I know there are two ways to do quantum gravity.

One can pick a background space-time (usually Minkowski flat space-time) and then at any time slice one can define the state of the universe as the probability of gravitons being at positions: $x_1, x_2, x_3, x_4, x_5,...$


Or one could specify the geometry given by the 3D metric tensor g. Then the wave function is


Then one either sums over all the Feynman diagrams for the graviton paths or alternatively one sums over all the different geometries.

What is the connection between the two ways of doing it? Is there a way to show that the same degrees of freedom are in both? How are the above wave functions related? Is one right and one wrong?

(As an extra, can one do a similar thing with the electromagnetic field. Is there a wave function $\psi(A)$ where A(x) is the electromagnetic field? Summing over fields instead of photons?)

What I don't get is that in QED there is no one electric field. Each electron has it's own electric field (minus it's own one). But in quantum gravity people talk as if there is one geometry. Surely each particle will feel a different geometry (minus its own self interaction?)


I don't think they are compatible since the first one is calculated by:

$$ Amp(In,Out) = \int e^{ i S[g,\phi] } In(g,\phi)Out(g,\phi) D[g]$$

where for example: $$ In(g) = a + \int g(x)\psi(x) dx^4 + \int g(x)g(y)\psi(x,y)dx^4 dy^4 +...$$

Where the second one is calculated as something like:

$$ Amp(g_{in},g_{out}) = \int_{g_{in}}^{g^{out}} e^{i S[g,\phi] } D[g]D[\phi] $$

Which seems to only be an amplitude for a single graviton field. Whereas it is believed that gravitons are quantised.

However, maybe just because gravity is also to do with space-time these two ways are somehow equivalent? Or that gravitons don't exist?

(Let's just assume for a moment that Supergravity is finite. Just to aid the discussion).

Edit 2

I think in the first description it is easy to describe entangled gravitons (which surely must exist), while in the second geometric description it is not possible to do this. So it is not a description of more than one graviton. Unless I am mistaken and entangled gravitons cannot exist. (Or entangled closed strings which amount to the same thing). So summing over geometries cannot be correct as it can't describe entangled gravitons. Or is something wrong with this argument? On the other hand if we believe the brane description of the Universe a single brane can describe multiple particles, so maybe a single geometry represents multiple gravitons?

Edit 3

I think in loop quantum gravity instead of expanding $\psi[g]$ in terms of positions of gravitons they are expanding it in terms of loops or spin networks. So an individual spin network would not correspond to one particular geometry but rather the superposition of various numbers of entangled gravitons. So I think this is fine. And probably also why it is difficult to reconstruct GR from LQG.

On the other hand I think the Hawking-Hartle type of summing over geometries cannot be true or otherwise only true in approximation.

  • $\begingroup$ Contrary to what you have written, there are zero ways to do quantum gravity. It is possible, however, to talk about QG in a context of string theory. In string theory string modes, corresponding to gravitons, have actually the same effect as if the worldsheet was embedded into the curved (instead of flat) background. $\endgroup$ Sep 28, 2015 at 20:46
  • $\begingroup$ That is true. For the aid of this discussion let us just assume that N=8 Supergravity is finite and can be quantised in terms of Feynman diagrams. $\endgroup$
    – user84158
    Sep 29, 2015 at 12:20
  • 2
    $\begingroup$ I think you are asking a more general question (not specific to gravity) about how to reconcile the field and particle ways of describing QFT. In fact the exact same issue shows up in QED (which has the advantage of being much simpler). One can talk about photons, or one can talk about the E/M field, but the two descriptions are complementary and one can't really talk about both simultaneously. Indeed, the number operator $N$ that counts photons does not commute with the photon field operator $A_\mu$ (or if you want something gauge invariant it doesn't commute with $F_{\mu\nu}$). $\endgroup$
    – Andrew
    Sep 29, 2015 at 19:30
  • $\begingroup$ Hi Andrew. Could you elaborate? Perhaps as a full answer? Is there REALLY such a thing as the EM field in QED? If the A turns from a field into an operator it is no longer a field but an operator for creation and annihilation of individual photons. Isn't the classical EM field simply an approximation to a full description for a large number of photons? $\endgroup$
    – user84158
    Sep 29, 2015 at 22:51
  • $\begingroup$ Also, I forgot to add, you can have entangled photons which has no description (as far as I know) in terms of an EM field. $\endgroup$
    – user84158
    Sep 29, 2015 at 23:18

3 Answers 3


There is a little confusion in your statements about the quantisation of the gravitational field, especially

then at any time slice one can define the state of the universe as the probability of gravitons being at positions:

is not correct. Let us start from the beginning instead.

