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What is a quantum computer?

A quantum computer is a computer that harnesses the properties of 'quantum' particles such as atoms and electrons, which make up matter, to perform information processing. In the future, it is expected that supercomputers will be able to perform calculations that would take an immense amount of time much more quickly, particularly for specific issues..


Bits in a Classical (conventional) Computer

In a computer, all information processing is performed as a string of 0s and 1s (called a bit string). By the way, magnetic materials and semiconductor materials are used as bits in classical computers, and in magnetic materials, the magnetic moment (you can think of it as a small magnet) is 0 to indicate whether it is facing up or down. or 1. MONOS type memory is a representative example of a semiconductor material, and 0 or 1 corresponds to whether or not an electron exists in a minute space. In this way, each bit stores a definite value of either 0 or 1 . It is an error if 0 is supposed to be saved, but it is 1 when the data is retrieved. In practice, how long the stored data can be retained correctly depends on the characteristics of the memory material. Now, let us introduce the symbols |0> and |1> as symbols representing the state of 0 or 1 of a classical bit, for correspondence with quantum mechanics later. Once again, a classical 1 bit can be in either the  |0> state or the |1> state.

Bits and Quantum Superposition States in Quantum Computers

Now let's move on to the bits of a quantum computer. Quantum computer bits are called  qubits from quantum bits. First, let's start with the case of 1 qubit. In the case of Qubit as well, when each bit is measured, either the |0> state or the |1> state is always observed. The |0.5> or |1.1> states are never observed. It is always either |0> or |1>. However, the decisive difference between a qubit and a classical bit is that it can take a state governed by quantum mechanics, and can take a mysterious state called a quantum superposition state. I will explain in detail in another commentary article what the quantum superposition state is, but here I will only explain mathematically that "the following state can be taken in the world of quantum mechanics". In general, a qubit is in a state such as  α0|0>+α1|1> (α0 and α1  are complex coefficients), and a state in which the |0> and |1> ), can be taken. In other words, a state in which the |0> state and the |1> state are mixed is generally allowed in the world of quantum mechanics. So, what is α0 here? Again, if you observe the state  α0|0>+α1|1> , it is either the |0> state or the  |1> be. The  |0.5> state between the |0> state and the  |1> state is not observed. Then, what is the physical meaning of α0 and α1? It has the physical meaning of being done. By the way, since either the |0> state or the |1> state is "always" measured, the condition |α0|^2+|α1|^2=1  conditions). By the way, for example, if you prepare a 1-qubit state of  α0=1 α1=0  (that is, the |0> state), the |0> state will be observed no matter how many times the measurement is repeated. This state is no different from the |0> state of a classical bit. On the other hand, if we prepare a 1-qubit state with α0=0 α1=1  (that is, the |1> state), the |1> state will be observed no matter how many times the measurement is repeated. This state is no different from the |1> state of the classical bit. Let's prepare another example, α0=1/√2, α1=1/√2  1 qubit state (that is, 1/√2(|0>+|1>) ), there is a 50% probability that the |0> state will be observed, and a 50% probability that the |1> state will be observed. This is a qubit note

It's not that the li is broken and causing a problem. Essentially, the quantum mechanical state is such a state. What makes me happy in such an ambiguous (?) state? As a little digression, the  1/√2(|0>+|1>)  state is the probability that the |0> and |1> are 50/50, and no one can predict in advance which state will appear in a single measurement. Usually, when we say that 0 or 1 appear at random, we imply that they appear to be random, either because of our ignorance or because they are too complex to predict too accurately. There are many times, but here ``When observing the 1/√2(|0>+|1>)  state, the |0> and |1> This probability is a "true" probability, meaning that the natural phenomenon itself contains stochastic elements that are not derived from human ignorance. Well, as I wrote at length, in order to understand quantum computers, it is necessary to understand the existence of these coefficients  α0 and α1 . And if we accept the existence of these coefficients  α0 and α1 , we will be able to see the outrageous benefits of quantum computers. For the time being, in the 1qubit state, the superposition state is generally  α0|0>+α1|1> (α0 and α1 are complex coefficients), and the coefficients  α0 and α1  Please know that it has the meaning of the probability of observing the state of .


How could a qubit state have such an enormous amount of memory?

What is the quantum entangled state at its essence?

