Semiclassical Quantization of Classical Field Theories

Abstract

These lectures are an introduction to formal semiclassical quantization of classical field theory. First we develop the Hamiltonian formalism for classical field theories on space time with boundary. It does not have to be a cylinder as in the usual Hamiltonian framework. Then we outline formal semiclassical quantization in the finite dimensional case. Towards the end we give an example of such a quantization in the case of Abelian Chern-Simons theory.

Appendix

1.1 A Discrete Time Quantum Mechanics

An example of a finite dimensional version of a classical field theory is a discrete time approximation to the Hamiltonian classical mechanics of a free particle on $\mathbb {R}$. We denote coordinates on this space $(p,q)$ where $p$ represents the momentum and $q$ represents the coordinate of the system.

In this case the space time is an ordered collection of $n$ points which represent the discrete time interval. If we enumerate these points $\{1,\ldots ,n\}$ the points $1,n$ represent the boundary of the space time. The space of fields is $\mathbb {R}^{n-1}\times \mathbb {R}^n$ with coordinates $p_i$ where $i=1,\ldots , n-1$ represents the “time interval” between points $i$ and $i+1$ and $q_i$ where $i=1,\ldots , n$. The coordinates $p_1, p_{n-1}, q_1, q_n$ are boundary fields.¹⁵ The action is

$$ S=\sum \limits _{i=1}^{n-1} p_i(q_{i+1}-q_i)-\sum \limits _{i=1}^{n-1} \frac{p_i^2}{2} $$

We have

$$ dS=\sum \limits _{i=1}^{n-2} (q_{i+1}-q_i-p_i)\,dp_i+\sum \limits _{i=2}^{n-1}(p_{i-1}-p_i)\,dq_i+ p_{n-1}dq_n- p_1dq_1 $$

From here we derive the Euler-Lagrange equations

$$ q_{i+1}-q_i=p_i, \quad i=1,\ldots n-1, $$

$$ p_{i-1}-p_i=0, \quad i=2,\ldots , n-1 $$

and the boundary 1-form

$$ \alpha =p_{n-1}dq_n-p_1dq_1 $$

This gives the symplectic structure on the space of boundary fields with

$$ \omega _\partial = dp_{n-1}\wedge dq_n-dp_1\wedge dq_1 $$

The boundary values of solutions of the Euler-Lagrange equations define the subspace

$$ L=\pi (EL)=\{(p_1,q_1,p_{n-1},q_n)| p_1=p_{n-1}, q_n=q_1+(n-1) p_1\} $$

It is clear that this a Lagrangian subspace.

1.2 B Feynman Diagrams

Let $\varGamma$ be a graph with vertices of valency $\ge$3 with one special vertex which may also have valency $0,1,2$. We define the weight $F_c(\varGamma )$ as follows.

Fig. 3

The “theta” diagram

A state on $\varGamma$ is a map from the set of half-edges of $\varGamma$ to the set $1,\ldots , n$, for an example see Fig. 3. The weight of $\varGamma$ is defined as

$$ F_c(\varGamma )=\sum \limits _{states} \left( \frac{\partial ^l v}{\partial x^{j_1},\ldots ,\partial x^{j_l}}(c) \prod _{vertices} \frac{\partial ^k S}{\partial x^{i_1},\ldots ,\partial x^{i_k}}(c)\prod _{edges} (B_c^{-1})_{ij}\right) $$

Here the sum is taken over all states on $\varGamma$, and $i_1,\ldots , i_k$ are states on the half-edges incident to a vertex. The first factor is the weight of the special vertex where $v$ is the density of the integration measure in local coordinates $\frac{dx}{db}=v(x) dx^1,\ldots ,dx^N$. The pair $(i,j)$ is the pair of states at the half-edges comprising an edge. Note that weights of vertices and the matrix $B_c$ are symmetric. This makes the definition meaningful.

