The Weyl and Wigner Transforms

Today I'd like to try to understand better how deformation quantization is related to the usual canonical quantization, and especially how the latter might be used to deduce the former, i.e., given an honest quantization (in the sense of operators), how might be reproduce the formula for the Moyal star product?

We'll fix our symplectic manifold once and for all to be \(\mathbb{R}^2\) with its standard symplectic structure, with Darboux coordinates \(x\) and \(p\). Let \(\mathcal{A}\) be the algebra of observables on \(\mathbb{R}^2\). For technical reasons, we'll restrict to those smooth functions that are polynomially bounded in the momentum coordinate (but of course the star product makes sense in general). Let \(\mathcal{D}\) be the algebra of pseudodifferential operators on \(\mathbb{R}\). We want to define a quantization map

\[ \Psi: \mathcal{A} \to \mathcal{D} \]

such that

\[ \Psi(x) = x \in \mathcal{D} \]

\[ \Psi(p) = -i\hbar \partial \]
Out of thin air, let us define
\[ \langle q| \Psi(f) |q' \rangle = \int e^{ik(q-q')} f(\frac{q+q'}{2}, k) dk \]
This is the Weyl transform. Its inverse is the Wigner transform, given by
\[ \Phi(A, q, k) = \int e^{-ikq'} \left\langle q+\frac{q'}{2} \right| A \left| q - \frac{q'}{2} \right\rangle dq' \]
Note: I am (intentionally) ignoring all factors of \(2\pi\) involved. It's not hard to work out what they are, but annoying to keep track of them in calculations, so I won't.

Theorem For suitably well-behaved \(f\), we have \( \Phi(\Psi(f)) = f\).

Proof Using the "ignore \(2\pi\)" conventions, we have the formal identities
\[ \int e^{ikx} dx = \delta(k), \ \ \int e^{ikx} dk = \delta(x). \]
The theorem is a formal result of these:
\begin{align} \Phi(\Psi(f))(q, k) &= \int e^{-ikq'} \left\langle q + \frac{q'}{2} \right| \Psi(f) \left| q - \frac{q'}{2} \right\rangle \\\
&= \int e^{-ikq'} e^{ik'q'} f(q, k) dk' dq' \\\
&= f(q,k).
\end{align}

One may easily check that \(\Psi(x) = x\) and \(Psi(k) = -i\partial\), so this certainly gives a quantization. But why is it particularly natural? To see this, let \(Q\) be the operator of multiplication by \(x\), and let \(P\) be the operator \(-i\partial\). We'd like to take \(f(q,p)\) and replace it by \(f(Q, P)\), but we can't literally substitute like this due to order ambiguity. However, we could work formally as follows:
\begin{align}
f(Q, P) &= \int \delta(Q-q) \delta(P - p) f(q,p) dq dp \\\
&= \int e^{ik(Q-q) + iq'(P-p)} f(q,p) dq dq' dp dk.
\end{align}
In this last expression, there is no order ambiguity in the argument of the exponential (since it is a sum and not a product), and furthermore the expression itself make sense since it is the exponential of a skew-adjoint operator. So let's check that this agrees with the Weyl transform. Using a special case of the Baker-Campbell-Hausdorff formula for the Heisenberg algebra, we have
\[ e^{ik(Q-q) + iq'(P-p)} = e^{ik(Q-q)} e^{iq'(P-p)} e^{-ikq'/2} \]
Let us compute the matrix element:
\begin{align}
\langle q_1 | P | q_2 \rangle &= \int \langle q_1 | p_1 \rangle
\langle p_1 | P | p_2 \rangle \langle p_2 | q_2 \rangle dp_1 dp_2 \\\
&= \int e^{iq_1p_1 - iq_2 p_2} p_2 \delta(p_2 - p_1) dp_1 dp_2 \\\
&= \int e^{i p(q_1-q_2)} p dp.
\end{align}
Hence we find that the matrix element for the exponential is
\begin{align} \langle q_1 |e^{ik(Q-q) + iq'(P-p)} | q_2 \rangle
&= e^{-ikq'/2 + ik(q_1-q)} \langle q_1 | e^{iq'(P-p)} | q_2 \rangle \\\
&=  \int e^{-ikq'/2 + ik(q_1-q) -iq'p} e^{iq'p'' + ip''(q_1-q_2)} dp'' \\\
&= \delta(q' + q_1 - q_2)  e^{-ikq'/2 + ik(q_1-q) -iq'p}
\end{align}
Plugging this back into the expression for \(f(Q, P)\) we find
\begin{align}
 \langle q_1 | f(Q. P) | q_2 \rangle &= \int \delta(q' + q_1 - q_2)  e^{-ikq'/2 + ik(q_1-q) -iq'p}
f(q,p) dq dq' dp dk \\\
&= \int  e^{ ik(q_1/2 +q_2/2-q) -ip(q_1-q_2)} f(q,p) dq dp dk \\\
&= \int e^{ip(q_1-q_2)} f(\frac{q_1+q_2}{2}, p) dp,
\end{align}
which is the original expression we gave for the Weyl transform.

