-increase of entanglement under LOCC. We also consider the asso-ciated quantum random walk and show that all unitary balanced coins give the sameasymmetric spatial probability distribution. It is further illustrated that unbalancedcoins, upon appropriate superposition, lead to new unbiased walks which have noclassical analogues.
Keywords:
Equal superposition ensemble; no-signalling; non-increase of entangle-ment; quantum random walk.
PACS:
Introduction
There has been considerable interest in the recent past to prove the non-existence of certainquantum unitary operations for arbitrary and unknown qubits. Some of the importantones are: the no-clonning theorem [1], the no-deleting principle [2], no-flipping operator[3] and the no-Hadamard operator [4]. These no-go theorems have been re-established byother physical fundamental principles, like the no-signalling condition and no-increase ofentanglement under LOCC [5]-[10]. It is then natural to ask that if these operations donot work universally (i.e., for all qubits), then for what classes of quantum states it wouldbe possible to perform a particular task by a single unitary operator. For example, the setof qubits which can be flipped exactly by the quantum NOT operator, lie on a great circleof the Bloch sphere [4, 11]. Likewise, the largest ensemble of states which can be rotatedby the Hadamard gate was obtained in [12].The Hadamard gate creates a superposition of qubit state and its orthogonal comple-ment with equal amplitudes. In the present work, we consider the most general trans-formation where the superposition is with amplitudes which are equal upto a phase. Inother words, the state and its orthogonal superimpose with equal probabilities but notnecessarily with exactly the same amplitudes. First, we obtain the largest class of quan-tum states which can be superposed via this transformation. Second, it is shown, byusing the no-signalling condition and non-increase of entanglement under LOCC, that thistransformation does not hold for an arbitrary qubit.The Hadamard transformation is known to be intimately connected to quantum randomwalks which were introduced in [13]. It has been used as a ‘coin flip’ transformation(balanced coin) to study the dynamics of such walks [14, 15]. In the same spirit, weconsider the quantum random walk associated with our general transformation and studythe probability distribution of the position of a particle. It is found that the entire familyof such walks gives the same asymmetric distribution. We have also considered a unitarytransformation with unequal amplitudes, serving as an unbalanced coin. It is shown that,after a suitable superposition, both types of coins lead to symmetric (unbiased) walks.However, in the case of unbalanced coin, we obtain new walks that have no classicalanalogues.The paper is organized as follows: In Sec. 2 we present the equal superposition en-semble. Sec. 3 and 4 pertain to the proving of the non-existence of equal superpositiontransformation for an arbitrary qubit. If the state and its orthogonal could be superposed,then it must belong to the ensemble presented in Sec. I. This is achieved by imposing thecondition of no-superluminal signalling and non-increase of entanglement under LOCC.Sec. 5 is devoted to the study of the associated quantum random walks. We end the paperwith some conclusions in Sec. 6. 2
The equal superposition ensemble
The computational basis (CB) states {| i , | i} of a qubit can be superposed most generallyvia the transformation U | i → α | i + β | i , U | i → γ | i + δ | i , (1) α, β, γ, δ being arbitrary non-zero complex numbers. We are, however, interested in equalsuperposition (upto a phase) of the basis vectors. So let β = e iθ α, δ = e iφ γ. (2)The transformed states are required to be normalized and orthogonal to each other. Thisimposes the following constraints αα ∗ = γγ ∗ = 1 / , φ = θ + π. (3)Eq.(1) then becomes U | i → α | i + e iθ α | i , U | i → γ | i − e iθ γ | i , (4)with the unitary matrix given by U = " α γe iθ α − e iθ γ . (5)This gives an infinite family of transformations since θ can take any value between 0 and2 π , and α, γ are c - numbers satisfying the constraint (3). One can get rid of the overallfactor by setting α = 1 = γ in Eq.