I wanted to see how the Fourier transform can turn field theory into many-particle mechanics. This is just silly fooling around, so you shouldn't take what follows too seriously (there are much better models of extra dimensions, to be sure!).
Take \(\phi(t, s)\) to be a field on a cylinder of radius \(R\). We consider the action
\[ S = \frac{1}{R} \int_{-\infty}^\infty \int_0^{R} |\nabla \phi|^2 ds dt \]
Expand \(\phi(t, s)\) in Fourier series:
\[ \phi(t,s ) = \sum_n \phi_n(t) e^{2 \pi i n s / R} \]
Then in Lorentzian signature, we have
\[ \int_0^{R} |\nabla \phi|^2 d\theta = R \sum_n \dot{\phi}_n^2 - \left(\frac{2\pi n}{R}\right)^2 \phi_n^2. \]
Putting this back into the action, we find
\[ S = \sum_n \int_{-\infty}^\infty \dot{\phi}_n^2- \left(\frac{2\pi n}{R}\right)^2 \phi_n^2 dt. \]
This is the action for infinitely many harmonic oscillators, with frequencies \(\omega_n = 2\pi |n| / R\). Recall that the energy levels of the harmonic oscillator are \(k\omega\) for \(k = 0, 1, \ldots\). So supposing that only a finite energy \(E\) is accessible in some particular experiment, we can only excite those modes \(\phi_n\) for which
\[ \frac{2\pi |n|}{R} < E. \]
In particular, only finitely many \(\phi_n\) may be excited at energies below \(E\), effectively reducing the field theory on the cylinder to many-particle quantum mechanics.
Monday, August 27, 2012
Thursday, July 26, 2012
Generating Functions
Method of Generating Functions
Let \(X\) and \(Y\) be two smooth manifolds, and let \(M = T^\ast X, N = T^\ast Y\) with corresponding symplectic forms \(\omega_M\) and \(\omega_N\).
Question: How can we produce symplectomorphisms \(\phi: M \to N\)?
The most important construction from classical mechanics is the method of generating functions. I will outline this method, shameless stolen from Ana Cannas da Silva's lecture notes.
Suppose we have a smooth function \(f \in C^\infty(X \times Y)\). Then its graph \(\Gamma\) is a submanifold of \(M \times N\): \( \Gamma = \{ (x,y, df_{x,y}) \in M \times N \}\). Since \(M \times N\) is a product, we have projections \(\pi_M, \pi_N\), and this allows us to write the graph as
\[ \Gamma = \{ (x, y, df_x, df_y) \}\]
Now there is a not-so-obvious trick: we consider the twisted graph \(\Gamma^\sigma\) given by
\[ \Gamma^\sigma = \{(x,y, df_x, -df_y) \} \]
Note the minus sign.
Proposition If \(\Gamma^\sigma\) is the graph of a diffeomorphism \(\phi: M \to N\), then \(\phi\) is a symplectomorphism.
Proof By construction, \(\Gamma^\sigma\) is a Lagrangian submanifold of \(M \times N\) with respect to the twisted symplectic form \(\pi_M^\ast \omega_M - \pi_N^\ast \omega_N\). It is a standard fact that a diffeomorphism is a symplectomorphism iff its graph is Lagrangian with respect to the twisted symplectic form, so we're done.
Now we have:
Modified question: Given \(f \in C^\infty(M \times N)\), when is its graph the graph of a diffeomorphism \(\phi: M \to N\)?
Pick coordinates \(x\) on \(X\) and \(y\) on \(Y\), with corresponding momenta \(\xi\) and \(\eta\). Then if \(\phi(x,\xi) = (y,\eta)\), we obtain
\[ \xi = d_x f, \ \eta = -d_y f \]
Note the simlarity to Hamilton's equations. By the implicit function theorem, we can construct a (local) diffeomorphism \(\phi\) as long as \(f\) is sufficiently non-degenerate.
Different Types of Generating Functions
We now concentrate on the special case of \(M = T^\ast \mathbb{R} = \mathbb{R} \times \mathbb{R}^\ast\). Note that this is a cotangent bundle in two ways: \(T^\ast \mathbb{R} \cong T^\ast \mathbb{R}^\ast\). Hence we can construct local diffeomorphisms \(T^\ast \mathbb{R} \to T^\ast \mathbb{R}\) in four ways, by taking functions of the forms
\[ f(x_1, x_2), \ f(x_1, p_2), \ f(p_1, x_2), \ f(p_1, p_2) \]
Origins from the Action Principle, and Hamilton-Jacobi
Suppose that we have two actions
\[ S_1 = \int p_1 \dot{q}_1 - H_1 dt, \ S_2 = \int p_2 \dot{q}_2 - H_2 dt \]
which give rise to the same dynamics. Then the Lagrangians must differ by a total derivative, i.e.
