Timeline for Measurement of quantum state
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Aug 14, 2015 at 4:10 | comment | added | Jahan Claes | @zeldredge but is there a measurement B could perform on the spins to distinguish between the two? I think in this case the answer is no. | |
| Mar 26, 2015 at 4:02 | comment | added | zeldredge | The answer is no, $| 1 \rangle + | 2 \rangle$ is measurably different than 50% $| 1 \rangle$ and 50% $|2 \rangle$. To see this formally you have to consider a measurement of a different observable. | |
| Feb 21, 2014 at 18:43 | comment | added | mcFreid | Ah, now I understand what you're asking. That is a good question and after thinking about it for a bit - I am not sure of the answer. I suggest making a new question on StackExchange titled something like "Is an ensemble of spin eigenstates equivalent to an ensemble of superimposed states?" Then ask the question you just asked in the comment above this one. | |
| Feb 21, 2014 at 17:10 | comment | added | user35122 | I mean, is the above ensemble (comprising of half of particles in definite up state and the rest in down) equivalent to a another ensemble of systems each of which contains particle in superposed state. Are these two ensembles equivalent? | |
| Feb 21, 2014 at 16:46 | comment | added | mcFreid | If I understand you correctly: you are asking whether an ensemble of spin eigenstates such that 50% of the ensemble is spin up and 50% of the ensemble is spin down is equivalent to a single particle in a superposition of spin up and and spin down? If that's your question, then the answer is no. An ensemble is many particles. A single particle is one particle. They are very different systems. | |
| Feb 21, 2014 at 15:57 | comment | added | user35122 | so each of the system would be in one particular state, not the superposition.I am just wondering if these two pictures are the same! | |
| Feb 21, 2014 at 14:41 | comment | added | mcFreid | Yes, I know what a statistical ensemble is - I just don't see how that changes what I wrote. | |
| Feb 21, 2014 at 14:39 | comment | added | user35122 | by statistical ensemble i mean a large collection of systems given by superposition of up spin and down spin becomes in this case a collection of large no of systems among which half of them are in up-spin state and the rest in down spin state! | |
| Feb 20, 2014 at 20:37 | comment | added | mcFreid | I don't understand your first sentence. As for a "hidden-variable"... That concept is in regards to the probablistic nature of quantum mechanics. Just because an observer lacks information in making a prediction does not mean there is some fundamental hidden variable in your theory. For example, if we tried to predict the orbit of Mercury yet didn't account for the small deviation due to Venus, this does not mean that the fundamental laws of gravity have a hidden variable. It just means we didn't account for all the interactions. | |
| Feb 20, 2014 at 16:25 | comment | added | user35122 | but does this not make the system an statistical ensemble of systems among which hafl are in up state and the other half in down state. and morever because there is an information which B has no knowledge of does it not make the system possess a 'hidden-variable' just out of reach of B? | |
| Feb 17, 2014 at 1:14 | history | answered | mcFreid | CC BY-SA 3.0 |