Prof. Sachindeo Vaidya (CHEP, IISc) | PDF
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We were discussing Grassmann integrals & differentiation.
We discussed that \(c\,a^{\dagger}\), \(c^{*}a\) behaves like ordinary operators. Note the order matters.
\[
\left[c\,a^{\dagger},\ c^{*}\right] = 0
\]
define \(|c\rangle\) & \(\langle c|\) as
\[
|c\rangle \;=\; e^{-ca^{\dagger}}|0\rangle
\]
We saw, \(a|c\rangle = c|0\rangle = c|c\rangle\)
&
\[
\begin{aligned}
\langle c| &= \langle 0|e^{-ac^{*}}\\
&= \langle 0|\left(1 - ac^{*}\right)\\
&= \langle 0| - \langle 0|a\,c^{*}\\
&= \langle 0| - \langle 1|c^{*}
\end{aligned}
\]
\[
\begin{aligned}
\langle c|a^{\dagger} &= \langle 0|a^{\dagger} - \langle 1|c^{*}a^{\dagger}\\
&= 0 + \langle 1|a^{\dagger}c^{*}\\
&= \langle 0|c^{*} \qquad\qquad {\text{-- verify!}}
\end{aligned}
\]
Overlap,
\[
\begin{aligned}
\langle c'|c\rangle &= \langle 0|e^{-ac'^{*}}\,e^{-ca^{\dagger}}|0\rangle\\
&= \langle 0|\left(1 - ac'^{*}\right)\left(1 - ca^{\dagger}\right)|0\rangle\\
&= \langle 0|1 - ca^{\dagger} - ac'^{*} + ac'^{*}ca^{\dagger}|0\rangle\\
&= \langle 0|1 + a^{\dagger}c + c'^{*}a - c'^{*}\,a\,c\,a^{\dagger}|0\rangle\\
&= \langle 0|0\rangle + \cancelto{0}{\langle 0|a^{\dagger}c|0\rangle} + \cancelto{0}{\langle 0|c'^{*}a|0\rangle} + \langle 0|c'^{*}c\,aa^{\dagger}|0\rangle
\end{aligned}
\]
\[
\langle c'|c\rangle \;=\; 1 + c'^{*}c \;\simeq\; e^{c'^{*}c}
\]
Remember coherent states of SHO \(\left(|z\rangle\right)\)
\[
I \;=\; C\int dz\,d\bar{z}\ e^{-\bar{z}z}\,|z\rangle\langle z|
\]
Similarly
\[
\begin{aligned}
I &= \int dc^{*}dc\ e^{-c^{*}c}\ |c\rangle\langle c|\\
&= \int dc^{*}dc\ \left(1 - c^{*}c\right)e^{-ca^{\dagger}}|0\rangle\langle 0|e^{-ac^{*}}\\
&= \int dc^{*}dc\ \left(1 - c^{*}c\right)\left(1 - ca^{\dagger}\right)|1\rangle\,\langle 0|\left(1 - ac^{*}\right)\\
&= \int dc^{*}dc\ \left(1 - c^{*}c\right)\left(|0\rangle - c|1\rangle\right)\left(\langle 0| - \langle 1|c^{*}\right)\\
&= \int dc^{*}dc\ \left(1 - c^{*}c\right)\Big[\,|0\rangle\langle 0| - c^{*}|0\rangle\langle 1| - c|0\rangle\langle 0|\\
&\hspace{12em} + c\,c^{*}|0\rangle\langle 1|\,\Big]\\
&= \int dc^{*}dc\ \left(1 - c^{*}c\right)|0\rangle\langle 0| - \int dc^{*}dc\ \left(1 - c^{*}c\right)c^{*}|0\rangle\langle 1|\\
&\quad - \int dc^{*}dc\ \left(1 - c^{*}c\right)c\,|0\rangle\langle 0| + \int dc^{*}dc\ \left(1 - c^{*}c\right)c\,c^{*}|0\rangle\langle 1|
\end{aligned}
\]
\[
\begin{aligned}
