Prof. Sachindeo Vaidya (CHEP, IISc) | PDF
Previous: Lecture 20
We were discussing canonical quantisation of gauge theories. We saw that \(\mathcal{L}\to H\) is
Full phase space: \(\left(\Pi_0,\Pi_i,A_0,A_i\right)\). We have \(\Pi_0 = 0\) and \(\dot{A}_0 = 0\),
We must have
\[
\dot{\Pi}_0 \;=\; \left\{\Pi_0,H\right\} \;=\; 0
\qquad\left(\Pi_0 = 0 \text{ at all times}\right)
\]
But \(\left\{\Pi_0,H\right\} = \partial_i\Pi_i\), so
\[
\partial_i\Pi_i \;=\; 0 \qquad(\text{to have dynamics}).
\]
i.e. not all \(\Pi_i\) are allowed, but only those \(\Pi_i\) which satisfy \(\partial_i\Pi_i = 0\).
For phase space functions we demand that only those functions which obey
\[
\left\{f,\ \partial_i\Pi^{i}\right\} \;=\; 0
\]
In \(\left\{A_i,\Pi^{j}\right\}\) space not all functions have well defined time evolution; only
\[
G\left(A_i,\Pi^{i}\right) \;=\; 0 \qquad\longrightarrow\qquad
\text{Gauss' law condition (or constraint).}
\]
Say we change variable from \(G\) to \(Z\) (another variable) that is conjugate to
\[
\text{Jacobian} \;=\; \det\left|\frac{\delta G}{\delta Z}\right|
\;=\; \det\left|\left\{G,\ \partial_i\Pi^{i}\right\}_{PB}\right| \;\neq\; 0
\]
(for an infinite dimensional basis). For finite dimensional bases \(A\to B\) with \(B = RA\), a
\[
\frac{\delta}{\delta Z(y)} \;=\; \int d^3x\
\frac{\delta\left(\partial_i\Pi^{i}\right)(y)}{\delta\Pi^{j}(x)}\,\frac{\delta}{\delta A^{j}(x)}
\;=\; \left\{\ \cdot\ ,\ \partial_i\Pi^{i}\right\}_{PB}
\]
Say we find a non-singular transformation \(G\to Z\). Then
\[
\left\{A_i,\Pi_i\right\} \;\xrightarrow{\ \ G\ \to\ Z\ \ }\;
\left\{\tilde{A}_i,\tilde{\Pi}_i\right\} \qquad(\text{change of basis})
\]
such that \(G\left(A_i,\Pi_i\right) = \tilde{A}_3\). Then
\[
\det\left|\left\{G,\partial_i\Pi^{i}\right\}_{PB}\right|
\;=\; \det\left\{\frac{\delta G}{\delta\tilde{A}_j}\frac{\delta\left(\partial_i\Pi^{i}\right)}{\delta\tilde{\Pi}_j}
- \frac{\delta G}{\delta\tilde{\Pi}_j}\frac{\delta\left(\partial_i\Pi^{i}\right)}{\delta\tilde{A}_j}\right\}
\;=\; \det\left|\frac{\delta\left(\partial_i\Pi^{i}\right)}{\delta\tilde{\Pi}_3}\right|
\]
i.e. the Jacobian for \(\partial_i\Pi^{i}\longrightarrow\tilde{\Pi}_3\).
Take \(G = \partial_i A^{i}\). Then
\[
\det\left|\left\{\partial_i A^{i},\ \partial_j\Pi^{j}\right\}_{PB}\right|
\;=\; \det\left|\partial_{i_x}\partial_{j_y}\,\delta(\vec{x}-\vec{y})\,\delta_{ij}\right|
\]
Look at the Laplacian in 3-space:
\[
\partial_i\partial^{i}F \;=\; \lambda F
\]
For zero eigen modes, \(\partial^2 F = 0\):
\[
\frac{\partial^2 F}{\partial x^2} + \frac{\partial^2 F}{\partial y^2} + \frac{\partial^2 F}{\partial z^2} \;=\; 0
\qquad\Longrightarrow\qquad F = \text{const.}
\]
which is non-normalizable. So \(F = 0\) (normalizable zero eigen modes).
Split
\[
\tilde{\Pi}_i \;=\; \tilde{\Pi}^{L}_{i} + \tilde{\Pi}^{T}_{i} \ ,\qquad
\tilde{A}_i \;=\; \tilde{A}^{L}_{i} + \tilde{A}^{T}_{i}
\]
into longitudinal (\(L\)) and transverse (\(T\)) parts, with the constraint
\[
\partial_i\tilde{\Pi}^{i} \;=\; 0 \qquad\Longrightarrow\qquad \partial_i\tilde{\Pi}^{L}_{i} \;=\; 0
\]
Let us go back and look at \(H\):
\[
H \;=\; \int d^3x\ \left(\frac{1}{4}F_{ij}F^{ij} + \frac{1}{2}\Pi^{T}_{i}\Pi^{T}_{i}\right)
\]
with \(\frac{1}{4}F_{ij}F^{ij}\to\frac{1}{2}B_iB_i\), so
\[
H \;=\; \frac{1}{2}\int d^3x\ \left(\Pi^{T}_{i}\Pi^{T}_{i} + B_iB_i\right)
\;=\; H_{\text{EM radiation}}
\]
In QED, here the gauge condition is \(A_3 = 0\). Then
\[
\det\left|\frac{\partial}{\partial x_3}\,\delta(\vec{x}-\vec{y})\right| \;\neq\; 0
\]
since \(\dfrac{\partial}{\partial x_3}\) is invertible (an inverse exists). So the determinant
We have
\[
\partial_i\Pi^{i} = 0 \qquad\Longrightarrow\qquad
\partial_3\Pi^{3} \;=\; -\left(\partial_1\Pi^{1} + \partial_2\Pi^{2}\right)
\]
\[
\Pi^{3}(\vec{x},t) \;=\; -\int_{-\infty}^{z}\left(\partial_1\Pi^{1} + \partial_2\Pi^{2}\right)dx_3
\]
So \(\Pi^{3} = \Pi^{3}\left[\Pi_1,\Pi_2\right]\), a functional of \(\Pi_1\) and \(\Pi_2\).
