# A simple gedankenexperiment

1. Apr 4, 2012

### tom.stoer

Think about an instable particle with exakt rotational symmetry, e.g. a neutral pion π°; let's measure its gravity on a very large sphere with the particle located at the center. Of course the initial gravitational field and the whole configuration has spherical symmetry. Describing the pion decay π° → 2γ by (quantum) fields it's obvious the the final 2γ-state has spherical symmetry, too. b/c no gravitational monopole radiation does exist, the initial and the final gravitational field must be identical; total energy and invariant mass of both initial and final state are identical as well (for a very large sphere we can define energy and invariant mass even in GR b/c we have a asymptotically flat field configuration).

Now let's modify the apparatus and let's detect the two photons on the large sphere. Detecting them of course breaks rotational invariance, may cause higher gravitational multipoles and therefore gravitational radiation. That means that the "collaps of the 2γ wave function" has a measurable effect, namely gravitational radiation.

What goes wrong?

2. Apr 4, 2012

### genneth

Is this fundamentally different from detecting an energy eigenstate which leads to "violation" of energy conservation? In any case, one has an entangled state between the outside of the sphere and the inside. The decay always breaks the rotational symmetry anyway, and everything detected should be a consequence of that.

3. Apr 5, 2012

### tom.stoer

My problem is that the symmetry breaking will result in gravitational waves. But they cannot both not exist (w/o detection) and exist (w/ detetction).

4. Apr 5, 2012

### tom.stoer

The idea for the solution is the following: before the detection we have a coherent superposition of 2γ states (with exact spehrical symmetry) and a coherent superposition of gravitational field states (with exact spehrical symmetry). When detecting the particles, the symmetry of both the particle and the gravitational field states is destroyed, both wave functions collapse, and we detect both localized photons in our detector and the corresponding gravitons.

So a solution requires a description in terms of quantum gravity.

5. Apr 5, 2012

### Haelfix

Hi Tom.

Hmm, why should the photons emit gravitational radiation in this example? It seems to me that won't be the case unless there is a nonideal interaction between the measuring device and the incoming photons.
If that is the case, we need to talk about response functions and ideal Von Neumann detectors, and there will be very large subtleties with quantum gravity.
For instance, typically we insist that such devices be very massive and classical (so that their quantum fluctuations can be made small). Of course, this is exactly what we don't want in the case of gravity, where we want the measuring device to not contribute significantly to the gravitational system in question (much less collapse into a blackhole if M --> infinity).

Anyway, suppose that you did jiggle an atom in the photomultiplier, such that there was a quadropole or higher moment and gravitational waves were emmitted. I still don't quite see what the contradiction is?

6. Apr 6, 2012

### Demystifier

Nothing goes wrong. In general, the collapse is a good phenomenological description of measurable quantum events, so there is nothing strange with the conclusion that it has measurable effects.

7. Apr 6, 2012

### Demystifier

You are right. In fact, an experiment that confirms the reality of collapse in systems involving gravity, and thus serving as an indirect experimental evidence that gravity is quantized, has already been performed:
http://prl.aps.org/abstract/PRL/v47/i14/p979_1

8. Apr 6, 2012

### tom.stoer

b/c after the collapse the spherical symmetry is broken and the energy-momentum tensor may have (in general) a non-zero energy-momentum quadrupole moment Q which causes a gravitational wave h.

$$h_{ik} = \frac{2}{r}\frac{d^2\langle Q_{ik}\rangle}{dt^2}$$

The contradiction is due to the description of the gravitational field following classical GR. Doing that you have perfect spherical symmetry before the collapse (no gravitational wave) and broken spherical symmetry after the collaps (with a gravitational wave). That means that the gravitational wave is caused by the detection (i.e. by the collapse), not by the decay itself.

The way out is to assume that the gravitational field is quantized, too, and that there is not only a collapse of the 2γ wave function but in addition a collapse of the gravitational field wave function; then the contradiction goes away, the gravitational field is described in the same quantum mechanical framework.

@Demystifier: fine, thanks, that confirmes my ideas

Last edited: Apr 6, 2012
9. Apr 6, 2012

### Haelfix

Yes, but the quadropole moment still vanishes, unless you have some sort of acceleration. That won't happen unless you have some sort of nonideal (in the Von Neumann sense) measurement process going, and there will be interaction terms that you cannot ignore.

More generally, you can never ignore the properties of the detector in quantum gravity (see the discussion in Birrel and Davies). You must specify the mass, and you must specify the state of motion.

Doing this carefully tends to make all conclusions rather ambiguous.

10. Jul 22, 2012

### ApplePion

What is going on is that "measurement" is a physical process, and in "measuring" the photons you are affecting their energies, etc.

I can construct a similar "paradox". Suppose with have an electron in a superposition state that is an even mix of an energy of 10 and an energy of 100. The "average energy is 55. Suppose you make a "measurement", and the electron is now in a pure state of 100. It might appear that energy was not conserved--the energy of the final state is not the same as the energy of the initial state. But what really happened is that the "measuring device" added enegy to the system.

11. Aug 1, 2012

### Prathyush

are u trying imply that detection photons + gravitons will some how save Rotational invariance

12. Aug 1, 2012

### tom.stoer

no, it will not save rotational invariance

the problem is that in QM rotational invariance is broken by the measurement = by the "collaps of the wave function" or by "splitting into many worlds" whereas in GR no such breaking of rotational invariance does exist; therefore this symmetry breaking must either be present on both soides of the Einstein equations or on neither side; this forces us to quantize gravity and to treat & interpret matter + gravity (geometry) consistently

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