None of the answers current explain an aspect of the question that I am interested in: what goes wrong if you try to construct a theory of physics where gravity is not quantized?

There are various arguments that strongly suggest that everything coupled to a quantum system should, fundamentally, also be quantum.

We know that the stress-energy tensor sources curvature for the gravitational field, $$G_{\mu\nu} \sim T_{\mu\nu}$$ but in a quantum theory the stress-energy tensor does not have a definite value, but rather may be in superposition. So then how do we describe the curvature? If you say the curvature may be in superposition too, so that $G_{\mu\nu} = T_{\mu\nu}$ holds for each branch of the superposition, then you've just quantized gravity -- quantization is exactly the process where we treat the set of classical physical states of a system as separate quantum states which may be superposed.

The only other option which reduces to the classical result when the matter is nearly classical is $$G_{\mu\nu} \sim \langle T_{\mu\nu} \rangle.$$ However, this is extremely strange for many reasons. For example, consider a particle of mass $m$ which is in an equal superposition of being here or in Andromeda. Then the classical gravitational field would be that of two masses $m/2$, each in one galaxy. If the particle is measured, the wavefunction collapses, and the gravitational field instantaneously changes, so the observed mass in Andromeda becomes either $m$ or zero. This nonlocal change in the field allows superluminal signalling by somebody in the Milky Way. (There's nothing special about gravity here; it would also hold if we insisted on a classical electromagnetic field. In either case, when the field is quantized, this problem is avoided by the usual way in quantum field theory.)

In addition, energy conservation may be violated. This is easier to see with the electromagnetic field. If one starts with an excited atom in an empty cavity, in state $|e \rangle$, after some time it will be in the superposition $(|e \rangle + |g \rangle) / \sqrt{2}$. If you insist the electromagnetic field have a definite classical configuration, then the branches of the wavefunction do not have equal energy. When you measure the energy, you'll generally find a different result than the initial energy; it can only match on average.

This is essentially the erroneous BKS theory which was rendered obsolete with the quantization of the electromagnetic field. In this case the wavefunction is $(|e\rangle \otimes |0 \rangle + |g \rangle \otimes |1 \rangle) / \sqrt{2}$ where the second factor indicates the number of photons, and the two branches of the wavefunction have exactly the same energy as they must. Similarly, if one couples to classical gravity, one must allow violations of energy conservation that only cancel out on the average, but there's no problem for quantized gravity.

I'm sure the mathematicians can come up with more sophisticated, complicated reasons that classical and quantum theories don't mesh, but these immediate issues are already bad enough.

None of the answers current explain an aspect of the question that I am interested in: what goes wrong if you try to construct a theory of physics where gravity is not quantized?

There are various arguments that strongly suggest that everything coupled to a quantum system should, fundamentally, also be quantum.

We know that the stress-energy tensor sources curvature for the gravitational field, $$G_{\mu\nu} \sim T_{\mu\nu}$$ but in a quantum theory the stress-energy tensor does not have a definite value, but rather may be in superposition. So then how do we describe the curvature? If you say the curvature may be in superposition too, so that $G_{\mu\nu} = T_{\mu\nu}$ holds for each branch of the superposition, then you've just quantized gravity -- quantization is exactly the process where we treat the set of classical physical states of a system as separate quantum states which may be superposed.

The only other option which reduces to the classical result when the matter is nearly classical is $$G_{\mu\nu} \sim \langle T_{\mu\nu} \rangle.$$ However, this is extremely strange for many reasons. For example, consider a particle of mass $m$ which is in an equal superposition of being here or in Andromeda. Then the classical gravitational field would be that of two masses $m/2$, each in one galaxy. If the particle is measured, the wavefunction collapses, and the gravitational field instantaneously changes, so the observed mass in Andromeda becomes either $m$ or zero. This nonlocal change in the field allows superluminal signalling by somebody in the Milky Way. (There's nothing special about gravity here; it would also hold if we insisted on a classical electromagnetic field. In either case, when the field is quantized, this problem is avoided by the usual way in quantum field theory.)

One could argue that collapse is really unphysical; all branches of the wavefunction exist and we should sum over all of them. If we take this interpretation, then non-quantized gravity is already ruled out experimentally. See Page and Geilker (1981), where the result of a radioactive decay is used to determine the positioning of a mass in a Cavendish experiment. If all branches of the wavefunction count, then the pendulum should point at the midpoint of the two possible positions for the mass, but it doesn't.

As a separate issue, energy conservation may be violated. This is easier to see with the electromagnetic field. If one starts with an excited atom in an empty cavity, in state $|e \rangle$, after some time it will be in the superposition $(|e \rangle + |g \rangle) / \sqrt{2}$. If you insist the electromagnetic field have a definite classical configuration, then the branches of the wavefunction do not have equal energy. When you measure the energy, you'll generally find a different result than the initial energy; it can only match on average.

