For a Deeper Understanding and Appreciation of Quantum Physics

BATTLING DECOHERENCE: THE FAULT-TOLERANT QUANTUM COMPUTER

EDWIN B. FABILLAR

Introduction

Information carried by a quantum system has notoriously weird properties. Physicists and engineers are now learning how to put that weirdness to work. Quantum computers, which manipulate quantum states rather than classical bits, may someday be able to perform tasks that would be inconceivable with conventional digital technology. A particularly daunting difficulty is that quantum computers are highly susceptible to making errors. The magical power of the quantum computer comes from its ability to process coherent quantum states; but such states are very easily damaged by uncontrolled interactions with the environment—a process called decoherence. In response to the challenge posed by decoherence, the new discipline of quantum error correction has arisen at the interface of physics and computer science. We have learned that quantum states can be cleverly encoded so that the debilitating effects of decoherence, if not too severe, can be resisted.

The power of the quantum computer

A classical computer executes a series of simple operations (often called “gates”), each of which acts on a single bit or pair of bits. By executing many gates in succession, the computer can evaluate any Boolean function of a set of input bits. Also a quantum computer executes a series of elementary quantum gates, each of which is a unitary transformation that acts on a single qubit or pair of qubits. By executing many such gates in succession, the quantum computer can apply a complicated unitary transformation to a particular initial state of a set of qubits. A classical computer can faithfully simulate a quantum computer, so that anything the quantum computer could do, the classical computer could also do. Still, there is a sense in which the quantum computer appears to be a more powerful device: In more physical terms, running a classical simulation of a quantum computer is hard because (as exemplified by John Bell’s famous inequalities) correlations among quantum bits are qualitatively different from correlations among classical bits. The exponential explosion in the size of Hilbert space as we increase the number of qubits arises because the correlations among qubits are too weird to be expressed easily in classical language.

The challenge of error correction

Are there any obstacles that might be fundamental matters of principle, that would prevent us from ever constructing a quantum computer? In fact, there is a problem of principle that is potentially very serious: decoherence. Unavoidable interactions with the environment will cause the quantum information stored in a quantum computer to decay, thus inducing errors in the computation. If quantum computers are ever to be capable of solving hard problems, a means must be found to control decoherence and other potential sources of error.

Quantum error-correcting codes

In 1995, Shor and Andrew Steane discovered that the obstacles were illusory— that quantum error correction really is possible. To appreciate the insights of Shor and Steane, lets first consider how to defend quantum information against a dragon who performs only bit flips . We are to protect the state

(1) $a|0> + b|1>$ ,

a coherent superposition of the red (|0>) and green (|1>) states of a single qubit, where the complex coefficients a and b are unknown. Were the dragon to attack, the bit flip would transform the state to

(2) $a|1> + b|0>$ ,

and damage would be inflicted unless a =±b. The beaver’s assignment is to diagnose and reverse bit flips, but without disturbing the delicate superposition state, that is without modifying a and b, the beaver applies the principle of redundant storage by encoding the qubit in a state of three qubits.

The red state is encoded as three red qubits, and the green state as three green qubits; that is,

(3.a) $|0> \rightarrow |0> = |000>$ ,

(3.b) $|1> \rightarrow |1> = |111>$ ,

Thus the unknown superposition state becomes

(4) $a|0> + b|1> \rightarrow a|0> + b|1> = a|000> + b|111>$ ,

This redundant state is not the same as three identical copies of the original unknown state, which would be

(5) $(a|0> + b|1>) (a|0> + b|1>) (a|0> + b|1>)$ ,

Although it is impossible to copy unknown quantum information, nothing prevents us from building a (unitary) machine that will execute the encoding transformation given as equation 4.

Now suppose that the dragon flips one of the three qubits, let’s say the first one, so that the state becomes

(6) $a|100> + b|011>$ ,

and the beaver is to detect and reverse the damage. His first impulse would be to open the boxes and look to see if one ball was a different color from the others, just as he would to diagnose errors in classical information, but he must resist that temptation. If he were to open door 1 of all three boxes, he would find either I100> (with probability |a|2), or I011 > (with probability IbI2); either way, the coherent quantum information (the values of a and b) would be irrevocably lost. But he is a clever beaver who knows he need not restrict his attention to single-qubit measurements. Instead, he performs collective measurements on two qubits at once. The beaver won that round, but now the dragon tries a more subtle approach. Rather than flipping the first qubit, he rotates it only slightly, so that the three-qubit state becomes

(7) $a|000> + b|111> \rightarrow a|000> + b|111>$

$+ e(a|100> + b|011>) + o (e2)$ ,

where |e| >> 1. What should the beaver do now? In fact, he can do the same thing as before. If he performs a collective measurement on the first two qubits, then most of the time

(with probability $1 - |e|^2$ ), the measurement well project the damaged state (equation 7) back to the completely undamaged state (equation 4). Only much more rarely (with probability $|e|^2$ ) will the measurement project onto the state given as equation 6 with a bit-flip error. But then the measurement outcome tells the beaver what action to take to repair the damage, just as in theprevious case.

