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Okay, this is a second part of my previous question. Again, I'm following Itzykson's book. The fermionic solution for the 2D Ising model is described in terms of a matrix $T = \theta \tilde{\theta}$, where: $$\theta(\beta) = e^{-i\beta \sum_{x}\Gamma_{x,x+\frac{1}{2}}\Gamma_{x+\frac{1}{2},x}} \quad \mbox{and} \quad \tilde{\theta}(\beta) = e^{-i\tilde{\beta}\sum_{x}\Gamma_{x-\frac{1}{2},x}\Gamma_{x,x+\frac{1}{2}}}$$ The idea now is to find the kernels of each of these operators in terms of Grassmann variables. To compute the kernel, Itzykson states that we have to use the unitary operator $U$ which shifts the indices by a half factor: $$U^{-1}\Gamma_{x,x+\frac{1}{2}}U = \Gamma_{x+\frac{1}{2},x+1} \quad \mbox{and} \quad U^{-1}\Gamma_{x-\frac{1}{2},x}U = \Gamma_{x,x+\frac{1}{2}}$$ The result should be: $$U(\bar{\xi},\xi) = 2^{-L/2}\exp\bigg{(}{{\sum_{x\ge x'}2i\bar{\xi}_{x}\xi_{x'}-\bar{\xi}_{x'}\bar{\xi}_{x}+\xi_{x}\xi_{x'}}}\bigg{)} = 2^{-L/2}\exp \int_{-\pi}^{\pi}\frac{dp}{2\pi}\bigg{[}\frac{2i}{1-e^{-ip}}\bar{\xi}_{p}\xi_{p}+\frac{1+e^{-ip}}{2(1-e^{-ip})}(\bar{\xi}_{-p}\bar{\xi}_{p}+\xi_{-p}\xi_{p})\bigg{]} .$$ As it became clear in the answer of my previous linked question, the kernel of such operators must use some Grassmann identities. But my problem is: how can I calculate the kernel of $U$, that is, $U(\bar{\xi},\xi)$ if I don't know what $U$ is? All I know is what $U$ does: it shifts the indices by a half factor. But I don't know $U$ (and $U^{-1}$) explicitly. How is the above expression for $U(\bar{\xi},\xi)$ obtained?

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  • $\begingroup$ What exactly do you mean by kernel? $\endgroup$ May 3, 2021 at 0:38
  • $\begingroup$ It is the integral kernel representation of the operator in the Grassmann algebra representation. My previous question (linked in the post) have more information. $\endgroup$
    – IamWill
    May 3, 2021 at 0:49

1 Answer 1

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To be honest, I'm not sure the $U$ operator exists. The book does not contain its complete consideration. I expect that there may be problems with boundaries and translational invariance. Nevertheless, here is the formal derivation of formula (65a).

Transformation (63) is linear. Therefore, the operator $U$ is probably the exponential of some quadratic fermion operator, and its kernel is probably a Gaussian function of the anticommuting variables. For this reason, let's look for the kernel in the following form $$ U(\overline{\xi},\xi) = C \exp\left(\sum_{x,x'}\ \frac12A_{x,x'}\overline{\xi}_x\overline{\xi}_{x'} + B_{x,x'}\overline{\xi}_x \xi_{x'} + \frac12 A'_{x,x'} \xi_{x'}\xi_x\right) \tag{O} $$ where $A$ and $A'$ are antisymmetric: $A_{x',x} = -A_{x,x'}$.

If $\langle 0|a^+_x = 0$ for any $x$, then by formulas (21) and (O) we have equality $$ \langle newvac| \equiv \langle 0|U = C\langle 0|\exp\left(\frac12\sum_{x,x'}\ A'_{x,x'}a_{x'}a_x\right). \tag{Ia} $$ At the same time we have another equality $$ \langle newvac|b^+_x = \langle 0|U b^+_x = \langle 0|a^+_x U = 0 \tag{Ib} $$ for any $x$. Substituting (63) and (Ia) into (Ib) gives the following equations for $A'$ $$ \langle newvac|\left(a_{x+1} + a_x +\sum_{x'}\ A'_{x+1,x'}a_{x'} - A'_{x,x'}a_{x'} \right) = 0, \tag{Ic} $$ whence follows $$ A'_{x+1,x} = -1,\quad A'_{x,x+1} = 1,\quad A'_{x+1,x'} = A'_{x,x'},\forall x'\neq x, x+1. \tag{Ic'} $$ The solution to these equations is $$ A'_{x,x'} = \left\{ \begin{array}{ccc} 1&\mbox{if}&x'>x\\ 0&\mbox{if}&x'=x\\ -1&\mbox{if}&x'<x \end{array}\right. . \tag{Id} $$ The vector $|newvac\rangle$ corresponding to the function (Id) has norm $C 2^{\frac{L-1}2}$. On the other hand, its norm must be equal to $1$ due to the unitarity of $U$. Therefore we get $$ C = 2^{-\frac{L-1}{2}}.\tag{Ie} $$

