August 2020
Beginner to intermediate
332 pages
8h 16m
Japanese
book
動かして学ぶ量子コンピュータプログラミング ―シミュレータとサンプルコードで理解する基本アルゴリズム
by Eric R. Johnston, Nic Harrigan, Mercedes Gimeno-Segovia, 北野 章, 丸山 耕司
Content preview from 動かして学ぶ量子コンピュータプログラミング ―シミュレータとサンプルコードで理解する基本アルゴリズム
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2.3
基本の
QPU
演算
21
対応する逆演算を持ち、このことを知っておくと役に立ちます。逆演算を持つ場合、その
QPU
演算は
可逆である言います。これは、演算を行うときに情報がまったく失われない、あるいは破棄されないこ
とを意味します。しかしながら、一部の
QPU
演算は不可逆です。つまり、その逆演算を持ちません（何
らかの理由で情報の損失が生じます）。演算が可逆的であるかどうかは、演算の使い方に重要な影響を
与えることを後で示します。
これらの
QPU
命令のいくつかは、奇妙で、有用性の面で疑わしいと思うかもしれません。ですが、
少し説明をした後で、実際にそれらを使ってみましょう。
2.3.1
QPU
命令：
NOT
NOT
は従来の
NOT
演算と等価な演算です。この演算では、
0
を
1
に、
1
を
0
に変換しま
す 。 し か し 、従 来 の
NOT
とは異なり、
QPU
の
NOT
演算は重ね合わせの状態にあるキュ
ビットにも作用します。
図2-6に示すように、円表示では、この結果は
∣
0
⟩
と
∣
1
⟩
の 円 の 交 換（
swapping
）で 表 さ れ ま す 。
図2-6
可逆性
デジタル論理と同様に、
NOT
演算はそれ自身が逆演算となっています。この演算を
2
回適用
するとキュビットは元に戻ります。
2.3.2
QPU
命令：
HAD
HAD
演 算（ ア ダ マ ー ル（
Hadamard
）の 略 ）は 、
∣
0
⟩
あるいは
∣
1
⟩
のどちらかの状態に対
して作用したときに、大きさが等しい重ね合わせを生成する演算です。これは量子的 ...
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I wanted to learn C and C++, but it didn't click for me until I picked up an O'Reilly book. When I went on the O’Reilly platform, I was astonished to find all the books there, plus live events and sandboxes so you could play around with the technology.Addison B.
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I'm always learning. So when I got on to O'Reilly, I was like a kid in a candy store. There are playlists. There are answers. There's on-demand training. It's worth its weight in gold, in terms of what it allows me to do.