\[ p_1 \dot{q}_1 - H_1 = p_2 \dot{q}_2 - H_2 + \frac{d f}{dt} \]
Suppose that \(f = -q_2 p_2 + g(q_1, p_2, t)\). Then we have
\[ p_1 \dot{q}_1 - H_1 = -q_2 \dot{p}_2 - H_2 + \frac{\partial g}{\partial t} + \frac{\partial g}{\partial q_1}\dot{q}_1 + \frac{\partial g}{\partial p_2} \dot{p_2} \]
Comparing coefficients, we find
\[ p_1 = \frac{\partial g}{\partial q_1}, \ q_2 = \frac{\partial g}{\partial p_2}, \ H_2 = H_1 + \frac{\partial g}{\partial t} \]
\[ p_1 = \frac{\partial g}{\partial q_1}, \ q_2 = \frac{\partial g}{\partial p_2}, \ H_2 = H_1 + \frac{\partial g}{\partial t} \]
Now suppose that the coordinates \((q_2, p_2)\) are chosen so that Hamilton's equations become
\[ \dot{q_2} = 0, \ \dot{p}_2 = 0 \]
Then we must have \(H_2 = 0\), i.e.
\[ H_1 + \frac{\partial g}{\partial t} = 0 \]
Now we also have \(\partial H_2 / \partial p_2 = 0\), so this tells us that \(g\) is independent of \(p_2\), i.e. \(g = g(q_1, t)\). Since \(p_1 = \partial g / \partial q_1\), we obtain
\[ \frac{\partial g}{\partial t} + H_1(q_1, \frac{\partial g}{\partial q_1}) = 0 \]
This is the Hamilton-Jacobi equation, usually written as
\[ \frac{\partial S}{\partial t} + H(x, \frac{\partial S}{\partial x}) = 0 \]
Note the similarity to the Schrodinger equation! In fact, one can derive the Hamilton-Jacobi equation from the Schrodinger equation by taking a wavefunction of the form
\[ \psi(x,t) = A(x,t) \exp({\frac{i}{\hbar} S(x,t)}) \]
and expanding in powers of \(\hbar\). This also helps to motivate the path integral formulation of quantum theory.
Monday, July 23, 2012
KAM I
In this post I want to sketch the idea of KAM, following these lecture notes.
I don't want to worry too much about details, so for now we'll define an integrable system to be a Hamiltonian system \((M, \omega, H)\) for which we can choose local Darboux coordinates \((I, \phi)\) with \(I \in \mathbb{R}^N\) and \(\phi \in T^N\), such that the Hamiltonian is a function of \(I\) only. Defining \(\omega_j := \partial H / \partial I_j\), Hamilton's equations then read
\begin{align}
\dot{I}_j &= 0, \\\
\dot{\phi}_j &= \omega_j(I).
\end{align}
Hence we obtain linear motion on the torus as our dynamics. Note in particular that the sets \(\{I = \mathrm{const}\}\) are tori, and that the dynamics are constrained to these tori. We call these tori "invariant".
Now suppose that our Hamiltonian \(H\) is of the form
\[ H(I, \phi) = h(I) + f(I, \phi) \]
with \(f\) "small". What can be said of the dynamics? Specifically, do there exist invariant tori? KAM theory lets us formulate this question in a precise way, and gives an explicit quantitative answer (as long as \(f\) is nice enough, and small enough).
I want to sketch the idea of the KAM theorem, completely ignoring analytical details.
Suppose we could find a symplectomorphism \(\Phi\): (I, \phi) \mapsto (\tilde{I}, \tilde{\phi})\) such that \(H(I, \phi) = H(\tilde{I}\). Then our system would still be integrable (just in new action-angle coordinates), and we'd be done. There are two relatively easy ways of constructing symplectomorphisms: integrating symplectic vector fields, and generating functions. In the lecture notes, generating functions are used, so let's take a minute to discuss them.
Proposition Let \(\Sigma(\tilde{I}, \phi)\) be a smooth function and suppose that the transformation
\[ I = \frac{\partial \Sigma}{\partial \phi}, \
\tilde{\phi} = \frac{\partial \Sigma}{\partial \tilde{I}}\]
can be inverted to produce a diffeomorphism \(\Phi: (I, \phi) \mapsto (\tilde{I}, \tilde{\phi})\). Then \(\Phi\) is a symplectomorphism.
