1.5 Path integral quantization of a Fermi particle

The particles observed in Nature are not necessarily Bose particles whose position and momentum operators obey the commutation relation $[p,x]=-\mathrm{i}$ . There are particles called fermions whose operators satisfy anti-commutation relations. A classical description of a fermion requires anti-commuting numbers called the Grassmann numbers.

1.5.1 Fermionic harmonic oscillator

The bosonic harmonic oscillator in the previous section is described by the Hamiltonian¹

$$H=\frac{1}{2}(a^{†}a+a{a}^{†})$$

where $a$ and $a^{†}$ satisfy the commutation relations

$$[a,a^{†}]=1[a,a]=[a^{†},a^{†}]=0.$$

The Hamiltonian has eigenvalues $(n+1/2)\omega$ ( $n\in \mathbb{N}$ ) with the eigenvector $|n⟩$ :

$$H|n⟩=(n+\frac{1}{2})\omega |n⟩.$$

Now suppose there is a Hamiltonian

$$H=\frac{1}{2}\left(c^{†}c-c{c}^{†}\right)\omega .\text{\hspace{1em}(1.166)}$$

This is called the fermionic harmonic oscillator, which may be regarded as a Fourier component of the Dirac Hamiltonian, which describes relativistic fermions. If the operators $c$ and $c^{†}$ should satisfy the same commutation relations as those satisfied by bosons, the Hamiltonian would be a constant $H=-\omega /2$ . Suppose, in contrast, they satisfy the anti-commutation relations

$$\{c,c^{†}\}\equiv c{c}^{†}+c^{†}c=1\{c,c\}=\left\{c^{†},c^{†}\right\}=0.\text{\hspace{1em}(1.167)}$$

The Hamiltonian takes the form

$$H=\frac{1}{2}[c^{†}c-(1-c{c}^{†})]\omega =(N-\frac{1}{2})\omega (1.168)$$

where $N=c^{†}c$ . It is easy to see that the eigenvalue of $N$ must be either 0 or 1. In fact, $N$ satisfies $N^2=c^{†}c{c}^{†}c=N$ , namely $N(N-1)=0$ . This is nothing other than the Pauli principle.

Let us study the Hilbert space of the Hamiltonian $H$ . Let $|n⟩$ be an eigenvector of $H$ with the eigenvalue $n$ , where $n=0,1$ as shown earlier. It is easy to verify the following relations;

$$H|0⟩=-\frac{\omega }{2}|0⟩H|1⟩=\frac{\omega }{2}|1⟩$$

$$c^{†}|0⟩=|1⟩c|0⟩=0c^{†}|1⟩=0c|1⟩=|0⟩.$$

It is convenient to introduce the component expressions

$$|0⟩=\left(\begin{array}{c}0\\ 1\end{array}\right)|1⟩=\left(\begin{array}{c}1\\ 0\end{array}\right).$$

Exercise 1.6

Suppose the basis vectors have this form. Show that the operators have the following matrix representations

$$\begin{array}{rl}& c=\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right),c^{†}=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right),\\ & N=\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right),H=\frac{\omega }{2}\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right).\end{array}$$

The commutation relation $[x,p]=\mathrm{i}$ for a boson has been replaced by $[x,p]=0$ in the path integral formalism of a boson. For a fermion, the anti-commutation relation $\{c,c^{†}\}=1$ should be replaced by $\{\theta ,{\theta }^{\ast }\}=0$ , where $\theta$ and ${\theta }^{\ast }$ are anti-commuting classical numbers called Grassmann numbers.

1.5.2 Calculus of Grassmann numbers

To distinguish anti-commuting Grassmann numbers from commuting real and complex numbers, the latter will be called the ‘c-number’, where $c$ stands for commuting. Let $n$ generators $\{{\theta }_1,\dots ,{\theta }_n\}$ satisfy the anti-commutation relations

$$\left\{{\theta }_i,{\theta }_j\right\}=0\forall i,j.\text{\hspace{1em}(1.169)}$$

Then the set of the linear combinations of $\{{\theta }_i\}$ with the c-number coefficients is called the Grassmann number and the algebra generated by $\{{\theta }_i\}$ is called the Grassmann algebra, denoted by ${\Lambda }^n$ . An arbitrary element $f$ of ${\Lambda }^n$ is expanded as

$$\begin{array}{rl}f(\theta )& =f_0+\sum _{i=1}^n{f}_i{\theta }_i+\sum _{i<j}{f}_{ij}{\theta }_i{\theta }_j+\dots \\ & =\sum _{0\le k\le n}\frac{1}{k!}\sum _{\{i\}}{f}_{i_1,\dots i_k}{\theta }_{i_1}\dots {\theta }_{i_k},\end{array}\text{\hspace{1em}(1.170)}$$

where $f_0,f_i,f_{ij},\dots$ and $f_{i_1,\dots ,i_k}$ are c-numbers that are anti-symmetric under the exchange of any two indices. The element $f$ is also written as

$$f(\theta )=\sum _{k_i=0,1}{\tilde{f}}_{k_1,\dots ,k_n}{\theta }_1^{k_1}\dots {\theta }_n^{k_n}.\text{\hspace{1em}(1.171)}$$

Take $n=2$ for example. Then

$$\begin{array}{rl}f(\theta )& =f_0+f_1{\theta }_1+f_2{\theta }_2+f_{12}{\theta }_1{\theta }_2\\ & ={\tilde{f}}_{00}+{\tilde{f}}_{10}{\theta }_1+{\tilde{f}}_{01}{\theta }_2+{\tilde{f}}_{11}{\theta }_1{\theta }_2.\end{array}$$