(4). However, the states then become unnormalized. Torestore normalization one could simply fix α = 1 / √ γ . With this choice, the states in(4) reduce to the specific form of states lying on the equatorial great circle. So for the sakeof generality, we shall refrain from assigning any particular value to these parameters.Now, we address the following question: Which other orthogonal pair of qubit states {| ψ i , | ψ i} would transform under U in a similar manner as {| i , | i} ? More precisely, wewish to find as to which class of qubits would satisfy U | ψ i → α | ψ i + e iθ α | ψ i , U | ψ i → γ | ψ i − e iθ γ | ψ i , αα ∗ = γγ ∗ = 1 / . (6)For this purpose, we start with an arbitrary qubit state | ψ i and its orthogonal complement | ψ i as a superposition of the CB states | ψ i = a | i + b | i , | ψ i = b ∗ | i − a ∗ | i , (7)where the non-zero complex numbers obey the normalization condition aa ∗ + bb ∗ = 1.Substituting the above states in the first expression of Eq.(6) gives U | ψ i = ( αa + e iθ αb ∗ ) | i + ( αb − e iθ αa ∗ ) | i . (8)3ssuming that U acts linearly on | ψ i , we have U | ψ i = aU | i + bU | i = ( αa + γb ) | i + ( e iθ αa − e iθ γb ) | i . (9)Equating the coefficients in (8) and (9) gives b = e iθ αγ b ∗ , a + a ∗ = ( e − iθ + γα ) b = e − iθ b + e iθ b ∗ (10)Thus, we can state our main result: The general equal superposition transformation (6) holds for all qubit pairs {| ψ i , | ψ i} whichsatisfy the constraint (10) .It can be explicitly checked that unitarity holds for these states. Consider two suchdistinct states {| ψ i , | ψ i} and their orthogonal complements {| ψ i , | ψ i} which transformaccording to (6). Taking the inner product, we have h ψ | ψ i → αα ∗ [ h ψ | ψ i + e iθ h ψ | ψ i + e − iθ h ψ | ψ i + h ψ | ψ i ] , h ψ | ψ i → γγ ∗ [ h ψ | ψ i − e iθ h ψ | ψ i − e − iθ h ψ | ψ i + h ψ | ψ i ] , (11)where αα ∗ = γγ ∗ = 1 /
2. To see that these states actually satisfy the inner productrelations, it is instructive to write the complex state parameters as a = x + iy, b = u + iv ,where x, y, u, v are all real. In this notation, a state from this ensemble reads as | ψ i = {
12 ( e − iθ + γα )( u + iv ) + iy }| i + ( u + iv ) | i , (12)while its orthogonal would be | ψ i = ( u − iv ) | i − {
12 ( e iθ + γ ∗ α ∗ )( u − iv ) − iy }| i = e − iθ γα ( u + iv ) | i − {
12 ( e − iθ + γα )( u + iv ) − iy }| i . (13)The inner product rules are then explicitly given as h ψ | ψ i = 14 (6 + e iθ γα + e − iθ γ ∗ α ∗ ) e − iθ γα ( u + iv )( u + iv ) + y y + i e − iθ + γα )[( u + iv ) y − ( u + iv ) y ] = h ψ | ψ i ∗ , h ψ | ψ i = ie − iθ γα [( u + iv ) y − ( u + iv ) y ] = −h ψ | ψ i ∗ . (14)Substituting these in (11), we find that the inner product relations are indeed preserved.Our result provides a very convenient unified framework to deduce any desired class ofequally superposable quantum states. If the CB states obey a particular transformation(out of the infinite family (4)), then in a single shot we can obtain the entire ensemble4f qubits which would satisfy the same transformation. To demonstrate its usefulness, wepresent below, two known examples as special cases of our result.1. Hadamard ensemble : Choose α = √ , γ = √ , θ = 0. Then ( U → U H ) U H | i = 1 √ | i + | i ] , U H | i = 1 √ | i − | i ] , (15)where U H = 1 √ " − . (16)This is the well known Hadamard gate with its corresponding transformation. Notice that U H = I since U H = [ σ x + σ z ] / √ σ x , σ z are Pauli matrices. However, in general U = I .Further, substituting the above choice of the parameters, the constraint (10) gives b = b ∗ , i.e., b is real, and a + a ∗ = 2 b , i.e., Re ( a ) = b . In terms of the real parameters x, y, u, v , the above deductions yield v = 0 and u = x . Therefore, the qubit states becomerestricted to | ψ i = ( x + iy ) | i + x | i , | ψ i = x | i − ( x − iy ) | i , x + y = 1 . (17)Hence, we have obtained a special class of states which transform under the action of theHadamard matrix U H via the transformation U H | ψ i = 1 √ | ψ i + | ψ i ] , U H | ψ i = 1 √ | ψ i − | ψ i ] . (18)In other words, this proves the existence of the Hadamard gate (16) for any qubit chosenfrom the ensemble (17).2. Invariant ensemble : Choose α = √ , γ = i √ , θ = π . Then ( U → U I ) U I | i = 1 √ | i + i | i ] , U I | i = 1 √ i | i + | i ] , (19)where U I = 1 √ " ii . (20)An interesting property of this transformation is that it goes into itself, i.e., U I | i ↔ U I | i under the interchange | i ↔ | i . For this reason we shall refer to it as being ‘invariant’.The matrix U I is symmetric but not hermitian and U I = iσ x (i.e., the NOT gate) since U I = [ I + iσ x ] / √ U I | ψ i = 1 √ | ψ i + i | ψ i ] , U I | ψ i = 1 √ i | ψ i + | ψ i ] (21)we substitute the above values of α, γ, and θ in (10). This yields b = b ∗ , i.e., b is real, and a + a ∗ = 0, i.e., Re ( a ) = 0, implying that a is purely imaginary. Again assuming a = x + iy b = u + iv , these constraints give v = 0 and x = 0. Therefore, the qubit states becomerestricted to | ψ i = iy | i + u | i , | ψ i = u | i + iy | i , y + u = 1 . (22)The above two ensembles were obtained in [12] by treating each one separately. Here wehave shown that they can be deduced from a single general ensemble of equally superposedqubits.The family of transformations which remain invaraint under the interchange of | i and | i is a subset of the general family (4), and every member is essentially of the type (19).To see this let us consider the general transformation (4). For this to be invariant we musthave α = − e iθ γ and γ = e iθ α which implies that θ = π/ , π/
2. Substituting γ = ± iα in(4) we obtain the general form of the invariant transformation ( U → U I ′ ) U I ′ | i = α [ | i ± i | i ] , U I ′ | i = α [ ± i | i + | i ] , αα ∗ = 1 / , (23)where U I ′ = α " ± i ± i . (24)Since α is an overall phase factor, it can be readily verified that every member of (23) wouldlead to exactly the same ensemble (22). Thus, (19) can be regarded as a representative ofthe invariant family (23). In what follows, we shall establish our main result in the contextof two other physical principles, namely; the no-superluminal signalling condition and thenon-increase of entanglement under LOCC. Let us consider the CB states transforming via Eq.(4), and a qubit state | ψ i transformingunder the same unitary matrix U via the first expression in (6). We first show that if | ψ i is completely arbitrary, then this would imply superluminal signalling. For this purpose,assume that Alice possesses a 3 d qutrit while Bob has a 2 d qubit and both share thefollowing entangled state: | φ i AB = 1 √ | i A | i B + | i A | ψ i B + | i A | i B ) . (25)The density matrix of the combined system is defined as ρ AB = | φ i AB h φ | . Alice’s reduceddensity matrix can be obtained by tracing out Bob’s part ρ A = tr B ( ρ AB ) = 13 [ | ih | + | ih | + | ih | + a | ih | + a ∗ | ih | + b | ih | + b ∗ | ih | ] . (26)Now Bob applies the above mentioned unitary transformation on his qubit states {| i , | i , | ψ i} in Eq.(25). But, he does not communicate any information to Alice regarding his operation.6he shared state then changes to( I ⊗ U ) | φ i AB = | φ ′ i AB = 1 √ α | i + e iθ α | i + α | ψ i + e iθ α | ψ i + γ | i− e iθ γ | i ] . (27)After this operation, Alice’s new reduced density matrix becomes ρ ′ A = 13 [ | ih | + | ih | + | ih | + 12 ( a + e − iθ b + e iθ b ∗ − a ∗ ) | ih | + 12 ( a ∗ + e iθ b ∗ + e − iθ b − a ) | ih | + αγ ∗ ( a − e − iθ b + e iθ b ∗ + a ∗ ) | ih | + α ∗ γ ( a ∗ − e iθ b ∗ + e − iθ b + a ) | ih | ] . (28)Comparing the coefficients of each term in (26) and (28), it is evident that ρ ′ A = ρ A forarbitrary choices of the parameters a and b . So, in principle, Alice can distinguish between ρ A and ρ ′ A , although Bob has not revealed anything to her about his operation. Thisimplies that, with the help of entanglement, superluminal communication has taken place.But faster-than-light communication is forbidden by special theory of relativity. Hence, weconclude that the equal superposition transformation does not exist for an arbitrary qubit.If, however, we impose that the no-signalling constraint should not be violated, then ρ A and ρ ′ A should be equal because the action of U is a trace preserving local operationperformed only at Bob’s side. Comparing coefficients of the term | ih | in (26) and (28) werecover the condition a + a ∗ = e − iθ b + e iθ b ∗ . From | ih | we have αγ ∗ ( a − e − iθ b + e iθ b ∗ + a ∗ ) = b which yields 2 αγ ∗ e iθ b ∗ = b . Substituting γ ∗ = γ , we get the other constraint b = e iθ αγ b ∗ .Thus, the no-signalling condition gives exactly the same class of states that was obtainedinitially from linearity. Here we shall first show the non-existence of the unitary operation (6) for