Chris's Senior Project

Helpful books and papers

Chris's Thesis and Work drafts

Quantum for mathematicians

Quantum Mechanics by Arno Bohm

Physicists' favorite quantum books

Shankar's Quantum Mechanics

Sakurai's Quantum Mechanics

A paper dealing with the factor-ordering problem in general

Ryan/Turbiner paper on factor ordering

A paper on factor ordering in quantum cosmology

(This is where the family of orderings we're considering comes from)

Steigl/Hinterleitner on factor ordering in FRW

2d Quantum Gravity

The 2d version of the Friedmann-Robertson-Walker model is a universe with one spatial dimension and one time dimension. We assume that the spatial part is just a circle whose radius can vary over time - analogous to the (3+1) FRW model where the spatial part of the universe is a 3d sphere with radius $a(t)$. What we need is the Lagrangian and Hamiltonian that describe the evolution of the radius of the (1+1) universe.

In this paper of Nakayama's, he works out that Lagrangian. Actually gravity in 2 dimensions needs to be modified from the usual Einstein general relativistic gravitational theory we use in 4 dimensions. If we just naively try to imitate Einstein gravity in 2d, we get something that doesn't have any dynamics. But the same general idea behind GR, that gravity is really spacetime curvature, can be adapted to the 2d case. Roughly, what people do is set up a Lagrangian action that includes some matter fields as well as gravity and then they integrate out the matter fields. I think it was Polyakov who first did this; we can go into that more sometime later. In any case what Nakayama does in his paper is to start with this 2d theory of gravity, without any symmetry assumptions on the shape of the universe (i.e. he assumes that topologically the spatial universe is a circle, but it could be a deformed circle; i.e. its curvature could vary from point to point). Then he proves (that's Eq (22) in his paper) that the metric can be factored into the product of two functions: one that only depends on time and one that only depends on space. This means that the space-dependent part does not evolve, so essentially we just care about the time-dependent part, and the problem of 2d quantum gravity becomes a quantum mechanics problem.

The most important part of Nakayama's paper for us starts at Eqs (35) and (36). Note that what Nakayama calls $x^0$ is the time coordinate. The coordinate $x^1$ was the space coordinate, essentially an angle parametrizing the spatial universe (recall that it is topologically a circle) which he just integrates out starting in Eq (29). By the time we get to Eq (35), we are looking at the Lagrangian of a one-dimensional mechanics problem. Note that Nakayama's $l \left( x^0 \right)$ is, up to a constant rescaling, the analog of our scale factor $a \left( t \right)$. After we talk more about metrics I can explain how to get this. To explain the potential terms in the Lagrangian (35), the lower-case $\lambda$ is the analog of the cosmological constant. In the term $\left( m + \frac{1}{2} \right)^2 / l$, $m \in \left\{ 0, 1, 2, \dots \right\}$. It comes from the fact that we integrated out the angular spatial coordinate (that's what's happening in Eq (33), which results from Eq (23)).

Now, the only complication right now is that in getting Eq (35), Nakayama treated time like a space variable. The technical name for this is to say that he was working in Euclidean signature, and after we talk about metrics I'll tell you what this means mathematically. But we actually want to treat time like time and space like space, which is called working in Lorentzian signature (will explain). Nakayama switches the signature when he goes to Eq (36) - that's what he means when he says "To quantize the model (35) we will switch to the Minkowski metric temporarily." The effect of the switch is to reverse the sign of the potential with respect to the kinetic term. For example, start with Eq (35) in Nakayama and do a Legendre transformation to get the Hamiltonian corresponding to his $S_m$. It will not be the $H_m$ in (36) - the potential terms will have the opposite sign. Nakayama switches the signature in (35) by negating the potential terms. If you do this and then do a Legendre transformation, you will get (36).

Next, notice that in (36) Nakayama has not quantized yet, but he has already indicated his choice of factor ordering by writing

\begin{align} H_m = \Pi_l l \Pi_l + \left( m+ \frac{1}{2} \right)^2 l^{-1} + \lambda l. \end{align}

When he promotes $\Pi_l$ (the momentum conjugate to $l$) to a quantum operator $-i \partial / \partial l$ (where note that he has set $\hbar = 1$, but I would recommend you reinstate $\hbar$), the kinetic term will become $- \partial_l l \partial_l$, so this is his chosen ordering. With this ordering he goes ahead and finds all the energy eigenfunctions of the Hamiltonian (see Eqs (38)-(40)). Note that in our problem we're actually just interested in the solving for the $E = 0$ case, but we want to vary the ordering.

