• #PDF
• #Path integral formulation of gauge theories ---
• #Hamiltonian formulation of gauge theories ---
• #Canonical momenta and Poisson brackets ---
• [[#The difficulty -- \(\Pi_0 = 0\) ---]]
• #Non uniqueness of the Hamiltonian ---
• #Consistency -- the secondary constraint ---
• #Dirac theory of constraints ---
• #The final Hamiltonian and Gauss' law ---

Prof. Sachindeo Vaidya (CHEP, IISc) | PDF

Previous: Lecture 19 | Next: Lecture 21

Recall the path integral in QM:

\[ \left\langle x',t'\,\middle|\,x,t\right\rangle \;=\; \left\langle x'\right|e^{-iH(t'-t)}\left|x\right\rangle \;\sim\; \text{propagator} \;=\; G(x',t';x,t) \]

Take \(T = t'-t\):

\[ \left\langle x',t'\,\middle|\,x,t\right\rangle \;=\; \left\langle x'\right|e^{-iHT}\left|x\right\rangle \]

We wrote

\[ e^{-iHT} \;=\; e^{-iH\Delta t}\,e^{-iH\Delta t}\cdots e^{-iH\Delta t} \]

and introduced a complete set of states at the time splitting points. We arrive at the formula
for the propagator as a sum over paths.

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In QFT we need to sum over different field configurations; also we need to use the Hamiltonian
of the field to get the correct PI formulation.

Look at QED (an abelian gauge theory with \(U(1)\) symmetry, i.e. local gauge invariance):

\[ \mathcal{L} \;=\; -\frac{1}{4}F_{\mu\nu}F^{\mu\nu} \ ,\qquad F_{\mu\nu} \;=\; \partial_\mu A_\nu - \partial_\nu A_\mu \] \[ \mathcal{L} \;=\; -\frac{1}{4}\left(F_{0\nu}F^{0\nu} + F_{i\nu}F^{i\nu}\right) \;=\; -\frac{1}{4}\left(F_{0i}F^{0i} + F_{i0}F^{i0} + F_{ij}F^{ij}\right) \] \[ \;=\; -\frac{1}{4}\left(2F_{0i}F^{0i} + F_{ij}F^{ij}\right) \;=\; \frac{1}{2}\left(E^2 - B^2\right) \]

Canonical momentum density:

\[ \Pi_\mu \;=\; \frac{\partial\mathcal{L}}{\partial\dot{A}^{\mu}} \;=\; \frac{\partial\mathcal{L}}{\partial\left(\partial_0 A^{\mu}\right)} \]

Postulate the Poisson bracket. We know \(\{x,p\} = 1\); we impose \(\{A_\mu,\Pi_\mu\} = 1\), i.e.

\[ \left\{A_\mu(\vec{x},t),\ \Pi_\nu(\vec{y},t)\right\} \;=\; -g_{\mu\nu}\,\delta(\vec{x}-\vec{y}) \ ,\qquad g_{\mu\nu} = (-1,1,1,1) \]

This blows up at \(\vec{x} = \vec{y}\) --- so integrate over?

Hamiltonian density:

\[ \mathcal{H} \;=\; \Pi^{\mu}\dot{A}_\mu - \mathcal{L} \;=\; \Pi^{\mu}\left(\partial_0 A_\mu\right) - \mathcal{L} \]

For any dynamical variable

\[ \dot{f} \;=\; \left\{f,H\right\} + \frac{\partial f}{\partial t} \]

where the last term is \(0\) if \(f\) is explicitly time independent.

Gauge theories present difficulties almost right from the beginning:

\[ \Pi_\mu \;=\; \frac{\partial\mathcal{L}}{\partial\dot{A}^{\mu}} \;=\; F_{0\mu} \qquad\Longrightarrow\qquad \Pi_0 \;=\; F_{00} \;=\; 0 \]

Proceed anyway:

\[ H_0 \;=\; \int d^3x\ \left(\Pi^{\mu}\partial_0 A_\mu - \mathcal{L}\right) \;=\; \int d^3x\ \left(\Pi^{\mu}\partial_0 A_\mu - \frac{1}{4}\left(2F_{0i}F_{0i} - F_{ij}F_{ij}\right)\right) \] \[ \;=\; \int d^3x\ \left(\underbrace{\Pi^{0}\partial_0 A_0}_{0} + \Pi^{i}\partial_0 A_i + \frac{1}{4}F_{ij}F_{ij} - \frac{1}{2}F_{0i}F_{0i}\right) \] \[ \;=\; \int d^3x\ \left(\frac{1}{4}F_{ij}F_{ij} + \Pi^{i}\partial_0 A_i - \frac{1}{2}F_{0i}F_{0i}\right) \]

We have 4 velocities \(\left(\dot{A}_\mu\right)\) but 3 conjugate momenta (\(\Pi_\mu\), with
\(\Pi_0 = 0\)). The transformation from velocities to momenta is \underline{singular}.

