Prof. Sachindeo Vaidya (CHEP, IISc) | PDF
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Read chapter 6, 7, 8 --- for path integrals
43, 44 --- path integral for fermions
Solve all problems in above chapters.
\[
\eta = \begin{pmatrix} \eta_1 \\ \eta_2 \end{pmatrix}
\qquad\qquad a_{ij} \in \mathbb{R} \ \text{or}\ \mathbb{C}
\]
\[
\bar{\eta}A\eta \;=\; \begin{pmatrix} \bar{\eta}_1 & \bar{\eta}_2 \end{pmatrix}
\begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}
\begin{pmatrix} \eta_1 \\ \eta_2 \end{pmatrix}
\]
\[
=\; \begin{pmatrix} \bar{\eta}_1 & \bar{\eta}_2 \end{pmatrix}
\begin{pmatrix} a_{11}\eta_1 + a_{12}\eta_2 \\ a_{21}\eta_1 + a_{22}\eta_2 \end{pmatrix}
\]
\[
\bar{\eta}A\eta \;=\; \bar{\eta}_1 a_{11}\eta_1 + \bar{\eta}_1 a_{12}\eta_2
+ \bar{\eta}_2 a_{21}\eta_1 + \bar{\eta}_2 a_{22}\eta_2
\]
\[
e^{-\bar{\eta}A\eta} \;=\; 1 - \bar{\eta}A\eta + \frac{\left(\bar{\eta}A\eta\right)^2}{2!}
- \frac{\left(\bar{\eta}A\eta\right)^3}{3!} + \cdots
\]
\[
=\; 1 - \underbrace{\left(\bar{\eta}_1 a_{11}\eta_1 + \bar{\eta}_1 a_{12}\eta_2
+ \bar{\eta}_2 a_{21}\eta_1 + \bar{\eta}_2 a_{22}\eta_2\right)}_{\enclose{circle}{\text{1}}}
\]
\[
\enclose{circle}{\text{2}}\left\{
\begin{aligned}
&+ \left(\bar{\eta}_1 a_{11}\eta_1 + \bar{\eta}_1 a_{12}\eta_2 + \bar{\eta}_2 a_{21}\eta_1 + \bar{\eta}_2 a_{22}\eta_2\right)\\
&\times \left(\bar{\eta}_1 a_{11}\eta_1 + \bar{\eta}_1 a_{12}\eta_2 + \bar{\eta}_2 a_{21}\eta_1 + \bar{\eta}_2 a_{22}\eta_2\right)\\
&+ \cdots
\end{aligned}\right.
\]
\(\hookrightarrow\) 2nd term contains \(\bar{\eta}_2^{\,2}\), \(\bar{\eta}_1^{\,2}\), \(\eta_1^2\eta_2^2\) which all vanishes.
\((2)\;=\;\) The terms which survive are
\[
\bar{\eta}_1 a_{11}\eta_1\,\bar{\eta}_2 a_{22}\eta_2 + \bar{\eta}_1 a_{12}\eta_2\,\bar{\eta}_2 a_{21}\eta_1
\]
\[
+\; \bar{\eta}_2 a_{21}\eta_1\,\bar{\eta}_1 a_{12}\eta_2 + \bar{\eta}_2 a_{22}\eta_2\,\bar{\eta}_1 a_{11}\eta_1
\]
\[
=\; a_{11}a_{22}\,\bar{\eta}_1\eta_1\bar{\eta}_2\eta_2 + a_{12}a_{21}\,\bar{\eta}_1\eta_2\bar{\eta}_2\eta_1
+ a_{21}a_{12}\,\bar{\eta}_2\eta_1\bar{\eta}_1\eta_2 + a_{22}a_{11}\,\bar{\eta}_2\eta_2\bar{\eta}_1\eta_1
\]
\[
=\; a_{11}a_{22}\left(\bar{\eta}_1\eta_1\bar{\eta}_2\eta_2 + \bar{\eta}_2\eta_2\bar{\eta}_1\eta_1\right)
+ a_{12}a_{21}\left(\bar{\eta}_1\eta_2\bar{\eta}_2\eta_1 + \bar{\eta}_2\eta_1\bar{\eta}_1\eta_2\right)
\]
\[
=\; \bar{\eta}_1\eta_1\bar{\eta}_2\eta_2\Big(a_{11}a_{22} + a_{11}a_{22} + a_{12}a_{21}(-1) + a_{12}a_{21}(-1)\Big)
\]
\[
=\; \bar{\eta}_1\eta_1\bar{\eta}_2\eta_2 \cdot 2\underbrace{\left(a_{11}a_{22} - a_{12}a_{21}\right)}_{\det A}
\]
So,
\[
\boxed{\ e^{-\bar{\eta}A\eta} \;=\; 1 - \bar{\eta}A\eta + \left(\det A\right)\bar{\eta}_1\eta_1\bar{\eta}_2\eta_2\ }
\]
\(_Z_ = 0 \rightsquigarrow\) \(\psi(x)\ \left\{\begin{array}{l}\psi_1\\ \psi_2\\ \psi_3\\ \psi_4\end{array}\right.\) \(\longrightarrow\) Grassmann co.
