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Visualise Hadamard gate as addition of $x$ and $z$ gate

Ichwerdennauchsonst
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⚡ Quantum Brief
A quantum computing enthusiast questioned whether the Hadamard gate can be visualized as a combination of Pauli-X and Pauli-Z rotations on the Bloch sphere, proposing a geometric interpretation where both rotations are added rather than applied sequentially. The user suggested the Hadamard gate’s effect might stem from summing X and Z gate matrices, with the 1/√2 factor normalizing the result via Pythagoras’ theorem, but noted this leads to an unexpected Y-axis outcome. The core confusion arises from adding rotation vectors: independent X and Z rotations cancel out X and Z components, leaving only a Y-axis displacement, contradicting the Hadamard’s known X-axis behavior. The post highlights a common misconception in visualizing quantum gates as classical vector additions, emphasizing the need for precise mathematical representation over intuitive geometric analogies. Experts were invited to clarify how the Hadamard gate’s superposition creation aligns with Bloch sphere dynamics without relying on simplistic vector addition models.
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Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Stack Overflow for Teams is now called Stack Internal. Bring the best of human thought and AI automation together at your work. Bring the best of human thought and AI automation together at your work. Learn more Stack InternalKnowledge at workBring the best of human thought and AI automation together at your work.I try to immagine the hadamard gate on a bloch sphere and it is obvious that it is the sum of the pauli $x$ and $z$ gate matrices.So that means we rotate the "vector" around the x axe and around the z axe (independently, not succesivily) and then add the results. The $\sqrt{2}$ then very likely norms it (Pythagoras).But herein lies the problem. If I add the vectors, the result lies indeed in the plane but on the y-axe instead of the x-axe. This is not how the hadamard gate is described to work.I hope you can understand how I mean it as I don't know how to visualise it. Immagine in your head, you rotate a point about the x axe and about the z axe, the resulting points will lie on the same y-point but on opposite x and z point. If you add these, only the y point is not zero but twice the original point with reversed sign divided through sqrt(2). So how it is possible?Thanks for contributing an answer to Quantum Computing Stack Exchange!But avoid …Use MathJax to format equations. MathJax reference.To learn more, see our tips on writing great answers.Required, but never shown By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy. Start asking to get answersFind the answer to your question by asking.Explore related questionsSee similar questions with these tags.To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Site design / logo © 2025 Stack Exchange Inc; user contributions licensed under CC BY-SA . rev 2025.11.27.37399

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