an arbitrary | ψ i by considering the fact that local operations and classical communication cannot increasethe entanglement content of a quantum system. It turns out that, ρ A and ρ ′ A above, haveequal eigenvalues (0 , / , / | Φ i AB = 1 √ b ∗ b [ | i A | i B | i B − | i B | i B √ | i A | i B | ψ i B − | ψ i B | i B √ , (29)where the first qubit is with Alice while the other two are at Bob’s side. Repeating theprotocol, we obtain Alice’s reduced density operator as ρ A = 11 + b ∗ b [ | ih | + b ∗ b | ih | + b | ih | + b ∗ | ih | ] . (30)7he amount of entanglement given by the von Neumann entropy is zero since the eigen-values of ρ A are 0 and 1. This means that the resource state (29) is a product state in theA:B cut. Now Bob applies the trace preserving general transformation on the last particle( B
2) in Eq.(29), which results in the state | Φ ′ i AB = 1 √ N [ γ | i − e iθ γ | i − α | i − e iθ α | i + α | ψ i + e iθ α | ψ i − α | ψ i − e iθ α | ψ i ] , (31)where N = 2 + { ( a − a ∗ ) − ( e − iθ b + e iθ b ∗ )( a + a ∗ ) } . Since a and b are arbitrary, soin general, the above state is entangled in the A:B cut. This implies that entanglementhas been created by local operation. However, we know that entanglement cannot beincreased by local operations even if classical communication is allowed. Therefore, theabove contadiction leads us to the conclusion that the unitary operator (5) cannot performthe same task for an arbitrary qubit, as it does for the CB states | i and | i .We now derive the conditions under which the entanglement in the state would remainzero even after the application of U . For this purpose we have to compare the eigenvaluesof the respective density matrices on Alice’s side. So after Bob’s operation ρ ′ A = 1 N [ | ih | + ( N − | ih | + D | ih | + D ∗ | ih | ] , (32)where D = { αγ ∗ ( a + a ∗ − e − iθ b + e iθ b ∗ ) + b } . The eigenvalue equation of the above matrixgives two roots, namely, λ ± = 12 ± q N − N − − DD ∗ )2 N , (33)In order to maintain the same amount of entanglement in the system before and after theunitary operation, we should equate these two roots λ ± of ρ ′ A to the eigenvalues 0 and 1of ρ A . This furnishes the constraint DD ∗ = N −
1. Substituting the expressions for
N, D and D ∗ , and rearranging the terms, this condition acquires the form[( a + a ∗ ) − ( e − iθ b + e iθ b ∗ )][ 14 ( a + a ∗ ) + 14 ( e − iθ b + e iθ b ∗ ) + γα ∗ b + αγ ∗ b ∗ ]+2( e − iθ γα ∗ b + e iθ αγ ∗ b ∗ + bb ∗ ) = 4 + ( a − a ∗ ) − ( e − iθ b + e iθ b ∗ )( a + a ∗ ) . (34)Using α ∗ = α , γ ∗ = γ on the L.H.S. and adding and subtracting 2 aa ∗ on the R.H.S., theabove relation is recast as[( a + a ∗ ) − ( e − iθ b + e iθ b ∗ )][ 14 ( a + a ∗ ) + 14 ( e − iθ b + e iθ b ∗ ) + 12 ( γα b + αγ b ∗ )]+[ e − iθ/ r γα b + e iθ/ s αγ b ∗ ] = 4 bb ∗ + [( a + a ∗ ) − ( e − iθ b + e iθ b ∗ )]( a + a ∗ ) (35)8hich can be written more compactly as[( a + a ∗ ) − ( e − iθ b + e iθ b ∗ )][ −
34 ( a + a ∗ ) + 14 ( e − iθ b + e iθ b ∗ ) + 12 ( γα b + αγ b ∗ )]= − [ e − iθ/ r γα b − e iθ/ s αγ b ∗ ] . (36)For convenience let us denote the two terms on L.H.S. by A and B . Then[ A ][ B ] = − [ e − iθ/ r γα b − e iθ/ s αγ b ∗ ] . (37)In the above, R.H.S. is either a real positive definite quantity or zero. For the L.H.S. tobe positive, there, however, exist two possibilities:(i) A > , B >
0: If we suppose that both terms in A are positive (they are already real),then a + a ∗ = ( e − iθ b + e iθ b ∗ ) + C , where C is a real positive constant. Thus B > γα b + αγ b ∗ ) > ( e − iθ b + e iθ b ∗ ) + C , which certainly is possible. Similarly if we suppose thatboth terms in A are negative, then A > | e − iθ b + e iθ b ∗ | = | a + a ∗ | + C . Thus B > γα b + αγ b ∗ ) + | a + a ∗ | > C .(ii) A < , B <
