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B Why wouldn't quantum gravity work this way?

  1. Mar 12, 2016 #1
    First of all, I would like to point out that I have zero background in general relativity, but due to my overwhelming curiosity I will post this question anyway.
    So, what's wrong actually with applying the correspondence principle, which is typical in quantum mechanics, to quantum gravity? Suppose a system of zero charges, why wouldn't the equation
    $$
    \left(\frac{p^2}{2m}+G\frac{mM}{r}\right)\psi = E\psi
    $$
    work as it does with the electric potential? Why is the search of a satisfying theory of quantum gravity still on going, what has been impeding the physicists in this field?
     
  2. jcsd
  3. Mar 12, 2016 #2

    jfizzix

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    For starters, the Schrodinger equation is only for non-relativistic systems. It treats time and space in a way that does not work well with special relativity. There are more advanced equations, like the Dirac equation or the Klein Gordon equation that do include special relativity, but there is no equation that includes General relativity in a way that gives accurate predictions.

    So, Gravity is a 1/r^2 force, and we could certainly model gravitational effects on a quantum system with this Schrodinger equation (like we can with the electric force), but that would only be in the low-mass weak field approximation (where Newton's description of gravity is good enough).
    When the mass and field are strong enough that the effects of warping spacetime need to be included, then we don't have complete answers yet.
     
  4. Mar 13, 2016 #3
    Alright that makes sense. But are there already proposals to realize a physical system where quantum effects can be prominent (i.e. the system is well below nanometer in dimension) but also involves very massive objects? If not, how do they plan to test whether their theory of quantum gravity works?
     
  5. Mar 13, 2016 #4

    mfb

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    To see nonlinear effects of gravity, you need something like a neutron star or a black hole. There is no way to bring a neutron star into a superposition of anything - even if we ignore the problem that we do not have a neutron star available. Well... if we would, we would have other much more serious problems...
    A black hole would give insight into quantum gravity, but we don't have that either. Some very exotic theories predict that the LHC could produce tiny black holes, that would help a lot.

    For linear gravity, things are better.
    • Neutrons can bounce above a surface, driven by Earth's gravity, and their energy levels are quantized. This system can be described with the formula you wrote in post 1.
    • Atom interferometers give very accurate measurements of the gravitational acceleration. So precise that their measurement is influenced by (still classical, ton-scale) nearby masses in the experiment, which is then used to measure the gravitational constant.
    • Instead of quantum systems as test masses, a great step would be quantum systems as source mass. Currently we are several orders of magnitude away from the required sensitivity, but the control of those systems improves rapidly. Such an experiment would entangle the position of the probe mass with the position of the source mass solely by gravity.
     
  6. Mar 13, 2016 #5

    Demystifier

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    There is nothing wrong with your equation above. Indeed, experiments with neutron interferometry show that this equation is correct. However, this equation cannot be the end of the story. For instance, what if the source of gravity is not in a state of some definite mass M but in a superposition of slightly different masses? The full theory of quantum gravity must be able to answer such a question, but the equation above does not answer it.
     
  7. Mar 13, 2016 #6
    Do you mean in the theory of quantum gravity, mass is represented as an operator having more than one eigenvalues? Why is it so? Its counterpart in the electric potential, the charge, is not associated to any operator in quantum mechanics, i.e. charge is a constant in QM.
     
  8. Mar 13, 2016 #7

    mfb

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    The formula works only if a quantum-mechanical particle couples to a classical external field. You cannot use it once the field itself can be in a superposition, or once your particles can absorb or emit it. This is not limited to gravity, it doesn't work with electromagnetism either. You need quantum field theory to quantize the field (surprise...). And in quantum fied theory, adding gravity is problematic.
     
  9. Mar 13, 2016 #8

    Demystifier

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    One way of explaining it at a heuristic level is by adding relativity into the picture. As you probably know, in relativity theory mass is closely related to energy. On the other hand, you already know that energy is an operator in quantum theory. This suggests that in relativistic quantum theory mass should be an operator too.

    Another, even better heuristic argument, involves particle creation/destruction. In QM the particles cannot be created or destructed. Yet, in reality they can. (When you turn on light, the electric bulb creates photons that did not exist a moment before.) So QM cannot be the whole story. Since the number of particles is not a constant, it must be a quantum operator. Moreover, when the number of particles changes, the mass does not need to remain constant. Therefore the mass must be an operator too. Furthermore, even charged particles can be created or destructed (such that the total charge remains constant), so even charge must be an operator. The theory that describes all this is called quantum field theory (QFT).
     
  10. Mar 13, 2016 #9
    I guess it has to do with that second quantization thing,
     
  11. Mar 14, 2016 #10

    Demystifier

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    Absolutely!
     
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