​By the way, I think that some of you who have read this far have a strange feeling as if you were pinched by a fox. Why did classical bits and qubits have such different memories? The reason for this disgust is that qubits skillfully use a state called quantum entanglement, which was not found in classical bits. Although it is a little advanced, I will explain what the quantum entangled state is below. Here, for simplicity, we will talk about the case of 2 qubits. Now consider the following situation with 2 qubits: 1/√2(|01>+|10>)  state. You know what this state is, right? It's like a 50% probability of |01> and a 50% probability of |10> being observed. If we actually measured this two-qubit state, we would measure |0> at 50% and |1> at 50%, considering the left qubit alone. The same is true if we consider only the right qubit. Measures |0>  at 50% and |1> at 50%. Considering the left and right alone, |0> However, when we focus on each qubit alone, it looks random, but when we consider both the left and right qubits, the measurement results are not actually random. They are correlated or coordinated (?) with each other, and are either |01> or |10>. It looks random, but if you look closely, states like |00> and |11> don't appear. In other words, the result of what is measured in the left qubit also affects the right qubit, and the measurement result of the left qubit determines the state of the right qubit. I'm here. Or vice versa, the result measured by the right qubit affects the state of the left qubit. Such a state is called a “quantum entangled state”. The situation is such that the two qubits are not “independent” of each other, but are interrelated with each other. This is an anomalous phenomenon from the classical bits. The left bit is measured, while the right bit is irrelevant whatever the left bit is. The measurement result of the left bit does not affect the right bit. Of course, in a real device, there is a kind of error (called cross-term) in which the electric or magnetic field when reading the left bit affects the right bit as a disturbance, but it is an error and should not occur. It's not a phenomenon, it's a malfunction, and it's a completely different story. It can be said that the qubit memory has a high degree of freedom of 2^N by considering the quantum entanglement state, which is impossible in the classical bit. By the way, as an aside, the left qubit state is  1/√2(|0>  left+|1>left) and the right qubit state is also  1/√2(|0>right+|1>right) state? Maybe, but it's not.  1/√2(|0>Left+|1>Left) 1/√2(|0>Right+|1>Right) = 1/2( |00>+|01>+|10> +|11>), which is different from the 1/√2(|01>+|10>)  state.  1/√2(|0>Left+|1>Left) 1/√2(|0> Right+|1>Right)  “It is a state in which we are in a quantum superposition state, and this state is not called a quantum entangled state. In fact, this state is just random, and you can see that the |00> and |11> states occur.

​Memory Explosion Problem and Effectiveness of Qubits in Quantum Chemical Computation

​The field of quantum chemical calculation is one of the fields that suffer from the small memory capacity of today's computers. Quantum chemical calculations have recently attracted a great deal of attention due to the trend of materials informatics (often abbreviated as MI). This is an attempt to predict in advance what kind of properties and characteristics the substance will have on a computer simulation before synthesizing the substance in an actual experiment. Quantum chemical calculations are intended to increase efficiency and accelerate the development of new materials.

The constituent elements of matter are atomic nuclei and electrons, and both atomic nuclei and electrons behave according to quantum mechanics. If so, it is possible to predict the properties of matter by simulating quantum mechanics on a computer. Simulation based on quantum mechanics without using experimental values is called first-principles calculation. The equation to be solved by the first-principles calculation is already known, and it is an equation called the Schrödinger equation. In principle, we can predict any material science problem by solving the Schrödinger equation.


Well, there is a wave function that appears in the Schrödinger equation. This wave function is actually one of the stragglers. Electrons are called fermions and obey the Pauli exclusion principle. The Pauli exclusion principle prohibits two electrons from occupying one state. Suppose now that there are Norb states. Then, each state has two states, occupied or not occupied by electrons. Normally, about  Norb=50  is necessary even for small molecules. The problem is that it works exponentially against Norb. It is easy to imagine that the memory will be pressed in no time. As Norb grows, exponential memory is required. This becomes the problem that plagues quantum chemical calculations. On the other hand, if this is calculated on a quantum computer, it will be solved if there are about Norb qubits. Details will be explained in the commentary article on quantum chemical calculations using quantum computers.


Quantum Computers have Powerful Memory!

Comparing 1 classical bit and 1 qubit is actually not very interesting. However, if you compare the 3 bits with each other, you will gradually get a glimpse of the anomaly. Now, let's compare 3 bits between classical bits and qubits. First of all, I will proceed with the story from the time of the classic bit. Each of the classical bits can be |0> or |1>, so the states of the three classical bits are |000>, |001>, |010> One of the eight states of |101>, |110>, and |111> will be taken as a fixed state. Here, the sequence of three numbers is the order of the three classical bits. For example, the |000> state is the state where all three classical bits are |0>. The |001> state is the state where the rightmost bit of the three classical bits is in the |1> state and the other two bits are in the |0> state. In any case, the 3-classical bit state takes ``only one'' of the above 8 states as its definite state. No matter how many times you measure, the same state will be repeated and returned.


But what about 3 qubits? Here's how things change. In the 3-qubit state, the qubit states generally assume a superposition state, as in the case of 1-qubit. In the 3-qubit state, we can take the superposition state of 8 states: |000>, |001>, |010>, |011>, |100>, |101>, |110>, |111>.