1.3 C Gauge Fixing in Maxwell’s Electromagnetism

In the special case of electromagnetism ($G = \mathbb {R}, \mathfrak {g} = \mathbb {R}$), the space of fields is $F_M = \varOmega ^1(M) \oplus \varOmega ^{n-2}(M)$ and similarly for the boundary. If $M$ has no boundary, the gauge group $G_M = \varOmega ^0(M)$ acts on fields as follows: $A \mapsto A + d \alpha , B \mapsto B$. We can construct a global section of the corresponding quotient using Hodge decomposition: we know that

$$\begin{aligned} \varOmega ^{\bullet }(M) \cong \varOmega _{\text {exact}}^{\bullet }(M) \oplus H^{\bullet }(M) \oplus \varOmega _{\text {coexact}}^{\bullet }(M) \end{aligned}$$

(69)

where the middle term consists of harmonic forms. In particular,

$$\begin{aligned} \varOmega ^1(M) = d \varOmega ^0(M) \oplus H^1(M) \oplus d^{*} \varOmega ^2(M) \end{aligned}$$

(70)

where the last two terms give a global section. In physics, choosing a global section is called gauge fixing, and this particular choice of gauge is called the Lorentz gauge, where $d^{*} A = 0$.

1.4 D Hodge Decomposition for Riemannian Manifolds With Boundary

1.4.1 D.1 Hodge Decomposition With Dirichlet and Neumann Boundary Conditions

Let $M$ be a smooth oriented Riemannian manifold with boundary $\partial M$. Recall some basic facts about the Hodge decomposition of differential forms on $M$. Choose local coordinates near the boundary in which the metric has the product structure with $t$ being the coordinate in the normal direction. Near the boundary any smooth form can be written as

$$ \omega =\omega _{tan}+\omega _{norm}\wedge dt $$

where $\omega _{tan}$ is the tangent component of $\omega$ near the boundary and $\omega _{norm}$ is the normal component.

We will denote by $\varOmega _D(M)$ the space of forms satisfying the Dirichlet boundary conditions $\iota ^*(\omega )=0$ where $\iota ^*$ is the pull-back of the form $\omega$ to the boundary. This condition can be also written as $\omega _{tan}=0$.

We will denote by $\varOmega _N(M)$ the space of forms satisfying the Neumann boundary conditions $\iota ^*(*\omega )=0$. Here $*: \varOmega ^i(M)\rightarrow \varOmega ^{n-i}(M)$ is the Hodge star operation, recall that $*^2=(-1)^{i(n-i)}\mathrm {id}$ on $\varOmega ^i(M)$. Because $\omega _{norm}=*^{\prime }\iota ^*(*\omega )$ the Neumann boundary condition can be written as $\omega _{norm}=0$.

Denote by $d^*=(-1)^i{*}^{-1} d *$ the formal adjoint of $d$, and by $\varDelta =dd^*+d^* d$ the Laplacian on $M$. Denote by $\varOmega _{cl}(M)$ closed forms on $M$, $\varOmega _{ex}(M)$ exact forms on, $\varOmega _{cocl}(M)$ the space of coclosed forms, i.e. closed with respect to $d^*$ and by $\varOmega _{coex}(M)$ the space of coexact forms.

Define subspaces:

$$ \varOmega _{cl,cocl}(M)=\varOmega _{cl}(M)\cap \varOmega _{cocl}(M), \ \ \varOmega _{cl,coex}(M)=\varOmega _{cl}(M)\cap \varOmega _{coex}(M) $$

and similarly $\varOmega _{ex,cocl}(M)$, $\varOmega _{cl,cocl, N}(M)$ and $\varOmega _{cl,cocl, D}(M)$.