The Moyal Product

Today I want to understand the Moyal product, as we will need to understand it in order to construct quantizations of symplectic quotients. (More precisely, to incorporate stability conditions.)



Let \(A\) be the algebra of polynomial functions on \(T^\ast \mathbb{C}^n\). This algebra has a natural Poisson bracket, given by
\[ \{p_i, x_j\} = \delta_{ij}. \]
We would like to define a new associative product \(\ast\) on \(A((\hbar))\) satisfying:

1. \(f  \ast g = fg + O(\hbar) \)
2. \(f \ast g - g \ast f = \hbar \{f, g\} + O(\hbar^2)\)
3. \(1 \ast f = f \ast 1 = f\)
4. \((f \ast g)^\ast = -g^\ast \ast f^\ast\)

In the last line, the map \((\cdot)^\ast\) takes \(x_i \mapsto x_i\) and \(p_i \mapsto -p_i\). To figure out what this new product should be, let's take \(f,g \in A\) and expand \(f \ast g\) in power series:

\[ f \ast g = \sum_{n=0}^\infty c_n(f,g) \hbar^n \]

Now, equations (1) and (2) will be satisfied by taking \(c_0(f,g) = fg\) and \(c_1(f,g) = \{f,g\}/2\). Let \(\sigma\) be the Poisson bivector defining the Poisson bracket. This defines a differential operator \(\Pi\) on \(A \otimes A\) by

\[ \Pi = \sigma^{ij} (\partial_i \otimes \partial_j) \]

Let \(B = \sum_{n=0}^\infty B_n \hbar^n\) and write the product as

\[ f \ast g = m \circ B(f \otimes g). \]
Now, condition (2) tells us that \(B(0) = 1\) and that
\[ \left. \frac{dB}{d\hbar} \right|_{\hbar=0} = \frac{\Pi}{2} \]
So
\[ B = 1 + \frac{\hbar \Pi}{2} + O(\hbar^2) \]
It is natural to guess that \(B\) should be built out of powers of \(\Pi\), and a natural guess is
\[ B = \exp(\frac{\hbar \Pi}{2}), \]
which certainly reproduces the first two terms of our expansion. Let's see that this choice actually works, i.e. defines an associative \(\ast\)-product. Let \(m: A \otimes A \to A\) be the multiplication, and
\(m_{12}, m_{23}: A \otimes A \otimes A \to A \otimes A\), \(m_{123}: A \otimes A \otimes A \to A\) the induced multiplication maps. Then
\begin{align}
f \ast (g \ast h) &= m \circ(B( f \otimes m \circ B(g \otimes h) ) ) \\\
&= m \circ B( m_{23} \circ (1 \otimes B)(f \otimes g \otimes h) ) \\\
&= m_{123} (B \otimes 1)(1 \otimes B)(f \otimes g \otimes h)
\end{align}
On the other hand, we have

\begin{align}
(f \ast g) \ast h) &= m \circ(B( m \circ B(f \otimes g) \otimes h) ) ) \\\
&= m \circ B( m_{12} \circ (B \otimes 1)(f \otimes g \otimes h) ) \\\
&= m_{123} (1 \otimes B)(B \otimes 1)(f \otimes g \otimes h)
\end{align}