The subset of ${\lambda }^n$ which is generated by monomials of even (resp. odd) power in ${\theta }_k$ is denoted by ${\Lambda }_{+}^n({\Lambda }_{-}^n)$ :

$${\Lambda }^n={\Lambda }_{+}^n\oplus {\Lambda }_{-}^n.\text{\hspace{1em}(1.172)}$$

The separation of ${\Lambda }^n$ into these two subspaces is called ${\mathbb{Z}}_2$ -grading. We call an element of ${\Lambda }_{+}^n({\Lambda }_{-}^n)$ G-even (G-odd). Note that $\dim {\lambda }^n=2^n$ while $\dim {\Lambda }_{+}^n=\dim {\Lambda }_{-}^n=2^{(n-1)}$ .

The generator ${\theta }_k$ does not have a magnitude and hence the set of Grassmann numbers is not an ordered set. Zero is the only number that is a c-number as well as a Grassmann number simultaneously. A Grassmann number commutes with a c-number. It should be clear that the generators satisfy the following relations:

$${\theta }_k^2=0$$

$${\theta }_{k_1}{\theta }_{k_2}\dots {\theta }_{k_n}={\epsilon }_{k_1{k}_2\dots k_n}{\theta }_1{\theta }_2\dots {\theta }_n\text{\hspace{1em}(1.173)}$$

$${\theta }_{k_1}{\theta }_{k_2}\dots {\theta }_{k_m}=0(m>n),$$

where

$${\epsilon }_{k_1\dots k_n}=\{\begin{array}{ll}+1& \text{i f}\{k_1\dots k_n\}\text{i s a n e v e n p e r m u t a t i o n o f}\{1\dots n\}\\ -1& \text{i f}\{k_1\dots k_n\}\text{i s a n o d d p e r m u t a t i o n o f}\{1\dots n\}\\ 0& \text{o t h e r w i s e .}\end{array}$$

A function of Grassmann numbers is defined as a Taylor expansion of the function. When $n=1$ , for example, we have

$${\mathrm{e}}^{\theta }=1+\theta$$

since higher-order terms in $\theta$ vanish identically.

1.5.3 Differentiation

It is assumed that the differential operator acts on a function from the left:

$$\frac{\partial {\theta }_j}{\partial {\theta }_i}=\frac{\partial }{\partial {\theta }_i}{\theta }_j={\delta }_{ij}.\text{\hspace{1em}(1.174)}$$

It is also assumed that the differential operator anti-commutes with ${\theta }_k$ . The Leibnitz rule then takes the form

$$\frac{\partial }{\partial {\theta }_i}\left({\theta }_j{\theta }_k\right)=\frac{\partial {\theta }_j}{\partial {\theta }_i}{\theta }_k-{\theta }_j\frac{\partial {\theta }_k}{\partial {\theta }_i}={\delta }_{ij}{\theta }_k-{\delta }_{ik}{\theta }_j.\text{\hspace{1em}(1.175)}$$

Exercise 1.7

Show that

$$\frac{\partial }{\partial {\theta }_i}\frac{\partial }{\partial {\theta }_j}+\frac{\partial }{\partial {\theta }_j}\frac{\partial }{\partial {\theta }_i}=0.\text{\hspace{1em}(1.176)}$$

It is easily shown from this exercise that the differential operator is nilpotent

$$\frac{{\partial }^2}{\partial {\theta }_i^2}=0.\text{\hspace{1em}(1.177)}$$

Exercise 1.8

Show that

$$\frac{\partial }{\partial {\theta }_i}{\theta }_j+{\theta }_j\frac{\partial }{\partial {\theta }_i}={\delta }_{ij}.\text{\hspace{1em}(1.178)}$$

1.5.4 Integration

Supprisingly enough, integration with respect to a Grassmann variable is equivalent to differentiation. Let $D$ denote differentiation with respect to a Grassmann variable and let $I$ denote integration, where integration is understood as a definite integral. Suppose they satisfy the relations

(1) $ID=0$
(2) $DI=0$
(3) $D(A)=0\Rightarrow I(BA)=I(B)A,$

where $A$ and $B$ are arbitrary functions of Grassmann variables. The first relation states that the integration of a derivative of any function yields the surface term and it is set to zero. The second relation states that a derivative of a definite integral vanishes. The third relation implies that $A$ is a constant if $D(A)=0$ and hence it can be taken out of the integral. These relations are satisfied if we take $I\propto D$ . Here we adopt the normalization $I=D$ and put

$$\int \mathrm{d}\theta f(\theta )=\frac{\partial f(\theta )}{\partial \theta }.\text{\hspace{1em}(1.179)}$$

We find from the previous definition that

$$\int \mathrm{d}\theta =\frac{\partial 1}{\partial \theta }=0\int \mathrm{d}\theta \theta =\frac{\partial \theta }{\partial \theta }=1.$$

If there are $n$ generators $\{{\theta }_k\}$ , equation (1.179) is generalized as

$$\int \mathrm{d}{\theta }_1\mathrm{d}{\theta }_2\dots \mathrm{d}{\theta }_{n}f\left({\theta }_1,{\theta }_2,\dots ,{\theta }_n\right)=\frac{\partial }{\partial {\theta }_1}\frac{\partial }{\partial {\theta }_2}\dots \frac{\partial }{\partial {\theta }_n}f\left({\theta }_1,{\theta }_2,\dots ,{\theta }_n\right).\text{\hspace{1em}(1.180)}$$

Note the order of $\mathrm{d}{\theta }_k$ and $\partial /\partial {\theta }_k$ .