Theorem 3

• (1) The space of forms decomposes as

  $$ \varOmega (M)=d^* \varOmega _N(M)\oplus \varOmega _{cl,cocl}(M) \oplus d\varOmega _D(M) $$

• (2) The space of closed, coclosed forms decomposes as

  $$ \varOmega _{cl, cocl}(M)=\varOmega _{cl,cocl, N}(M)\oplus \varOmega _{ex,cocl}(M) $$
  $$ \varOmega _{cl, cocl}(M)=\varOmega _{cl,cocl, D}(M)\oplus \varOmega _{cl,coex}(M) $$

We will only outline the proof of this theorem. For more details and references on the Hodge decomposition for manifolds with boundary and Dirichlet and Neumann boundary conditions see [17]. Riemannian structure on $M$ induces the scalar product on forms

$$\begin{aligned} (\omega , \omega ^{\prime })=\int \limits _M \omega \wedge *\omega ^{\prime } \end{aligned}$$

(71)

For two forms of the same degree we have $\omega (x)\wedge *\omega ^{\prime }(x)= \langle \omega (x),\omega ^{\prime }(x)\rangle \,dx$ where $dx$ is the Riemannian volume form and $\langle .,. \rangle$ is the scalar product on $\wedge ^kT^*_xM$ induced by the metric. This is why (71) is positive definite.

Lemma 1

With respect to the scalar product (71)

$$ (d\varOmega _D(M))^\perp =\varOmega _{cocl} $$

Proof

By the Stokes theorem for any form $\theta \in \varOmega ^{i-1}_D(M)$ we have

$$ (\omega , d\theta )=\int \limits _M \omega \wedge *d\theta = (-1)^{(i+1)(n-i)}(\int \limits _{\partial M} \iota ^*(*\omega )\wedge \iota ^*(\theta )+ \int \limits _M d*\omega \wedge \theta ) $$

The boundary integral is zero because $\theta \in \varOmega _D(M)$. Thus $(\omega , d\theta )=0$ for all $\theta$ if and only if $d*\omega =0$ which is equivalent to $\omega \in \varOmega _{cocl}(M)$.

Corollary 1

Because $d\varOmega _D(M)\subset \varOmega _{cl}(M)$, we have $\varOmega _{cl}(M)=\varOmega _{cl}(M)\cap (d\varOmega _D(M))^\perp \oplus d\varOmega _D(M)$. i.e.

$$ \varOmega _{cl}(M)=\varOmega _{cl,cocl}(M)\oplus d\varOmega _D(M) $$

Here we are sketchy on the analytical side of the story. If $U\subset V$ is a subspace in an inner product space, in the infinite dimensional setting more analysis might be required to prove that $V=U\oplus U^\perp$. Here and below we just assume that this does not create problems. Similarly to Lemma 1 we obtain

$$ (d^*\varOmega _N(M))^\perp =\varOmega _{cl}(M) $$

This completes the sketch of the proof of the first part. The proof of the second part is similar.

Note that the spaces in the second part of the theorem are harmonic forms representing cohomology classes:

$$ \varOmega _{cl,cocl, N}(M)=H(M), \ \ \varOmega _{cl,cocl, D}(M)=H(M,\partial M) $$

1.4.2 D.2 More General Boundary Conditions

1.4.3 D.2.1 General setup

Assume that $M$ is a smooth compact Riemannian manifold, possibly with non-empty boundary $\partial M$. Let $\pi : \varOmega ^i(M)\rightarrow \varOmega ^i(\partial M)$, $i=0,\ldots , n-1$ be the restriction map (the pull-back of a form to the boundary) and $\pi (\varOmega ^n(M))=0$.

The Riemannian structure on $M$ induces the metric on $\partial M$. Denote by $*$ the Hodge star for $M$, and by $*_\partial$ the Hodge star for the boundary $*_\partial : \varOmega ^i(\partial M)\rightarrow \varOmega ^{n-1-i}(\partial M)$. Define the map $\widetilde{\pi }: \varOmega (M)\rightarrow \varOmega (\partial M)$, $i=1,\ldots , n$ as the composition $\widetilde{\pi }(\alpha )=*_\partial \pi (*\alpha )$. Note that $\tilde{\pi }(\varOmega ^0(M))=0$.