Collective measurement and fault tolerance

Topological ideas arise naturally in the theory of fault tolerance. The topological properties of an object remain invariant when we smoothly deform the object. Similarly, how a fault-tolerant gate acts on encoded information should remain unchanged when we deform the gate by introducing a small amount of noise. In seeking fault-tolerant implementations of quantum logic, we are led to contemplate physical interactions with a topological character. What comes quickly to mind is the Aharonov-Bohm effect. When an electron is transported around a magnetic flux tube, its wave function acquires a phase that depends only on the winding number of the electron about the solenoid; it is unmodified if the electron’s trajectory is slightly deformed. If we could perform quantum logic by means of topological interactions, then we would be able to give the beaver a rest! We could protect encoded information not by vigilantly checking for errors and reversing them, but rather by weaving fault tolerance into the design of our hardware.

Outlook

For a quantum computer to compete with a state-of-the-art classical computer, we will need machines with hundreds or thousands of qubits capable of performing millions or billions of operations. The technology clearly has far to go before quantum computers can assume their rightful place as the world’s fastest machines. But now that we know how to protect quantum information from errors, there are no known insurmountable obstacles blocking the path. Quantum computers of the 21st century may well unleash the vast computational power woven into the fundamental laws of physics. Apart from enabling a new technology, the discovery of fault-tolerant methods for quantum error correction and quantum computation may have deep implications for the future of physics. What else might coherent quantum systems be capable of? In what ways will they surprise, baffle, and delight us? Armed with new tools for maintaining and controlling intricate quantum states, physicists of the next century will seek the answers.

References

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124.

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Zurek, Nature 299, 802 (1982).

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77, 198 (1996). C. Bennett, D. DiVincenzo, <L Smolin, W.

Wootters, Phys. Rev. A 54, 3824 (1996).

8. A. R. Calderbank, P. W. Shor, Phys. Rev. A 54, 1098 (1996).

A. M. Steane, Proc. Roy. Soc. London, Ser. A452, 2551 (1996).

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Computer Science, Los Alamitos, Calif, IEEE Press (1996), p.

56.

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9608012.

11. E. Knill, R. Laflamme, W. Zurek, Proc. Roy. Soc. London, Ser.

A454, 365 (1998). A. Yu. Kitaev, Russ. Math. Surveys 52,1191

(1997). D. Aharonov, M. Ben-Or, in Proc. of the 29th Annual

ACM Symp. on the Theory of Computing, New York, ACM

(1997), p. 176. J. Preskill, Proc. Roy. Soc. London, Ser. A 454,

385(1998).

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9707021. J. Preskill, preprint, http://xxx.lanl.gov/

abs/quant-ph/9712048.

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abs/quant-ph/9811068.

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Wineland, Phys. Rev. Lett. 75, 4714 (1995). Q. A. Turchette,

C. J. Hood, W. Lange, H. Mabuchi, H. J. Kimble, Phys. Rev.

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AN UNDULATORY THEORY OF THE MECHANICS OF ATOMS AND MOLECULES by E. Schrodinger

EDMAR G. PANTOHAN

This report is based on the very interesting researches of L. de Broglie on what he called “phase waves”. The advantages of the wave theory,

The laws of motion and quantum condition can be derived from Hamiltonian principle.
The discrepancy between the frequency of motion and frequency of emission disappears when the latter frequencies coincide with the difference of the former.
It possible to pursue the so called transitions.
This wave theory is better supported by experiment

Consider a point mass m moving in a conservative field of force in q-space. Using the well known Hamiltonian principle and the kinetic energy of the particle, we can have the Hamiltonian partial differential equation. To solve this equation we put W=Et+S(x,y,z), geometrically we described W as a system of surfaces. By allowing this to vary with time, the phase velocity u of the wave is solve. But this velocity u is not the velocity of the particle which proportional to the square root of the difference between energy and potential.

