silvercats Posted July 10, 2012 Posted July 10, 2012 >."OBSERVED"< . how does the observing of a particle happen in here exactly? If we look at a mobile phone,light electrons hit the phone and return it to eyes.like that how does the observation of a particle happen(so it collapses the wave function!)?
swansont Posted July 10, 2012 Posted July 10, 2012 >."OBSERVED"< . how does the observing of a particle happen in here exactly? If we look at a mobile phone,light electrons hit the phone and return it to eyes.like that how does the observation of a particle happen(so it collapses the wave function!)? If an observation, or measurement, is of a particular state of the system (e.g. you measure/observe the energy), then it must be in one state, rather than being in a superposition.
Widdekind Posted July 18, 2012 Posted July 18, 2012 So, the pluri-potentiality, embodied in the wave-function of some particle, is like holding a "hand of cards". And, when that particle interacts with its external environment, in a sufficiently intense "measurement" way, then the "non-conscious particle" must "make a decision". And, doing the best a "dumb particle" can do, the particle, accommodating its environment, randomly "picks a card", "plays the card", and "discards the rest of the hand" (thereby becoming a "hand of one card") ?
Bill Angel Posted July 26, 2012 Posted July 26, 2012 So, the pluri-potentiality, embodied in the wave-function of some particle, is like holding a "hand of cards". And, when that particle interacts with its external environment, in a sufficiently intense "measurement" way, then the "non-conscious particle" must "make a decision". And, doing the best a "dumb particle" can do, the particle, accommodating its environment, randomly "picks a card", "plays the card", and "discards the rest of the hand" (thereby becoming a "hand of one card") ? That is an interesting way of looking at the situation. What if one had a shuffled deck of cards sitting on a table? The wavefunction for the deck would be the superposition of all possible arrangements of the cards. The observer now turns over the top card and places it on the table. The system now consists of the collapsed wave function for one card, and the deck whose wave function is the superposition of all possible arrangements of the remaining 51 cards.
Ronald Hyde Posted August 3, 2012 Posted August 3, 2012 If an observation, or measurement, is of a particular state of the system (e.g. you measure/observe the energy), then it must be in one state, rather than being in a superposition. That statement is only meaningful in the past tense. If an observation, or measurement, is of a particular state of the system (e.g. you measure/observe the energy), then it must have been in one state. Measurement either destroys the state, or makes a new one with different observables.
MigL Posted August 3, 2012 Posted August 3, 2012 Its been almost 90 yrs and people still try to analyse quantum situations by comparing then to everyday situations or objects like decks of cards ( although I thought Bill's example was an excellent interpretation ). 1
swansont Posted August 4, 2012 Posted August 4, 2012 If an observation, or measurement, is of a particular state of the system (e.g. you measure/observe the energy), then it must have been in one state. Measurement either destroys the state, or makes a new one with different observables. No, there have been EPR experiments that show that the system in an indeterminate state prior to measurement. 1
Ronald Hyde Posted August 4, 2012 Posted August 4, 2012 No, there have been EPR experiments that show that the system in an indeterminate state prior to measurement. You do realize that that is what I said? That measurement destroys the state that it was in, and makesa new one that is determined, or simply destroys it. A really good model of all these ideas is in the polarization states of the photon as they pass through filters that separate them into plane and circular polarization states. A plane polarized state is a superposition of circular polarized states and vice versa. But to actually measure whether a particular photon was in a particular state it has to be detected in some way, running it through filters only selects its state, detecting it destroys it, so the statement is only meaningful in the past tense.
swansont Posted August 4, 2012 Posted August 4, 2012 You do realize that that is what I said? That measurement destroys the state that it was in, and makes a new one that is determined, or simply destroys it. No, you said it must have been in one state. Experiment shows that not to be the case. A really good model of all these ideas is in the polarization states of the photon as they pass through filters that separate them into plane and circular polarization states. A plane polarized state is a superposition of circular polarized states and vice versa. But to actually measure whether a particular photon was in a particular state it has to be detected in some way, running it through filters only selects its state, detecting it destroys it, so the statement is only meaningful in the past tense. A system in a superposition can have different behavior than a system in a single state, prior to detecting the state. 1
Ronald Hyde Posted August 4, 2012 Posted August 4, 2012 No, you said it must have been in one state. Experiment shows that not to be the case. A system in a superposition can have different behavior than a system in a single state, prior to detecting the state. All the states of a photon are superpositions of other states, that's the way it is, all the time. If we separate it into a particular state of one kind of polarization, it's still a superposition of the other kinds of polarization. To find what state it was in we have to detect it. That's the crazy way that QM works.
swansont Posted August 5, 2012 Posted August 5, 2012 All the states of a photon are superpositions of other states, that's the way it is, all the time. If we separate it into a particular state of one kind of polarization, it's still a superposition of the other kinds of polarization. To find what state it was in we have to detect it. That's the crazy way that QM works. But that's not really the way it works in this context. Saying that the particle is or isn't in one state doesn't involve changing the basis in the middle of the explanation. You pick a basis and stay with it.
Ronald Hyde Posted August 5, 2012 Posted August 5, 2012 (edited) What am I missing here? I'm thinking of the polarization states of the photon in the context of the Poincare' sphere. You can use sugar water to rotate the plane of the polarization, but you can't actually measure the plane of polarization either before or after it passes through the filter, without destroying the photon. And whether you use a calcite crystal or sugar water as a filter it just moves the state from one 'position' on the sphere to another, it doesn't destroy it, but it also doesn't provide any information on it's actual state. The Poincare' sphere provides all the information that there is about the polarization, but when we measure it, it's either plane or circularly polarized, so that much of the 'sphere information' is lost or changed. Does this make sense? It does to me, the sphere is the 'basis' that I use when I think about the possible states of the photon. Edited August 5, 2012 by Ronald Hyde
swansont Posted August 6, 2012 Posted August 6, 2012 The basis is the orthogonal system you use to describe the states, often dictated by how you do a measurement. If you use linear polarization, then it's the two perpendicular directions. But by making that choice, you are deciding to do a measurement of linear polarization. If you want to use circular, then that's your basis.
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