Hence, associativity is the condition
\[ m_{123} \circ [1\otimes B, B \otimes 1] = 0. \]

On \(A \otimes A \otimes A\), write \(\partial_i^1\) for the partial derivative acting on the first factor, \(\partial_i^2\) on the second, etc. Then
\[ 1 \otimes B = \sum_n \frac {\hbar^n}{2^n n!}
 \Pi^{i_1 j_1} \cdots \Pi^{i_n j_n} \partial^2_{i_1} \partial^3_{j_1} \cdots
\partial^2_{i_n} \partial^3_{j_n} \]
and similarly for \(B \otimes 1\). So we have
\begin{align}

m_{123} (B\otimes 1)(1 \otimes B) &= \sum_n \sum_{k=0}^n \frac {\hbar^n}{2^n k! (n-k)!}

 \Pi^{k_1 l_1} \cdots \Pi^{k_k l_k} \partial_{k_1} \partial_{l_1} \cdots
\partial_{k_k} \partial_{l_k} \\\
 & \ \times  \Pi^{i_1 j_1} \cdots \Pi^{i_{n-k} j_{n-k}} \partial_{i_1} \partial_{j_1} \cdots

\partial_{i_{n-k}} \partial_{j_{n-k}} \\\
&= m_{123}(1 \otimes B)(B \otimes 1)
\end{align}
Hence we obtain an associative \(\ast\)-product. This is called Moyal product.

Sheafifying the Construction

Now suppose that \(U\) is a (Zariski) open subset of \(X = T^\ast \mathbb{C}^n\). Then the star product induces a well-defined map
\[ \ast: O_X(U)((\hbar)) \otimes_\mathbb{C} O_X(U)((\hbar)) \to O_X(U)((\hbar)) \]
In this way we obtain a sheaf \(\mathcal{D}\) of \(O_X\) modules with a non-commutative \(\ast\)-product defined as above.

Define a \(\mathbb{C}^\ast\) action on \(T^\ast \mathbb{C}^n\) by acting on \(x_i\) and \(p_i\) with weight 1. Extend this to an action on \(\mathcal{D}\) by acting on \(hbar\) with weight -1.

Proposition: The algebra \(C^\ast\)-invariant global sections of \(\mathcal{D}\) is naturally identified with the algebra of differential operators on \(\mathbb{C}^n\).

Proof: The \(\mathbb{C}^\ast\)-invariant global sections are generated by \(\hbar^{-1} x_i\) and \(\hbar^{-1} p_i\). So define a map \(\Gamma(\mathcal{D})^{\mathbb{C}^\ast} \to \mathbb{D}\) by
\[ \hbar^{-1} x_i \mapsto x_i \]
\[ \hbar^{-1} p_i \mapsto \partial_i \]
From the definition of the star product, it is clear that this is an algebra map, and that it is both injective and surjective.