Proof
\[ dI = \frac{\partial^2 \Sigma}{\partial \phi \partial \tilde{I}} d \tilde{I} \]
\[ d\tilde{\phi} = \frac{\partial^2 \Sigma}{\partial \phi \partial \tilde{I}} d\phi \]
Hence
\[ dI \wedge d\phi = \frac{\partial^2 \Sigma}{\partial \phi \partial \tilde{I}} d \tilde{I} \wedge d\phi = d\tilde{I} \wedge d\tilde{\phi}. \]
We want a symplectomorphism \(\Phi\) such that
\[ H \circ \Phi(\tilde{I}, \tilde{\phi}) = \tilde{h}(\tilde{I} \]
If \(\Phi\) came from a generating function \(\Sigma\), then we have
\[ H(\frac{\partial \Sigma}{\partial \phi}, \phi) = \tilde{h}(\tilde{I}) \]
Expanding things, we have
\[ h(\frac{\partial \Sigma}{\partial \phi}) + f(\frac{\partial \Sigma}{\partial \phi}, \phi) = \tilde{h}(\tilde{I}). \]
If \(f\) is small, then we might expect \(\Phi\) to be close to the identity, and hence \(\Sigma\) ought to be close to the generating function for the identity (which is \(\langle I, \phi \rangle\)). So we take
\[ \Sigma(\tilde{I}, \phi) = \langle \tilde{I}, \phi \rangle + S(\tilde{I}, \phi) \]
where \(S\) should be "small". So we linearize the equation in \(S\):
\[ \langle \omega(\tilde{I}), \frac{\partial S}{\partial \phi} \rangle
+ f(\tilde{I}, \phi)
= \tilde{h}(\tilde{I}) - h(\tilde{I}) \]
Now we can expand \(S\) and \(f\) in Fourier series and solve coefficient-wise. This gives a formal solution \(S(\tilde{I}, \phi)\) of the equation
\[ \langle \omega, \frac{\partial S}{\partial \phi} \rangle + f(\tilde{I}, \phi) = 0. \]
Unfortunately, the Fourier series for \(S\) has no chance to converge, so instead we take a finite truncation. If we assume \(f\) is analytic, its Fourier coefficients decay exponentially fast, so this provides a very good approximate solution to the linearized equation (and we can give an explicit bound in terms of a certain norm of \(f\)). Call this function \(S_1\). We then use \(S_1\) to construct a symplectomorphism \(\Phi_1\).
Now we take
\[ H_1(I, \phi) = H \circ \Phi_1(I, \phi) = h_1(I) + f_1(I, \phi). \]
Some hard analysis then shows that \(h - h_1\) is small, and \(f_1\) is much smaller than f.
The above arguments sketch a method to put the system "closer" to an integrable form. By carefully controlling \(\epsilon\)'s and \(\delta\)'s, one then shows that iterated sequence \(\Phi_1, \Phi_2 \circ \Phi_1, \ldots\) converges to some limiting symplectomorphism \(\Phi_\infty\).
Integrable Systems
I don't want to worry too much about details, so for now we'll define an integrable system to be a Hamiltonian system \((M, \omega, H)\) for which we can choose local Darboux coordinates \((I, \phi)\) with \(I \in \mathbb{R}^N\) and \(\phi \in T^N\), such that the Hamiltonian is a function of \(I\) only. Defining \(\omega_j := \partial H / \partial I_j\), Hamilton's equations then read
\begin{align}
\dot{I}_j &= 0, \\\
\dot{\phi}_j &= \omega_j(I).
\end{align}
Hence we obtain linear motion on the torus as our dynamics. Note in particular that the sets \(\{I = \mathrm{const}\}\) are tori, and that the dynamics are constrained to these tori. We call these tori "invariant".
Now suppose that our Hamiltonian \(H\) is of the form
\[ H(I, \phi) = h(I) + f(I, \phi) \]
with \(f\) "small". What can be said of the dynamics? Specifically, do there exist invariant tori? KAM theory lets us formulate this question in a precise way, and gives an explicit quantitative answer (as long as \(f\) is nice enough, and small enough).
I want to sketch the idea of the KAM theorem, completely ignoring analytical details.
Constructing the Symplectomorphism
Suppose we could find a symplectomorphism \(\Phi\): (I, \phi) \mapsto (\tilde{I}, \tilde{\phi})\) such that \(H(I, \phi) = H(\tilde{I}\). Then our system would still be integrable (just in new action-angle coordinates), and we'd be done. There are two relatively easy ways of constructing symplectomorphisms: integrating symplectic vector fields, and generating functions. In the lecture notes, generating functions are used, so let's take a minute to discuss them.
Proposition Let \(\Sigma(\tilde{I}, \phi)\) be a smooth function and suppose that the transformation
\[ I = \frac{\partial \Sigma}{\partial \phi}, \
\tilde{\phi} = \frac{\partial \Sigma}{\partial \tilde{I}}\]
can be inverted to produce a diffeomorphism \(\Phi: (I, \phi) \mapsto (\tilde{I}, \tilde{\phi})\). Then \(\Phi\) is a symplectomorphism.
Proof
\[ dI = \frac{\partial^2 \Sigma}{\partial \phi \partial \tilde{I}} d \tilde{I} \]
\[ d\tilde{\phi} = \frac{\partial^2 \Sigma}{\partial \phi \partial \tilde{I}} d\phi \]
Hence
\[ dI \wedge d\phi = \frac{\partial^2 \Sigma}{\partial \phi \partial \tilde{I}} d \tilde{I} \wedge d\phi = d\tilde{I} \wedge d\tilde{\phi}. \]
We want a symplectomorphism \(\Phi\) such that
\[ H \circ \Phi(\tilde{I}, \tilde{\phi}) = \tilde{h}(\tilde{I} \]
If \(\Phi\) came from a generating function \(\Sigma\), then we have
\[ H(\frac{\partial \Sigma}{\partial \phi}, \phi) = \tilde{h}(\tilde{I}) \]
Expanding things, we have
\[ h(\frac{\partial \Sigma}{\partial \phi}) + f(\frac{\partial \Sigma}{\partial \phi}, \phi) = \tilde{h}(\tilde{I}). \]
If \(f\) is small, then we might expect \(\Phi\) to be close to the identity, and hence \(\Sigma\) ought to be close to the generating function for the identity (which is \(\langle I, \phi \rangle\)). So we take
\[ \Sigma(\tilde{I}, \phi) = \langle \tilde{I}, \phi \rangle + S(\tilde{I}, \phi) \]
where \(S\) should be "small". So we linearize the equation in \(S\):
\[ \langle \omega(\tilde{I}), \frac{\partial S}{\partial \phi} \rangle
+ f(\tilde{I}, \phi)
= \tilde{h}(\tilde{I}) - h(\tilde{I}) \]
Now we can expand \(S\) and \(f\) in Fourier series and solve coefficient-wise. This gives a formal solution \(S(\tilde{I}, \phi)\) of the equation
\[ \langle \omega, \frac{\partial S}{\partial \phi} \rangle + f(\tilde{I}, \phi) = 0. \]
Getting it to Work
Unfortunately, the Fourier series for \(S\) has no chance to converge, so instead we take a finite truncation. If we assume \(f\) is analytic, its Fourier coefficients decay exponentially fast, so this provides a very good approximate solution to the linearized equation (and we can give an explicit bound in terms of a certain norm of \(f\)). Call this function \(S_1\). We then use \(S_1\) to construct a symplectomorphism \(\Phi_1\).