0: In a similar manner, restrictions can be obtained for this case.When R.H.S. is identically zero, then Eq.(37) would be satisfied uniquely if A = 0 , B = 0.This gives a + a ∗ = e − iθ b + e iθ b ∗ and b = e iθ αγ b ∗ , which are exactly the constraints that wehave earlier obtained by linearity and no-signalling. It can be easily checked that the othercases A = 0 , B = 0 and A = 0 , B = 0 cannot exist due to the constraint fixed by R.H.S.being zero.The above analysis demonstrates the possibility of existence of more solutions from theprinciple of non-increase of entanglement under LOCC. In the case of Hadamard operation,we had obtained a unique solution [10] from linearity, no-signalling and non-increase ofentanglement under LOCC. Here we get a larger set of states with zero entanglement,from the last method. However, we must remember that we are looking for orthogonalpairs of states {| ψ i , | ψ i} which transform under the unitary operation defined by (6). Inthe above, we have considered only | ψ i . Therefore, we must now carry out a similar analysiswith | ψ i . More precisely, we take the set of qubit states {| i , | i , | ψ i} and the shared stateas | Ψ i AB = 1 √ a ∗ a [ | i A | i B | i B − | i B | i B √ | i A | i B | ψ i B − | ψ i B | i B √ . (38)Then Alice’s reduced matrix reads as ρ A = 11 + a ∗ a [ | ih | + a ∗ a | ih | − a ∗ | ih | − a | ih | ] . (39)Bob now applies U on the states {| i , | i , | ψ i} of his last qubit, thereby changing the sharedstate to | Ψ ′ i AB = 1 √ N [ γ | i − e iθ γ | i − α | i − e iθ α | i + γ | ψ i − e iθ γ | ψ i − α | ψ i − e iθ α | ψ i ] , (40)9here N = 2 − { γα ∗ b ( a + a ∗ + e − iθ b − e iθ b ∗ ) + αγ ∗ b ∗ ( a + a ∗ − e − iθ b + e iθ b ∗ ) } . Thecorresponding reduced density matrix at Alice’s end becomes ρ ′ A = 1 N [ | ih | + ( N − | ih | + D| ih | + D ∗ | ih | ] , (41)where D = { ( a − a ∗ − e − iθ b − e iθ b ∗ ) − a ∗ } . Like the previous case, this matrix has thefollowing two eigenvalues λ ± = 12 ± q N − N − − DD ∗ )2 N . (42)Equating λ ± to the eigenvalues 0 and 1 of ρ A gives the constraint DD ∗ = N − a + a ∗ ) − ( e − iθ b + e iθ b ∗ )][ −
38 ( a + a ∗ ) −
18 ( e − iθ b + e iθ b ∗ ) + 12 ( γα b + αγ b ∗ )]= − [ e − iθ/ r γα b − e iθ/ s αγ b ∗ ] . (43)Interestingly, this is a new restriction on the expression on R.H.S. This has to be consistentwith the earlier restriction (36). Therefore equating (43) with (36) renders a + a ∗ = e − iθ b + e iθ b ∗ . Substituting this in either (36) or (43) yields b = e iθ αγ b ∗ . Thus we finally obtaina unique solution which is exactly the constraint (10) that defines our equal superpositionensemble.We remark that such a situation was not encountered in the case of the Hadamardoperation [10]. The reason is that the Hadamard transformation on | ψ i is not independentsince it can be obtained from the Hadamard transformation on the states {| i , | i , | ψ i} byusing the special property of the Hadamard operator, namely, U H = I . However, in thepresent scenario (and in general), U | ψ i cannot be deduced from { U | i , U | i , U | ψ i} . So itis necessary to take U | ψ i into consideration, although whether this would provide somenew restriction or not depends on the particular situation. For example, if we proceed with | ψ i , then linearity and no-signalling give nothing new but the same constraint (10) whichwas obtained from | ψ i . However, in the framework of non-increase of entanglement underLOCC, this indeed yields a different condition (43), thereby forcing the set of solutions toa single unique solution. In view of the above, we are now in a position to make a strongerstatement regarding our main result: Any pair of qubit states {| ψ i , | ψ i} can be equally superposed via the unitary operation(6) if and only if they satisfy the constraint (10) . In the previous sections, we have obtained by different methods the class of qubit stateswhich transform under the action of the unitary matrix U in a manner similar to Eq.104). As an application of this transformation (4), we are now going to study the quantumrandom walk associated with it. A particularly nice detailed survey of quantum walkshas been given by Kempe [16], while [17] is a short review devoted to their applications toalgorithms. The Hadamard matrix U H has been widely used as a balanced coin (translationto the left or to the right with equal probability) to study the properties of a discrete-timequantum random walk (QRW)[15]. For example, the probability of finding the particle ata particular site after T steps of the walk have been investigated in detail. The Hadamardcoin gives an asymmetric probability distribution for the QRW on a 1 d line. This is becausethe Hadamard coin treats the two CB states differently; it multiplies the phase by − | i . It has also been pointed out [16, 17] that if the Hadamard coin is replacedwith the more symmetric coin U I , then the probability distribution becomes symmetric.However, our analysis shows that this is not the case, even though U I treats both | i and | i in a symmetrical way. This also motivates us to investigate the discrete-time QRWfrom a more general point of view. We shall study the behaviour of the walk by takingthe general unitary matrix U given by Eq.