Here's where I hand it over to you! Go ahead and construct a quantum Hamiltonian with ordering varying analogously to what I showed you for FRW in 4 dimensions. Note, though, that there I wrote the varied orderings all in terms of the preferred ordering $a^{-1/2} \partial_a a^{-1/2} \partial_a$. In your case, since the kinetic term is different, the preferred ordering would be $l^{1/2} \partial_l l^{1/2} \partial_l$. I'll tell you why we prefer these orderings after we talk about metrics!

The reasoning behind our preferred ordering

To explain why we prefer the ordering $a^{-1/2} \partial_a a^{-1/2} \partial_a$ in the 3d FRW case or the ordering $l^{1/2} \partial_l l^{1/2} \partial_l$ in 2d gravity, we need a quick discussion of what Lagrangian and Hamiltonian mechanics looks like on a curved background. Basically, on a flat background Lagrangians look like

\begin{align} L = \frac{1}{2} \mathbf{\dot x} \cdot \mathbf{\dot x} - V \left( \mathbf{x} \right) \end{align}

and Hamiltonians look like

\begin{align} H = \frac{1}{2} \mathbf{p} \cdot \mathbf{p} + V \left( \mathbf{x} \right), \end{align}

where $\mathbf{x}$ is the position vector, $\mathbf{\dot x}$ is its time derivative, $\mathbf{p}$ is the momentum vector, and $\cdot$ represents the standard dot product. On the other hand in a curved background, we have to use the metric to take inner products.

First, we establish a splitting into a time coordinate and three spatial coordinates, and we call the spatial part of the metric $g_{ij} , i,j = 1,2,3$. As usual, we let an overdot denote a derivative with respect to our chosen time coordinate. (Note: this is what I was mentioning yesterday about how a canonical formalism forces us to split spacetime into space and time. It's not actually restrictive since we can always do this, but for aesthetic reasons it would be nice not to have to, and this is one of the arguments why the path integral approach is an important complement to canonical quantization).

So the Lagrangian for a particle moving on a curved background described by the spatial metric $g_{ij}$ and experiencing a potential $V \left( \mathbf{x} \right)$ is

\begin{align} L = \frac{1}{2} g_{ij} \dot{x}^i \dot{x}^j - V \left( \mathbf{x} \right), \end{align}

the momentum is given by the 1-form $p_i = g_{ij} \dot{x}^j$, and the Hamiltonian is

\begin{align} H = \frac{1}{2} g^{ij} p_i p_j + V \left( \mathbf{x} \right) \end{align}

Here $g^{ij}$ is defined to be the inverse matrix of $g_{ij}$, so that $g_{ij} g^{jk} = \delta_i^k$ (where we are using Einstein summation convention as always here and $\delta_i^k$ is the Kronecker delta. It's just matrix multiplication).

Now if we look again at or 1 in light of these definitions, we notice that both can be interpreted as Hamiltonians for a particle moving on a curved one-dimensional manifold (parametrized by $a$ in the case of and by $l$ in the case of 1), experiencing a potential. In , we see that the metric on the curved one-dimensional manifold should be taken as $\left( a \right)$ and in 1 as $\left( l^{-1} \right)$. Notice these are just matrices with only one component since the space is 1-dimensional so the metric is a 1x1 matrix. In the case of , we can compute the Laplace-Beltrami operator to be $a^{-1/2} \partial_a a^{-1/2} \partial_a$, and in 1 it is $l^{1/2} \partial_l l^{1/2} \partial_l$. * Exercise: Work these out.

This is why we prefer these orderings. In quantum mechanics on a curved background, as you can see in the D'Olivo and Torres paper, people want the Schrodinger equation to be covariant, so they want the kinetic term to be a Laplace-Beltrami operator. And this is the ordering for which people actually do know the corresponding path integral measure.


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