Now we take
\[ H_1(I, \phi) = H \circ \Phi_1(I, \phi) = h_1(I) + f_1(I, \phi). \]
Some hard analysis then shows that \(h - h_1\) is small, and \(f_1\) is much smaller than f.
The Induction Step
The above arguments sketch a method to put the system "closer" to an integrable form. By carefully controlling \(\epsilon\)'s and \(\delta\)'s, one then shows that iterated sequence \(\Phi_1, \Phi_2 \circ \Phi_1, \ldots\) converges to some limiting symplectomorphism \(\Phi_\infty\).
Friday, July 13, 2012
Circle Diffeomorphisms I
This is the first of a series of posts based on these lecture notes on KAM theory. For now I just want to outline section 2, which is a toy model of KAM thoery.
We consider a map \(\phi: \mathbb{R} \to \mathbb{R}\) defined by
\[ \phi(x) = x + \rho + \eta(x) \]
where \(\rho\) is its rotation number and \(\eta(x)\) is "small".
Define \(S_\sigma\) to be the strip \(\{ |\mathrm{Im} z|<\sigma\} \subset \mathbb{C}\) and let \(B_\sigma\) be the space of holomorphic functions bounded on \(S_\sigma\) with sup norm \(\|\cdot\|_\sigma\).
Goal: Show that if \(\|\eta\|_\sigma\) is sufficiently small, then there exists some diffeomorphism \(H(x)\) such that
\[ H^{-1} \circ \phi \circ H (x) = x + \rho \]
i.e. that \(\phi\) is conjugate to a pure rotation.
The idea is that if \(\eta\) is small, then \(H\) should be close to the identity, so we suppose that
\[ H(x) = x + h(x) \]
where \(h(x)\) is small. Plugging this into the equation above and discarding higher order terms yields
\[ h(x+\rho) - h(x) = \eta(x) \]
Since \(\eta\) is periodic, we Fourier transform both sides to obtain an explicit formula for the Fourier coefficients of \(h(x)\). We have to show several things:
1. The Fourier series defining \(h(x)\) converges in some appropriate sense.
2. The function \(H(x) = x + h(x)\) is a diffeomorphism.
3. The composition \(\tilde{\phi} = H^{-1} \circ \phi \circ H\) is closer to a pure rotation than \(\phi\), in the sense that
\[ \tilde{\phi}(x) = x + \rho + \tilde{\eta}(x) \]
where \(\|\tilde{\eta}\| \ll \|\eta\|\).
\[ H_\infty^{-1} \circ \phi \circ H_\infty (x) = x + \rho, \]
as desired.
So in fact the idea of the proof is extremely simple, and all of the hard work is in proving some estimates.
Circle Diffeomorphisms
We consider a map \(\phi: \mathbb{R} \to \mathbb{R}\) defined by
\[ \phi(x) = x + \rho + \eta(x) \]
where \(\rho\) is its rotation number and \(\eta(x)\) is "small".
Define \(S_\sigma\) to be the strip \(\{ |\mathrm{Im} z|<\sigma\} \subset \mathbb{C}\) and let \(B_\sigma\) be the space of holomorphic functions bounded on \(S_\sigma\) with sup norm \(\|\cdot\|_\sigma\).
Goal: Show that if \(\|\eta\|_\sigma\) is sufficiently small, then there exists some diffeomorphism \(H(x)\) such that
\[ H^{-1} \circ \phi \circ H (x) = x + \rho \]
i.e. that \(\phi\) is conjugate to a pure rotation.
