High School Quantum Computing for Beginners: Understanding Double Hadamard Gates

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SUMMARY

This discussion focuses on the mechanics of double Hadamard gates in quantum computing, specifically addressing the cancellation of basis vectors |1⟩ while |0⟩ remains unchanged. The participants clarify that |0⟩ and |1⟩ are labels for basis vectors in a vector space, not numerical values. The mathematical representation demonstrates how the application of double Hadamard gates leads to the resultant state |0⟩ through vector addition and normalization, emphasizing the importance of understanding vector components in quantum mechanics.

PREREQUISITES
  • Basic understanding of quantum mechanics terminology
  • Familiarity with quantum states represented as kets (|0⟩, |1⟩)
  • Knowledge of vector spaces and vector addition
  • Understanding of normalization in quantum states
NEXT STEPS
  • Study the principles of quantum superposition and entanglement
  • Learn about the mathematical representation of quantum gates, particularly Hadamard gates
  • Explore the concept of quantum state normalization in detail
  • Investigate the implications of vector space theory in quantum computing
USEFUL FOR

Beginner quantum computing enthusiasts, students of quantum mechanics, and anyone interested in the mathematical foundations of quantum gates.

Quark Effect
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TL;DR
Cannot entirely understand why some of the terms cancel each other out while double Hadamard gates are applied.
Hi guys, I am an absolute beginner in quantum computing. I am really curious about its mechanics but my lack of knowledge in mathematics makes me struggle. I cannot entirely understand why numbers |1> cancel each other out while double Hadamard gates are applied. If the rule is to cancel out the same numbers, why then |0> stays the same, as well as the √2? Thanks

cancelling out0.png
 
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Quark Effect said:
I cannot entirely understand why numbers |1> cancel each other out while double Hadamard gates are applied.

The ##0## and ##1## inside the kets ##|0\rangle## and ##|1\rangle## are not numbers; they're labels for basis vectors in a vector space. You could just as well call them ##|A\rangle## and ##|B\rangle##, or ##|\text{Laurel}\rangle## and ##|\text{Hardy}\rangle## for that matter, and everything would be the same.

The rule for adding vectors is that you add them component by component, i.e., you add up all the terms involving each basis vector separately. So you have:

$$
\frac{1}{\sqrt{2}} \left( \frac{|0\rangle + |1\rangle}{\sqrt{2}} + \frac{|0\rangle - |1\rangle}{\sqrt{2}} \right)
$$

equals

$$
\frac{1}{\sqrt{2}} \left( \frac{|0\rangle}{\sqrt{2}} + \frac{|1\rangle}{\sqrt{2}} + \frac{|0\rangle}{\sqrt{2}} - \frac{|1\rangle}{\sqrt{2}} \right)
$$

equals

$$
\frac{1}{\sqrt{2}} \left( (1 + 1) \frac{|0\rangle}{\sqrt{2}} + (1 - 1) \frac{|1\rangle}{\sqrt{2}} \right)
$$

equals

$$
\frac{1}{\sqrt{2}} \left( 2 \frac{|0\rangle}{\sqrt{2}} \right)
$$

equals

$$
|0\rangle
$$
 
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PeterDonis said:
The ##0## and ##1## inside the kets ##|0\rangle## and ##|1\rangle## are not numbers; they're labels for basis vectors in a vector space. You could just as well call them ##|A\rangle## and ##|B\rangle##, or ##|\text{Laurel}\rangle## and ##|\text{Hardy}\rangle## for that matter, and everything would be the same.

The rule for adding vectors is that you add them component by component, i.e., you add up all the terms involving each basis vector separately. So you have:

$$
\frac{1}{\sqrt{2}} \left( \frac{|0\rangle + |1\rangle}{\sqrt{2}} + \frac{|0\rangle - |1\rangle}{\sqrt{2}} \right)
$$

equals

$$
\frac{1}{\sqrt{2}} \left( \frac{|0\rangle}{\sqrt{2}} + \frac{|1\rangle}{\sqrt{2}} + \frac{|0\rangle}{\sqrt{2}} - \frac{|1\rangle}{\sqrt{2}} \right)
$$

equals

$$
\frac{1}{\sqrt{2}} \left( (1 + 1) \frac{|0\rangle}{\sqrt{2}} + (1 - 1) \frac{|1\rangle}{\sqrt{2}} \right)
$$

equals

$$
\frac{1}{\sqrt{2}} \left( 2 \frac{|0\rangle}{\sqrt{2}} \right)
$$

equals

$$
|0\rangle
$$
Ahh, I see! I really appreciate your help, it's not the first time you are spending your time and patiently explaining me things. Thanks a lot man
 

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