questionposter Posted March 24, 2012 Author Share Posted March 24, 2012 They're all in the ground state. So they aren't technically entangled, but they exist in a sort of undefined superposition as to also instantaneously respond to various state changes like entanglement? I guess instead of the superposition being with energy, it's with...location? Link to comment Share on other sites More sharing options...
swansont Posted March 24, 2012 Share Posted March 24, 2012 So they aren't technically entangled, but they exist in a sort of undefined superposition as to also instantaneously respond to various state changes like entanglement? I guess instead of the superposition being with energy, it's with...location? How can you have a superposition of states when they are in the same state? BTW, entanglement doesn't respond to state changes. In entanglement, the state is undetermined. The system responds to a measurement. Prior to that, you can't separate the states. http://en.wikipedia.org/wiki/Quantum_entanglement#Quantum_mechanical_framework Link to comment Share on other sites More sharing options...
questionposter Posted March 24, 2012 Author Share Posted March 24, 2012 (edited) How can you have a superposition of states when they are in the same state? BTW, entanglement doesn't respond to state changes. In entanglement, the state is undetermined. The system responds to a measurement. Prior to that, you can't separate the states. http://en.wikipedia....nical_framework I didn't mean superposition of energy states in superfluid, I meant superposition of general location, since the location of any particular atom can't be defined. Edited March 24, 2012 by questionposter Link to comment Share on other sites More sharing options...
swansont Posted March 25, 2012 Share Posted March 25, 2012 I didn't mean superposition of energy states in superfluid, I meant superposition of general location, since the location of any particular atom can't be defined. Then how is position an eigenstate, that you could have a superposition? Link to comment Share on other sites More sharing options...
questionposter Posted March 25, 2012 Author Share Posted March 25, 2012 Then how is position an eigenstate, that you could have a superposition? Doesn't the uncertainty principal say that if you know the momentum, i.e. we know the ground state, then it's position in undetermined? Link to comment Share on other sites More sharing options...
derek w Posted March 25, 2012 Share Posted March 25, 2012 (edited) In a sense all particles within the universe are entangled,as they must interact with and be effected by each other.The only way to completely disentangle a particle is to take it outside the universe. Edited March 25, 2012 by derek w Link to comment Share on other sites More sharing options...
questionposter Posted March 25, 2012 Author Share Posted March 25, 2012 In a sense all particles within the universe are entangled,as they must interact with and be effected by each other.The only way to completely disentangle a particle is to take it outside the universe. No, they aren't entangled, and their probability fields are practically 0 at macroscopic distances, so it doesn't do much. Link to comment Share on other sites More sharing options...
derek w Posted March 25, 2012 Share Posted March 25, 2012 (edited) Yes you say practically zero,but never zero,which means the probability of particle entanglement is greater the smaller the distance of separation,and lesser the greater the distance of separation. But if 2 particles are separated by a large distance with nothing else to interact or effect them,then they are only effected by each other.(but the probability of that is practically zero) Edited March 25, 2012 by derek w Link to comment Share on other sites More sharing options...
swansont Posted March 25, 2012 Share Posted March 25, 2012 Doesn't the uncertainty principal say that if you know the momentum, i.e. we know the ground state, then it's position in undetermined? Ground state does not imply there is a known momentum. Ground state is not zero energy. Link to comment Share on other sites More sharing options...
questionposter Posted March 26, 2012 Author Share Posted March 26, 2012 Ground state does not imply there is a known momentum. Ground state is not zero energy. That doesn't make sense to me, the ground state is the lowest possible energy state for an atom, and we can calculate the energy that a particle has at that state as well as it's average radius, so...how does knowing that every atom is in the ground state mean we don't know the momentum? If we can say they are in the ground state, doesn't that mean we have to know their momentum is a specific thing in order to know that they are at their lowest possible energy state? Link to comment Share on other sites More sharing options...
swansont Posted March 26, 2012 Share Posted March 26, 2012 That doesn't make sense to me, the ground state is the lowest possible energy state for an atom, and we can calculate the energy that a particle has at that state as well as it's average radius, so...how does knowing that every atom is in the ground state mean we don't know the momentum? If we can say they are in the ground state, doesn't that mean we have to know their momentum is a specific thing in order to know that they are at their lowest possible energy state? Even if you know the kinetic energy exactly, how does that tell you the momentum? Link to comment Share on other sites More sharing options...
questionposter Posted March 26, 2012 Author Share Posted March 26, 2012 Even if you know the kinetic energy exactly, how does that tell you the momentum? Can't you figure out the momentum based on the energy level? Link to comment Share on other sites More sharing options...
swansont Posted March 26, 2012 Share Posted March 26, 2012 Can't you figure out the momentum based on the energy level? No. If I tell you a particle has an energy E, which you can write as p^2/2m, you might think that you know the momentum as [math]\sqrt{2mE}[/math] But momentum is a vector and the energy doesn't tell you which direction the particle is moving. Thus for a particle of energy E in a potential well, the average momentum is zero, and the uncertainty is p. Link to comment Share on other sites More sharing options...
questionposter Posted March 26, 2012 Author Share Posted March 26, 2012 (edited) No. If I tell you a particle has an energy E, which you can write as p^2/2m, you might think that you know the momentum as [math]\sqrt{2mE}[/math] But momentum is a vector and the energy doesn't tell you which direction the particle is moving. Thus for a particle of energy E in a potential well, the average momentum is zero, and the uncertainty is p. So in both instances, the location of an individual particle is undefined so that they both act as a single entity, but with entanglement the vector state is undefined where-as in liquid helium it isn't? Edited March 26, 2012 by questionposter Link to comment Share on other sites More sharing options...
swansont Posted March 27, 2012 Share Posted March 27, 2012 So in both instances, the location of an individual particle is undefined so that they both act as a single entity, but with entanglement the vector state is undefined where-as in liquid helium it isn't? In entanglement, you have two particles each in one of two states, but you don't know which (e.g. one is spin up and the other down, or orthogonal polarizations of photons). In a BEC/superfluid, all the particles are in the ground state. In a BEC generally it's a magnetic trap, so all the spins are the same. Link to comment Share on other sites More sharing options...
questionposter Posted March 27, 2012 Author Share Posted March 27, 2012 In entanglement, you have two particles each in one of two states, but you don't know which (e.g. one is spin up and the other down, or orthogonal polarizations of photons). In a BEC/superfluid, all the particles are in the ground state. In a BEC generally it's a magnetic trap, so all the spins are the same. Ok, well can this these types of things happen in degenerate matter? Link to comment Share on other sites More sharing options...
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