This is essentially the erroneous BKS theory which was rendered obsolete with the quantization of the electromagnetic field. In this case the wavefunction is $(|e\rangle \otimes |0 \rangle + |g \rangle \otimes |1 \rangle) / \sqrt{2}$ where the second factor indicates the number of photons, and the two branches of the wavefunction have exactly the same energy as they must. Similarly, if one couples to classical gravity, one must allow violations of energy conservation that only cancel out on the average, but there's no problem for quantized gravity.

I'm sure the mathematicians can come up with more sophisticated, complicated reasons that classical and quantum theories don't mesh, but these immediate issues are already bad enough.

None of the answers current explain an aspect of the question that I am interested in: what goes wrong if you try to construct a theory of physics where gravity is not quantized?

There are various arguments that strongly suggest that everything coupled to a quantum system should, fundamentally, also be quantum.

We know that the stress-energy tensor sources curvature for the gravitational field, $$G_{\mu\nu} \sim T_{\mu\nu}$$ but in a quantum theory the stress-energy tensor does not have a definite value, but rather may be in superposition. So then how do we describe the curvature? If you say the curvature may be in superposition too, so that $G_{\mu\nu} = T_{\mu\nu}$ holds for each branch of the superposition, then you've just quantized gravity -- quantization is exactly the process where we treat the set of classical physical states of a system as separate quantum states which may be superposed.

The only other option, which reduces to the classical result when the matter is nearly classical, is $$G_{\mu\nu} \sim \langle T_{\mu\nu} \rangle.$$ However, this is extremely strange for many reasons. For example, consider a particle of mass $m$ which is in an equal superposition of being here or in Andromeda. Then the classical gravitational field would be that of two masses $m/2$, each in one galaxy. If the particle is measured, the wavefunction collapses, so the gravitational field instantaneously changes. This affects the (physically measurable) gravitational field nonlocally and hence allows superluminal signaling.

Furthermore, energy conservation may be violated. This is easier to see with the electromagnetic field. If one starts with an excited atom in an empty cavity, in state $|e \rangle$, after some time it will be in the superposition $(|e \rangle + |g \rangle) / \sqrt{2}$. If you insist the electromagnetic field have a definite configuration, then the branches of the wavefunction do not have equal energy. When you measure the energy, you'll generally find a different result than the initial energy; it will only match on average. This is essentially the erroneous BKS theory which was rendered obsolete with the quantization of the electromagnetic field. Similarly, if one does not quantize gravity, one must allow violations of energy conservation that only cancel out on the average.

I'm sure the mathematicians can come up with more sophisticated, complicated reasons that classical and quantum theories don't mesh, but these immediate issues are already bad enough.

None of the answers current explain an aspect of the question that I am interested in: what goes wrong if you try to construct a theory of physics where gravity is not quantized?

There are various arguments that strongly suggest that everything coupled to a quantum system should, fundamentally, also be quantum.

We know that the stress-energy tensor sources curvature for the gravitational field, $$G_{\mu\nu} \sim T_{\mu\nu}$$ but in a quantum theory the stress-energy tensor does not have a definite value, but rather may be in superposition. So then how do we describe the curvature? If you say the curvature may be in superposition too, so that $G_{\mu\nu} = T_{\mu\nu}$ holds for each branch of the superposition, then you've just quantized gravity -- quantization is exactly the process where we treat the set of classical physical states of a system as separate quantum states which may be superposed.

The only other option which reduces to the classical result when the matter is nearly classical is $$G_{\mu\nu} \sim \langle T_{\mu\nu} \rangle.$$ However, this is extremely strange for many reasons. For example, consider a particle of mass $m$ which is in an equal superposition of being here or in Andromeda. Then the classical gravitational field would be that of two masses $m/2$, each in one galaxy. If the particle is measured, the wavefunction collapses, and the gravitational field instantaneously changes, so the observed mass in Andromeda becomes either $m$ or zero. This nonlocal change in the field allows superluminal signalling by somebody in the Milky Way. (There's nothing special about gravity here; it would also hold if we insisted on a classical electromagnetic field. In either case, when the field is quantized, this problem is avoided by the usual way in quantum field theory.)

In addition, energy conservation may be violated. This is easier to see with the electromagnetic field. If one starts with an excited atom in an empty cavity, in state $|e \rangle$, after some time it will be in the superposition $(|e \rangle + |g \rangle) / \sqrt{2}$. If you insist the electromagnetic field have a definite classical configuration, then the branches of the wavefunction do not have equal energy. When you measure the energy, you'll generally find a different result than the initial energy; it can only match on average. 

This is essentially the erroneous BKS theory which was rendered obsolete with the quantization of the electromagnetic field. In this case the wavefunction is $(|e\rangle \otimes |0 \rangle + |g \rangle \otimes |1 \rangle) / \sqrt{2}$ where the second factor indicates the number of photons, and the two branches of the wavefunction have exactly the same energy as they must. Similarly, if one couples to classical gravity, one must allow violations of energy conservation that only cancel out on the average, but there's no problem for quantized gravity.

I'm sure the mathematicians can come up with more sophisticated, complicated reasons that classical and quantum theories don't mesh, but these immediate issues are already bad enough.