I am a little confused about a few papers I read on the Einstein-Klein-Gordon (EKG) equations.

From what I understood one takes the energy-stress-tensor of the scalar field:

$$T_{\mu\nu } = −\partial_\mu \varphi ∂_\nu \varphi − \frac12 g_{\mu\nu}∂_\alpha\varphi∂_\alpha\varphi − V(φ )$$

$$V(φ) = −\frac12 (mφ)^2 + \fracκ4φ^4$$

Where $κ$ is the usual self-interaction coupling constant .

Then this stress-tensor is plugged into Einstein's equation and solved usually with the Schwarzschild or other convenient metrics.

Now from what I know about QFT, isn't $φ$ an operator that maps the Hilbert space $H$ of particle states to $H$ itself? Doesn't that make the components of the energy-stress-tensor observables and thus operators as well? If so, then how can one equate the components of the Einstein tensor (which are purely geometric tensor fields) to operators?

I once read that people (as of today) usually plug in < $T_{μν} $ > in Einstein's equation, but in the papers I read they directly used the operator itself and not it's expectation value.

I also wondered how people plug the Maxwell stress tensor into Einstein's equations in a similar way? I get that the EM and scalar field are real valued fields, but shouldn't we use the expectation values in Einstein's equation?

What am I missing? Here is a link to one such paper: http://arxiv.org/abs/0805.3211

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    $\begingroup$ This seems to be a semantic issue. People sometimes use the phrase Klein-Gordon equation to refer to the classical wave equation with a mass term. There is nothing quantum about it. So $\phi$ is not an operator, its just a normal classical field. $\endgroup$ Aug 21, 2013 at 17:56

1 Answer 1


This is not how I would usually go about formulating this problem. When i think of the Einstein Klein-gordon equation, I start from an action principle:

$$S = \int d^{4}x\sqrt{|g|}\left(\frac{1}{16\pi G}R -\left[\nabla_{a}\phi\nabla^{a}\phi + V(\phi)\right]\right)$$

Which will then yield EOM:

$$R_{ab} - \frac{1}{2}Rg_{ab} = 8\pi G\left(\nabla_{a}\phi\nabla_{b}\phi -\frac{1}{2}g_{ab}\left[\nabla_{c}\phi\nabla^{c}\phi + V(\phi)\right]\right)$$


$$\nabla^{c}\nabla_{c}\phi - V'(\phi) = 0$$

From here, the question is what are you doing with these equations?

Are you looking at general relativity in the context of a classical Klein-Gordon source? If so, you just solve these equations.

Are you trying to do semi-classical gravity? Well, then, you set your metric to a fixed background metric, and just analyze the Klein-Gordon EOM using the appropriate $\nabla$ for this background metric, quantizing the field using a scheme like you'll find in Wald's book.

Are you looking to work through the back-reaction of semi-classical effects on the background metric? Well, then you need to write down $g_{ab} = g^{0}_{ab} + g^{1}_{ab}$ where $g^{1}_{ab} \ll g^{0}_{ab}$, assume that $\phi$ is first-order, and substitute the expectation value of your solved $\phi$ in on the right hand side, and solve for $g^{1}_{ab}$ in this limit.

Or are you trying to do something else? If you want to treat this as a fully quantum problem, you're going to need to first quantize gravity.

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    $\begingroup$ @dj_mummy: I do agree that the approach seems a little confused. I tell them to use the expectation value if they're looking to solve the semi-classical problem. And it's a little pedantic, but $\phi_{a}\phi_{a}$, as notation, seems to imply the use of a Minkowski background, and starting with an action was an attempt on my part to say "no, there really is an enveloping 4-geometry, and this affects the KG EOM, and needs to be thought about throughout the problem." And, if you just treat the whole thing classically, of course, you can just think of the KG equation as a classical equation. $\endgroup$ Aug 22, 2013 at 13:17
  • $\begingroup$ I think a (1/2) factor is missing from the scalar field kinetic term, in order for the KG equation to be the one you wrote. $\endgroup$
    – ApolloRa
    Apr 10, 2020 at 20:09

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