# B Does gravity affect quantum transition amplitudes?

1. Dec 18, 2017

### Aidyan

I suppose the answer is no, since there is no reason to believe that it does. Or is there any? Has this been tested experimentally? Or is there an obvious reason that it does or does not?

2. Dec 18, 2017

### phyzguy

It certainly does. The classic "COW experiment" used neutron interferometry to show that the gravitational potential affects quantum phases just as one would expect. The paper is attached.

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• ###### COW_PhysRevLett.34.1472.pdf
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3. Dec 18, 2017

### Aidyan

Hmm... sorry my question was ill posed. I mean the transition probability which should not depend from the quantum phase (squared modulus of the amplitude).

4. Dec 18, 2017

### George Jones

Staff Emeritus
I am not quite sure what you mean, so I apologise if I have misinterpreted things.

An overall phase does not change the squared modulus, and thus is not observable, but a relative phase usually causes interference effects that are observable. The latter is the case in the famous paper that @phyzguy posted.

For example, if $\psi' = e^{i\phi} \psi$, then $\left| \psi' \right|^2 = \left| \psi \right|^2$.

If $\psi' = \psi + e^{i\phi} \psi$, then
\begin{align} \left| \psi' \right|^2 &= \left| \psi \right|^2 + e^{-i\phi} \left| \psi \right|^2 + e^{i\phi} \left| \psi \right|^2 + \left| \psi \right|^2\\ &= 2 \left( 1 + \cos \phi \right) \left| \psi \right|^2 \end{align}
and
$$\frac{\left| \psi' \right|^2}{\left| \psi' \right|^2_{\phi = 0}} = \frac 1 2 \left( 1 + \cos \phi \right) ,$$
which varies between 0 and 1.

Roughly, in the above paper, the relative phase in the above paper depends on relative gravitational potential (due to height difference) of two arms of a neutron interferometer, and change in gravitational potential is very much obsevable.

Not Beyond the Standard Model, though.

5. Dec 19, 2017

### Aidyan

What I mean are the transition probabilities in QM, i.e. the probability between the initial and final state of a quantum system (atomic or nuclear transitions, scattering, etc.). That one that are calculated with time-dependent perturbation theory in the interaction picture which gives:

$P_{if}(t)=\left|\left<\Psi_{f}|\Psi(t)\right>_{I}\right|^2 = \left|\left<\Psi_{f}|\hat{U_{I}}(t,t_{i})| \Psi_{i}\right>\right|^2,$

which is the transition probability from an unperturbed initial state $\Psi_{i}$ to a final state $\Psi_{f}$ with $\hat{U_{I}}(t,t_{i})$ the evolution operator in the interaction picture. And, when expressed back in the Schrödinger picture it is:

$P_{if}(t)=|\left<\Psi_{f}|\Psi(t)\right>_{S}|^2 = |\left<\Psi_{f}|\sum_{n} c_{n}(t) e^{-iE_{n}t/\hbar}|\Psi_{n}\right>|^2 = |c_{f}(t) e^{-iE_{n}t/\hbar}|^2=|c_{f}(t)|^2,$

with $c_{n}$ an $c_{f}$ complex amplitudes which squared modulus gives the transition probabilities (which value can be calculated from matrix elements of the perturbing potential) and the latter equality showing that phases are irrelevant (I skipped lots of stuff here... please see every textbook on QM for more details).

Since this is classical QM, my doubt was that this does no longer hold in the presence of a curved spacetime background and needs extension (not just with SR, as in QFT of the SM, but with GR). If so, transition probabilities change in the presence of gravity, which means that the radiative transition spectrum of matter changes, for instance that of falling into a BH (it is not just redshift, it is about the structure of the spectrum). I suppose it is something already extensively analyzed (especially in quantum gravity theories) but could not find a reference to that. I ask because eventually that should not be too complicate to check experimentally. Or is there an obvious reason to dismiss this altogether?

6. Dec 19, 2017

### Mentz114

The Unruh effect applies to accelerating frames and by the equivalence principle one could expect something similar for a system at rest in a gravitational field. This paper could be of interest

https://arxiv.org/abs/quant-ph/0509151

[ On reflection this is probably irrelevant. The neutrons in the COW experiment were falling, and the effect is attributed to the gravtational potential. No accelerations there.]