An Exercise in Quantum Hamiltonian Reduction

Semiclassical Setup

Let the group \(GL(2)\) act on \(V = \mathrm{Mat}_{2\times n}\) and consider the induced symplectic action on \(T^\ast V\). If we use variables \((x,p)\) with \(x\) a \(2 \times n\) matrix and \(p\) an \(n \times 2\) matrix, then the classical moment map \(\mu\) is given by
\[ \mu(x,p) = xp \]
This is equivariant with respect to the adjoint action, so we can form the \(GL(2)\)-invariant functions
\[ Z_1 = \mathrm{Tr} \mu \]
\[ Z_2 = \mathrm{Tr} (\mu)^2 \]
If we think of \(x\) as being made of column vectors
\[ x = ( x_1 \cdots x_n ) \]
and similarly think of \(p\) as being made of row vectors, then there are actually many more \(GL(2)\) invariants, given by
\[ f_{ij} = \mathrm{Tr} x_i p_j = p_j x_i \]
In terms of the invariants, the \(Z\) functions are
\[ Z_1 = \sum_k f_{kk} \]
\[ Z_2 = \sum_{jk} f_{jk} f_{kj} \]
Let us compute Poisson brackets:
\begin{align}
 \{f_{ij}, f_{kl}\} &= \{p_j^\mu x_i^\mu, p_l^\nu x_k^\nu\} \\\
&= x_i^\mu p_l^\nu \delta_{jk} \delta^{\mu\nu} - p_j^\mu x_k^\nu \delta_{il} \delta^{\mu\nu} \\\
&= f_{il} \delta_{jk} - f_{kj} \delta_{il}.
\end{align}
So we see that the invariants form a Poisson subalgebra (as they should!). Let's compute:
\begin{align}
\{Z_1, f_{ij} &= \sum_k \{ f_{kk}, f_{ij} \} \\\
&= \sum_k \left( f_{kj} \delta_{ki} - f_{ik} \delta_{kj} \right) \\\
&= f_{ij} - f_{ij} = 0.
\end{align}
Hence \(Z_1\) is central with respect to the invariant functions \(f_{ij}\). Similarly,
\begin{align}
\{Z_2, f_{kl}\} &= \sum_{ij} \{f_{ij} f_{ji}, f_{kl}\} \\\
&= \sum_{ij} f_{ij} \left(f_{jl} \delta_{ik} - f_{ki} \delta_{jl} \right) + f_{ji} \left(f_{il} \delta_{jk} - f_{kj} \delta_{il} \right) \\\
&= \sum_j f_{kj} f_{jl} - \sum_i f_{il} f_{ki} + \sum_i f_{ki} f_{il} - \sum_j f_{jl} f_{kj} \\\
&= 0.
\end{align}
So we see that the \(Z_i\) are in the center of the invariant algebra. In fact, they generate it, so we'll denote by \(Z\) the algebra generated by \(Z_1, Z_2\). Let \(A\) be the algebra generated by the \(f_{ij}\). The inclusion \(Z \hookrightarrow A\) can be thought of as a purely algebraic version of the moment map. In particular, given any character \(\lambda: Z \to \mathbb{C}\), we can define the Hamiltonian reduction of \(A\) to be
\[ A_\lambda := A / A\langle \ker \lambda \rangle \]
The corresponding space is of course \(\mathrm{Spec} A\).

The Cartan Algebra and the Center

Define functions

\[ h_1 = Z_1 = \sum_i f_{ii} \]
\[ h_2 = Z_2 = \sum_{ij} f_{ij} f_{ji} \]
\[ h_3 = \sum_{ijk} f_{ij} f_{jk} f_{ki} \]
\[ h_k = \sum_{i_1, i_2, \ldots, i_k} f_{i_1 i_2} f_{i_2 i_3} \cdots f_{i_k i_1} \]

These are just the traces of various powers of the \(n \times n\) matrix \(px\). In particular, \(h_k\) for \(k>n\) may be expressed as a function of the \(h_i\) for \(i \leq n\). The algebra generated by the \(H\) plays the role of a Cartan subalgebra. So we have inclusions
\[ Z \subset H \subset A \]

Quantization

Now we wish to construct a quantization of \(A\) and \(A_\lambda\). The quantization of \(A\) is obvious: we quantize \(T^\ast V\) by taking the algebraic differential operators on \(V\). Denote this algebra by \(\mathbb{D}\). It is generated by \(x_i\) and (\partial_i\) satisfying the relation
\[ [\partial_i, x_j] = \delta_{ij} \]
Then we simply the subalgebra of \(GL(2)\)-invariant differential operators as our quantization of \(A\). Call this subalgebra \(U\). We can define Hamiltonian reduction analogously by taking central quotients. So we need to understand the center \(Z(U)\), but this is just the subalgebra generated by quantizations of \(Z_1\) and \(Z_2\), i.e. the subalgebra of all elements whose associated graded lies in \(Z(A)\).

More to come: stability conditions, \(\mathbb{D}\)-affineness, and maybe proofs of some of my claims.

A Toy Model for Effective Field Theory and Extra Dimensions

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.

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.

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.

Path Integrals 3: Recovering the Spectrum from Asymoptotics

In my previous posts on path integrals, I described (rather tersely) how the path integral, suitably defined and interpreted, can be used to compute the Schwartz kernel of the operators \(e^{iHt}\) (Lorentzian signature) and \(e^{-Ht}\) (Euclidean signature).