\[
\psi_1^2 = 0 \ , \qquad \underline{\psi_1\psi_2 = -\psi_2\psi_1}\ , \quad \psi_2^2 = 0
\]
\[
Z = \text{Tr}\, e^{-\beta H} = \sum_{n=0}^{\infty}\langle n|e^{-\beta H}|n\rangle
\]
\[
=\; \sum_{n=0}^{\infty} e^{-\beta n\left(1+\frac{1}{2}\right)\hbar\omega}
\;=\; \frac{1}{2\sinh\frac{\hbar\omega}{2}}
\]
If we use position basis instead of energy eigenvector basis.
\[
Z \;=\; \int dx_0\, \langle x_0|e^{-\beta H}|x_0\rangle
\qquad\qquad \text{where,}\quad \epsilon = \frac{\beta}{N}
\]
\[
=\; \int dx_0\, \langle x_0| e^{-\epsilon H}\ \underbrace{\phantom{x}}_{I}\ e^{-\epsilon H}\ \underbrace{\phantom{x}}_{I}\ \cdots\, e^{-\epsilon H}|x_0\rangle
\]
Inserting \(N-1\) complete set of \(|x\rangle\) states
\[
I = \int dx\, |x\rangle\langle x|
\]
\[
=\; \int dx_0\, \langle x_0|e^{-\epsilon H}\left|\int dx_1 |x_1\rangle\langle x_1|\right. e^{-\epsilon H}\int dx_2 |x_2\rangle\langle x_2| \ \cdots\ \int dx_{N-1}|x_{N-1}\rangle\langle x_{N-1}|x_0\rangle
\]
\[
Z \;=\; \int dx_0\,dx_1\,dx_2 \cdots dx_{N-1}\ \prod_{i=0}^{N-1}\langle x_i|e^{-\epsilon H}|x_{i+1}\rangle
\qquad \text{with}\quad \left(x_N = x_0\right)
\]
let us find \(\langle x|e^{-\epsilon H}|y\rangle\)
\[
\hat{H} = \frac{\hat{p}^2}{2m} + V(\hat{x})
\]
\[
e^{-\left(\epsilon\frac{p^2}{2m} + \epsilon V(x)\right)} \;=\; e^{-\epsilon\frac{\hat{p}^2}{2m}}\, e^{-\epsilon V(\hat{x})} + O(\epsilon^2)
\]
So,
\[
\langle x|e^{-\frac{\epsilon p^2}{2m}}\ e^{-\epsilon V(\hat{x})}|y\rangle
\qquad\qquad \text{inserting}\quad I = \int\frac{dp}{2\pi}|p\rangle\langle p|
\]
\[
=\; \int\frac{dp}{2\pi}\ \langle x|e^{-\frac{\epsilon \hat{p}^2}{2m}}|p\rangle\langle p|e^{-\epsilon V(\hat{x})}|y\rangle
\]
\[
=\; \int\frac{dp}{2\pi}\ e^{-\frac{\epsilon p^2}{2m}}\, e^{-\epsilon V(x)}\, e^{ip\cdot(x-y)}
\]
\[
=\; \frac{1}{(2\pi)^2}\, e^{-\epsilon V(x)}\int dp\ e^{-\frac{\epsilon p^2}{2m} + ip\cdot(x-y)}
\]
\[
=\; \frac{1}{(2\pi)^2}\, e^{-\epsilon V(x)}\int dp\ e^{-\frac{\epsilon}{2m}\left(p^2 - \frac{2mi(x-y)}{\epsilon}p + \frac{4m^2(x-y)^2}{\epsilon^2} - \frac{4m^2(x-y)^2}{\epsilon^2}\right)}
\]
\[