Though in the above we are dealing with wave surfaces and calculating phase velocity, the whole established analogy is more on geometrical optics than real physical or undulatory optics. The fundamental mechanical conception is that of the path or the orbit of the particle and it corresponds to the conception of rays in optical analogy. But the concept of rays loses its significance in real physical optics as soon as the dimensions of the beam or of material become comparable with the wavelength . Considering this striking fact, the ordinary mechanics is really not applicable to mechanical system of a very small atomic dimensions. The same kind as the non-applicability of geometrical optics to the phenomena of diffraction or interference.

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Perturbation of a 3-dimensional infinite cubical well

Karl Patrick S. Casas

Consider a three-dimensional infinite cubical well

Cubical Well with perturation

$V(x,y,z)=\left\{ \begin{array} {cccccc} 0, & if &0<x<a,& 0<y<a,& and& 0<z<a \\ \infty, &otherwise& & & & \\ \end{array}$

The stationary states are

$\psi^{0}_{n_x,n_y,n_z}(x,y,z)=\left(2/a\right)^{3/2}\sin\left(\frac{n_x \pi}{a}x\right)\sin\left(\frac{n_y \pi}{a}y\right)\sin\left(\frac{n_z \pi}{a}z\right)$

and the allowed ground state energy is given by

$E^0_0=3\frac{\pi^2\hbar^2}{2ma^2}$

The first excited state is triply degenerate,

$E^0_1=3\frac{\pi^2\hbar^2}{ma^2}$

and we denote each degenerate state as

$\psi_a=\psi_{112}, \psi_b=\psi_{121}, \psi_c=\psi_{211}$

Now, we introduce the perturbation $H'$ (shown in the figure above),

$H'=\left\{ \begin{array} {cccccc} V_0, & if &0<x<a/2,& a/2<y<a,& and& 0<z<a \\ 0, &otherwise& & & & \\ \end{array}$

We can get the first-order correction to the ground state energy by

$E^1_0=\langle\psi_{111}|H'|\psi_{111}\rangle$

$=\left(2/a\right)^3V_0\int_0^{a/2}\sin^2\left(\frac{\pi}{a}x\right)dx\int_{a/2}^{a}\sin^2\left(\frac{\pi}{a}y\right)dy\int_0^{a}\sin^2\left(\frac{\pi}{a}z\right)dz$

Using the integral formula,

$\int{\sin^2\alpha xdx}=\frac{x}{2}-\frac{\sin2\alpha x}{4\alpha}$ .

And so, solving each integral,

$\int_0^{a/2}\sin^2\left(\frac{\pi}{a}x\right)dx=\left[\frac{x}{2}-\frac{\sin2\pi x/a}{4\pi/a}\right]_0^{a/2}=\frac{x}{2}\left|_0^{a/2}=\frac{a}{4}$

$\int_{a/2}^{a}\sin^2\left(\frac{\pi}{a}y\right)dy=\left[\frac{y}{2}-\frac{\sin2\pi y/a}{4\pi/a}\right]_{a/2}^a=\frac{y}{2}\left|_{a/2}^a=\frac{a}{4}$

$\int_0^{a}\sin^2\left(\frac{\pi}{a}z\right)dz=\left[\frac{z}{2}-\frac{\sin2\pi z/a}{4\pi/a}\right]_0^{a}=\frac{z}{2}\left|_0^{a}=\frac{a}{2}$

Thus,

$E_0^1=\left(\frac{2}{a}\right)^3V_0\left(\frac{a}{2}\right)^3\left(\frac{1}{4}\right)=\frac{V_0}{4}$ .

To solve for the eigenvalue $E_1^1$ , we consider the components of

$W=\left( \begin{array} {ccc} W_{aa} & W_{ab} & W_{ac}\\ W_{ba} & W_{bb} & W_{bc}\\ W_{ca} & W_{cb} & W_{cc} \end{array} \right)$

where we can then solve the corresponding eigenvalue equation,

$\left( \begin{array} {ccc} W_{aa} & W_{ab} & W_{ac}\\ W_{ba} & W_{bb} & W_{bc}\\ W_{ca} & W_{cb} & W_{cc}\\ \end{array} \right)$

$\times\left( \begin{array} {c} \alpha \\ \beta\\ \gamma \end{array} \right) = E' \left( \begin{array} {c} \alpha \\ \beta\\ \gamma \end{array} \right)$