The canonical variables are \(A_1, A_2, \Pi^{1}, \Pi^{2}\), and
\[
H \;=\; \frac{1}{2}\int d^3x\ \left(\Pi_i\Pi_i + B_iB_i\right)
\]
\[
H \;=\; \frac{1}{2}\int d^3x\ \left(\Pi_1^2 + \Pi_2^2 + \Pi_3^2\left[\Pi_1,\Pi_2\right]
+ B_1^2 + B_2^2 + B_3^2\right)
\]
\(\Pi_3\) is not a local function, so \(H\) is \underline{non-local}. That is the problem with the
\[
B_i \;=\; \frac{1}{2}\epsilon_{ijk}F_{jk}
\]
\[
B_1 \;=\; \frac{1}{2}\epsilon_{1jk}\left(\partial_j A_k - \partial_k A_j\right)
\;=\; \frac{1}{2}\left(\epsilon_{123}\left(\partial_2 A_3 - \partial_3 A_2\right)
+ \epsilon_{132}\left(\partial_3 A_2 - \partial_2 A_3\right)\right)
\;=\; -\partial_3 A_2
\]
\[
B_2 \;=\; -\partial_1\underbrace{A_3}_{0} + \partial_3 A_1 \;=\; \partial_3 A_1
\ ,\qquad
B_3 \;=\; \partial_1 A_2 - \partial_2 A_1
\]
With \(A^{T}\equiv A_{\perp}\),
\[
\text{PI} \;=\; \int\mathcal{D}\tilde{A}_{\perp}\int\mathcal{D}\tilde{\Pi}_{\perp}\
\exp\left(i\int\left(\tilde{\Pi}_{\perp}\cdot\dot{\tilde{A}}_{\perp} - \mathcal{H}\right)d^4x\right)
\]
Compare with the non-relativistic path integral
\[
\text{PI} \;=\; \int\mathcal{D}q\,\mathcal{D}p\ \exp\left(\int\left(p\dot{q} - H\right)dt\right)
\]
So
\[
\text{PI} \;=\; \int\mathcal{D}\tilde{A}_{\perp}\,\mathcal{D}\tilde{\Pi}_{\perp}
\underbrace{\int\mathcal{D}\tilde{A}_3\,\delta\!\left(\tilde{A}_3\right)\mathcal{D}\tilde{\Pi}_3\,
\delta\!\left(\tilde{\Pi}_3 - \tilde{\Pi}_3(\Pi_{\perp})\right)}_{1}\
\exp\left(i\int\left(\tilde{\Pi}_i\dot{\tilde{A}}_i - \mathcal{H}\right)d^4x\right)
\]
with
\[
\delta\!\left(\tilde{\Pi}_3 - \tilde{\Pi}_3(\Pi_{\perp})\right)
\;=\; \delta\!\left(\partial_i\tilde{\Pi}^{i}\right)
\det\left|\left\{\partial_i\tilde{\Pi}^{i},\ \tilde{A}_3\right\}_{PB}\right|
\]
Write
\[
\delta\!\left(\partial_i\tilde{\Pi}^{i}\right) \;=\; \int\mathcal{D}A_0\ e^{\,i\int A_0\partial_i\tilde{\Pi}^{i}}
\]
using \(\delta(x) = \int da\ e^{iax}\). (Note that here \(A_0\neq\phi\): it is a variable introduced
\[
\text{PI} \;=\; \int\mathcal{D}\tilde{A}_{\perp}\,\mathcal{D}\tilde{A}_3\,\mathcal{D}\tilde{\Pi}_{\perp}\,
\mathcal{D}A_0\,\mathcal{D}\tilde{\Pi}_3\
\det\left|\left\{\partial_i\Pi_i,\ \tilde{A}_3\right\}\right|
\exp\left[i\int d^4x\left\{\tilde{\Pi}_{\perp}\cdot\dot{\tilde{A}}_{\perp}
+ \tilde{\Pi}_3\dot{\tilde{A}}_3 - \mathcal{H} + A_0\partial_i\tilde{\Pi}^{i}\right\}\right]
\]
Here
\[
\mathcal{H} \;=\; \frac{1}{2}\left(\tilde{\Pi}_{\perp}\cdot\tilde{\Pi}_{\perp}
+ \tilde{\Pi}_3\tilde{\Pi}_3 + \tilde{B}_i\tilde{B}_i\right)
\]
and in compact notation
\[
\text{PI} \;=\; \int\mathcal{D}A_\mu\,\mathcal{D}\Pi_i\ \delta(\mathcal{G})
\det\left|\left\{\partial_i\Pi^{i},\ \mathcal{G}\right\}_{PB}\right|\
e^{\,i\int d^4x\left(\Pi_i\dot{A}_i - \mathcal{H} + A_0\partial_i\Pi^{i}\right)}
\]
Now do the integral over the \(\Pi_i\)'s.
Ch. 7 --- all problems.
Ch. 8.1 --- A, B, C; and Ch. 8.2 --- A, B.