According to formula (20), the operator with kernel (O) must satisfy the following relations $$ a_xU = \sum_{x'}\ A_{x,x'}a_{x'}^+ U + B_{x,x'} U a_{x'} \tag{IIa}, $$ $$ Ua^+_x = \sum_{x'}\ A'_{x,x'}Ua_{x'} + B_{x',x}a^+_{x'}U \tag{IIb} $$ Or equivalently $$ U^{-1}a_xU = \sum_{x'}\ A_{x,x'} U^{-1}a_{x'}^+ U + B_{x,x'} a_{x'} \tag{IIa'}, $$ $$ a^+_x = \sum_{x'}\ A'_{x,x'}a_{x'} + B_{x',x} U^{-1} a^+_{x'}U \tag{IIb'} $$ Substituting of transformation (63) into these equations gives the equations for $B$ and $A$. From (IIb') and (Id) it follows $$ a^+_x = \sum_{x'>x} a_{x'} - \sum_{x'<x}a_{x'} + \sum_{x'}B_{x,x'}\frac{i}2\left( a^+_{x'+1} - a^+_{x'} + a_{x'+1} + a_{x'} \right),\forall x.\tag{IIb''} $$ Analysis of equation (IIb'') leads to the following expression for $B$: $$ B_{x,x'} = \left\{ \begin{array}{ccc} i&\mbox{if}&x\geq x\\ -i&\mbox{if}&x<x \end{array}\right. . \tag{IIc} $$ Similarly, from (IIa') it follows $$ \frac{i}2\left(-a^+_{x+1} - a^+_x - a_{x+1} + a_x \right) = $$ $$ = \sum_{x'}\ B_{x,x'}a_{x'} + A_{x,x'}\frac{i}2\left(a^+_{x'+1} - a^+_{x'} + a_{x'+1} + a_{x'} \right),\forall x. \tag{IIa''} $$ Equation (IIa'') is consistent with (IIc) and leads to $$ A_{x,x'} = A'_{x,x'}. \tag{Id'} $$

Now, substituting (Id), (Ie), (IIc) and (Id') into (O) gives $$ U(\overline{\xi},\xi) = 2^{-\frac{L-1}{2}}\exp\left(i\sum_{x\geq x'} \overline{\xi}_x \xi_{x'} - i \sum_{x < x'} \overline{\xi}_x \xi_{x'} - \sum_{x>x'} (\overline{\xi}_x \overline{\xi}_{x'} - \xi_x \xi_{x'})\right).\tag{R} $$ This result does not coincide with the central expression in the line of equalities (65a). I consider the difference in normalization constants to be insignificant; due to boundary effects, much more significant corrections can arise. I don't know how to explain the difference in exponent arguments. But, to my surprise, substituting transformation (64) into (R) leads exactly to the final expression in (65a). To obtain this result it is necessary to use equalities like the following $$ \sum_x e^{i(p\pm p')x} = 2\pi\delta(p \pm p'),\quad \sum_{n=1}^\infty e^{\pm ipn} = \frac{e^{\pm ip}}{1- e^{\pm ip}}, \tag{E} $$ $$ \sum_{x>x'} \xi_x \xi_{x'} = \sum_x\sum_{n=1}^\infty \xi_x \xi_{x-n},\quad \sum_{x<x'} \xi_x \xi_{x'} = \sum_x\sum_{n=1}^\infty \xi_x \xi_{x+n}. \tag{E'} $$

Due to unitarity, the kernel of the inverse operator $U^{-1}$ is conjugate to the kernel of the operator $U$.

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