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B Subsystem has no wave function

  1. Jan 27, 2017 #1
    It is mentioned that subsystems don't have a wave function, in general. If two subsystems are entangled, there can be a wave function for the composite system, but not for each subsystem.

    Let's say you have two entangled photon pair.. it has a wave function but not for each separate photon.. but then a photon as quantum stuff always have a wave function.. you can entangled one of the photons to another quantum object.. so how can you state the subsystem has no wave function?
     
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  3. Jan 27, 2017 #2
    I wouldn't say that a system has a wavefunction or doesn't have a wavefunction - the wavefunction is just a mathematical formalism that we developed to describe a quantum mechanical system. However, it happens that the wavefunction formalism can only be used to describe what we call pure states. It cannot be used to describe mixed states, which behave as classical mixtures (i.e. not superpositions) of pure states.

    Mixed states occur in a variety of contexts, such as coupling to a noisy environment etc. Let's consider your example of a maximally entangled pair of qubits that can be either spin up or spin down. If we consider the pair as a whole, then it exists in a pure state. However, if we look at only one qubit and ignore the existence of the other, when we make many measurements on an ensemble of identically prepared pairs, we will find that qubit to be spin up 50% of the time and spin down the other 50% of the time. Note however that this is not a superposition of up and down. In order to describe this classical mixture behaviour, the wavefunction formalism therefore fails, and we have to introduce the notion of a density matrix or operator to describe the system.
     
  4. Jan 27, 2017 #3
    But if we used Bohmian Mechanics where the particles really have a wave function. Then when we just looked at the subsystem of an entangled system, how do you force it to have a wave function? and what kind of wave function would it be (theoretically Bohmian wise)?
     
  5. Jan 27, 2017 #4
    Unfortunately I'm not familiar enough with Bohmian mechanics to be able to describe subsystems of an entangled system within that framework. Perhaps @Demystifier can help us out here?
     
  6. Jan 27, 2017 #5

    A. Neumaier

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    Bohmian mechanics ascribes a wave function only to the state of the universe, which determines the dynamics of all Bohmian particles.

    It does not say anything at all about how the Bohmian dynamics of the universe relates to a putative Bohmian dynamics of a subsystem. The subsystem is described statistically by ordinary quantum mechanics only, which is claimed to be identical with the common formalsim, since the effect of the environment has been traced out.
     
  7. Jan 27, 2017 #6

    Demystifier

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    Bohmian mechanics can define a wave function of the subsystem as a conditional wave function. For instance, if ##\Psi(x_1,x_2,t)## is a full wave function, then the conditional wave function is
    $$\psi_c(x_1,t)=\Psi(x_1,X_2(t),t)$$
    where ##X_2(t)## is the Bohmian trajectory. In general, ##\psi_c(x_1,t)## does not satisfy the Schrodinger equation. This is actually good, because it can explain the effective "collapse" of the conditional wave function.

    Furthermore, when the entanglement is absent, i.e. when
    $$\Psi(x_1,x_2,t)=\psi_1(x_1,t)\psi_2(x_2,t)$$
    then
    $$\psi_c(x_1,t)=\psi_1(x_1,t)f(t)$$
    where ##f(t)=\psi_2(X_2(t),t)##. The factor ##f(t)## does not influence the trajectory ##X_1(t)##, so the physical content of ##\psi_c(x_1,t)## does not differ from the physical content of standard subsystem wave function ##\psi_1(x_1,t)##.
     
  8. Jan 27, 2017 #7

    A. Neumaier

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    For a system consisting of two noninteracting subsystems that are entangled with each other, do the subsystems propagate under the conditional Bohmian dynamics in the same way as under the full Bohmian dynamics?
     
    Last edited: Jan 27, 2017
  9. Jan 27, 2017 #8

    Demystifier

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    I'm not sure I understand the question. Did you perhaps mean or rather than as?
    Or if you meant what you wrote, the answer is a trivial yes.
     
    Last edited: Jan 27, 2017
  10. Jan 27, 2017 #9

    A. Neumaier

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    ''in the same way as''
     
  11. Jan 27, 2017 #10

    A. Neumaier

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    How does it explain the effective "collapse"?
     
  12. Jan 28, 2017 #11

    PeterDonis

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    This is not correct. A quantum system always has a wave function, but a single photon is only a quantum system (instead of just a subsystem) if it is isolated, i.e., if it does not interact with anything else. But if it does not interact with anything else, it can't be entangled with anything else.
     
  13. Jan 30, 2017 #12

    Demystifier

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    I will explain if you promise that you will not complain that "it is not a rigorous proof". :wink:
     
  14. Jan 30, 2017 #13

    A. Neumaier

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    I don't expect any rigorous proof, so I won't complain about that.
     
  15. Jan 30, 2017 #14

    Demystifier

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    Fine. Let ##x_2## represent the collective coordinate for positions of the macroscopic pointer of the measuring apparatus. Then, as I explained to you several times, the total wave function at some fixed time ##t## after the measurement is of the form
    $$\Psi(x_1,x_2)=\sum_A c_A \psi_A(x_1) \varphi_A(x_2)$$
    where
    $$\varphi_A(x_2)\varphi_B(x_2)\simeq 0.........(1)$$
    for ##A\neq B##. The corresponding conditional wave function is
    $$\psi_c(x_1)=\sum_A c_A \psi_A(x_1) \varphi_A(X_2)............(2)$$
    where ##X_2## represents the actual particle positions at time ##t##. But from (1) we see that if ##\varphi_A(X_2)## is non-negligible for one value of ##A##, then it is negligible for all other values. Hence (2) reduces to
    $$\psi_c(x_1) \simeq c_A \psi_A(x_1) \varphi_A(X_2)\equiv \tilde{c}_A \psi_A(x_1)$$
    which corresponds to an effective collapse.
     
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