+ α010|010> + α011|011> + α100|100> + α101|101> + α110|110> + α111|111> All superposition states can be obtained. Here, the coefficient  |α000|^2  means the probability that the |000> state is measured, and |α001|^2  means the probability that the |001> state is measured. Also, since states other than the eight states are never observed, the relation |α000|^2+ |α001|^2+ |α010|^2+....+ |α111|^2=1 It holds (normalization condition).


Now, in the case of classical bits, calculations etc. are carried out by controlling the arrangement of three strings of numbers (bit strings) sandwiched between ket symbols (pairs of | and >). Quantum computers, however, take a completely different approach. Do not tamper with the string of numbers inside the brackets. The coefficients (α000, α001, etc.) that are applied before it will be manipulated. Why? Because it gives you a huge amount of freedom. In the case of the current 3 qubits, it means that we have 8 degrees of freedom that can be arbitrarily manipulated (exactly speaking, there are normalization conditions, so there are some restrictions...). In classical bit, the calculation proceeds in the form of what to do with the number string in the ket symbol. On the other hand, in a quantum computer, we do not manipulate the contents of the ket symbol, but the coefficients in front of it. Because that gives you a lot of freedom. An illustration of the difference between the two using a slightly more concrete example will be given after the generalizations below. If you look at them, you will be able to feel the difference more strongly.

Generalize the discussion above. That is, in the N-bit case, we compare classical bits and qubits. When it's N-classic bit, the story is simple. One of 2^N states |0...00>, |0...01>, ..., |1...11>  It will be. Qubits, on the other hand, are a very different story. For N qubits, the superposition state of 2^N states |0...00>, |0...01>, ..., |1...11>  can take. That is, α_{0...00}|0...00> + α_{0...01}|0...01> + ... + α_{1...11}|1.. .11>  where |α_{0...00}|^2 is the probability that |0...00> state is observed, |α_{0...01}|^2 is |0...01>  The probability that a state is observed, and so on. With N classical bits, the calculation proceeds in the form of what to do with the number string in the ket symbol. On the other hand, in a quantum computer, you do not manipulate the contents of the ket symbol, but the coefficients in front of it. This is because you can obtain a huge degree of freedom of 2^N . By freely substituting numbers for the coefficients, we can obtain a memory that can input 2^N pieces of information.


01. Example of 3-bit state

02. Generalization for N-bit states

03. Non-Obvious Example

​Now, let's show a non-trivial example to get a feel for the qualitative difference between the amount of memory that classical bits and quantum bits have. Here is an example of handwritten digit characters from MNIST. This handwritten numeral character is a black and white image of 28×28=784 pixel data. Consider storing this MNIST digit handwriting in memory. By associating the color of each pixel of this black-and-white image data with 0 for white and 1 for black, the image data of 784 pixels can be stored as equivalent information on 784 classical bits as a 784-bit string.

On the other hand, consider storing this MNIST handwritten digit character on a qubit. How many qubits do we need? Actually, 10 qubits turns out to be enough. Why? Because 10 qubits have 2^10=1024 freedoms. In fact, when we consider 10 qubits, the qubit states are represented by 1024 superposition states of |0...0>, ..., |1...1>.

In other words, there are 1024 coefficients, and assuming that normalization conditions will be considered later, and considering that bit strings of MNIST handwritten numeric characters are put in order into coefficients {α}, a square that can accommodate 1024 bit strings , so we know that we can store data on 10 qubits well enough. In this way, 784 bits were required for classical bits, but 10 bits are enough for quantum bits. This is due to the exponential increase in the memory capacity of  2^N derived from the quantum superposition state. Please understand that qubits have an enormous amount of memory behind them that cannot be imagined with classical bits. In fact, there are many cases where the lack of memory is more serious than the calculation speed itself in today's computers. Alternatively, even if memory is distributed and stored on a huge supercomputer, there are many cases where data communication itself takes an enormous amount of time.


I would like you to understand that the problem with computers today is not only the computation time but also the problem of memory shortage, and that quantum computers can solve the problem of memory shortage. One example of a problem where memory shortage is serious is quantum chemical calculation. At the end of the commentary article, I will explain why memory shortage is so serious in quantum chemical calculations. If you are not interested in quantum chemical calculations, you can skip it.

2 reasons why Quantum Computers are so Powerful

In October 2019, the news of quantum supremacy from Google circulated around the world. The news is that quantum computers have solved in just 200 seconds a calculation that would take 10,000 years even with today's supercomputers. This news means that a next-generation computer called a quantum computer is becoming a reality. In today's advanced information society, it is no exaggeration to say that it is impossible to live without a computer. With the advent of computers with higher computing power than today, it is expected that various social problems will be solved by deriving various optimal solutions that could not be derived with current computer systems. In fact, there are complex and intertwined issues that cannot be processed by today's computers. First, in this commentary article, I will explain why quantum computers have such powerful computing power. Broadly speaking, quantum computers are amazing in two ways. Those two points are exponential memory and massively parallel algorithms. Below, I'll explain how each one works.


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