&= \cancelto{0}{\int dc^{*}dc\ |0\rangle\langle 0|} - \underline{\int dc^{*}dc\ c^{*}c\ |0\rangle\langle 0|} - \cancelto{0}{\int dc^{*}dc\ |0\rangle\langle 1|}\\
&\quad + \cancelto{0}{\int dc^{*}dc\ c^{*}c\,c^{*}|0\rangle\langle 1|} - \cancelto{0}{\int dc^{*}dc\,c|0\rangle\langle 0|} + \cancelto{0}{\int dc^{*}dc\,c\,c^{*}|0\rangle\langle 0|}\\
&\quad + \underline{\int dc^{*}dc\ c\,c^{*}|0\rangle\langle 1|} - \cancelto{0}{\int dc^{*}dc\ c^{*}c\,c\,c^{*}|0\rangle\langle 1|}
\end{aligned}
\]
$\left\{\begin{aligned} &c2 = c{2} = 0\ &\int dc c = \int dc{}c{} = 1\ &\{c, c^{}\} = 0
\end{aligned}\right.$
We will get,
\[
\underline{I} \;=\; |0\rangle\langle 0| + |1\rangle\langle 1|
\]
Similarly, we can find
\[
\text{Tr}\, A \;=\; \langle 0|A|0\rangle + \langle 1|A|1\rangle
\]
\[
=\; \int dc^{*}dc\ e^{-c^{*}c}\ \langle -c|A|c\rangle
\]
\[
=\; \langle 0|A|0\rangle + \langle 1|A|1\rangle
\]
\[
=\; \int dc^{*}dc\ e^{-c^{*}c}\ \langle -c|e^{-\beta H}|c\rangle
\qquad\qquad \text{let}\quad \epsilon = \frac{\beta}{N}
\]
\[
=\; \int dc^{*}dc\ e^{-c^{*}c}\ \langle -c|\,e^{-\epsilon H}\cdot e^{-\epsilon H}\cdot e^{-\epsilon H}\cdots e^{-\epsilon H}\,|c\rangle
\]
Insert
\[
I \;=\; \int dc_i^{*}\,dc_i\ e^{-c_i^{*}c_i}\ |c_i\rangle\langle c_i|
\]
We gonna need matrix elements like...
\[
e^{-c_{i+1}^{*}c_{i+1}}\ \langle c_{i+1}|e^{-\epsilon H(a, a^{\dagger})}|c_i\rangle
\]
\[
\simeq\; e^{-c_{i+1}^{*}c_{i+1}}\ \langle c_{i+1}|c_i\rangle\ e^{-\epsilon H\left(c_{i+1}^{*},\, c_i\right)}
\qquad\left(\longrightarrow e^{c_{i+1}^{*}c_i}\right)
\]
\[
=\; \exp\left[-\epsilon\left(c_{i+1}^{*}\ \frac{\left(c_{i+1} - c_i\right)}{\epsilon} - H\left(c_{i+1}^{*}, c_i\right)\right)\right]
\]
\[
Z \;=\; \int dc_N^{*}\,dc_N\ dc_{N-1}^{*}\,dc_{N-1}\cdots dc_1^{*}\,dc_1\ e^{-S_E/\hbar}
\]
where,
\[
S_E \;=\; \epsilon\sum_{i=1}^{N}\left[c_{i+1}^{*}\left(\frac{c_{i+1} - c_i}{\epsilon}\right) + H\left(c_{i+1}^{*}, c_i\right)\right]
\]
\[
c_{N+1} = -c_1
\]
\[
c_{N+1}^{*} = -c_1^{*}
\]
Finally we take \(N \to \infty\), \(\epsilon \to 0\) with \(\beta = \epsilon N\)
\[
Z \;=\; \int \mathcal{D}c^{*}(\tau)\ \mathcal{D}c(\tau)\ e^{-\int_{0}^{\beta\hbar} d\tau\ \left(c^{*}\dot{c} - H\right)}
\]
\[
\left.\begin{aligned}
c(\beta\hbar) &= -c(0)\\
c^{*}(\beta\hbar) &= -c^{*}(0)
\end{aligned}\right\}
\begin{array}{l}\text{Antiperiodic boundary cond}^{\text{n}}.\\ \text{in imaginary time axis or}\\ \text{axis. (inv. Temp)}\end{array}
\]
Consider \(H = \dfrac{p^2}{2m} + \dfrac{1}{2}m\omega^2 x^2 + gx\)
Solve by ``completing the square''.