Solving for the components,

$W_{aa}=\langle\psi_a|H'|\psi_a\rangle$

$=\left(2/a\right)^3V_0\int_0^{a/2}\sin^2\left(\frac{\pi}{a}x\right)dx\int_{a/2}^{a}\sin^2\left(\frac{\pi}{a}y\right)dy\int_0^{a}\sin^2\left(\frac{2\pi}{a}z\right)$

we already have the first and second integral. Now, consider the z-part

$\int_0^a\sin^2(2\pi z/a )dz=\left[\frac{z}{2}-\frac{\sin2\left(2\pi/a\right)z}{4\left(2\pi/a\right)}\right]_0^a=\frac{z}{2}\left|_0^a=\frac{a}{2}$

hence,

$W_{aa}=\left(\frac{2}{a}\right)^3V_0\left(\frac{a}{4}\right)\left(\frac{a}{4}\right)\left(\frac{a}{2}\right)=\frac{V_0}{4}$

Now,

$W_{bb}=\langle\psi_b|H'|\psi_b\rangle$

$=\left(2/a\right)^3V_0\int_0^{a/2}\sin^2\left(\frac{\pi}{a}x\right)dx\int_{a/2}^{a}\sin^2\left(\frac{2\pi}{a}y\right)dy\int_0^{a}\sin^2\left(\frac{\pi}{a}z\right)dz$

solving the y-integral

$\int_{a/2}^{a}\sin^2\left(\frac{2\pi}{a}y\big)dy$

$=\left[\frac{y}{2}-\frac{\sin2\left(2\pi/a\right)y}{4\left(2\pi/a\right)}\right]_0^a=\frac{y}{2}\left|_{a/2}^a =\frac{a}{2}-\frac{a}{4}=\frac{a}{4}$ .

So,

$W_{bb}=\left(\frac{2}{a}\right)^3V_0\left(\frac{a}{4}\right)\left(\frac{a}{4}\right)\left(\frac{a}{2}\right)=\frac{V_0}{4}$ .

And then,

$W_{cc}=\langle\psi_c|H'|\psi_c\rangle$

$=\left(2/a\right)^3V_0\int_0^{a/2}\sin^2\left(\frac{2\pi}{a}x\right)dx\int_{a/2}^{a}\sin^2\left(\frac{\pi}{a}y\right)dy\int_0^{a}\sin^2\left(\frac{\pi}{a}z\right)dz$ .

Considering the x-part,

$\int_{a/2}^{a}\sin^2\left(\frac{2\pi}{a}x\right)dx=\frac{x}{2}-\frac{\sin2\left(2\pi/a\right)x}{4\left(2\pi/a\right)}\left]_0^a=\frac{x}{2}\left|_{a/2}^a =\frac{a}{2}-\frac{a}{4}=\frac{a}{4}$

Thus,

$W_{cc}=\left(\frac{2}{a}\right)^3V_0\left(\frac{a}{4}\right)\left(\frac{a}{4}\right)\left(\frac{a}{2}\right)=\frac{V_0}{4}$

Next,

$W_{ab}=\langle\psi_a|H'|\psi_b\rangle=\left(2/a\right)^3V_0\int_0^{a/2}\sin^2\left(\frac{\pi}{a}x\right)dx$

$\times\int_{a/2}^{a}\sin\left(\frac{\pi}{a}y\right)\sin\left(\frac{2\pi}{a}y\right)dy\int_0^{a}\sin\left(\frac{2\pi}{a}z\right)\sin\left(\frac{\pi}{a}z\right)dz$

Here, it’s different from the previous. So we use the following integral,

$\int\sin px\sin qxdx = \frac{\sin(p-q)x}{2(p-q)}-\frac{\sin(p+q)x}{2(p+q)}$

Solving for the y- and z- integrals,

$\int_{a/2}^{a}\sin\left(\frac{\pi}{a}y\right)\sin\left(\frac{2\pi}{a}y\right)dy=\frac{\sin(1-2)(\pi/a)y}{2(1-2)(\pi/a)}\left|_{a/2}^a-\frac{\sin(1+2)(\pi/a)y}{2(1+2)(\pi/a)}\right|_{a/2}^a$

$=-\frac{2}{2\pi}-\frac{a}{6\pi}=-\frac{4a}{6\pi}=-\frac{2a}{3\pi}$

$\int_0^{a}\sin\left(\frac{2\pi}{a}z\right)\sin\left(\frac{\pi}{a}z\right)dz$

$=\frac{\sin(2-1)(\pi/a)y}{2(2-1)(\pi/a)}\left|_{0}^a-\frac{\sin(2+1)(\pi/a)y}{2(2+1)(\pi/a)}\right|_{0}^a=0$