Denote by $\varOmega _D(M, L)$ and $\varOmega _N(M,L)$ the following subspaces:

$$ \varOmega _D(M, L)=\pi ^{-1}(L), \ \ \varOmega _N(M, L)=\widetilde{\pi }^{-1}(L) $$

where $L\subset \varOmega (\partial M)$ is a subspace.

Denote by $L^\perp$ the orthogonal complement to $L$ with respect to the Hodge inner product on the boundary. The following is clear:

Lemma 2

$$ (*L^{(i)})^\perp =*(L^{(i)})^\perp , *(L^\perp )=L^{sort} $$

Here $L^{sort}$ is the space which is symplectic orthogonal to $L$.

Proposition 3

$(d^*\varOmega _N(M,L))^\perp =\varOmega _D(M,L^\perp )_{cl}$

Proof

Let $\omega$ be an $i$-form on $M$ such that

$$ \int \limits _M \omega \wedge d*\alpha =0 $$

for any $\alpha$. Applying Stocks theorem we obtain

$$ \int \limits _M \omega \wedge d*\alpha = (-1)^i \int \limits _{\partial M} \pi (\omega )\wedge *_\partial \tilde{\pi }(\alpha )+ (-1)^{i+1} \int \limits _M d\omega \wedge *\alpha $$

The boundary integral is zero for any $\alpha$ if and only if $\pi (\omega )\in L^\perp$ and the bulk integral is zero for any $\alpha$ if and only if $d\omega =0$.

As a corollary of this we have the orthogonal decomposition

$$ \varOmega (M)=\varOmega _D(M,L^\perp )_{cl}\oplus d^*\varOmega _N(M,L) $$

Similarly, for each subspace $L\subset \varOmega (\partial M)$ we have the decomposition

$$ \varOmega (M)=\varOmega _N(M,L^\perp )_{cocl}\oplus d\varOmega _D(M,L) $$

Now, assume that we have two subspaces $L, L_1\subset \varOmega (\partial M)$ such that

$$\begin{aligned} d_\partial (L_1^{\perp })\subset L^\perp , \end{aligned}$$

(72)

Note that this implies $d^*_\partial L\subset L_1$. Indeed, fix $\alpha \in L$, then (72) implies that for any $\beta \in L_1^\perp$ we have

$$ \int \limits _{\partial M} \alpha \wedge *d_\partial \beta =0 $$

This is possible if and only if

$$ \int \limits _{\partial M} *d_{\partial } *\alpha \wedge *\beta =0 $$

Thus, $d_\partial ^*\alpha \in L_1$. Here we assumed that $(L_1^\perp )^\perp =L_1$.

Because $\pi d=d_\partial \pi$ and $\tilde{\pi } d^*=d^*_\partial \tilde{\pi }$ we also have

$$ d\varOmega _D(M,L_1^\perp )\subset \varOmega _D(M,L^\perp )_{cl}, \ \ d^*\varOmega _N(M,L)\subset \varOmega _N(M,L_1)_{cocl} $$

Theorem 4

Under assumption (72) we have

$$\begin{aligned} \varOmega (M)=d^*\varOmega _N(M,L)\oplus \varOmega _D(M,L^\perp )_{cl}\cap \varOmega _N(M,L_1)_{cocl}\oplus d\varOmega _D(M,L_1^\perp ) \end{aligned}$$

(73)

Indeed, if $V, W\subset \varOmega$ are liner subspaces in the scalar product space $\varOmega$ such that $W\subset V^\perp$ and $V\subset W^\perp$ then $\varOmega =V\oplus V^\perp =W\oplus W^\perp$ and

$$ \varOmega =V\oplus W^\perp \cap V^\perp \oplus W $$

We will call the identity (73) the Hodge decomposition with boundary conditions. The following is clear:

Theorem 5

The decomposition (73) agrees with the Hodge star operation if and only if

$$ *L_1^\perp =L $$

Remark 13

In the particular case $L=\{0\}$ and $L_1^\perp =\{0\}$ we obtain the decomposition from the previous section:

$$ \varOmega (M)=d^*\varOmega _N(M)\oplus \varOmega _{cl,cocl}(M)\oplus d\varOmega _D(M) $$

Lemma 3

If $L\subset \varOmega (\partial M)$ is an isotropic subspace then $*L\subset \varOmega (\partial M)$ is also an isotropic subspace.