(5) as our balanced coin. Subsequently, we shallcomment on some interesting features that these walks share.Consider a particle localized at position z on a 1 d line. The Hilbert space H P is spannedby basis states | z i , where z is an integer. This position Hilbert space is augmented by acoin space H C spanned by the two CB states | i and | i . To avoid confusion with theposition states, we now introduce a change of notation, and instead denote the CB statesas | ↑i and | ↓i . The total state of the particle lies in the Hilbert space H = H C ⊗ H P .The first step of the random walk is a rotation in the coin space. We follow a proceduresimilar to what was adopted for the Hadamard walk [16]. In our general scenario, thematrix U given by (5) serves as the coin, with the following action (cf. Eq.(4)) U | ↑i = α | ↑i + e iθ α | ↓i , U | ↓i = γ | ↑i − e iθ γ | ↓i . (44)The rotation is followed by translation with the application of the unitary operator S = | ↑ih↑ | ⊗ X z | z + 1 ih z | + | ↓ih↓ | ⊗ X z | z − ih z | (45)in the position space H P . Note that S is a ‘conditional’ translation operator since it movesthe particle by one unit to the right if the coin state is | ↑i , and to the left if it is | ↓i S | ↑i ⊗ | z i = | ↑i ⊗ | z + 1 i , S | ↓i ⊗ | z i = | ↓i ⊗ | z − i . (46)The particle is subjected to these two alternating unitary transformations. Therefore, theQRW of T steps is defined as the transformation A T , where A acts on the total Hilbertspace H and is given by A = S ( U ⊗ I ) (47)To start with, let the particle be in the | ↑i coin state and located at the position 0. Thusthe total initial state is denoted by | φ i = | ↑i ⊗ | i . Let us now evolve the walk, for a few11teps, under successive action of the operator A : | φ i → α | ↑i ⊗ | i + e iθ α | ↓i ⊗ | − i→ α | ↑i ⊗ | i + e iθ ( α | ↓i + αγ | ↑i ) ⊗ | i − e iθ αγ | ↓i ⊗ | − i→ α | ↑i ⊗ | i + e iθ ( α | ↓i + 2 α γ | ↑i ) ⊗ | i − e iθ αγ | ↑i ⊗ | − i + e iθ αγ | ↓i ⊗ | − i→ α | ↑i ⊗ | i + e iθ ( α | ↓i + 3 α γ | ↑i ) ⊗ | i + e iθ ( α γ | ↓i − α γ | ↑i ) ⊗ | i− e iθ ( α γ | ↓i − αγ | ↑i ) ⊗ | − i − e iθ αγ | ↓i ⊗ | − i (48)After T iterations, the particle is in an entangled state, say | φ T i . The probability of findingthe particle at a particular site z is given by P z = | ( h↑ | ⊗ h z | ) | φ T i| + | ( h↓ | ⊗ h z | ) | φ T i| . (49)Let us analyze, step by step, the spatial probability distribution of the walk.After T = 1: If we measure the position space after the first step, then the particle canbe found at the site 1 with probability αα ∗ and at the site − αα ∗ = 1 / T = 2: The probabilities of finding the particle at positions 2, − P = α α ∗ = ( αα ∗ ) = 1 / , P − = αα ∗ γγ ∗ = 1 / , P = α α ∗ + αα ∗ γγ ∗ = 1 / . (50)This step is also similar to the case of classical walk since the P ′ s are symmetricallydistributed.After T = 3: The distribution is P = α α ∗ = ( αα ∗ ) = 1 / , P − = αα ∗ γ γ ∗ = 1 / ,P = ( αα ∗ ) + 4( αα ∗ ) γγ ∗ = 5 / , P − = αα ∗ ( γγ ∗ ) = 1 / . (51)After the third step, the quantum walk begins to deviate from its classical counterpart.Although P = P − , note that P = P − . So the walk starts to be asymmetric, driftingtowards the right since the site 1 has greater probalility.After T = 4: Similarly, upon measuring the position space after four iterations, we getthe following asymmetric distribution P = 1 / , P − = 1 / , P = 5 / , P − = 1 / , P = 1 / . (52)Again, this differs from the symmetric classical probability distribution P = 1 / , P − =1 / , P = 1 / , P − = 1 / , P = 3 /