The Hamiltonian is not unique --- we can add a term proportional to \(\Pi_0\) to \(H_0\):

\[ H \;=\; H_0 + \int d^3x\ C\,\Pi_0 \ ,\qquad C:\ \text{arbitrary function.} \]

Then

\[ \dot{A}_0 \;=\; \left\{A_0,H\right\}_{PB} \;=\; \left\{A_0,H_0\right\} + \left\{A_0,\int C\Pi_0\,d^3x\right\} \]

with

\[ H_0 \;=\; \int d^3x\left(\frac{1}{4}F_{ij}F^{ij} - \frac{1}{2}\Pi_i\Pi^{i} + A_0\partial_i\Pi^{i}\right) \]

which does not contain \(\Pi^{0}\). So \(\left\{A_0,H_0\right\} = 0\) and
\(\left\{A_0,\int C\Pi_0 d^3x\right\} = C\), giving

\[ \dot{A}_0 \;=\; C(x) \]

But this is effectively the same as

\[ A_0 \;\longrightarrow\; A_0 + \dot{\lambda} \ ,\qquad A_\mu \;\longrightarrow\; A_\mu + \partial_\mu\lambda \]

where \(\lambda(\vec{x},t_0) = 0\) but \(\dot{\lambda}(\vec{x},t_0)\neq 0\) --- i.e. a gauge
transformation.

We require \(\dot{A}_0 = 0\), i.e. \(\dot{\Pi}_0 = 0\) for all times:

\[ \dot{\Pi}_0 \;=\; \left\{\Pi_0,H\right\}_{PB} \;=\; \left\{\Pi_0,\int d^3x\left(\frac{1}{4}F_{ij}F^{ij} - \frac{1}{2}\Pi_i\Pi^{i} + A_0\partial_i\Pi^{i}\right)\right\} + \underbrace{\left\{\Pi_0,\int C\Pi_0 d^3x\right\}}_{0} \] \[ \;=\; -\,\partial_i\Pi^{i} \]

So

\[ \dot{\Pi}_0 = 0 \qquad\Longrightarrow\qquad \partial_i\Pi^{i} = 0 \quad\forall\,t \qquad\text{i.e.}\qquad \vec{\nabla}\cdot\vec{\Pi} = 0 \quad\forall\,t \]

To summarize: first we had \(\Pi_0 = 0\); then the requirement that \(\Pi_0 = 0\ \forall t\), i.e.
\(\dot{\Pi}_0 = 0\), gave \(\partial_i\Pi^{i} = 0\).

\[ H_{\text{extra}} \;=\; \int d^3x\ G(x,t)\,\partial_i\Pi^{i} \]

Look at \(\dot{A}_0\), or alternatively

\[ \delta A_0 \;=\; \left\{A_0,H_{\text{extra}}\right\} \;=\; \left\{A_0,\int d^3x\ G(x,t)\,\partial_i\Pi^{i}\right\} \;=\; 0 \qquad\left(\left\{A_0,\Pi^{i}\right\} = 0\right) \]

Similarly,

\[ \delta A_i \;=\; \left\{A_i,H_{\text{extra}}\right\}_{PB} \;=\; \left\{A_i,\int d^3x\ G\,\partial_j\Pi^{j}\right\} \;=\; \partial_i G \]

using \(\left\{A_i,\Pi^j\right\}\sim\delta_{ij}\delta(\vec{x}-\vec{y})\) and throwing away the
surface term \(\int d^3x\ \partial_j\left(G\Pi^{j}\right) = 0\).

Recall that \(A_i' = A_i + \partial_i\lambda\), so \(\delta A_i = \partial_i\lambda\). Hence
\(H_{\text{extra}}\) generates a gauge transformation on \(A_i\).

So the final form of \(H\) is

\[ H \;=\; \int d^3x\ \left(\frac{1}{4}F_{ij}F^{ij} - \frac{1}{2}\Pi_i\Pi^{i} + G\,\partial_i\Pi^{i}\right) \]

The new Hamiltonian no longer depends on \(A_0\).

\(\dot{f} = \left\{f,H\right\}\) depends on arbitrary terms \(\sim\left(\partial_i\Pi^{i}\right)\).
We demand that we consider only those functions that obey

\[ \left\{f,\ \partial_i\Pi^{i}\right\} \;=\; 0 \]

i.e. we only consider \underline{gauge invariant functions}.

In phase space \(\left(A_i,\Pi^{j}\right)\) not all functions have well defined time evolution;
only functions defined on the surface \(G\left(A_i,\Pi^{j}\right) = 0\) are well defined
physically:

\[ G\left(A_i,\Pi^{j}\right) \;=\; 0 \qquad\longrightarrow\qquad \text{Gauss' law (constraint)} \] \[ \partial_i\Pi^{i} \;=\; 0 \qquad\Longleftrightarrow\qquad G \;=\; 0 \]