Indeed, if $L$ is isotropic then for any $\alpha , \beta \in L$ we have $\int \limits _{\partial M} \alpha \wedge *\beta =0$, but

$$ \int \limits _{\partial M} *\alpha \wedge *^2\beta =\pm \int \limits _{\partial M} \alpha \wedge *\beta $$

therefore $*L$ is also isotropic.

Remark 14

We have

$$ *\varOmega _N(M)=\varOmega _D(M), \ \ *H(M)=H(M,\partial M) $$

In the second formula $H(M)$ is the space of closed-coclosed forms with Neumann boundary conditions and $H(M, \partial M)$ is the space of closed-coclosed forms with Dirichlet boundary conditions. They are naturally isomorphic to corresponding cohomology spaces. Note that as a consequence of the first identity we have $*d^*\varOmega _N(M)=d\varOmega _D(M)$. We also have more general identity

$$ *\varOmega _N(M, L)=\varOmega _D(M, *_\partial L) $$

and consequently $*\varOmega _D(M,L)=\varOmega _N(M,*_\partial L)$.

Let $\pi$ and $\tilde{\pi }$ be maps defined at the beginning of this section. Because $\pi$ commutes with de Rham differential and $\tilde{\pi }$ commutes with its Hodge dual, we have the following proposition

Proposition 4

Let $H_M(\partial M)$ be the space of harmonic forms on $\partial M$ extendable to closed forms on $M$, then

$$ \pi (\varOmega _{cl}(M))=H_M(\partial M)\oplus d\varOmega (\partial M), \ \ \widetilde{\pi }(\varOmega _{cocl}(M))=H_M(\partial M)^\perp \oplus d^*\varOmega (\partial M) $$

Here is an outline of the proof. Indeed, let $\theta \in \varOmega _{cl}(M)$ and $\sigma \in \varOmega _{cocl}(M)$. Then

$$ \int \limits _{\partial M} \pi (\theta )\wedge *_\partial \widetilde{\pi }(\sigma )= \int \limits _{\partial M} \pi (\theta )\wedge \pi (*\sigma )= \int \limits _M d(\theta \wedge *\sigma ) $$

The last expression is zero because by the assumption $\theta$ and $*\sigma$ are closed. The proposition follows now from the Hodge decomposition for forms on the boundary and from $\pi (\varOmega _{cl}(M))\subset \varOmega _{cl}(\partial M)$, $\widetilde{\pi }(\varOmega _{cocl}(M))\subset \varOmega _{cocl}(\partial M)$.

D.2.2 $\varvec{\dim M=3}$

Let us look in details at the 3-dimensional case. In order to have the Hodge decomposition with boundary conditions we required

$$ d L_{1}^{\perp } \subset L^{\perp } $$

If we want it to be invariant with respect to the Hodge star we should also have $*L_{1}^{\perp } = L$. Together these two conditions imply that $L$ should satisfy $d*L\subset L^\perp$ or

$$ \int \limits _{\partial M} d*\alpha \wedge *\beta =0 $$

for any $\alpha , \beta \in L$. This condition is equivalent to

$$ \int \limits _{\partial M} d^* \alpha \wedge \beta =0 $$

for any $\alpha \in L^{(1)}$ and any $\beta \in L^{(2)}$.