Linearization
The idea is that if \(\eta\) is small, then \(H\) should be close to the identity, so we suppose that
\[ H(x) = x + h(x) \]
where \(h(x)\) is small. Plugging this into the equation above and discarding higher order terms yields
\[ h(x+\rho) - h(x) = \eta(x) \]
Since \(\eta\) is periodic, we Fourier transform both sides to obtain an explicit formula for the Fourier coefficients of \(h(x)\). We have to show several things:
1. The Fourier series defining \(h(x)\) converges in some appropriate sense.
2. The function \(H(x) = x + h(x)\) is a diffeomorphism.
3. The composition \(\tilde{\phi} = H^{-1} \circ \phi \circ H\) is closer to a pure rotation than \(\phi\), in the sense that
\[ \tilde{\phi}(x) = x + \rho + \tilde{\eta}(x) \]
where \(\|\tilde{\eta}\| \ll \|\eta\|\).
Newton's Method
Carrying out the analysis, one finds that for appropriate epsilons and deltas, if \(\eta \in B_\sigma\) then \(H \in B_{\sigma - \delta}\) and that \(\|\tilde{\eta}\|_{\sigma-\delta} \leq C \|\eta\|_\sigma^2\). By carefully choosing the deltas, we can iterate this procedure (composing the \(H\)'s) to obtain a well-defined limit \(H_\infty \in B_{\sigma/2}\) such that\[ H_\infty^{-1} \circ \phi \circ H_\infty (x) = x + \rho, \]
as desired.
So in fact the idea of the proof is extremely simple, and all of the hard work is in proving some estimates.
Saturday, March 3, 2012
Gaussian Integrals: Wick's Theorem
We saw in the last update that the generating function \(Z[J]\) can be expressed as
\[ Z[J] = e^{\frac{1}{2} J \cdot A^{-1} J} \]
(at least as long as we've normalize things so that \(Z[0] = 1\). Now the wonderful thing is that this is something we can compute explicitly:
\[ Z[J] = \sum_{n = 0}^{\infty} \frac{(\frac{1}{2} A^{-1}_{ij} J^i J^j)^n}{n!}
= \sum_{n=0}^\infty \frac{(A^{-1}_{ij} J^i J^j)^n}{2^n n!} \]
For example, in the one-dimensional case (taking \(A = 1\)) we get
\[ Z[J] = \sum_{n=0}^\infty \frac{J^{2n}}{2^n n!} \]
On the other hand, by the definition of the generating function we have
\[ Z[J] = \sum_{n=0}^\infty \frac{\langle x^n \rangle}{n!} J^n \]
Comparing coefficients, we find
\[ \frac{\langle x^{2n} \rangle}{(2n)!} = \frac{1}{2^n n!} \]
so that
\[ \langle x^{2n} \rangle = \frac{(2n)!}{2^n n!}. \]
Let's give a combinatorial description. Given \(2n\) objects, in how many ways can we divide them into pairs? If we care about the order in which we pick the pairs, then we have
\[ {2n \choose 2}{2n - 2 \choose 2} \cdots {2n-(2n-2) \choose 2} = \frac{(2n)!}{2^n} \]
Of course, there are \(n!\) ways of ordering the \(n\) pairs, so after dividing by this (to account for the overcounting) we get exactly the expression for \(\langle x^{2n} \rangle\). This is the first case of Wick's theorem.
Now consider the general multidimensional case. Given \(I = (i_1, \cdots, i_{2n})\), we define a contraction to be
\[ \langle x^{j_1} x^{k_1} \rangle \cdots \langle x^{j_n} x^{k_n} \rangle \]
where \(j_1, k_1, \cdots, j_n, k_n\) is a choice of parition of \(I\) into pairs.
Theorem (Wick's theorem, Isserlis' theorem) The expectation value
\[ \langle x^{i_1} \cdots x^{i_{2n}} \rangle \]
is the sum over all full contractions. There are \((2n)!/ 2^n n!\) terms in the sum.
Proof This follows from our formula for the power series of the generating function. The reason is that the coefficient of \(J^I\) in \((\frac{1}{2} A^{-1}_{ij} J^i J^k)^n\) is exactly given by summing products of \(A^{-1}_{ij}\) over partitions of \(I\) into pairs, and the \(n!\) in the denominator takes care of the overcounting.
Next up: perturbation theory and Feynman diagrams.
\[ Z[J] = e^{\frac{1}{2} J \cdot A^{-1} J} \]
(at least as long as we've normalize things so that \(Z[0] = 1\). Now the wonderful thing is that this is something we can compute explicitly:
\[ Z[J] = \sum_{n = 0}^{\infty} \frac{(\frac{1}{2} A^{-1}_{ij} J^i J^j)^n}{n!}
= \sum_{n=0}^\infty \frac{(A^{-1}_{ij} J^i J^j)^n}{2^n n!} \]
For example, in the one-dimensional case (taking \(A = 1\)) we get
\[ Z[J] = \sum_{n=0}^\infty \frac{J^{2n}}{2^n n!} \]
On the other hand, by the definition of the generating function we have
\[ Z[J] = \sum_{n=0}^\infty \frac{\langle x^n \rangle}{n!} J^n \]
Comparing coefficients, we find
\[ \frac{\langle x^{2n} \rangle}{(2n)!} = \frac{1}{2^n n!} \]
so that
\[ \langle x^{2n} \rangle = \frac{(2n)!}{2^n n!}. \]
Let's give a combinatorial description. Given \(2n\) objects, in how many ways can we divide them into pairs? If we care about the order in which we pick the pairs, then we have
\[ {2n \choose 2}{2n - 2 \choose 2} \cdots {2n-(2n-2) \choose 2} = \frac{(2n)!}{2^n} \]
Of course, there are \(n!\) ways of ordering the \(n\) pairs, so after dividing by this (to account for the overcounting) we get exactly the expression for \(\langle x^{2n} \rangle\). This is the first case of Wick's theorem.