8. Proceeding in a similar way one can check theveracity of the foregoing conclusions by considering more steps of iterations. Clearly, theparameter θ which appears in the phase factor does not contribute to the probabilities. Also12ince αα ∗ = γγ ∗ , so for the purpose of probability distribution, only one of the parametersmay be regarded as independent.It is observed that the spatial probability distribution of the QRW corresponding tothe general matrix U is asymmetrical and coincides exactly with that of the already knownHadamard walk. This means that every unitary transformation in which the qubit CBstates are equally weighted, leads to the same probability distribution if the particle istaken in the same initial state. Therefore, we infer that even the symmetric coin U I induces an asymmetrical walk. In fact, it can be argued easily as to why a symmmetricprobability distribution for the initial state | ↑i ⊗ | i (or | ↓i ⊗ | i ) is impossible. Let usrefer to the distribution (51) after three iterations. If we want to make it symmetric, wemust have P = P − . This implies that αα ∗ [( αα ∗ ) + 4 αα ∗ γγ ∗ − ( γγ ∗ ) ] = 0 . (53)This equation cannot be satisfied since we know that αα ∗ = 1 /
2, and the term in thebracket equals 1. So L.H.S. can never be zero.The direction of drift in the walk depends on the initial coin state and the bias is theresult of quantum interference. So phases play a very crucial role in inducing asymmetry.This bias can, however, be taken care of if we again allow interference, so that the effectof the earlier superposition is negated. Thus, in order to make the walk symmetric orunbiased, we must take a superposition of | ↑i and | ↓i as the initial coin state. However,we shall not assume apriori that | ↑i and | ↓i are superimposed with equal probability. Forour general approach, we shall rather superimpose them with arbitrary amplitudes andobtain restrictions under which we can get a symmetric distribution. So we start the walkin the state | φ ′ i = ( x | ↑i + y | ↓i ) ⊗ | i , xx ∗ + yy ∗ = 1 (54)( x and y are, in general, complex numbers) and let it evolve under the repeated action ofthe operator A , as was done earlier.After T = 1, the state becomes | φ i = ( xα + yγ ) | ↑i ⊗ | i + e iθ ( xα − yγ ) | ↓i ⊗ | − i = A | ↑i ⊗ | i + e iθ B | ↓i ⊗ | − i (55)We now demand that the particle should be found at sites 1 and − AA ∗ = BB ∗ = 1 / xy ∗ αγ ∗ + x ∗ yα ∗ γ = 0 . (57)Clearly, this holds only if xy ∗ αγ ∗ is purely imaginary.13fter T = 2, the state of the particle becomes | φ i = αA | ↑i ⊗ | i + e iθ ( αA | ↓i + γB | ↑i ) ⊗ | i − e iθ γB | ↓i ⊗ | − i (58)and the probabilities are P = αα ∗ AA ∗ = 1 / , P − = γγ ∗ BB ∗ = 1 / , P = αα ∗ AA ∗ + γγ ∗ BB ∗ = 1 / . (59)After T = 3, the state evolves into | φ i = α A | ↑i ⊗ | i + e iθ ( α A | ↓i + 2 xα γ | ↑i ) ⊗ | i− e iθ ( γ B | ↑i + 2 yγ α | ↓i ) ⊗ | − i + e iθ γ B | ↓i ⊗ | − i (60)The probabilities for the odd sites are P = ( αα ∗ ) AA ∗ = 1 / , P − = ( γγ ∗ ) BB ∗ = 1 / P = ( αα ∗ ) AA ∗ + 4 xx ∗ ( αα ∗ ) γγ ∗ = 1 / xx ∗ / P − = ( γγ ∗ ) BB ∗ + 4 yy ∗ ( γγ ∗ ) αα ∗ = 1 / yy ∗ / P = P − which, in turn, implies that xx ∗ = yy ∗ . But from normaliza-tion, we have xx ∗ + yy ∗ = 1. This restricts the value to xx ∗ = yy ∗ = 1 /
2. So the twoamplitudes are equal, upto a phase factor, leading to an equal superposition of | ↑i and | ↓i . We have thus found that in order to make the quantum walk associated with thematrix U symmetric, it is necessary to take an ‘equally’ superposed coin state in such away that xy ∗ αγ ∗ is purely imaginary. We present a new example to illustrate this situation. Example:
Choose α = γ = i and θ = π . The transformation (44) becomes U | ↑i = 1 + i | ↑i − i | ↓i ) , U | ↓i = 1 + i | ↑i + i | ↓i ) (62)This has the features of a ‘hybrid’ between the Hadamard and the Invariant transforma-tions discussed earlier. As expected, this gives an asymmetric walk. However, if we takethe coin state in a superposition with amplitudes x = √ and y = − i √ , then the conditionthat xy ∗ αγ ∗ is purely imaginary is satisfied. Thus upon evolving the walk with the initialstate √ ( | ↑i − i | ↓i ) ⊗ | i , we do get the symmetric probability distribution which coincideswith the classical one. Unbalanced coin:
We have seen above that unitary balanced coins lead to a symmetric walk after appropriatesuperposition. Now we shall show that even unbalanced coins can yield an unbiased walkunder similar restrictions.Consider the unequal superposition transformation given in [12]
