Leonard
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Registered User
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Dec 25 |
awarded | ● Critic |
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Dec 24 |
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A novice question on Quantum Mechanics Hence, you are right. As Professor Andreas Blass has mentioned in his comment below the wording of your question, superposition is a binary operation on non-zero vectors, not on states, which are equivalence classes of non-zero vectors. |
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Dec 24 |
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A novice question on Quantum Mechanics Hi Ryan. We always have $ a_{1} |A \rangle \sim a_{2} |A \rangle $ and $ b_{1} |B \rangle \sim b_{2} |B \rangle $, yet we may still end up with $ |R_{1} \rangle \nsim |R_{2} \rangle $. Superposition of vectors is not a quantum-state-preserving binary operation on $ \mathcal{H} \setminus \lbrace 0_{\mathcal{H}} \rbrace $, unless $ |A \rangle \sim |B \rangle $. You can prepare a myriad of quantum states from just two distinct quantum states. This principle is important in quantum computing, in which one can produce infinitely many states from a fixed basis of a finite-dimensional Hilbert space. |
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Dec 24 |
revised |
A novice question on Quantum Mechanics deleted 67 characters in body |
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Dec 24 |
answered | A novice question on Quantum Mechanics |
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Dec 22 |
asked | On the definition of ‘smooth vectors’ in Rieffel’s “Deformation Quantization for Actions of $ \mathbb{R}^{d} $”. |
