Timeline for A novice question on Quantum Mechanics
Current License: CC BY-SA 3.0
5 events
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| Dec 24, 2012 at 21:32 | comment | added | Leonard | 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. | |
| Dec 24, 2012 at 21:22 | comment | added | Leonard | 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. | |
| Dec 24, 2012 at 21:10 | history | edited | Leonard | CC BY-SA 3.0 |
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| Dec 24, 2012 at 18:15 | comment | added | Ryan | So, would it be correct to say that even though $a_{1}|A\rangle$ is the same state as $a_{2}|A\rangle$ and $b_{1}|B\rangle$ is the same state as $b_{2}|B\rangle$, $|R_{1}\rangle$ and $|R_{2}\rangle$ represent two different ways of superposing these states, so that the probabilities of being in the state represented by $|A\rangle$ and $|B\rangle$ are different for each? | |
| Dec 24, 2012 at 7:12 | history | answered | Leonard | CC BY-SA 3.0 |