Hence,

$W_{ab}=W_{ba}=0$

Moving on,

$W_{ac}=\langle\psi_a|H'|\psi_c\rangle=\left(2/a\right)^3V_0\int_0^{a/2}\sin\left(\frac{\pi}{a}x\right)\sin\left(\frac{2\pi}{a}x\right)dx$

$\times\int_{a/2}^{a}\sin^2\left(\frac{\pi}{a}y\right)dy\int_0^{a}\sin\left(\frac{2\pi}{a}z\right)\sin\left(\frac{\pi}{a}z\right)dz$

We have seen in the previous section that the z-part is zero, and so

$W_{ac}=W_{ca}=0$

Lastly,

$W_{bc}=\langle\psi_b|H'|\psi_c\rangle=\left(2/a\right)^3V_0\int_0^{a/2}\sin\left(\frac{\pi}{a}x\right)\sin\left(\frac{2\pi}{a}x\right)dx$

$\int_{a/2}^{a}\sin\left(\frac{2\pi}{a}y\right)\sin\left(\frac{\pi}{a}y\right)dy\int_0^{a}\sin^2\left(\frac{\pi}{a}z\right)dz$

Solving for the x-integral,

$\int_0^{a/2}\sin\left(\frac{\pi}{a}x\right)\sin\left(\frac{2\pi}{a}x\right)dx=\frac{\sin(1-2)(\pi/a)y}{2(1-2)(\pi/a)}\left|_{0}^a-\frac{\sin(1+2)(\pi/a)y}{2(1+2)(\pi/a)}\right|_{0}^a$

$=\frac{a}{2\pi}+\frac{a}{6\pi}=\frac{4a}{6\pi}=\frac{2a}{3\pi}$

Therefore,

$W_{bc}=\left(\frac{2}{a}\right)^3V_0\left(\frac{2a}{3\pi}\right)\left(\frac{2a}{3\pi}\right)\left(\frac{a}{2}\right)=-\frac{16}{9\pi^2}V_0$

$=\left(-\frac{V_0}{4}\right)\left(\frac{8}{3\pi}\right)^2=-\frac{V_0}{4}\kappa=W_{cb}$ ,

where $\kappa=(8/3\pi)^2$

Finally,

$W=\frac{V_0}{4}\left( \begin{array} {ccc} 1 & 0 & 0 \\ 0 & 1 & -\kappa \\ 0 & -\kappa & 1 \end{array} \right)$

We can now solve for the characteristic equation given by

$\left| \begin{array} {ccc} \frac{V_0}{4}-\lambda & 0 & 0 \\ 0 & \frac{V_0}{4}-\lambda & -\kappa\frac{V_0}{4} \\ 0 & -\kappa\frac{V_0}{4} & \frac{V_0}{4}-\lambda \end{array} \right|$

$= \left(\frac{V_0}{4}-\lambda\right)\left[\left(\frac{V_0}{4}-\lambda\right)^2-\left(-\kappa\frac{V_0}{4}\right)^2\right]=0$

$\Rightarrow\frac{V_0}{4}-\lambda=0\Rightarrow \lambda=\frac{V_0}{4}$

$\Rightarrow\left(\frac{V_0}{4}-\lambda\right)=\pm\left(\kappa\frac{V_0}{4}\right)$

$\Rightarrow\lambda=\frac{V_0}{4}\left(1\mp\kappa\right)$

Therefore,

$E'_{\lambda}=\left\{ \begin{array} {c} E_1^0 + \lambda(V_0/4) \\ E_1^0 + \lambda(1+\kappa)(V_0/4) \\ E_1^0 + \lambda(1-\kappa)(V_0/4) \end{array}$

This result is exactly the same as in the sample problem related to this given in Grifftihs “Introduction to quantum mechanics”.

About the author: Karl Patrick S. Casas is a masters student of Mindanao State University-Iligan Institute of Technology. He hopes to finish his degree as soon as possible.