Note that if $L^{(2)}=\{0\}$ we have no conditions on the subspace $L^{(1)}$. In this case for any choice of $L^{(0)}$ and $L^{(1)}$ the $*$-invariant Hodge decomposition is:

$$ \varOmega ^0(M)=d^*\varOmega ^1_N(M,L^{(0)})\oplus \varOmega ^0_D(M, {L^{(0)}}^\perp )_{cl} $$

$$ \varOmega ^1(M)=d^*\varOmega ^2_N(M,L^{(1)})\oplus \varOmega ^1_D(M, {L^{(1)}}^\perp )_{cl}\cap \varOmega ^1_N(M, L^{(0)})_{cocl}\oplus d\varOmega _D^0(M) $$

Here we used $\varOmega _N^i(M,L_1)=\varOmega ^i_N(M, L_1^{(i-1)})=\varOmega ^i_N(M, (*L^{(3-i)})^\perp )$. The condition $L^{(2)}=\{0\}$ implies that $\varOmega ^1_N(M,(*L^{(2)})^\perp )= \varOmega ^1(M)$. We also used $\varOmega ^0_D(M, L^\perp _1)=\varOmega ^0(M, *L^{(2)})=\varOmega ^0_D(M)$.

The decomposition of $2$- and $3$-forms is the result of application of Hodge star to these formulae.

1.4.4 D.2.3 The gauge-fixing subspace

Consider the bilinear form

$$\begin{aligned} B(\alpha , \beta )=\int \limits _M \beta \wedge d\alpha \end{aligned}$$

(74)

on the space $\varOmega ^\bullet (M)$.

Let $I\subset \varOmega ^\bullet (\partial M)$ be an isotropic subspace.

Proposition 5

The form $B$ is symmetric on the space $\varOmega _D(M,I)$.

Indeed

$$ \int \limits _M(\beta \wedge d\alpha )=(-1)^{|\beta |+1}\int \limits _{\partial M}\pi (\beta )\wedge \pi (\alpha )+ \int \limits _M d\beta \wedge \alpha =(-1)^{(|\alpha |+1)(|\beta |+1)}B(\alpha , \beta ) $$

The boundary term vanishes because boundary values of $\alpha$ and $\beta$ are in an isotropic subspace $I$.

Proposition 6

Let $I\subset \varOmega (\partial M)$ be an isotropic subspace, then $B$ is nondegenerate on $d^*\varOmega _N(M,I^\perp )\cap \varOmega _D(M,I)$.

Proof

If $I$ is isotropic, $\beta \in \varOmega _D(M,I)$ and $B(\beta , \alpha )=0$ for any $\alpha \in \varOmega _D(M,I)$, we have:

$$ B(\beta , \alpha )=B(\alpha , \beta )=\int \limits _M \alpha \wedge d\beta $$

and therefore $d\beta =0$. Therefore $\varOmega _D(M,I)_{cl}$ is the kernel of the form $B$ on $\varOmega _D(M,I)$. But we have the decomposition

$$ \varOmega (M)=\varOmega _D(M,I)_{cl}\oplus d^*\varOmega _N(M,I^\perp ) $$

This implies

$$ \varOmega _D(M,I)= \varOmega _D(M,I)_{cl}\oplus d^*\varOmega _N(M,I^\perp )\cap \varOmega _D(M,I) $$

This proves the statement.

In particular, the restriction of the bilinear form $B$ is nondegenerate on $\Lambda _I=d^*\varOmega _N^2(M,{I^{(1)}}^\perp )\,\cap \,\varOmega _D^1(M,I^{(1)})$. For the space of all 1-forms with boundary values in $I^{(1)}$ we have:

$$ \varOmega ^1_D(M,I^{(1)})= \varOmega ^1_D(M,I^{(1)})_{cl}\oplus d^*\varOmega ^2_N(M,{I^{(1)}}^\perp )\cap \varOmega _D^1(M,I^{(1)}) $$

The first part is the space of solutions to the Euler-Lagrange equations with boundary values in $I^{(1)}$.