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I Can Energy be Entangled?

  1. Dec 13, 2016 #1

    referframe

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    Can two particles be entangled in the energy property/observable? If so, what kind of experiment could verify that the two particles were energy-entangled?

    Thanks in advance.
     
  2. jcsd
  3. Dec 13, 2016 #2
    I am not an expert, but I think you need to define 'energy entanglement'.
     
  4. Dec 14, 2016 #3

    hilbert2

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    Suppose we have two replicas of a two-state system with available states ##\left| 1 \right>## and ##\left| 2 \right>## corresponding to energies ##E_1## and ##E_2##. Now the state of the combined system is something like:

    ##\left| \psi \right> = a \left| 1 \right>\left| 1 \right> + b\left| 1 \right>\left| 2 \right>+c\left| 2 \right>\left| 1 \right>+d\left| 2 \right>\left| 2 \right>## .

    An entangled state of this system could be something like ##\frac{1}{\sqrt{2}}\left(\left| 1 \right>\left| 1 \right>+\left| 2 \right>\left| 2 \right>\right)##, where you can be certain that the both subsystems have equal energies.
     
  5. Dec 14, 2016 #4

    DrChinese

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    Since energy is conserved in entanglement generation (say of photon pairs), I would say there is entanglement on that basis. For photons pairs created via Parametric Down Conversion, for example: frequency is proportional to energy and so frequency is conserved. If you knew the input frequency fairly precisely (which you normally would), then the relationship between the 2 output photons would be well defined.

    How would test you that? Usually you have an inequality of some type involving (partially) non-commuting observables. The experimental realization is using energy-time entanglement which I am not so familiar with. Here is a reference, I have not read it but one of the authors (Larsson) writes on this a lot and I have read some of his other work.

    https://arxiv.org/abs/1103.6131
    "In the spirit of Einstein, Podolsky, and Rosen it is possible to ask if the quantum-mechanical description (of this setup) can be considered complete. This question will be answered in detail in this paper, by delineating the quite complicated relation between energy-time entanglement experiments and Einstein-Podolsky-Rosen (EPR) elements of reality."
     
  6. Dec 14, 2016 #5
    As far as I understand, entanglement has no understood physical mechanism, it just "pops out" of the math. I like to explain entanglement using way simpler algebra.

    Say you have a wave function f(x) which describes two particles. Now when you want to split them, you define one of the particles as g(x). You have now automatically defined the other particle as f(x) - g(x), there is nothing else it can be.
     
  7. Dec 14, 2016 #6

    DrChinese

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    Your comment has nothing to do with this thread.

    Yes, it is true that entanglement is often associated with a conservation rule. However, the connections between entangled systems are more complicated than the algebra you present would imply. For one thing, not every system of 2 particles is entangled. For another, you have the 2 independent functions as being a sum when it might be a Product state. Entangled systems cannot be described by a Product state.
     
  8. Dec 15, 2016 #7

    Zafa Pi

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    That's true. Thus if |ψ⟩ = √½(|00⟩ + |11⟩), which is an entangled state, then |ψ⟩⊗|ψ⟩ is not entangled. Kind of weird.
     
  9. Dec 15, 2016 #8

    referframe

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    Conservation laws help experimenters prove that the two output particles are correlated but do not cause entanglement directly. Agree?
     
  10. Dec 15, 2016 #9

    atyy

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    Would this qualify as energy entanglement? https://arxiv.org/abs/quant-ph/0507189
     
  11. Dec 15, 2016 #10

    referframe

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    In your example, both subsystems would have equal probabilities, but the two energy eigenvalues need not be equal. Right?
     
  12. Dec 15, 2016 #11

    hilbert2

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    If you measure the energy of one of the subsystems and find out thst it's ##E_1##, then you can be certain that the energy of the second subsystem is ##E_1##, too.
     
  13. Dec 15, 2016 #12

    Nugatory

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    That is correct. Even the equal probabilities are not fundamental to entanglement: ##\frac{1}{2}|1\rangle|1\rangle+\frac{\sqrt{3}}{2}|2 \rangle|2\rangle## is an entangled state with different probabilties of the two possible outcomes. The essential property if the entangled state is that it can be written as a superposition of two states such that collapsing the superposition determines the value of both observables. (This can be rephrased to conform to your favorite interpretation, and note that the two observables must have have common eigenfunctions to superimpose).
     