Suppose that we understand the spectrum of \(H\) completely (nb: for a given system described by \(H\), this is the goal). For example, suppose we know that the spectrum of \(H\) consists of discrete eigenvalues \(E_n, n = 0, \cdots\) with corresponding eigenvectors \(|n\rangle\),
\[ H|n\rangle = E_n|n\rangle. \]
(For simplicity, I assume there is no continuous spectrum and that the eigenvalues are nondegenerate.) Then we have
\[ e^{-iHt} = \sum_n e^{-i E_n t} \langle n|n\rangle \]
and
\[ e^{-Ht} = \sum_n e^{-E_n t} \langle n|n\rangle \]
Now the second expression turns out to be very useful. Assume the eigenvalues are ordered so that
\[ E_0 < E_1 < \cdots \]
Then we can write
\[ e^{-Ht} = e^{-E_0 t} |0\rangle + \sum_{n \geq 1} e^{-(E_n-E_0)t}|n\rangle \]
Now suppose that \(v\) is some vector which is close to the ground state, in the sense that
\[ \langle v|0\rangle \neq 0 \]
(This is obviously a generic condition, so if we just pick \(v\) randomly we can expect this to be true.) Then we can consider
\[ e^{-Ht} v = e^{-E_0 t} v_0 |0\rangle + \sum_{n \geq 1} e^{-(E_n-E_0)t} v_n|n\rangle \]
Now for \(n \geq 1\), \(E_n-E_0\) is strictly positive, and so for large \(t\) all of the higher terms are exponentially damped. So, we have the asymptotic
\[ e^{-Ht} v \sim e^{-E_0 t }v_0|0\rangle \]
Next comes the really interesting part. Multiply on the right by a position-representation eigenbra \(\langle x|\):
\[ \langle x|e^{-Ht} v \sim e^{-E_0 t} v_0 \langle x|0\rangle \]
Now \(v_0\) is an irrelevant constant, so we might as well take it to be 1 (rescale \(v\) as necessary). The expression \(\langle x|0\rangle\) is exactly the ground state wavefunction in the position representation! Call it \(\psi_0(x)\). So to conclude: the large-t asymptotic of the expression \(\langle x|e^{-Ht}v\) is (up to an overall constant) given by \(e^{-E_0 t} \psi_0(x)\), hence we can recover both the ground state energy and the ground state wavefunction. But the value of this expression is exactly given by the Euclidean path integral. So we have a correspondence:

Asymptotics of Euclidean path integral \(\leftarrow\rightarrow\) The spectrum of \(H\).

Coming next: instantons.

Path Integrals 2: Euclidean Path Integrals and Heat Kernels

Consider the heat equation
\[ \frac{\partial \psi}{\partial t} = -\hat{H} \psi \]
We find similarly
\[ \langle x_N|U_t|x_0\rangle = \int \exp \sum_{j=0}^{N-1} ik_j(x_j - x_{j+1}) -\Delta t H(x_j, k_j) dx dk. \]

Now take \(H(x,k) = k^2/2m + V(x)\) and complete the square:

\[ ik_j(x_j - x_{j+1}) - \Delta t k_j^2/2m - \Delta tV(x_j) + ik_{j+1}(x_{j+1} - x_{j+2}) - k_{j+1}^2/2m - V(x_{j+1}) \]

\begin{align}
 ik_j a_j - \Delta t k_j^2/2m &= -(-2mi k_j a_j / \Delta t + k_j^2) \Delta t/2m \\
&= -(k_j^2 -2mi k_j a_j/\Delta t -m^2 a_j^2/(\Delta t)^2 + m^2 a_j^2/(\Delta t)^2) \Delta t/2m \\
&= -(k_j -mia_j/\Delta t)^2 \Delta t/2m -ma_j^2/2\Delta t
\end{align}

Combining things together, we have that the heat kernel is given by
\[ \langle y|e^{-t \hat{H}}|x\rangle = \int e^{-S_{\textrm{euc}}} \mathcal{D}x \]

That is, Schwartz kernel of time evolution operator is given by the oscialltory Lorentzian signature path integral, whereas the heat kernel is given by the exponentially decaying path integral (better chance of being well-defined). Most importantly, the heat kernel contains most of the essential information about the spectrum of \(\hat{H}\), which is really all we need in order to understand the dynamics.

