Gauge fixing in canonical quantum gravity - Physics Stack Exchange most recent 30 from physics.stackexchange.com 2019-09-23T17:50:09Z https://physics.stackexchange.com/feeds/question/477376 https://creativecommons.org/licenses/by-sa/4.0/rdf https://physics.stackexchange.com/q/477376 2 Gauge fixing in canonical quantum gravity Michael Angelo https://physics.stackexchange.com/users/63967 2019-05-02T12:58:04Z 2019-05-02T13:59:25Z <p>In analogy with QFT, the partition function in canonical quantum gravity is defined as a functional integral over the metric tensor (which is now the quantum field),</p> <p><span class="math-container">$$\int \mathcal{D} g \mathcal{D}\phi \exp{I_E(g,\phi)}$$</span> where <span class="math-container">$\phi$</span> are all matter fields and <span class="math-container">$I_E$</span> is the Euclidean Einstein-Hilbert action.</p> <p>This path integral can be seen as a way to generate solutions to the Wheeler-deWitt equation, which is a canonically quantised version of the Hamiltonian constraint of GR; <span class="math-container">$$\hat{H}\Psi=0$$</span></p> <p>I am confused that there never seems to be a discussion on gauge-fixing, which is usually essential if you quantise something. In GR the gauge transformations are basically diffeomorphism, and I know superspace factors out all diffeomorphisms, but I don't see how this is implemented in either the path integral nor the WdW equation.</p> <p>Passing over to the minisuperspace approximation, my worry becomes clearer. Let's only include homogeneous and isotropic metrics in the path integral. A general <span class="math-container">$0(4)$</span> metric in the 3+1 split is of the form <span class="math-container">$$ds^2 = N^2(\lambda)\text{d} \lambda ^2 + a^2(\lambda)\text{d} \Omega^2_3,$$</span> where <span class="math-container">$a$</span> is the scale factor and <span class="math-container">$N$</span> is the lapse. Importantly, the lapse <span class="math-container">$N$</span> incorporates all gauge freedom (it ensures that in making the 3+1 split we do not kill the reparametrisation invariance of GR). The path integral in this case is simply given by</p> <p><span class="math-container">$$\int \mathcal{D} a \mathcal{D}N \exp{I_E(a,N)}$$</span></p> <p>In other words, we <em>integrate</em> over the gauge freedom? What does this mean? How does this correspond to the usual gauge fixing paradigm in QFT?</p>