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2nd-Order Correction

Rommel J. Jagus

Find the 2^nd-order correction to the energies $(E_n^{2})$ for the potential $H=\alpha \delta (x-\frac{a}{2})$

Solution:

$<\Psi_{m}^{0} | H | \Psi_{n}^{0}> = \frac{2} {a} \alpha \int_0^a sin(\frac{m x \pi }{a} ) \delta (x-\frac{a}{2}) sin(\frac{n x \pi }{a})$

$<\Psi_{m}^{0} | H | \Psi_{n}^{0}> =\frac{2} {a} \alpha [a sin(\frac{m \pi }{a} ) sin(\frac{n \pi }{a})]$

This is zero unless both m and n are odd in which case it is $\pm\frac{2\alpha}{a}$

$E_{n}^{2}=\sum_{m \ne n, odd} (\frac{2\alpha}{a})^2$

But

$E_{n}^{o}=\frac{\pi^2 \hbar^2 n^2}{2ma^2}$

So,

$E_{n}^{2}=2m(\frac{2\alpha}{\pi m})^2\sum_{m \ne n, odd} \frac{1}{(n^2-m^2)}$

Since $\frac{1}{n^2-m^2}=\frac{1}{2n}(\frac{1}{m+n}-\frac{1}{m-n})$

For n=1

$\sum= \frac{1}{2} \sum_{3,5,7} (\frac{1}{m+1}-\frac{1}{m-1})$

$\sum=\frac{1}{2}(\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+$ … $-\frac{1}{2}-\frac{1}{4}-\frac{1}{6} - \frac{1}{8})$

$\sum= \frac{1}{2}(\frac{-1}{2}) = - \frac{1}{4}$

For n=3

$\sum= \frac{1}{6} \sum_{1,5,7} (\frac{1}{m+3}-\frac{1}{m-3})$

$\sum=\frac{1}{6}(\frac{1}{4}+\frac{1}{8}+\frac{1}{10}+$ … $-\frac{1}{2}-\frac{1}{4}-\frac{1}{8} - \frac{1}{10})$

$\sum= \frac{1}{6}(\frac{-1}{6}) = - \frac{1}{36}$

Therefore

$E_{n}^{2}=0$ if n is even

$E_{n}^{2}=-2m (\frac{\alpha}{\pi\hbar^2\n})^2$ if n is odd

Source: Problem 6.4 A in “Introduction to Quantum Mechanics” by David J. Griffiths

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Quantum Information Processing with Atoms and Photons

Michael J. Jabines

Quantum information science(QIS) is a new field of science and technology, combining and drawing on the disciplines of physical science, mathematics, computer science and engineering. Here, Quantum information processors exploit the quantum features of superposition and entanglement for application not possible in classical devices. It offers the potential for significant improvements in the communication and processing of information.

For an isolated quantum system, the fundamental unit of information is the quantum bit or qubit. Qubits are just quantum two-level system such as the spin of an electron or the polarization of a photon and can be prepared in a coherent superposition state. For a quantum information processor, there are three (3) requirements to have a good quantum hardware:

1. The quantum system must be initialized in a well-defined state.

2. Arbitrary unitary operators must be available and controlled to launch the initial state to an arbitrary entangled state.

3. Measurements of the qubits must be performed with high quantum efficiency.

The first requirements demand that the qubits are well isolated from the environment to ensure pure initial quantum states and to preserve their superposition character, but they must also interact strongly between one another to become entangled. Atomic gases and single photon are among promising candidates to implement quantum information technology because they can be well isolated from their environment. Despite this advantage it is challenging to design controllable interaction between these particles and to store or manipulate quantum information in a reliable way. Quantum information processing requires qubits to behave as quantum memories for long-storage and for many applications to behave as quantum transmitters for long-distance communication. Cold and localized individual atoms are the natural choice for qubit memories and sources of local entanglement for quantum information processing. Photons, on the other hand are the natural source for the communication of quantum information, as they can traverse large distance through the atmosphere or optical fibers with minimal distance.

Quantum information technology is likely to have an important role in information processing after the demise of Moore’s Law. Current devices come primarily from the areas of quantum optics and atomic physics, usually involving laser-cooled and trapped atoms. But perhaps the most exciting feature of this field is that the first-large scale quantum computer will probably be built from a physical system that is not currently known. Current experiments that control the individual atoms and photons will continue to lead the bizzare features of quantum-mechanical foundations to the forefront. With this new language of information, hope we can gain more insight in the underlying quantum-physical principle, exactly as Shannon’s theory of classical information ushered advance physics responsible for the current digital age.

*A summary of Christopher Monroe’s “Quantum Information Processing with Atoms and Photons”

Michael Jabines is a graduate student in Physics of Mindanao State University-Iligan Institute of technology (MSU-IIT) Iligan City Philippines.

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