See ABC of Instantons. (I never understood the title of Nekrasov's "ABCD of Instantons" until I found this classic).

Path Integrals 1: Feynman's Derivation

Consider the Hilbert space \(\mathcal{H} = L^2(\mathbb{R})\) with Lebesgue measure and a Hamiltonian \(H = T(k) + V(x)\) (a sum of kinetic and potential energy). Then the quantum hamiltonian \(\hat{H}\) acts as

\begin{align}
(\hat{H}\psi)(y) &= \frac{1}{2\pi} \int e^{ik(y-x)} T(k) \psi(x) dx dk + V(y)\psi(y) \\
&= \frac{1}{2\pi}\int e^{ik(y-x)} H(k,x) \psi(x) dx dk \\
\end{align}

Now consider the Schrodinger equation
\[ \frac{\partial \psi}{\partial t} = i \hat{H} \psi. \]

We would like to obtain a formula for the solution operator \(U_t = e^{-i \hat{H} t}\). Let us consider its Schwartz kernel \(\langle{y}|U_t|{x}\rangle\). Let \(N\) be a large integer so that \(\Delta t = t/N\) is "small". Then we can write

\[ U_t = U_{\Delta t}^N, \]

Now consider a single term:

\begin{align}
 \langle{y}|U_{\Delta t}|x\rangle &\simeq \langle{y}|1-i\Delta{t}\hat{H}|x\rangle \\
&= \delta(y-x) -i \Delta t \langle x|\hat{H}|y\rangle \\
&= \frac{1}{2\pi}\int e^{ik(y-x)}( 1 - i \Delta t H(k,x)) dk \\
&\simeq \frac{1}{2\pi}\int e^{ik(y-x) - i \Delta t H(k,x)} dk\\
\end{align}
Now we have (taking \(x_0 = x\) and \(x_N = y\))
\begin{align}
\langle y|U_t|x\rangle &= \int dx_1 \cdots dx_{N-1} \\
& \ \times \langle x_N|U_{\Delta t}|x_{N-1}\rangle \cdots \langle x_1|U_{\Delta t}|x_0\rangle \\

&= \frac{1}{(2\pi)^N} \int dx_1 \cdots dx_{N-1} dk_0 \cdots dk_{N-1} \\
& \ \times \ e^{ik_N(x_N-x_{N-1}) - i \Delta t H(k_{N-1},x_{N-1})} \cdots
e^{-ik_N(x_1-x_0) - i \Delta t H(k_0,x_0)} \\
&= \frac{1}{(2\pi)^N} \int dx_1 \cdots dx_{N-1} dk_0 \cdots dk_{N-1} \\
& \ \times \ \exp \sum_{j=0}^{N-1} ik_j(x_{j+1} - x_j) - i \Delta t H(k_j,x_j)

\end{align}
Note this kind of integral seems like it should have a natural interpretation in symplectic or contact geometry, since the expression \(dxdk/(2\pi)^N\) is very nealy the Liouville measure. This is the most general form of the path integral.

Now assume that \(H(k,x) = k^2/2m + V(x)\). Then the \(k\)-dependent terms have the form
\[ \int e^{ik(y-x) - i\Delta t k^2/2m}. \]
Complete the square
\begin{align}
 ik(y-x) -i\Delta t k^2/2m &= -\frac{i \Delta t}{2m} (k^2 - \frac{2m}{\Delta t}k(y-x)) \\
&= -\frac{i \Delta t}{2m} (k^2 - \frac{2m}{\Delta t}k(y-x) + \frac{m^2}{\Delta t^2}(y-x)^2 -\frac{m^2}{\Delta t^2}(y-x)^2) \\
&= -\frac{i \Delta t}{2m}(k - \frac{m}{\Delta t}(y-x))^2 + \frac{i m}{2 \Delta t}(y-x)^2
\end{align}
Now, using that
\[ \int e^{-ak^2} = \sqrt{\frac{\pi}{a}} \]
We have
\[ \int e^{-\frac{i \Delta t}{2m}(k-\frac{m}{\Delta t}(y-x))^2} = \sqrt{\frac{2\pi m}{i \Delta t}} \]
Putting it altogether, we get the more familiar version of the path integral,
\begin{align} \langle y|U_t|x\rangle &\simeq C^N \int dx_1 \cdots dx_{N-1} \\
& \ \times \ \exp i\sum_{j=0}^{N-1} \frac{m}{2\Delta t}(x_{j+1}-x_j)^2 - V(x_j) \Delta t
\end{align}
where
\[ C = \frac{1}{2\pi} \sqrt{\frac{2 \pi m}{i \Delta t}} = \sqrt{\frac{m}{2\pi i \Delta t}} \]