The equivalence of differentiation and integration leads to an odd behaviour of integration under the change of integration variables. Let us consider the case $n=1$ first. Under the change of variable ${\theta }^{\prime }=a\theta$ ( $a\in \mathbb{C}$ ), we obtain

$$\int \mathrm{d}\theta f(\theta )=\frac{\partial f(\theta )}{\partial \theta }=\frac{\partial f({\theta }^{\prime }/a)}{\partial {\theta }^{\prime }/a}=a\int \mathrm{d}{\theta }^{\prime }f({\theta }^{\prime }/a)$$

which leads to $\mathrm{d}{\theta }^{\prime }=(1/a)\mathrm{d}\theta$ . This is readily extended to the case of $n$ variables. Let ${\theta }_i\to {\theta }_i^{\prime }=a_{ij}{\theta }_j$ . Then

$$\begin{array}{rl}\int \mathrm{d}{\theta }_1\dots {\theta }_{n}f(\theta )& =\frac{\partial }{\partial {\theta }_1}\dots \frac{\partial }{\partial {\theta }_n}f(\theta )\\ & =\sum _{k_i=1}^n\frac{\partial {\theta }_{k_1}^{\prime }}{\partial {\theta }_1}\dots \frac{\partial {\theta }_{k_n}^{\prime }}{\partial {\theta }_n}\frac{\partial }{\partial {\theta }_{k_1}^{\prime }}\dots \frac{\partial }{\partial {\theta }_{k_n}^{\prime }}f(a^{-1}{\theta }^{\prime })\\ & =\sum _{k_i=1}^n{\epsilon }_{k_1\dots k_n}{a}_{k_{1}1}\dots a_{k_{n}n}\frac{\partial }{\partial {\theta }_{k_1}^{\prime }}\dots \frac{\partial }{\partial {\theta }_{k_n}^{\prime }}f(a^{-1}{\theta }^{\prime })\\ & =\det a\int \mathrm{d}{\theta }_1^{\prime }\dots {\theta }_n^{\prime }f\left(a^{-1}{\theta }^{\prime }\right).\end{array}$$

Accordingly, the integral measure transforms as

$$\mathrm{d}{\theta }_1\mathrm{d}{\theta }_2\dots {\theta }_n=\det a\mathrm{d}{\theta }_1^{\prime }\mathrm{d}{\theta }_2^{\prime }\dots \mathrm{d}{\theta }_n^{\prime }.\text{\hspace{1em}(1.181)}$$

1.5.5 Delta-function

The $\delta$ -function of a Grassmann variable is introduced as

$$\int \mathrm{d}\theta \delta (\theta -\alpha )f(\theta )=f(\alpha )\text{\hspace{1em}(1.182)}$$

for a single variable. If we substitute the expansion $f(\theta )=a+b\theta$ into this definition, we obtain

$$\int \mathrm{d}\theta \delta (\theta -\alpha )(a+b\theta )=a+b\alpha$$

from which we find that the $\delta$ -function is explicitly given by

$$\delta (\theta -\alpha )=\theta -\alpha .\text{\hspace{1em}(1.183)}$$

Extension of this result to $n$ variables is easily verified to be (note the order of variables)

$${\delta }^n(\theta -\alpha )=\left({\theta }_n-{\alpha }_n\right)\dots \left({\theta }_2-{\alpha }_2\right)\left({\theta }_1-{\alpha }_1\right).\text{\hspace{1em}(1.184)}$$

The integral form of the $\delta$ -function is obtained from

$$\int \mathrm{d}\xi {\mathrm{e}}^{\mathrm{i}\xi \theta }=\int \mathrm{d}\xi (1+\mathrm{i}\xi \theta )=\mathrm{i}\theta$$

as

$$\delta (\theta )=\theta =-\mathrm{i}\int \mathrm{d}\xi {\mathrm{e}}^{\mathrm{i}\xi \theta }.\text{\hspace{1em}(1.185)}$$

1.5.6 Gaussian integral

Let us consider the integral

$$I=\int \mathrm{d}{\theta }_1^{\ast }\mathrm{d}{\theta }_1\dots \mathrm{d}{\theta }_n^{\ast }\mathrm{d}{\theta }_n{\mathrm{e}}^{-{\sum }_{ij}{\theta }_i^{\ast }{M}_{ij}{\theta }_j}\text{\hspace{1em}(1.186)}$$

where $\{{\theta }_i\}$ and $\{{\theta }_i^{\ast }\}$ are two sets of independent Grassmann variables. The $n\times n$ c-number matrix $M$ is taken to be anti-symmetric since ${\theta }_i$ and ${\theta }_i^{\ast }$ anti-commute. The integral is evaluated with the help of the change of variables ${\theta }_i^{\prime }={\sum }_j{M}_{ij}{\theta }_j$ as

$$\begin{array}{rl}I& =\det M\int \mathrm{d}{\theta }_1^{\ast }\mathrm{d}{\theta }_1^{\prime }\dots \mathrm{d}{\theta }_n^{\ast }\mathrm{d}{\theta }_n^{\prime }{\mathrm{e}}^{-{\sum }_i{\theta }_i^{\ast }{\theta }_i^{\prime }}\\ & =\det M{\left[\int d{\theta }^{\ast }d\theta \left(1+{\theta }^{\prime }{\theta }^{\ast }\right)\right]}^n\\ & =\det M.\end{array}\text{\hspace{1em}(1.187)}$$

We prove an interesting formula as an application of the Gaussian integral.