=\; \frac{1}{(2\pi)^2}\, e^{-\epsilon V(x)}\int dp\ e^{-\frac{\epsilon}{2m}\left(p - \frac{2mi(x-y)}{\epsilon}\right)^2}\ e^{-\frac{2m}{\epsilon}(x-y)^2}
\]
\[
=\; \frac{1}{(2\pi)^2}\ e^{-\left(\epsilon V(x) + \frac{2m}{\epsilon}(x-y)^2\right)}\ \sqrt{\frac{\pi\,2m}{\epsilon}}
\]
So,
\[
Z \;=\; \int dx_0\,dx_1\cdots dx_{N-1}\ \prod_{i=0}^{N-1}\frac{1}{(2\pi)^2}\
e^{-\left(\epsilon V(x_i) + \frac{2m}{\epsilon}(x_i - x_{i+1})^2\right)}\cdot\sqrt{\frac{2m\pi}{\epsilon}}
\]
\[
=\; \frac{\sqrt{m}}{(2\pi)^{3/2}}\left(\frac{1}{\sqrt{\epsilon}}\right)^{N}
\int dx_0\,dx_1\cdots dx_{N-1}\
e^{-\underbrace{\epsilon\left(\sum\limits_{i=0}^{N-1}\left(V(x_i) + 2m\left(\frac{dx_i}{dt}\right)^2\right)\right)}_{\displaystyle \downarrow\ S_E}}
\]
\[
=\; \sqrt{\frac{m}{\epsilon}}\left(\frac{1}{\sqrt{(2\pi)^3}}\right)^{N}\int \mathcal{D}x\ e^{-S_E}
\]
Note : Path integral approach is difficult / lengthy for Q.M. but for field theory this is much helpful approach.
For bosons we have commutator relations but for fermions we have anti commutation relations
\[
\{a, a^{\dagger}\} = 1 \qquad \longrightarrow\qquad 2a^{\dagger}a^{\dagger} = 0\ ,\qquad \left(a^{\dagger}\right)^2 = 0
\]
\[
\{a, a\} = \{a^{\dagger}, a^{\dagger}\} = 0
\]
we also define fermionic vacuum \(|0\rangle\) as
\[
a|0\rangle = 0
\]
\[
a^{\dagger}|0\rangle = \text{some state} \ldots |1\rangle
\]
\[
\left(a^{\dagger}\right)\left(a^{\dagger}\right)|0\rangle = 0|0\rangle = 0
\]
So, \(a^{\dagger}|1\rangle = a^{\dagger}a^{\dagger}|0\rangle = 0\)
So, we have only two states \(|0\rangle\) & \(|1\rangle\).
if we try to add one more particle by \(\left(a^{\dagger}\right)^2|0\rangle\) which becomes zero. (obeys Pauli-exclusion principle; does not allow 2 particles).
\[
H \;=\; \frac{\hbar\omega}{2}\left(a^{\dagger}a - aa^{\dagger}\right) \;=\; \hbar\omega\left(a^{\dagger}a - \frac{1}{2}\right)
\]
Such that,
\[
H|0\rangle = -\frac{\hbar\omega}{2}|0\rangle
\]
\[
H|1\rangle = \frac{\hbar\omega}{2}|1\rangle
\]
So, \(H = \dfrac{\hbar\omega}{2}\,\sigma_3 = \dfrac{\hbar\omega}{2}\begin{pmatrix}1&0\\0&-1\end{pmatrix}\)
we just have 2 eigen states of \(H\).