Now consider the general multidimensional case. Given \(I = (i_1, \cdots, i_{2n})\), we define a contraction to be
\[ \langle x^{j_1} x^{k_1} \rangle \cdots \langle x^{j_n} x^{k_n} \rangle \]
where \(j_1, k_1, \cdots, j_n, k_n\) is a choice of parition of \(I\) into pairs.
Theorem (Wick's theorem, Isserlis' theorem) The expectation value
\[ \langle x^{i_1} \cdots x^{i_{2n}} \rangle \]
is the sum over all full contractions. There are \((2n)!/ 2^n n!\) terms in the sum.
Proof This follows from our formula for the power series of the generating function. The reason is that the coefficient of \(J^I\) in \((\frac{1}{2} A^{-1}_{ij} J^i J^k)^n\) is exactly given by summing products of \(A^{-1}_{ij}\) over partitions of \(I\) into pairs, and the \(n!\) in the denominator takes care of the overcounting.
Next up: perturbation theory and Feynman diagrams.
Introduction to Gaussian Integrals
As a warm-up for more serious stuff, I'd like to discuss Gaussian integrals over \(\mathbb{R}^d\). Gaussian integrals are the main tool for perturbative quantum field theory, and I find that understanding Gaussian integrals in finite dimensions is an immense aid to understanding how perturbative QFT works. So let's get started.
The Basics
Let \(A\) be some \(d \times d\) symmetric positive definite matrix. We are interested in the integral
\[ \int_{-\infty}^\infty \exp(-\frac{x \cdot Ax}{2}) dx. \]
Out of laziness, I will suppress the limits of integration and just write this as
\[ \int e^{-S(x)} dx. \]
where \(S(x) = x \cdot Ax / 2\). Now for a function \(f(x)\), we define the expectation value \(\langle f(x) \rangle\) to be
\[ \langle f(x) \rangle_0 = \int f(x) e^{-S(x)} dx \]
Occasionally, we might care about the normalized expectation value
\[ langle f(x) \rangle = \frac{\langle f(x) \rangle_0}{\langle 1 \rangle_0} = \frac{1}{\langle 1 \rangle_0} \int f(x) e^{-S(x)} dx. \]
We mostly care about asymptotics, so we will typically think of a function \(f(x)\) as being a polynomial (or Taylor series). So what we're really interested in is
\[ \langle x^I \rangle = c\int x^I e^{-S(x)} dx, \]
where \(I\) is a multi-index.
The Partition Function
Let us define \(Z[J]\) by
\[ Z[J] = \int e^{-S(x) + J \cdot x} dx. \]
Now the great thing is that
\[ \langle x^I \rangle = \left. \frac{d^I}{dJ^I} \right|_{J = 0} Z[J], \]
so that once we know \(Z[J]\), we can calculate anything. So let's try to compute it. We have
\begin{align}
(Ax - J) \cdot A^{-1} (Ax - J) &= (Ax - J) \cdot (x - A^{-1} J) \\\
&= x \cdot Ax - x \cdot J - J \cdot x + J \cdot A^{-1} J \\\
&= x \cdot Ax - 2 x \cdot J + J \cdot A^{-1} J.
\end{align}
So we see that
\[ -\frac{1}{2} x \cdot A x + J \cdot x = \frac{1}{2} J \cdot A^{-1} J -\frac{1}{2} (x-A^{-1}J) \cdot A(x - A^{-1} J). \]
So, after a change of variales \(x \mapsto x - A^{-1} J\) we find
\[ Z[J] = e^{\frac{1}{2} J \cdot A^{-1} J} Z[0]. \]
Now the argument in the exponential is
\[ \frac{1}{2} A^{-1}_{ij} J^i J^j \]
So we find that
\[ \langle x^i x^j \rangle = \frac{d^2}{dx^i dx^j} Z[J]|_{J = 0} = A^{-1}_{ij}. \]
Now we are ready to prove Wick's theorem and discuss Feynman diagrams, which we'll do in the next post.
The Basics
Let \(A\) be some \(d \times d\) symmetric positive definite matrix. We are interested in the integral
\[ \int_{-\infty}^\infty \exp(-\frac{x \cdot Ax}{2}) dx. \]
Out of laziness, I will suppress the limits of integration and just write this as
\[ \int e^{-S(x)} dx. \]
where \(S(x) = x \cdot Ax / 2\). Now for a function \(f(x)\), we define the expectation value \(\langle f(x) \rangle\) to be
\[ \langle f(x) \rangle_0 = \int f(x) e^{-S(x)} dx \]
Occasionally, we might care about the normalized expectation value
\[ langle f(x) \rangle = \frac{\langle f(x) \rangle_0}{\langle 1 \rangle_0} = \frac{1}{\langle 1 \rangle_0} \int f(x) e^{-S(x)} dx. \]
We mostly care about asymptotics, so we will typically think of a function \(f(x)\) as being a polynomial (or Taylor series). So what we're really interested in is
\[ \langle x^I \rangle = c\int x^I e^{-S(x)} dx, \]
where \(I\) is a multi-index.