Proposition 1.3

Let $a$ be an anti-symmetric matrix of order $2n$ and define the Pfaffian of $a$ by

$$\mathrm{Pf}(a)=\frac{1}{2^{n}n!}\sum _{\begin{array}{c}\text{Permutations of}\\ \{i_1,\dots ,i_{2n}\}\end{array}}\mathrm{sgn}(P)a_{i_1{i}_2}\dots a_{i_{2n-1}{i}_{2n}}.\text{\hspace{1em}(1.188)}$$

Then

$$\det a=\mathrm{Pf}(a{)}^2.\text{\hspace{1em}(1.189)}$$

Proof

Observe that

$$\begin{array}{rl}I& =\int \mathrm{d}{\theta }_{2n}\dots \mathrm{d}{\theta }_1\exp \left[\frac{1}{2}\sum _{ij}{\theta }_i{a}_{ij}{\theta }_j\right]=\frac{1}{2^{n}n!}\int \mathrm{d}{\theta }_{2n}\dots \mathrm{d}{\theta }_1{\left(\sum _{ij}{\theta }_i{a}_{ij}{\theta }_j\right)}^n\\ & =\mathrm{Pf}(a).\end{array}$$

Note also that

$$I^2=\int \mathrm{d}{\theta }_{2n}\dots \mathrm{d}{\theta }_1\mathrm{d}{\theta }_{2n}^{\prime }\dots \mathrm{d}{\theta }_1\exp \left[\frac{1}{2}\sum _{ij}({\theta }_i{a}_{ij}{\theta }_j+{\theta }_i^{\prime }{a}_{ij}{\theta }_j^{\prime })\right].$$

Under the change of variables

$${\eta }_k=\frac{1}{\sqrt{2}}({\theta }_k+{\theta }_k^{\prime }),{\eta }_k^{\ast }=\frac{1}{\sqrt{2}\mathrm{i}}({\theta }_k-{\theta }_k^{\prime }),$$

we obtain the Jacobian $=(-1{)}^n$ and

$${\theta }_i{\theta }_j+{\theta }_i^{\prime }{\theta }_j^{\prime }={\eta }_i{\eta }_j^{\ast }-{\eta }_j^{\ast }{\eta }_i$$

$$\mathrm{d}{\eta }_{2n}\dots \mathrm{d}{\eta }_i\mathrm{d}{\eta }_{2n}^{\ast }\dots \mathrm{d}{\eta }_1^{\ast }=(-1{)}^{n^2}\mathrm{d}{\eta }_1\mathrm{d}{\eta }_1^{\ast }\dots \mathrm{d}{\eta }_{2n}\mathrm{d}{\eta }_{2n}^{\ast },$$

from which we verify that

$$\mathrm{Pf}(a{)}^2=\int \mathrm{d}{\eta }_1\mathrm{d}{\eta }_1^{\ast }\dots \mathrm{d}{\eta }_{2n}\mathrm{d}{\eta }_{2n}^{\ast }\exp \left[\sum _{ij}{\eta }_i^{\ast }{a}_{ij}{\eta }_j\right]=\det a.$$

Exercise 1.9

(1) Let $M$ be a skewsymmetric matrix and $K_i$ be Grassmann numbers. Show that

$$\int \mathrm{d}{\theta }_1\dots \mathrm{d}{\theta }_n{\mathrm{e}}^{-\frac{1}{2}t\theta \cdot M\cdot \theta +tK\cdot \theta }=2^{n/2}\sqrt{\det M}{\mathrm{e}}^{-tK\cdot M^{-1}\cdot K/4}.\text{\hspace{1em}(1.190)}$$

(2) Let $M$ be a skew-Hermitian matrix and $K_i$ and $K_i^{\ast }$ be Grassmann numbers. Show that

$$\int \mathrm{d}{\theta }_1^{\ast }\mathrm{d}{\theta }_1\dots \mathrm{d}{\theta }_n^{\ast }\mathrm{d}{\theta }_n{\mathrm{e}}^{-{\theta }^{†}\cdot M\cdot \theta +K^{†}\cdot \theta +{\theta }^{†}\cdot K}=\det M{\mathrm{e}}^{K^{†}\cdot M^{-1}\cdot K}.\text{\hspace{1em}(1.191)}$$

1.5.7 Functional derivative

The functional derivative with respect to a Grassmann variable can be defined similarly to that for a commuting variable. Let $\psi (t)$ be a Grassmann variable depending on a c-number parameter $t$ and $F[\psi (t)]$ be a functional of $\psi$ . Then we define

$$\frac{\delta F[\psi (t)]}{\delta \psi (s)}=\frac{1}{\epsilon }\left\{F[\psi (t)+\epsilon \delta (t-s)]-F[\psi (t)]\right\},\text{\hspace{1em}(1.192)}$$

where $\epsilon$ is a Grassmann parameter. The Taylor expansion of $F[\psi (t)-\epsilon \delta (t-s)]$ with respect to $\epsilon$ is linear in $\epsilon$ since ${\epsilon }^2=0$ . Accordingly, the limit $\epsilon \to 0$ is not necessary. A word of caution: division by a Grassmann number is not well defined in general. Here, however, the numerator is proportional to $\epsilon$ and division by $\epsilon$ simply means picking up the coefficient of $\epsilon$ in the numerator.

