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Physics & Astronomy International Journal

Conceptual Paper Volume 2 Issue 2

Standard quantum algorithms and the property of a certain function

Koji Nagata,1 Tadao Nakamura2

1Department of Physics, Korea Advanced Institute of Science and Technology, Korea
2Department of Information and Computer Science, Keio University, Japan

Correspondence: Koji Nagata, Department of Physics, Korea Advanced Institute of Science and Technology, Daejeon 34141, Korea

Received: March 12, 2017 | Published: April 18, 2018

Citation: Nagata K, Nakamura T. Standard quantum algorithms and the property of a certain function. Phys Astron Int J. 2018;2(2):146-147. DOI: 10.15406/paij.2018.02.00076

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Abstract

We discuss a new mathematical structure for standard quantum algorithms. It says a certain property in case of a special function f that the relationf(x)=f(x) holds. That is, the particular mathematical structure of the special function is thatf(x) should be even if we assume|x=|x .

Keywords: quantum algorithms, quantum computation

Introduction

In 1985, the Deutsch–Jozsa algorithm was discussed.1–3 In 1993, the Bernstein–Vazirani algorithm was published.4,5 This work can be considered an extension of the Deutsch–Jozsa algorithm. In 1994, Simon’s algorithm6 and Shor’s algorithm7 were discussed. In 1996, Grover et al.8 provided the highest motivation for exploring the computational possibilities offered by quantum mechanics.

In this short contribution, we discuss a new mathematical structure for standard quantum algorithms. They say a certain property in case of a special functionf that the relationf(x)=f(x) holds.

Mathematical structure for standard quantum algorithms

We discuss a new mathematical structure for standard quantum algorithms in case of a special functionf . Let us suppose that we are given the following function

f:{(2N1),(2N2),,2N2,2N1}{0,1,,2N2,2N1}. (1)

We shall assume thatf(y)0 . Let us introduce a functiong(x) that transforms binary strings into positive integers. We also defineg1(f(g(x)))=F(x) . We shall assume, for the time being, that the given function is even. Thus, we have
F(x)=F(x){0,1}Nx{0,1}N.    (2)

We see that the condition (2) holds in standard quantum algorithms.

What the functionf(x) does in (1) is to map a set of discrete values onto another one. In (2), we assume thatx is the binary representation of one element. x will be given by a binary string belonging to the Cartesian productN{0,1}×{0,1}××{0,1} , for instance,x=(0,1,1,0,0,1,,1) . We then definex  as(0,1,1,0,0,1,,1) .

Throughout the discussion, we omit any normalization factor. Let us suppose|x=|x . The input state is

|ψ1=|N0,0,,0,1|N1,1,,1.            (3)

The functionF is evaluated by using the following unitary2N qubits gate

UF:|x,z|x,z+F(x)    (4)

with

UF:|x,z|x,z+F(x)

|x,z|x,z+F(x)

|x,z|x,z+F(x)

|x,z|x,z+F(x)                (5)

And employing the fact thatF(x)=F(x) . Here,z+F(x)=(z1F1(x),z2F2(x),,zNFN(x)) (the symbol indicates addition modulo 2).

We have the following fact

UF|N0,0,...,0,1|N1,1,...,1=  |N0,0,...,0,1|¯F(0,0,...,0,1). (6)

Here, for example, if we haveF(0,0,...,0,1)=(0,1,1,0,0,1,,1) , then¯F(0,0,...,0,1)=(1,0,0,1,1,0,,0) . Surprisingly the relation F(x)=F(x)  is necessary for the fundamental relation (6) as shown below. From the definition in (5), we have

UF|x|N1,1,...,1=|x|¯F(x).            (7)

 This implies forxx , wit x0

UF|x|N1,1,...,1=|x|¯F(x).    (8)

We state that|x=|x . Then it follows that the minus sign on left and right hand side of (8) drop off. This implies

UF|x|N1,1,...,1=|x|¯F(x).          (9)

We furthermore assume such that

|P=|QP=Q.             (10)

Comparing (7) with (9) we see|¯F(x)=|¯F(x) . Hence, we cannot avoid the following property of the function in order to maintain consistency for the fundamental relation (6)

¯F(x)=¯F(x).      (11)

That is, the function under study is even

F(x)=F(x).      (12)

Conclusion

In conclusion, we have discussed a new mathematical structure for standard quantum algorithms. They have said a certain property in case of a special functionf that the relationf(x)=f(x) holds.

Acknowledgements

We thank Professor Han Geurdes for valuable discussions.

Conflict of interest

Authors declare there is no conflict of interest.

References

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