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EPR reasoning with triplet state

  1. Oct 2, 2013 #1
    EPRB is a know reasoning about the singlet state.

    However if we apply this to the triplet state [tex]\frac{1}{\sqrt{2}}(\mid+-\rangle+\mid-+\rangle)[/tex].

    we find the following :

    suppose we measure spin A with result +, then spin B is - (we see this by the projection on the possible state), however it's a state with total spin 1, since this triplet state is eigenvector of the sum operator (S1+S2)^2.

    So we fing that the following is different : measuring spin A and then spin B and making the sum,

    or making the measurement of the sum. How does the system know in advance the sum will be taken ?
     
  2. jcsd
  3. Oct 2, 2013 #2

    jfizzix

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    Though the triplet state gives perfect negative-correlations when you measure in the spin-z (+-) basis, it doesn't do this if you measure in the spin-x or spin-y bases. In those bases, you can get perfect positive correlations.

    It's only with the singlet state, that you have perfect anti correlations in all cases.
     
  4. Oct 3, 2013 #3
    Thanks
     
  5. Oct 3, 2013 #4
    Does this mean that, even if the angle between the direction of measurement is 0, the correlation depends on the angle of the measurement direction, and is not involving only the relative angle ?

    I remind also : does anyhow the triplet state exist, since those are fermions and that the wf has to be antisymmetric ? Or does the fact that this representation does not include space, that the 2 particle can have same spin because they are located at different places ?
     
  6. Oct 3, 2013 #5

    jfizzix

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    Generally, the correlations depend on both the angle of the measurement direction and the relative angle. There are mixed states whose correlations are maximum in the spin-z basis which exhibit no correlations at all in the spin-x or spin-y bases.

    The triplet state does exist for fermions. The quantum state as a whole has to be antisymmetric, which means that if the spinor is symmetric, the wavefunction has to be antisymmetric, so that the total state is antisymmetric (i.e. even function times odd function = odd function).
     
  7. Oct 5, 2013 #6
    Thanx. Coming back to my original post question about difference between EPR and QM.

    For the triplet state, if we measure (S1+S2)^2 we obtain S=1 with certainty in quantum mechanics.

    Does this mean that we cannot "think" locally by measuring only S1 and deducing S2 ?
     
  8. Oct 5, 2013 #7

    jfizzix

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    If you prepare a pair of particles to be in the triplet [itex]S=1[/itex] eigenstate of the joint measurement operator [itex](\vec{S}_{1}+\vec{S}_{2})^{2}[/itex], then you already know what you expect to find (in probability) of whatever you measure of this state. That is, once you know the state of a system, you can figure out the probabilities of every conceivable measurement, though what measurement outcome you happen to see is completely random (so far as we have been able to see).

    That being said, this state is (quite) entangled, and with the mathematical formalism, you can work out what you would expect to find if you were to also measure [itex]\vec{S}_{2}[/itex] given that you just found out the outcome of measuring [itex]\vec{S}_{1}[/itex]. Experimentally, entanglement is something that can only be seen if you look at both particles.

    On their own, single particles are just that, single. Without any outside information, there is no way in principle of ever knowing whether the particle you are looking is single or half of an entangled pair. Because of this (side note) there is no way of using entanglement to communicate faster than light.
     
  9. Oct 9, 2013 #8
    Yes we cannot transmit information FTL since we don't know which measurement to apply on the other side.

    The question was rather this : If we measure (S1+S2)^2 we get aligned spin, S=1. If we measure S1 we get S2 is in the opposite direction, S=0.

    This is a quantum mechanical effect, since the measurement of S1 disturbs the system.

    However is it not the case that some delay always happen between the measurement of S1 and S2, I suppose the measurement of (S1+S2)^2 implies a "simultaneous" measurement of S1 and S2, but due to the uncertainty principle (on the source) S1 could happen a bit before S2 (hence giving S=0) ?
     
  10. Oct 9, 2013 #9

    jfizzix

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    If you find the total spin to be 1, and the total z-component of the angular momentum to be zero, you have the entangled triplet state. Measuring in the Z-basis, if we find that the z-comoponents of the spin have opposite signs whenever we look. This does not mean though, that the x and y-components will also have opposite signs whenever we look, in particular, because different components of the angular momentum don't commute with one another.

    The issue of what sort of delay there is between measuring one particle, and updating the state of the second particle is still unresolved as it is a point that distinguishes different interpretations of quantum mechanics. Experimentally, the effect seems to be instantaneous, or at least, no one has been able to find an upper speed limit yet (10,000x the speed of light is a lower limit at least)
     
  11. Oct 9, 2013 #10
    thanks for the answer.
    I had one more point to ask : is it possible to make a rotation of the triplet space basis to get [tex]\mid ++\rangle +\mid--\rangle[/tex] ? But then the ms quantum number won't be defined ?
     
  12. Oct 9, 2013 #11

    jfizzix

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    You can perform a basis rotation to get to the state you mention. The ms quantum number given as the z-component of the total spin will not be defined, but either the x or y-component of the total spin will be well-defined.
     
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