1.5.8 Complex conjugation

Let $\{{\theta }_i\}$ and $\{{\theta }_i^{\ast }\}$ be two sets of the generators of Grassmann numbers. Define the complex conjugation of ${\theta }_i$ by $({\theta }_i{)}^{\ast }={\theta }_i^{\ast }$ and $({\theta }_i^{\ast }{)}^{\ast }={\theta }_i$ . We define

$${\left({\theta }_i{\theta }_j\right)}^{\ast }={\theta }_j^{\ast }{\theta }_i^{\ast }.\text{\hspace{1em}(1.193)}$$

Otherwise, the real c-number ${\theta }_i{\theta }_i^{\ast }$ does not satisfy the reality condition $({\theta }_i{\theta }_i^{\ast }{)}^{\ast }={\theta }_i{\theta }_i^{\ast }$ .

1.5.9 Coherent states and completeness relation

The fermion annihilation and creation operators $c$ and $c^{†}$ satisfy the anti-commutation relations $\{c,c\}=\{c^{†},c^{†}\}=0$ and $\{c,c^{†}\}=1$ and the number operator $N=c^{†}c$ has the eigenvectors $|0⟩$ and $|1⟩$ . Let us consider the Hilbert space spanned by these vectors

$$\mathcal{H}=\mathrm{Span}\{|0⟩,|1⟩\}.$$

An arbitrary vector $|f⟩$ in $\mathcal{H}$ may be written in the form

$$|f⟩=|0⟩f_0+|1⟩f_1,$$

where $f_0,f_1\in \mathbb{C}$ .

Now we consider the states

$$|\theta ⟩=|0⟩+|1⟩\theta \text{\hspace{1em}(1.194)}$$

$$⟨\theta |=⟨0|+{\theta }^{\ast }⟨1|\text{\hspace{1em}(1.195)}$$

where $\theta$ and ${\theta }^{\ast }$ are Grassmann numbers. These states are called the coherent states and are eigenstates of $c$ and $c^{†}$ respectively,

$$c|\theta ⟩=|0⟩\theta =|\theta ⟩\theta ,⟨\theta |c^{†}={\theta }^{\ast }⟨0|={\theta }^{\ast }⟨\theta |.$$

Exercise 1.10

Verify the following identities;

$$⟨{\theta }^{\prime }|\theta ⟩=1+{\theta }^{\prime \ast }\theta ={\mathrm{e}}^{{\theta }^{\prime \ast }\theta },$$

$$⟨\theta |f⟩=f_0+{\theta }^{\ast }{f}_1,$$

$$⟨\theta |c^{†}|f⟩=⟨\theta |1⟩f_0={\theta }^{\ast }{f}_0={\theta }^{\ast }⟨\theta |f⟩,$$

$$⟨\theta |c|f⟩=⟨\theta |0⟩f_1=\frac{\partial }{\partial {\theta }^{\ast }}⟨\theta |f⟩.$$

Let

$$h(c,c^{†})=h_{00}+h_{10}{c}^{†}+h_{01}c+h_{11}{c}^{†}c{h}_{ij}\in \mathbb{C}$$

be an arbitrary function of $c$ and $c^{†}$ . Then the matrix elements of $h$ are

$$⟨0|h|0⟩=h_{00}⟨0|h|1⟩=h_{01}⟨1|h|0⟩=h_{10}⟨1|h|1⟩=h_{00}+h_{11}.$$

It is easily found from these matrix elements that

$$⟨\theta |h|{\theta }^{\prime }⟩=\left(h_{00}+{\theta }^{\ast }{h}_{10}+h_{01}{\theta }^{\prime }+{\theta }^{\ast }{\theta }^{\prime }{h}_{11}\right)e^{{\theta }^{\ast }{\theta }^{\prime }}.\text{\hspace{1em}(1.196)}$$

Lemma 1.3

Let $|\theta ⟩$ and $⟨\theta |$ be defined as before. Then the completeness relation takes the form

$$\int \mathrm{d}{\theta }^{\ast }\mathrm{d}\theta |\theta ⟩⟨\theta |{\mathrm{e}}^{-{\theta }^{\ast }\theta }=I.\text{\hspace{1em}(1.197)}$$

Proof

Straightforward calculation yields

$$\begin{array}{rl}& \int \mathrm{d}{\theta }^{\ast }\mathrm{d}\theta |\theta ⟩⟨\theta |{\mathrm{e}}^{-{\theta }^{\ast }\theta }\\ & =\int d{\theta }^{\ast }d\theta (|0⟩+|1⟩\theta )(⟨0|+{\theta }^{\ast }⟨1|)(1-{\theta }^{\ast }\theta )\\ & =\int d{\theta }^{\ast }d\theta (|0⟩⟨0|+|1⟩\theta ⟨0|+|0⟩{\theta }^{\ast }⟨1|+|1⟩\theta {\theta }^{\ast }⟨1|)(1-{\theta }^{\ast }\theta )\\ & =|0⟩⟨0|+|1⟩⟨1|=I.\end{array}$$

$\square$

1.5.10 Partition function of a fermionic oscillator

We obtain here the partition fuction of a fermionic harmonic oscillator as an application of the path integral formalism of fermions. The Hamiltonian is $H=(c^{†}c-1/2)\omega$ , which has eigenvalues $\pm \omega /2$ . The partition function is then

$$Z(\beta )=\mathrm{Tr}{\mathrm{e}}^{-\beta H}=\sum _{n=0}^1⟨n|{\mathrm{e}}^{-\beta H}|n⟩={\mathrm{e}}^{\beta \omega /2}+{\mathrm{e}}^{-\beta \omega /2}=2\cosh (\beta \omega /2).\text{\hspace{1em}(1.198)}$$

Now we evaluate $Z(\beta )$ in two different ways using a path integral. We start our exposition with the following lemma.