Also do variation of above
\[
H \;=\; H_{SHO} + \underbrace{g(t)\,x}_{\text{position}} + \underbrace{p\,f(t)}_{\text{momentum}}
\]
\[
H \;=\; H_{SHO} + a\,F^{*}(t) + a^{\dagger}F(t)
\qquad \left(\downarrow\ \frac{f - ig}{\sqrt{2}}\right)
\]
-- compute the propagator
-- compute the vacuum to vacuum transition amplitude.
\[
\langle 0; t_{-\infty}\,|\,0, t_{\infty}\rangle
\qquad \text{where}\ \left\{ F, g \to 0 \ \text{as}\ t \to \pm\infty \right.
\]
(margin working)
\[
H' = a\left(\frac{f+ig}{\sqrt{2}}\right) + a^{\dagger}\left(\frac{f-ig}{\sqrt{2}}\right)
\qquad
a = \frac{x+ip}{\sqrt{2}}\ ,\quad a^{\dagger} = \frac{x-ip}{\sqrt{2}}
\]
\[
\left(\frac{x+ip}{\sqrt{2}}\right)\left(\frac{f+ig}{\sqrt{2}}\right) + \left(\frac{x-ip}{\sqrt{2}}\right)\left(\frac{f-ig}{\sqrt{2}}\right)
\]
\[
= \frac{1}{2}\left(xf + ixg + ipf - pg + xf - ixg - ipf - pg\right)
\]
\[
= \frac{1}{2}\Big(x(f + ig + f - ig) + p(if - g - if - g)\Big)
\]
\[
= \frac{1}{2}\left(2xf - 2pg\right) \;=\; \underline{xf} - \underline{pg}\ .
\]
These are not features of QFT alone.
ex : Electrostatics :
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\[
\phi \;=\; \frac{q}{\epsilon_0}
\qquad\Rightarrow\qquad E\,(2\pi x \ell) \;=\; \frac{\lambda \ell}{\epsilon_0}
\]
\[
E(x) \;=\; \frac{\lambda}{2\pi x \epsilon_0}
\]
\[
\phi(x) - \phi(y) \;=\; \int_{x}^{y}\vec{E}\cdot \vec{dx}
\]
using \(E(x)\).
\[
\phi(x) - \phi(y) \;=\; \frac{\lambda}{2\pi\epsilon_0}\,\ln\frac{y}{x}
\]
But we know
\[
\phi(r) \;=\; \int \frac{k\,dq}{r}
\qquad \left(\text{here } x = \rho\right)
\]
\[
=\; \frac{1}{4\pi\epsilon_0}\int_{-\infty}^{\infty}\frac{\lambda\,dy}{\sqrt{x^2+y^2}}
\]
\[
\left.\begin{aligned} y &= x\tan\theta\\ dy &= x\sec^2\theta\,d\theta \end{aligned}\right\}
\quad\Rightarrow\quad
\phi(r) = k\lambda\int\frac{\cancel{x}\sec^2\theta\,d\theta}{\cancel{x}\sec\theta} = k\lambda\int \sec\theta\,d\theta
\]
\[
\Rightarrow\quad \phi(r) = k\lambda\int \frac{\sec\theta\left(\sec\theta + \tan\theta\right)d\theta}{\sec\theta + \tan\theta}
\]
\[
\left.\begin{aligned} \sec\theta + \tan\theta &= t\\ \left(\sec\theta\tan\theta + \sec^2\theta\right)d\theta &= dt \end{aligned}\right\}
\qquad
\phi(r) = k\lambda\int\frac{dt}{t} = k\lambda\ln t
= k\lambda \ln\left(\sec\theta + \tan\theta\right)\Big|_{-\pi/2}^{\pi/2}
\]
\[
=\; \infty - \infty
\]
But answer shall be finite!
