Timeline for Definition of Entanglement
Current License: CC BY-SA 4.0
19 events
| when toggle format | what | by | license | comment | |
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| Feb 1, 2019 at 9:08 | answer | added | lcv | timeline score: 0 | |
| Jan 31, 2019 at 18:46 | history | edited | Qmechanic♦ |
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| Jan 31, 2019 at 11:27 | answer | added | anna v | timeline score: 0 | |
| Jan 31, 2019 at 10:57 | answer | added | Andrew Steane | timeline score: 5 | |
| Jan 31, 2019 at 10:13 | history | edited | Norbert Schuch | CC BY-SA 4.0 |
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| Jan 31, 2019 at 10:12 | comment | added | Norbert Schuch | @WillO One could also argue that people are used to think of mixed states when they see $\rho$ rather than $\psi$. | |
| Jan 31, 2019 at 3:55 | answer | added | PhysicsTeacher | timeline score: 1 | |
| Jan 31, 2019 at 2:10 | history | edited | user353840 | CC BY-SA 4.0 |
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| Jan 31, 2019 at 2:08 | comment | added | user353840 | @WillO fair enough, I will edit my question. | |
| Jan 31, 2019 at 2:08 | comment | added | user353840 | @DanYand Well, the premise is that I have to explain to a bunch of experimentalists how this is a good definition for quantum entanglement and I have no idea how to do that. | |
| Jan 31, 2019 at 1:59 | comment | added | WillO | Ah. Your post as written is quite misleading, then. It would help a lot if you clarified that you are talking about mixed states; to many readrers (including me) the unmodified word "state" refers to a pure state by default. | |
| Jan 31, 2019 at 1:13 | comment | added | user353840 | @WillO That's true, but in that case you're only talking about pure states (= vector states). This definition deals with mixed states, i.e. density matrices. They are not elements of a Hilbert space, but rather operators acting on a Hilbert space. They are already normalised, so your point does not hold in this context. | |
| Jan 31, 2019 at 1:09 | comment | added | WillO | A state is an equivalence class of vectors. Every linear combination is equivalent to a convex combination. | |
| Jan 31, 2019 at 1:02 | comment | added | user353840 | @WillO That's not true. Every element is a LINEAR combination of $A \otimes B$, but I am talking about a CONVEX combination of tensors, i.e. $\sum_i \lambda_i = 1$. | |
| Jan 31, 2019 at 0:56 | comment | added | WillO | It is nearly trivial to check that the tensor product consists entirely of states of the form you're calling separable. If you've found an article that suggests otherwise, you've found an article with a very elementary error. | |
| Jan 31, 2019 at 0:55 | review | Close votes | |||
| Feb 1, 2019 at 11:09 | |||||
| Jan 31, 2019 at 0:46 | comment | added | user353840 | @WillO no, this is definitely the definition found in the literature, for example in this article definition 3 | |
| Jan 31, 2019 at 0:41 | comment | added | WillO | By this definition, no state is entangled. The correct definition is that a state is entangled if it is not a product state. | |
| Jan 30, 2019 at 23:53 | history | asked | user353840 | CC BY-SA 4.0 |