Lemma 1.4

Let $H$ be the Hamiltonian of a fermionic harmonic oscillator. Then the partition function is written as

$$\mathrm{Tr}{\mathrm{e}}^{-\beta H}=\int \mathrm{d}{\theta }^{\ast }\mathrm{d}\theta ⟨-\theta |{\mathrm{e}}^{-\beta H}|\theta ⟩{\mathrm{e}}^{-{\theta }^{\ast }\theta }.\text{\hspace{1em}(1.199)}$$

Proof

Let us insert the completeness relation (1.197) into the definition of a partition function to obtain

$$\begin{array}{rl}& Z(\beta )=\sum _{n=0,1}⟨n|{\mathrm{e}}^{-\beta H}|n⟩\\ & =\sum _n\int \mathrm{d}{\theta }^{\ast }\mathrm{d}\theta {\mathrm{e}}^{-{\theta }^{\ast }\theta }⟨n|\theta ⟩⟨\theta |{\mathrm{e}}^{-\beta H}|n⟩\\ & =\sum _n\int \mathrm{d}{\theta }^{\ast }\mathrm{d}\theta (1-{\theta }^{\ast }\theta )(⟨n|0⟩+⟨n|1⟩\theta )(⟨0|{\mathrm{e}}^{-\beta H}|n⟩+{\theta }^{\ast }⟨1|{\mathrm{e}}^{-\beta H}|n⟩)\\ & =\sum _n\int \mathrm{d}{\theta }^{\ast }\mathrm{d}\theta (1-{\theta }^{\ast }\theta )[⟨0|e^{-\beta H}|n⟩⟨n|0⟩\\ & -{\theta }^{\ast }\theta ⟨1|e^{-\beta H}|n⟩⟨n|1⟩+\theta ⟨0|e^{-\beta H}|n⟩⟨n|1⟩+{\theta }^{\ast }⟨1|e^{-\beta H}|n⟩⟨n|0⟩].\end{array}$$

The last term of the last line does not contribute to the integral and hence we may change ${\theta }^{\ast }$ to $-{\theta }^{\ast }$ . Then

$$\begin{array}{rl}& Z(\beta )=\sum _n\int \mathrm{d}{\theta }^{\ast }\mathrm{d}\theta (1-{\theta }^{\ast }\theta )[⟨0|{\mathrm{e}}^{-\beta H}|n⟩⟨n|0⟩\\ & -{\theta }^{\ast }\theta ⟨1|e^{-\beta H}|n⟩⟨n|1⟩+\theta ⟨0|e^{-\beta H}|n⟩⟨n|1⟩-{\theta }^{\ast }⟨1|e^{-\beta H}|n⟩⟨n|0⟩]\\ & =\int \mathrm{d}{\theta }^{\ast }\mathrm{d}\theta {\mathrm{e}}^{-{\theta }^{\ast }\theta }⟨-\theta |{\mathrm{e}}^{-\beta H}|\theta ⟩.\end{array}$$

Accordingly, the coordinate in the trace is over anti-periodic orbits. The Grassmann variable is $\theta$ at $\tau =0$ while $-\theta$ at $\tau =\beta$ and we have to impose an anti-periodic boundary condition over $[0,\beta ]$ in the trace.

Use the expression

$${\mathrm{e}}^{-\beta H}=\underset{N\to \infty }{\lim }(1-\beta H/N{)}^N$$

and insert the completeness relation at each time step to find

$$\begin{array}{rl}& Z(\beta )=\underset{N\to \infty }{\lim }\int \mathrm{d}{\theta }^{\ast }\mathrm{d}\theta {\mathrm{e}}^{-{\theta }^{\ast }\theta }⟨-\theta |(1-\beta H/N{)}^N|\theta ⟩\\ & =\underset{N\to \infty }{\lim }\int \mathrm{d}{\theta }^{\ast }\mathrm{d}\theta \prod _{k=1}^{N-1}\int \mathrm{d}{\theta }_k^{\ast }\mathrm{d}{\theta }_k{\mathrm{e}}^{-{\sum }_{n=1}^{N-1}{\theta }_n^{\ast }{\theta }_n}\\ & \times ⟨-\theta |(1-\epsilon H)|{\theta }_{N-1}⟩⟨{\theta }_{N-1}|\dots |{\theta }_1⟩⟨{\theta }_1|(1-\epsilon H)|\theta ⟩\\ & =\underset{N\to \infty }{\lim }\int \prod _{k=1}^{N}d{\theta }_k^{\ast }d{\theta }_k{e}^{-{\sum }_{n=1}^N{\theta }_n^{\ast }{\theta }_n}\\ & \times ⟨{\theta }_N|(1-\epsilon H)|{\theta }_{N-1}⟩⟨{\theta }_{N-1}|\dots |{\theta }_1⟩⟨{\theta }_1|(1-\epsilon H)|-{\theta }_N⟩\end{array}$$

where we have put $\epsilon =\beta /N$ and $\theta =-{\theta }_N={\theta }_0,{\theta }^{\ast }=-{\theta }_N^{\ast }={\theta }_0^{\ast }$