\[
Z = \text{Tr}\,\left(e^{-\beta H}\right) = \underbrace{\langle 0|e^{-\beta H}|0\rangle}_{\enclose{circle}{\text{1}}} + \underbrace{\langle 1|e^{-\beta H}|1\rangle}_{\enclose{circle}{\text{2}}}
\]
\[
=\; \langle 0|1 - \beta H + \frac{\beta^2 H^2}{2!} - \frac{\beta^3 H^3}{3!} + \cdots|0\rangle
+ \langle 1|1 - \beta H + \frac{\beta^2 H^2}{2!} - \cdots|1\rangle
\]
\[
\enclose{circle}{\text{1}} \;=\; \langle 0|0\rangle - \beta\langle 0|H|0\rangle + \frac{\beta^2}{2!}\langle 0|H^2|0\rangle - \cdots
\]
\[
=\; 1 + \beta\frac{\hbar\omega}{2} + \frac{\beta^2}{2!}\frac{\hbar^2\omega^2}{4} + \cdots
\;=\; e^{\frac{\beta\hbar\omega}{2}}
\]
Similarly \((2) = e^{-\frac{\beta\hbar\omega}{2}}\)
\[
Z \;=\; e^{\frac{\beta\hbar\omega}{2}} + e^{-\frac{\beta\hbar\omega}{2}} \;=\; 2\cosh\frac{\beta\hbar\omega}{2}
\]
\underline{Grassmann variables \(c, c^{*}\)}
Axioms :
\(c\) & \(c^{*}\) are independent
\(c^2 = \left(c^{*}\right)^2 = 0\) \(\{c, c^{*}\} = 0\)
Integrals --
\(\displaystyle\int dc = 0 = \int dc^{*}\)
\(\displaystyle\int dc\ c = 1 = \int dc^{*}\,c^{*}\)
\(0 = \{c, dc\} = \{c^{*}, dc^{*}\} = \{c, dc^{*}\} = \{c^{*}, dc\}\) [2mm]
\(\Rightarrow\quad \displaystyle\int c\,dc = -\int dc\,c = -1\)
\(\{dc, dc\} = \{dc^{*}, dc^{*}\} = \{dc, dc^{*}\} = \{dc^{*}, dc\} = 0\)
Convension : write `\(*\)' variables to extreme left.
\[
\int dc^{*}\,dc \qquad \text{while doing mult}^{\text{n}} \quad \text{-- Grassmann variable integrals.}
\]
How fermionic operators & Grassmann var. are related --
\(c, c^{*}\) anticommute with \(a, a^{\dagger}\)
\(c\,a^{\dagger}\) (or) \(c^{*}a\) behave like ordinary operator
ex \(\left[c\,a^{\dagger},\, c^{*}\right] = 0\)
Note : A state \(|c\rangle\) which is labelled by Grassmann variable `\(c\)' is \(e^{-ca^{\dagger}}|0\rangle\)
\[
e^{-ca^{\dagger}}|0\rangle \;=\; \left(1 - ca^{\dagger}\right)|0\rangle
\]
and,
\[
\begin{aligned}
a|c\rangle &= a\,e^{-ca^{\dagger}}|0\rangle\\
&= a\left(1 - ca^{\dagger}\right)|0\rangle\\
&= \cancelto{0}{a|0\rangle} - a\,c\,a^{\dagger}|0\rangle\\
&= -\underbrace{a\,c}_{\text{anticommute}}a^{\dagger}|0\rangle\\
&= +\,c\,a\,a^{\dagger}|0\rangle
\end{aligned}
\]
\[
a|c\rangle = c|0\rangle \qquad\qquad (1)
\]
also
\[
\begin{aligned}
c|c\rangle &= c\left(1 - ca^{\dagger}\right)|0\rangle\\
&= c|0\rangle - \cancelto{0}{c^2}a^{\dagger}|0\rangle\\
&= c|0\rangle \qquad\qquad (2)
\end{aligned}
\]
(1) & (2) implies \(\longrightarrow\)
\[
\boxed{\ a|c\rangle = c|c\rangle\ }
\]