The Partition Function
Let us define \(Z[J]\) by
\[ Z[J] = \int e^{-S(x) + J \cdot x} dx. \]
Now the great thing is that
\[ \langle x^I \rangle = \left. \frac{d^I}{dJ^I} \right|_{J = 0} Z[J], \]
so that once we know \(Z[J]\), we can calculate anything. So let's try to compute it. We have
\begin{align}
(Ax - J) \cdot A^{-1} (Ax - J) &= (Ax - J) \cdot (x - A^{-1} J) \\\
&= x \cdot Ax - x \cdot J - J \cdot x + J \cdot A^{-1} J \\\
&= x \cdot Ax - 2 x \cdot J + J \cdot A^{-1} J.
\end{align}
So we see that
\[ -\frac{1}{2} x \cdot A x + J \cdot x = \frac{1}{2} J \cdot A^{-1} J -\frac{1}{2} (x-A^{-1}J) \cdot A(x - A^{-1} J). \]
So, after a change of variales \(x \mapsto x - A^{-1} J\) we find
\[ Z[J] = e^{\frac{1}{2} J \cdot A^{-1} J} Z[0]. \]
Now the argument in the exponential is
\[ \frac{1}{2} A^{-1}_{ij} J^i J^j \]
So we find that
\[ \langle x^i x^j \rangle = \frac{d^2}{dx^i dx^j} Z[J]|_{J = 0} = A^{-1}_{ij}. \]
Now we are ready to prove Wick's theorem and discuss Feynman diagrams, which we'll do in the next post.
Saturday, February 25, 2012
Geometry of Curved Spacetime 5: Bianchi Identity and Einstein Equations
Background
Following last time, we are almost ready to write down the Einstein equations. Before doing any math, let's understand what we're trying to do. Minkowski realized that Einstein's special relativity was best understood by combining space and time into 4-dimensional spacetime, with Lorentzian metric
\[ ds^2 = -dt^2 + dx^2 + dy^2 + dz^2. \]
The spacetime approach works wonderfully and even explains the Lorentz invariance of Maxwell's equations (indeed, it was Maxwell's equations that motivated Einstein to postulate his principle of relativity). However, (for reasons that I may discuss later) gravity is not a "force" but rather the geometry of spacetime itself.
By mass-energy equivalence (which is one of the most basic consequences of relativity), the gravitational field, whatever it is, must couple to the stress-energy tensor \(T_{ij}\). I won't get into details, but the stress-energy tensor is a familiar object from physics that roughly tells you what the energy-momentum density/flux is in each direction at every point in spacetime. If the matter is completely static, then it is ok to think of this as measuring the mass density, but for nonstatic matter it also takes things like pressure into account.
Now, as I said above, the gravitational field is just the geometry of spacetime, which is measured by the metric tensor \(g_{ij}\). Mass-energy equivalence says that it must couple to the stress-energy tensor \(T_{ij}\). The simplest field equation then would be
\[ G_{ij} = c T_{ij} \]
where \(G_{ij}\) is some tensor built out of \(g_{ij}\) and its derivatives, and \(c\) is some constant. The equations of Newtonian gravity are 2nd order in the gravitational field, so if we want these equations to reduce to Newton's in the appropriate limit, \(G_{ij}\) should only depend on the metric and its first two derivatives. Now there is an obvious 2nd rank tensor satisfying these constraints: \(R_{ij}\), the Ricci tensor. However, this turns out to be completely wrong (except in the vacuum).
Any reasonable matter will satisfy local energy-momentum conservation,
\[ \nabla_j T^{ij} = 0. \]
It turns out that the Ricci tensor does not satisfy this condition in general. So to look for the right tensor \(G_{ij}\), we turn to the Bianchi identity.
The Bianchi Identity
As discussed in the previous post, the curvature of a connection is the endomorphism-valued 2-form
\[ F = d\Omega - \Omega \wedge \Omega \]
where \(\Omega\) is the matrix of 1-forms telling us how to take the covariant derivative of a frame, i.e.
\[ \nabla_i e_j = \Omega_{ij} \otimes e_j. \]
Since a connection can be extended to all tensor powers in a natural way, we can consider the covariant derivative of the curvature \(F\) (thought of as a section of the appropriate bundle). Quick calcluation:
\begin{align}
\nabla F &= \nabla(d\Omega - \Omega \wedge \Omega) \\
&= d^2 \Omega - d\Omega \wedge \Omega + \Omega \wedge d\Omega \\
& \ \ + d\Omega \wedge \Omega - \Omega \wedge \Omega \wedge \Omega \\
& \ \ - \Omega \wedge d\Omega + \Omega \wedge \Omega \wedge \Omega \\
&= 0.
