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Measurements of a superposition: Conservation laws?

  1. Jan 14, 2014 #1
    Hi,

    Assume we have a source that emits several copies of the same quantum state which is a superposition of several eigenstates of the Hamilton operator with different energies. We can calculate the expectation value of the energy of this state and therefore also the energy the source releases, which is the same for each state. However, if we measure the energy, the system collapses to one eigenstate with a specific energy.
    How does this comply with conservation of energy? I guess there must be some energy exchanged with the measurement device, but in standard quantum mechanics there is no framework for the measurement process itself that could explain this exchange.
     
  2. jcsd
  3. Jan 14, 2014 #2

    DrChinese

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    Let's take an example where 2 entangled particles A and B have conserved total energy K. Thus e(A)+e(B)=K. We measure e(A), which makes e(B) take on the value K-e(A). So is your question about a possible transfer of energy to/from B? Or is it about the transfer of energy from A to the measuring device that determined the value of e(A)?
     
  4. Jan 14, 2014 #3
    Hi,

    Assume we have a source that continuously emits copies of the state $$\left|\psi\right\rangle=\alpha\left| E_1\right\rangle+\beta\left| E_2\right\rangle$$
    where [itex]\left| E_1\right\rangle[/itex] and [itex]\left| E_2\right\rangle[/itex] are eigenstates of the Hamilton operator to different energies. Since all copies are exactly alike, the source must consume the exact same amount of energy for emitting any of the states.

    However, if we now measure the energy of one state, we'll find either [itex]E_1[/itex] or [itex]E_2[/itex]. Since energy is preserved, there has to be some exchange with the measurement device. But how can we model that? In the standard formulation of quantum mechanics measurement devices don't even appear (a measurement is just a projection).
     
  5. Jan 14, 2014 #4

    jcsd

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    I think the problem is that you are conflating the expectation value of the energy with its actual energy, which is not the case. We should not expect the expectation value to be conserved by the measurement process as it is merely an expectation value. You can see in this example (assuming E1 and E2 are non-degenerate) sometimes the expectation value isn't even an allowed value for the measurement and so it will not be conserved by a measurement.
     
  6. Jan 14, 2014 #5
    Hi,

    Yes I thought of that argument, that's why I talked about the source. I think we can safely assume that the source consumes exactly the same amount for producing any of the copies (since it just repeats the same process over and over). Some of this energy will be lost in the machine, some will be emitted with the state, and this energy is the same for each copy (I'm trying to avoid using the expectation value here).
    When we now measure the energy of several different copies, we will get different energies, however we know that the had the same when they left the source. Hence there has to be exchange with the measurement device, and that's what's unclear to me.
     
  7. Jan 14, 2014 #6
    I don't think a source that consumed the same amount of energy each time could produce the sort of superposition you outline.
     
  8. Jan 14, 2014 #7

    jcsd

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    The point is then that if the energy of each particle is known before the measurement (even if not by a direct measurement on the particle) then it can't be in a superposition of states.
     
  9. Jan 14, 2014 #8
    Ok, let's do this differently. Assume we have a Stern-Gerlach apparatus and we take a single atom that is in the "up" state. The expectation value of the horizontal component of the angular momentum is zero.
    We now measure this atom in a second Stern-Gerlach apparatus that is rotated by 90°. In this basis, the state of the atom is in a perfect superposition of "left" and "right". Assume we measure "left". To be consistent with conservation of angular momentum, some exchange with the measurement device must have taken place. However, in the standard formulation of QM the measurement is just a projection, the measurement device is not modeled in any way that would explain how this exchange is mediated.
     
  10. Jan 14, 2014 #9
    It's a common mistake to assume that a particle in an up state has no angular momentum in a perpendicular direction. It's not true. Again you are mixing up expectation values with the actual value.
     
  11. Jan 14, 2014 #10
    So what's the true story? And what is conserved in QM if not expectation values?
     
  12. Jan 14, 2014 #11

    jcsd

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    You're still describing the non-conservation of an expectation value, however after making our 2nd measurement we might measure the particle again with our first set of apparatus and this time we might get a "down" result and then we might be entitled to ask ourselves what happened to cause the sign of the spin in that direction to reverse. This time in between making measurements of the up-down component of the spin we have disturbed the state of the system by making a measurement of the left-right component. Exactly how this came to be disturbed depends very much on the ontology that you choose to apply/details of how the measurement was made. As the formalism doesn't describe either it won't answer that question.
     
  13. Jan 14, 2014 #12

    DrChinese

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    You could say that the left-right superposition is transferred (swapped) to the measuring device. It being macroscopic, you wouldn't really notice anything. On the other hand, there are situations in which that superposition CAN be explicitly transferred to another quantum object (and measured as such). Although not the same physical setup, that happens with entanglement swapping experiments. There is a measurement device that facilitates this as in your setup.
     
  14. Jan 14, 2014 #13
    This led me to another question which could be related, but I'm not sure.

    Let's assume we have a box containing an observer with two Stern-Gerlach-apparatuses, the first horizontal, the second vertical. I put an "up"-state into the box, the observers measures twice, and, say, gets a "down" state which he outputs. From my point of view, the outside observer who can only measure the initial and final state, is there a way to tell this situation from a box where the "up"-"down" transition takes place unitarily (i.e. there is a Hamilton operator), for example by a magnetic field?
     
  15. Jan 14, 2014 #14

    DrChinese

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    If I recall correctly, this problem is very similar to the one that led Everett to the Many Worlds interpretation. Basically observers at different levels... not sure that has any bearing on your question, but it shows how difficult it is to draw a firm line between observer and that being observed.
     
  16. Jan 15, 2014 #15

    Demystifier

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    That part is correct ...

    ... but this part is not. In standard QM measurement process is described as entanglement with the measuring apparatus, as first described by von Neumann in 1932.
     
  17. Jan 15, 2014 #16

    Demystifier

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  18. Mar 18, 2015 #17
    I have been studying Ion Traps for quantum computing and was also wondering about conservation of energy in the measurement of superimposed states of the ion.
    If | ##\psi\rangle = a\cdot |0\rangle + b\cdot |1\rangle## is the state of the ion,
    then the expected energy would be
    |a^2| ##E_0## + |b^2|##E_1## = ## E_(ex)##
    According to Caltech notes on this subject, if one shines a laser beam of the resonance frequency, ## Omega = (E_1 - E_2)/h ##, on an ion and the ion is in state |##1\rangle##, the photon will not be absorbed. Now the energy of the ion state is ##E_1##, which is greater than the ##E_(ex)##. Where does this energy come from? The photon doesn't seem to have actually interacted with the ion, so I fail to see how it would be imparting energy to the ion's state.

    Can anyone clarify this for me?
     
  19. Mar 18, 2015 #18

    bhobba

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    Exactly.

    And it has deepened our knowledge of what the measurement issue in QM is. We have an explanation for apparent collapse - but actual collapse - that's another matter.

    Thanks
    Bill
     
  20. Mar 18, 2015 #19

    bhobba

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  21. Mar 19, 2015 #20
    This statement is wrong: "I think we can safely assume that the source consumes exactly the same amount for producing any of the copies (since it just repeats the same process over and over)."
    The source consumes more energy to produce copies which have more energy. Before measurement, the copies and the source both exist in a superposition of energies. When you measure a copy's energy, you are indirectly measuring the source energy.
    This is fully analogous to a common example of quantum entanglement where a particle decays into two particles with opposite spin. Let's say particle A is emitted and particle B stays in the source. A can either have spin up or spin down. Is the angular momentum conserved? Of course. If you measure particle A to have spin up, then you are also measuring particle B to have spin down.
     
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