Lattice Quantum Mechanics in 1D

For some reason I've been interested in lattice QFT recently, especially lattice gauge theory (note to self: a miniproject for the Christmas break is to understand the paper by Kogut and Susskind http://prd.aps.org/abstract/PRD/v11/i2/p395_1). As a warm-up, I thought I would try understanding plain-old 1D QM on the lattice, and writing some code to see if I got results that are at all reasonable.

The Setup: We will take as our space of states \(\mathscr{H} = L^2([0,1])\) and hamiltonian
$$
H = -\frac{d^2}{dx^2} + V(x).
$$
for some real function \(V(x)\) defined on \([0,1]\).

Now fix some large positive integer \(N\). Let \(\epsilon = 1/N\). We will consider the subspace \(\mathscr{H}_N\) of \(\mathscr{H}\) spanned by those functions that are constant on the subintervals \(i\epsilon, (i+1)\epsilon)\). Such a function is defined (a.e.) by the \(N\) values it takes on these intervals, so we may identify \(\mathscr{H}_N \cong \mathbb{C}^N\) as vector spaces. Let us denote elements of \(\mathscr{H}_N\) by \(\psi_i\) for \(i = 0, \ldots, N-1\). Thinking of these as the values of a function \(\psi(x)\) at \(x = i\epsilon\), we see that the inner product on \(\mathscr{H}_N\) is given by
$$
(\phi, \psi) = \int_0^1 \bar{\phi}(x) \psi(x) dx = \sum_{i=0}^{N-1} \bar{\phi}_i \psi(i)
$$
So we see that with these identifications, \(\mathscr{H}_N\) is just \(\mathbb{C}^N\) with the usual hermitian inner product.

Now, we can approximate \(d/dx\) with the forward and backward difference operators
\begin{align}
(D_+ \psi)_i &= \frac{\psi_{i+1} - \psi_i}{\epsilon} \\
(D_-\psi)_i &= \frac{\psi_i - \psi_{i-1}}{\epsilon}
\end{align}
Note: throughout I will assume periodic boundary conditions to make life easy. In this case we have \(\psi_{i+N} = \psi_i\).
Now consider
\begin{align}
(D_+\phi, \psi) &= \epsilon^{-1} \sum_i \left( \bar{\phi}_{i+1}\psi_i - \bar{\phi}_i \psi_i \right) \\
&= \epsilon^{-1} \sum_i \left( \bar{\phi}_i \psi_{i-1} - \bar{\phi}_i \psi_i \right) \\
&= -(\phi, D_-\psi).
\end{align}
Thus we have \(D_+^\ast = -D_-\). Similarly, \(D_-^\ast = -D_+\). Then the operator \(D = (D_+ + D_-)/2\) approximates \(d/dx\) and satisfies \(D^\ast = -D\) (as it should!), and the discrete Laplacian \(D^2\) is self-adjoint.

Finally, we can form the discrete hamiltonian \(H_N\) by taking \(H_N = -D^2 + \hat{V}\), where \(\hat{V}\) is the operator \(\psi_i \mapsto V(x_i) \psi_i\), where \(x_i = i\epsilon\).

Note: typically one further imposes Dirichlet or Neumann boundary conditions. This corresponds to projecting to a smaller subspace of \(\mathscr{H}_N\).

I wrote some Sage code to test this. With \(V(x) = 0\), and \(N = 500\), here is one of the lowest-energy states:

Rather encouraging.