Each matrix element is evaluated as

$$\begin{array}{rl}⟨{\theta }_k|(1-\epsilon H)\mid {\theta }_{k-1}⟩& =⟨{\theta }_k\mid {\theta }_{k-1}⟩[1-\epsilon \frac{⟨{\theta }_k|H\mid {\theta }_{k-1}⟩}{⟨{\theta }_k\mid {\theta }_{k-1}⟩}]\\ & \simeq ⟨{\theta }_k|{\theta }_{k-1}⟩e^{-\epsilon ⟨{\theta }_k|H|{\theta }_{k-1}⟩/⟨{\theta }_k|{\theta }_{k-1}⟩}\\ & ={\mathrm{e}}^{{\theta }_k^{\ast }{\theta }_{k-1}}{\mathrm{e}}^{-\epsilon \omega ({\theta }_k^{\ast }{\theta }_{k-1}-1/2)}\\ & ={\mathrm{e}}^{\epsilon \omega /2}{\mathrm{e}}^{(1-\epsilon \omega ){\theta }_k^{\ast }{\theta }_{k-1}}.\end{array}$$

The partition function is now expressed in terms of the path integral as

$$\begin{array}{rl}Z(\beta )& =\underset{N\to \infty }{\lim }{\mathrm{e}}^{\beta \omega /2}\prod _{k=1}^N\int \mathrm{d}{\theta }_k^{\ast }\mathrm{d}{\theta }_k{\mathrm{e}}^{-{\sum }_{n=1}^N{\theta }_n^{\ast }{\theta }_n}{\mathrm{e}}^{(1-\epsilon \omega ){\sum }_{n=1}^N{\theta }_n^{\ast }{\theta }_{n-1}}\\ & ={\mathrm{e}}^{\beta \omega /2}\underset{N\to \infty }{\lim }\prod _{k=1}^N\int \mathrm{d}{\theta }_k^{\ast }\mathrm{d}{\theta }_k{\mathrm{e}}^{-{\sum }_{n=1}^N\left[{\theta }_n^{\ast }\left({\theta }_n-{\theta }_{n-1}\right)+\epsilon \omega {\theta }_n^{\ast }{\theta }_{n-1}\right]}\\ & ={\mathrm{e}}^{\beta \omega /2}\underset{N\to \infty }{\lim }\prod _{k=1}^N\int \mathrm{d}{\theta }_k^{\ast }\mathrm{d}{\theta }_k{\mathrm{e}}^{-{\theta }^{†}\cdot B\cdot \theta },\end{array}\text{\hspace{1em}(1.200)}$$

where

$$\theta =\left(\begin{array}{c}{\theta }_1\\ {\theta }_2\\ ⋮\\ {\theta }_N\end{array}\right){\theta }^{†}=\left({\theta }_1^{\ast },{\theta }_2^{\ast },\dots ,{\theta }_N^{\ast }\right)$$

$$B_N=\left(\begin{array}{ccccc}1& 0& \dots & 0& -y\\ y& 1& 0& \dots 0& \\ 0& y& 1& \dots 0& \\ ⋮& & \ddots & & ⋮\\ 0& 0& \dots & y& 1\end{array}\right)$$

with $y=-1+\epsilon \omega$ in the last line. We finally find from the definition of the Gaussian integral of Grassmann numbers that

$$\begin{array}{rl}Z(\beta )& ={\mathrm{e}}^{\beta \omega /2}\underset{N\to \infty }{\lim }\det B_N={\mathrm{e}}^{\beta \omega /2}\underset{N\to \infty }{\lim }[1+(1-\beta \omega /N{)}^N]\\ & ={\mathrm{e}}^{\beta \omega /2}\left(1+{\mathrm{e}}^{-\beta \omega }\right)=2\cosh \frac{1}{2}\beta \omega .\end{array}\text{\hspace{1em}(1.201)}$$

This should be compared with the partition function (1.151) of the bosonic harmonic oscillator.

This partition function is also obtained by making use of the $\zeta$ -function regularization. It follows from the second line of equation (1.200) that

$$\begin{array}{rl}Z(\beta )& ={\mathrm{e}}^{\beta \omega /2}\underset{N\to \infty }{\lim }\prod _{k=1}^N\int \mathrm{d}{\theta }_k^{\ast }\mathrm{d}{\theta }_k{\mathrm{e}}^{-{\sum }_n[(1-\epsilon \omega ){\theta }_n^{\ast }({\theta }_n-{\theta }_{n-1})/\epsilon +\omega {\theta }_n^{\ast }{\theta }_n]}\\ & ={\mathrm{e}}^{\beta \omega /2}\int \mathcal{D}{\theta }^{\ast }\mathcal{D}\theta \exp \left[-{\int }_0^{\beta }\mathrm{d}\tau {\theta }^{\ast }\left((1-\epsilon \omega )\frac{\mathrm{d}}{\mathrm{d}\tau }+\omega \right)\theta \right]\\ & ={\mathrm{e}}^{\beta \omega /2}{\mathrm{Det}}_{\mathrm{APBC}}\left((1-\epsilon \omega )\frac{\mathrm{d}}{\mathrm{d}\tau }+\omega \right).\end{array}$$