The standard way to quantise fields is to write down their path integral, which formally represents the vacuum-vacuum transition amplitude given the field being there in the between. In order to do so one needs an action of the field $S[\phi]$ and a well defined measure (details are complicated though) to be able to define some sort of integral thereof as $$ Z[\phi] = N\int \mathcal{D}[\phi]\,\textrm{e}^{iS_J(\phi)}\propto\langle 0|0\rangle _J.$$ The above expression, after having been manipulated, produces the correlation functions which are the corner stones from which one can reconstruct back the entire field theory (provided some assumptions to hold) and calculate the scattering amplitudes, which in turn contain the "physics". So far so good, only that when one goes explicitly down to calculate all the pieces, infinities suddenly appear and they must be renormalised back to produce finite measurable quantities. However, techniques at hand, this can be done for all the interactions present in the universe except for quantum gravity, reason being the fact that the physical dimensions of the force do not allow it to be renormalisable (in 4 dimensions) following the standard aforementioned approach.

Therefore we need to find out something else.

A good new class of approaches to deal with the problem is given by loop quantum gravity together with the spin foam models (you can find some references in my other answer here): essentially, the idea is to reduce the complexity of the infinite degrees of freedom carried by the field describing the space-time itself as a finite triangulated network rather than a continuum; the states of the system are represented in terms of functions on the nodes of such triangulations. Evolution between states (i. e. transition amplitudes) can be calculated, turning out to be expressed in terms of irreducible representations of some particular Lie groups. Again, infinities appear in 4 dimensions mostly due to the fact that the Lorentz group does not contain finite dimensional representations. A new area of study, the Group Field Theory (links therein), has been almost ad hoc introduced to deal with such mathematical refinements.

Another way to approach the problem is to perform some sort of canonical quantisation based on a linear Hamiltonian that can be derived from a first order (Palatini) Lagrangian, plus some diffeomorphisms contraints to be implemented. So far so good again, only such additional constraints fail to be renormalisable when implemented.

Since you mentioned the graviton, let us see how that can emerge from these approaches. Actually, it most of the times does not, because in quantum field theory field carriers emerge as basis of the gauge groups you model the fields on your manifold with; here, instead, since the approach is a little different, there are non-standard ways to pull it out. Most of them rely on calculations of vacuum-vacuum transition amplitudes for the second order terms appearing in the Lagrangian (which are associated to the propagator of the fields).

Then there is string theory, which is a strong candidate as well, but that's a completely different story.

Then the wave function is $\psi[g]$

It is not completely clear to me what you mean by wave function of the field. However, regarding a possible way to do the same job with the electromagnetic field: there is really no need to do so because the quantisation of the electromagnetic field is maybe the only thing in nature whose formulation we have found to perfectly work as it is, and QED is perhaps the most complete, correct and experimentally proven amongst the theories. You may of course re-invent the wheel from scratch, should you wish to, bus as far as I am aware there are no research areas in this direction, at the moment.

  • $\begingroup$ Hello. I don't think you really got what I was saying. Say the first approach is like doing supergravity calculations with Feynman diagrams. Therefor the input and output sates would be amplitudes $\psi(x_1,x_2,...)$ corresponding to terms g(x_1)g(x_2).. in the path integral. The second approach is like the Wheeler-de-Witt approach where the input and output is a specific 3-geometry. You have a 'wave-function' of the Universe. Are the two approaches compatible? $\endgroup$
    – user84158
    Sep 29, 2015 at 12:02
  • $\begingroup$ I am not aware of such approaches, nor am I about supergravity or wave functions of the universe, but maybe someone else has better insights. $\endgroup$
    – gented
    Sep 29, 2015 at 12:30

I see here in the memoirs of Bryce De Witt that he says the Wheeler-de-Witt equation is WRONG and is only an approximation good for cosmology but not as a basis of quantum gravity.

I will take that to mean the Wheeler-de-Witt equation is an approximation in the same way that the single-particle Shrodinger equation is an approximation to full quantum electrodynamics. The history of the WdW equation is that it is in analogy with the Shrødinger equation.

Since one metric defines one geometry, anything with more than one graviton field will not be described by a single geometry. Although on the cosmological scale one might approximate a vast number of gravitons to a single geometry.


I have found the answer.

A wave functional of a field $\Psi[\phi]$ can be sharply peaked at a particular function of $\phi$. And hence it will correspond to a classical field f. For example:

$$\Psi[\phi] = \exp \left( -\int (\phi(x)-f(x))^2 dx^3 \right)$$

is maximum at $\Psi[f]$.

Just like a normal wave function $\psi(x)= \exp(-(x-a)^2)$ can be sharply peaked at a and hence correspond to a particle at a.


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