  14. Dec 16, 2016 #13

    vanhees71

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    Sure, concerning the above cited paper I didn't understand, why one could claim that the given state is not entangled. By definition it is. I also don't need two particles to have entanglement. The classic example is the Stern-Gerlach experiment, where after the magnets you have an entanglement between position and spin of one particle (take a neutron instead of the original version with an Ag atom).
     
  15. Dec 16, 2016 #14

    Demystifier

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    |ψ⟩⊗|ψ⟩ is entangled. If we think of |ψ⟩ as a 2-particle state, then |ψ⟩⊗|ψ⟩ is a 4-particle state. It is entangled because it cannot be written as a product of four 1-particle states.
     
  16. Dec 16, 2016 #15

    referframe

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    All very interesting. It's almost as if the central players in entanglement are not individual particles but individual sets where each set is a complete eigenbasis (with corresponding eigenvalues) for some observable, for the same or different particle.
     
  17. Dec 16, 2016 #16

    Nugatory

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    As far as the math is concerned, there aren't two particles. There is a single quantum system with a single wave function evolving according to Schrodinger's equations. There are various observables on this system, such "the result at instrument A" and "the result at instrument B"; but it's residual classical thinking that leads us to sort these observables into groups and call one of the groups "the properties of particle A" and another "the properties of particle B".
     
  18. Dec 16, 2016 #17

    Zafa Pi

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    In responding to my post #7: That's true. Thus if |ψ⟩ = √½(|00⟩ + |11⟩), which is an entangled state, then |ψ⟩⊗|ψ⟩ is not entangled. Kind of weird.
    This is great, some more conflict - a big part of what makes science fun, educational + prizes.
    Well, I certainly agree that |ψ⟩⊗|ψ⟩ cannot be written as a product of four 1-particle states. However, several sources say a state in a tensor product space is entangled if it is not a (tensor) product. But |ψ⟩⊗|ψ⟩ sits as a product of |ψ⟩ with itself, and furthermore |ψ⟩ is a pure state, for what that's worth.

    Suppose we have 4 photons P1, P2 from |ψ⟩, and another two P3, P4 from a different |ψ⟩. It seems that there will be no weird correlations when measuring both P1 and P3, as there would be in measuring both P1 an P2. This may not prove anything, but does lead me to think that |ψ⟩⊗|ψ⟩ is not entangled. Say, as opposed to the state √½(|0000⟩ + |1111⟩).

    The "That's true." in my #7 post refers to the DrChinese post #6 where he says, "Entangled systems cannot be described by a Product state." I wonder what he would say?

    OK, your turn.:rolleyes:
     
  19. Dec 17, 2016 #18

    referframe

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    How does the Planck energy equation, E=hf, fit into the above picture? Does the "single quantum system with a single wave function" have a single frequency?
     
  20. Dec 17, 2016 #19

    Nugatory

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    It doesn't. That equation relates the frequency of an infinite monochromatic plane wave to the energy of the photons associated with it (and be warned that the word "associated" must not be taken too literally - a more accurate but still very simplified picture of the relationship can be found here). You won't find such waves anywhere; we study them because any real waveform can be written and analyzed as a mathematical superposition of these waves.

    That equation also belongs to the "old" quantum mechanics that was largely abandoned after 1925 or thereabouts; it's not part of the modern treatment that we're using in this thread and that is needed to analyze entanglement situations. You'll see this equation used sometimes because under the right conditions (a single particle isolated well enough to be treated as a quantum system in its own right and prepared with a definite energy) you can still get good results out of it - but you have to be aware of its limitations.

    It depends on whether the system is in an energy eigenstate or a superposition of energy eigenstates.
     
  21. Dec 18, 2016 #20

    Demystifier

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    Either the sources are sloppy or you are a sloppy reader. Take for example wikipedia
    https://en.wikipedia.org/wiki/Quantum_entanglement#Meaning_of_entanglement
    which says (my bold):
    "An entangled system is defined to be one whose quantum state cannot be factored as a product of states of its local constituents; that is to say, they are not individual particles but are an inseparable whole."
     
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