\end{align}
Thus the endomorphism valued 3-form \(\nabla F\) is identically 0. Writing \(F\) in components as \(R_{ijkl}\), this is equivalent to
\[ R_{ijkl|m} + R_{ijlm|k} + R_{ijmk|l} = 0. \]
Now let's contract:
\begin{align}
0 &= g^{ik} g^{jl} R_{ijkl|m} + g^{ik} g^{jl} R_{ijlm|k} + g^{ik} g^{jl} R_{ijmk|l}\\
&= g^{ik} R_{ik|m} - g^{ik}R_{im|k} - g^{jl} R_{jm|l} \\
&= \nabla_m S - 2 \nabla^k R_{mk}
\end{align}
So we see that the tensor
\[ G_{ij} = R_{ij} - \frac{S}{2} g_{ij} \]
is divergence free. This yields the Einstein field equations:
\[ R_{ij} - \frac{S}{2} g_{ij} = c T_{ij}. \]
Actually, there is another obvious divergence free tensor: \(g_{ij}\) itself. So a more general form is
\[ G_{ij} + \Lambda g_{ij} = c T_{ij} \]
where \(\Lambda\) is a constant called the cosmological constant.
Following last time, we are almost ready to write down the Einstein equations. Before doing any math, let's understand what we're trying to do. Minkowski realized that Einstein's special relativity was best understood by combining space and time into 4-dimensional spacetime, with Lorentzian metric
\[ ds^2 = -dt^2 + dx^2 + dy^2 + dz^2. \]
The spacetime approach works wonderfully and even explains the Lorentz invariance of Maxwell's equations (indeed, it was Maxwell's equations that motivated Einstein to postulate his principle of relativity). However, (for reasons that I may discuss later) gravity is not a "force" but rather the geometry of spacetime itself.
By mass-energy equivalence (which is one of the most basic consequences of relativity), the gravitational field, whatever it is, must couple to the stress-energy tensor \(T_{ij}\). I won't get into details, but the stress-energy tensor is a familiar object from physics that roughly tells you what the energy-momentum density/flux is in each direction at every point in spacetime. If the matter is completely static, then it is ok to think of this as measuring the mass density, but for nonstatic matter it also takes things like pressure into account.
Now, as I said above, the gravitational field is just the geometry of spacetime, which is measured by the metric tensor \(g_{ij}\). Mass-energy equivalence says that it must couple to the stress-energy tensor \(T_{ij}\). The simplest field equation then would be
\[ G_{ij} = c T_{ij} \]
where \(G_{ij}\) is some tensor built out of \(g_{ij}\) and its derivatives, and \(c\) is some constant. The equations of Newtonian gravity are 2nd order in the gravitational field, so if we want these equations to reduce to Newton's in the appropriate limit, \(G_{ij}\) should only depend on the metric and its first two derivatives. Now there is an obvious 2nd rank tensor satisfying these constraints: \(R_{ij}\), the Ricci tensor. However, this turns out to be completely wrong (except in the vacuum).
Any reasonable matter will satisfy local energy-momentum conservation,
\[ \nabla_j T^{ij} = 0. \]
It turns out that the Ricci tensor does not satisfy this condition in general. So to look for the right tensor \(G_{ij}\), we turn to the Bianchi identity.
The Bianchi Identity
As discussed in the previous post, the curvature of a connection is the endomorphism-valued 2-form
\[ F = d\Omega - \Omega \wedge \Omega \]
where \(\Omega\) is the matrix of 1-forms telling us how to take the covariant derivative of a frame, i.e.
\[ \nabla_i e_j = \Omega_{ij} \otimes e_j. \]
Since a connection can be extended to all tensor powers in a natural way, we can consider the covariant derivative of the curvature \(F\) (thought of as a section of the appropriate bundle). Quick calcluation:
\begin{align}
\nabla F &= \nabla(d\Omega - \Omega \wedge \Omega) \\
&= d^2 \Omega - d\Omega \wedge \Omega + \Omega \wedge d\Omega \\
& \ \ + d\Omega \wedge \Omega - \Omega \wedge \Omega \wedge \Omega \\
& \ \ - \Omega \wedge d\Omega + \Omega \wedge \Omega \wedge \Omega \\
&= 0.
\end{align}
Thus the endomorphism valued 3-form \(\nabla F\) is identically 0. Writing \(F\) in components as \(R_{ijkl}\), this is equivalent to
\[ R_{ijkl|m} + R_{ijlm|k} + R_{ijmk|l} = 0. \]
Now let's contract:
\begin{align}
0 &= g^{ik} g^{jl} R_{ijkl|m} + g^{ik} g^{jl} R_{ijlm|k} + g^{ik} g^{jl} R_{ijmk|l}\\
&= g^{ik} R_{ik|m} - g^{ik}R_{im|k} - g^{jl} R_{jm|l} \\
&= \nabla_m S - 2 \nabla^k R_{mk}
\end{align}
So we see that the tensor
\[ G_{ij} = R_{ij} - \frac{S}{2} g_{ij} \]
is divergence free. This yields the Einstein field equations:
\[ R_{ij} - \frac{S}{2} g_{ij} = c T_{ij}. \]
Actually, there is another obvious divergence free tensor: \(g_{ij}\) itself. So a more general form is
\[ G_{ij} + \Lambda g_{ij} = c T_{ij} \]
where \(\Lambda\) is a constant called the cosmological constant.
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