Similarly when we calculate things in QFT, we sometimes get \(\infty\), but we know output must be some finite no.
put \(\eta = y/x_0\)
\[
\phi(x) - \phi(x_0) \;=\; \frac{\lambda}{4\pi\epsilon_0}\int du\left[\left(\left(\frac{x}{x_0}\right)^2 + u^2\right)^{-1/2} - \left(1+u^2\right)^{-1/2}\right]
\]
use,
\[
\displaystyle -\frac{1}{2}\int_{1}^{b} d\alpha\,\left(u^2+\alpha\right)^{-3/2} \;=\; \left(u^2+b\right)^{-1/2} - \left(u^2+1\right)^{-1/2}
\]
\[
\phi(x) - \phi(x_0) \;=\; \frac{-\lambda}{8\pi\epsilon_0}\int_{-\infty}^{\infty} du \int_{1}^{(x/x_0)^2} d\alpha\ \left(u^2+\alpha\right)^{-3/2}
\]
put,
\[
z \;=\; u\,\alpha^{-1/2}\,\left(x/x_0\right)^2
\]
\[
\phi(x) - \phi(x_0) \;=\; -\frac{\lambda}{8\pi\epsilon_0}\int_{1}^{(x/x_0)^2}\frac{d\alpha}{\alpha}\int_{-\infty}^{\infty}\frac{dz}{\left(1+z^2\right)^{3/2}}
\]
\[
=\; \frac{-\lambda}{2\pi\epsilon_0}\,\ln\frac{x}{x_0}
\]
Above way is called dimensional Regularization (DR). DR is faster.
Now we will perform same calculation with DR :
Notice that \(\phi(x)\) is scale invariant. \(\left\{\begin{aligned} x &\to kx\\ y &\to ky\end{aligned}\right.\)
\[
\phi(kx) = \phi(x)
\]
\[
\phi(kx) \;=\; \int_{-\infty}^{\infty}\frac{k\,dy}{k\sqrt{y^2+x^2}}
\;=\; \underbrace{\int_{-\infty}^{\infty}\frac{dy}{\sqrt{y^2+x^2}}}_{{\text{This integral is divergent}}} \;=\; \phi(x)
\]
For \(d\)-dimension \(\phi\) is not scale invariant!
\[
\phi_d(x) \;=\; \frac{\lambda}{4\pi\epsilon_0}\ \mu^{d-1}\int_{-\infty}^{\infty}\frac{d^d y}{\sqrt{y^2+x^2}}
\]
\(\downarrow\) \(d\) -- is dimension of charge confn.
\(\mu\) has dimension of \(\dfrac{1}{\ell}\) i.e. \(\mu\) has mass dimension.
Because
\[
\phi_d(kx) \;=\; k^{d-1}\,\phi(x)
\]
to make it scale invariant we multiply by \(\mu^{d-1}\). \((\mu = \text{const.})\)
\[
\phi_d(x) \;=\; \frac{\lambda}{4\pi\epsilon_0}\ \mu^{d-1}\int\frac{\rho^{d-1}\,d\rho}{\sqrt{\rho^2+x^2}}\ \int d\Omega_{d-1}
\]
i.e. we will have \(d-1\) angular variables & 1 radial variable, in \(d\)-dimensions.
\[
=\; \frac{\lambda}{4\pi\epsilon_0}\ \mu^{d-1}\ {\frac{2(\pi)^{d/2}}{\Gamma\!\left(d/2\right)}}\ \underbrace{\int_{0}^{\infty}\frac{\rho^{d-1}\,d\rho}{\left(\rho^2+x^2\right)^{1/2}}}_{I}
\qquad {\text{check!}}
\]
\[
I \;=\; \frac{\frac{1}{2}\,\Gamma\!\left(d/2\right)\,\Gamma\!\left(\frac{1-d}{2}\right)}{\sqrt{\pi}}
\qquad {\longleftarrow \text{should include `}x\text{'!}}
\]
\[
(z-1)! = \Gamma(z) = \int_{0}^{\infty} e^{-t}\,t^{z-1}\,dt
\qquad
{m! = \Gamma(m+1) = \int_{0}^{\infty}t^{m}\left(e^{-t}\,dt\right)}
\]