U | ↑i = p | ↑i + q | ↓i , U | ↓i = q ∗ | ↑i − p ∗ | ↓i , pp ∗ + qq ∗ = 1 , (63)14here, in general, pp ∗ = qq ∗ and the unitary matrix U is given by U = " p q ∗ q − p ∗ . (64)Following the same procedure, let us start the walk in the superposed state | Φ ′ i = ( r | ↑i + s | ↓i ) ⊗ | i , rr ∗ + ss ∗ = 1 , (65) r, s being non-zero c − numbers. After T = 1, the state becomes | Φ i = ( rp + sq ∗ ) | ↑i ⊗ | i + ( rq − sp ∗ ) | ↓i ⊗ | − i = E | ↑i ⊗ | i + F | ↓i ⊗ | − i . (66)For a symmetric probability distribution, we must have EE ∗ = F F ∗ = 1 / , (67)which leads to the constraint rs ∗ pq + r ∗ sp ∗ q ∗ = 0 , rr ∗ = ss ∗ . (68)This implies that like the previous case, here also the coin state must be in equal superpo-sition (upto a phase) of | ↑i and | ↓i .After T = 2, the particle is in the entangled state | Φ i = pE | ↑i ⊗ | i + ( qE | ↓i + q ∗ F | ↑i ) ⊗ | i − p ∗ F | ↓i ⊗ | − i (69)with the probability distribution P = pp ∗ EE ∗ = 12 pp ∗ , P − = pp ∗ F F ∗ = 12 pp ∗ , P = qq ∗ EE ∗ + qq ∗ F F ∗ = qq ∗ , (70)After T = 3, the entangled state is read as | Φ i = p E | ↑i ⊗ | i + pqE | ↓i ⊗ | i + ( qq ∗ E + pq ∗ F ) | ↑i ⊗ | i + ( − p ∗ qE + qq ∗ F ) | ↓i ⊗ | − i − p ∗ q ∗ F | ↑i ⊗ | − i + p ∗ F | ↓i ⊗ | − i , (71)and the associated probabilities are P = p p ∗ EE ∗ = ( pp ∗ ) EE ∗ = 12 ( pp ∗ ) , P − = ( pp ∗ ) F F ∗ = 12 ( pp ∗ ) ,P = pp ∗ qq ∗ + 12 ( qq ∗ ) , P − = pp ∗ qq ∗ + 12 ( qq ∗ ) . (72)Clearly, the above distribution is symmetric. One can continue like this for large times.15 xample: Let p = √ , q = , r = √ and s = i √ . It can be checked that the constraint(68) holds for this choice. The associated probabilities are T = 1 : P = P − = 1 / ,T = 2 : P = P − = 3 / , P = 1 / ,T = 3 : P = P − = 9 / , P = P − = 7 / . (73)Hence we get a new symmetric distribution which is different from the classical one. Thisexample demonstrates new possiblities in the quantum world which have no classical analo-ques. The distribution depends only on the values of p and q , after the initial constraint(68) is satisfied. For p = q = √ , we recover the Hadamard walk. So this walk can bethought of as a ‘generalized’ Hadamard walk for unequal amplitudes p and q . In this work, we have established that it is not possible to create a superposition with equalprobabilities, of an arbitrary qubit state and its orthogonal. The class of states for whichthis can be achieved is presented. In addition, by using the principles of no-superluminalsignalling and non-increase of entanglement under LOCC we have shown that this is theonly set of qubits which would satisfy the equal superposition transformation. In otherwords, a qubit state and its complement can be equally superposed if and only if theybelong to the aforementioned ensemble.The quantum random walk associated with this general unitary equal superpositiontransformation has been investigated from the point of view of probability distribution ofa particle. We have found that the entire family leads to the same asymmetric distribution.This implies that even the symmetric transformation (19) gives an asymmetric walk. Itmay be mentioned that apart from the CB vectors, any state from our ensemble specifiedby (10), can be used as a coin state to study the evolution of the walk. The measurementon the coin register would then have to be carried out in the {| ψ i , | ψ i} basis. We have alsoobtained conditions under which equal and unequal superpositions would yield unbiasedwalks. To illustrate this, a few examples have been presented. We have analysed theevolution explicitly only upto four iterations. It would be interesting to simulate the walkassociated with the unbalanced coin for large times and study the mixing time and otherproperties. Acknowledgement:
I thank G. Kar for useful discussions.
References [1] W.K. Wootters and W.H. Zurek,
Nature , 802 (1982).[2] A.K. Pati and S. Braunstein,
Nature , 164 (2000).163] V. Buzek, M. Hillery and R.F. Werner,
Phys. Rev. A , R2626 (1999).[4] A.K. Pati, Phys. Rev. A , 062319 (2002).[5] N. Gisin, Phys. Lett. A , 1 (1998).[6] L. Hardy and D.D. Song,
Phys. Lett. A , 331 (1999).[7] A.K. Pati,
Phys. Lett. A , 103 (2000).[8] A.K. Pati and S. Braunstein,
Phys. Lett. A , 208 (2003).[9] I. Chattopadhyay, S.K. Choudhary, G. Kar, S. Kunkri and D. Sarkar,
Phys. Lett.A , 384 (2006).[10] P. Parashar, quant-ph/0606231, (2006).[11] S. Ghosh, A. Roy and U. Sen,
Phys. Rev. A , 014301 (2000).[12] A. Maitra and P. Parashar, Int. J. Quant. Info. , 653 (2006).[13] Y. Aharonov, L. Davidovich and N. Zagury, Phys. Rev. A , 1687 (1993).[14] D. Meyers, J. Statistical Phys. , , 551 (1996).[15] A. Nayak and A. Vishwanath, quant-ph/0010117, (2000).[16] J. Kempe, Contemporary Physics ,44