Here the subscript APBC implies that the eigenvalue should be evaluated for the solutions that satisfy the anti-periodic boundary condition $\theta (\beta )=-\theta (0)$ . It

might seem odd that the differential operator contains $\epsilon$ . We find later that this gives a finite contribution to the infinite product of eigenvalues. Let us expand the orbit $\theta (\tau )$ in the Fourier modes. The eigenmodes and the corresponding eigenvalues are

$$\exp \left(\frac{\pi \mathrm{i}(2n+1)\tau }{\beta }\right),(1-\epsilon \omega )\frac{\pi \mathrm{i}(2n+1)}{\beta }+\omega ,$$

where $n=0,\pm 1,\pm 2,\dots$ . It should be noted that the coherent states are overcomplete and that the actual number of degrees of freedom is $N$ , which is related to $\epsilon$ as $\epsilon =\beta /N$ . Then we have to truncate the product at $-N/4\le k\le N/4$ since one complex variable has two real degrees of freedom. Accordingly, the partition function takes the form

$$\begin{array}{rl}Z(\beta )& ={\mathrm{e}}^{\beta \omega /2}\underset{N\to \infty }{\lim }\prod _{k=-N/4}^{N/4}\left[\mathrm{i}(1-\epsilon \omega )\frac{\pi (2n-1)}{\beta }+\omega \right]\\ & ={\mathrm{e}}^{\beta \omega /2}{\mathrm{e}}^{-\beta \omega /2}\prod _{k=1}^{\infty }\left[{\left(\frac{2\pi (n-1/2)}{\beta }\right)}^2+{\omega }^2\right]\\ & =\prod _{k=1}^{\infty }{\left[\frac{\pi (2k-1)}{\beta }\right]}^2\prod _{n=1}^{\infty }\left[1+{\left(\frac{\beta \omega }{\pi (2n-1)}\right)}^2\right].\end{array}$$

The first infinite product, which we call $P$ , is divergent and requires regularization. Note, first, that

$$\log P=\sum _{k=1}^{\infty }2\log \left[\frac{2\pi (k-1/2)}{\beta }\right].$$

Define the corresponding $\zeta$ -function by

$$\tilde{\zeta }(s)=\sum _{k=1}^{\infty }{\left[\frac{2\pi (k-1/2)}{\beta }\right]}^{-s}={\left(\frac{\beta }{2\pi }\right)}^s\zeta (s,1/2)$$

with which we obtain $P={\mathrm{e}}^{-2{\zeta }^{\prime }(0)}$ .Here

$$\zeta (s,a)=\sum _{k=0}^{\infty }\frac{1}{(k+a{)}^s}(0<a<1)\text{\hspace{1em}(1.202)}$$

is the generalized $\zeta$ -function (the Hurwitz $\zeta$ -function). The derivative of $\tilde{\zeta }(s)$ at $s=0$ yields

$${\tilde{\zeta }}^{\prime }(0)=\log \left(\frac{\beta }{2\pi }\right)\zeta (0,1/2)+{\zeta }^{\prime }(0,1/2)=-\frac{1}{2}\log 2,$$

where use has been made of the values ²

$$\zeta (0,1/2)=0{\zeta }^{\prime }(0,1/2)=-\frac{1}{2}\log 2.$$

Finally we obtain

$$P={\mathrm{e}}^{-2{\bar{\zeta }}^{\prime }(0)}={\mathrm{e}}^{\log 2}=2.\text{\hspace{1em}(1.203)}$$

Note that $P$ is independent of $\beta$ after regularization.

Putting them all together, we arrive at the partition function

$$Z(\beta )=2\prod _{n=1}^{\infty }\left[1+{\left(\frac{\beta \omega }{\pi (2n-1)}\right)}^2\right].\text{\hspace{1em}(1.204)}$$

By making use of the well-known formula

$$\cosh \frac{x}{2}=\prod _{n=1}^{\infty }\left[1+\frac{x^2}{{\pi }^2(2n-1{)}^2}\right]\text{\hspace{1em}(1.205)}$$

we obtain

$$Z(\beta )=2\cosh \frac{\beta \omega }{2}.\text{\hspace{1em}(1.206)}$$

Suppose, alternatively, we are ignorant about the formula (1.205). Then, by equating equation (1.201) with equation (1.204), we have proved the formula (1.205) with the help of path integrals. This is a typical application of physics to mathematics: evaluate some physical quantity by two different methods and equate the results. Then we often obtain a non-trivial relation which is mathematically useful.

Endnotes

来源与版本

正文：英文 · OCR 机器稿 · 待校对

核对状态：OCR 机器稿 · 待校对

原书 PDF

来源版本：second-edition-local

来源 PDF SHA-256：fcd24c3371d092a8c699ccf5cf0765462924b6c48412804231076e8084b1940d

OCR 来源 SHA-256：0b6324f01172a3548cd34382e82fd8bfae574602b6b2cfaa357ba96bc2c35478

OCR 产物 SHA-256：75dbaea4adf4b126a7c55263f185e1503a69483afaeebb7b4642969dde0e4b14

Footnotes

1. We will drop $\hat{}$ on operators from now on unless this may cause confusion. ↩

2. The first formula follows from the relation $\zeta (s,1/2)=(2^s-1)\zeta (s)$ , which is derived from the identity $\zeta (s,1/2)+\zeta (s)=2^s{\sum }_{n=1}^{\infty }[1/(2n-1{)}^s+1/(2n{)}^s]=2^s\zeta (s)$ . The second formula is obtained by differentiating $\zeta (s,1/2)=(2^s-1)\zeta (s)$ with respect to $s$ and using the formula $\zeta (0)=-1/2$ . ↩