On Correlation Functions in the Coordinate and the Algebraic Bethe Ansatz

Abstract

The Bethe ansatz, both in its coordinate and its algebraic version, is an exceptional method to compute the eigenvectors and eigenvalues of integrable systems. However, computing correlation functions using the eigenvectors thus constructed complicates rather fast. In this article, we will compute some simple correlation functions for the isotropic Heisenberg spin chain to highlight the shortcomings of both Bethe ansätze. In order to compare the results obtained from each approach, a discussion on the normalization of states in each ansatz will be required. We will show that the analysis can be extended to the long-range spin chain governing the spectrum of anomalous dimensions of single trace operators in four-dimensional Yang-Mills with maximal supersymmetry.

Appendices

A General Form of $\mathcal{F}^L_n$

In this appendix we are going to obtain the general expression of the function $\mathcal{F}^L_n$. All along the calculation, the limit $\alpha \rightarrow 0$ will be assumed. Using the first recurrence relation in (4.39) and setting both d and $\frac{\partial ^{} d}{\partial \lambda ^{}}$ to zero we find

$$\begin{aligned} \mathcal{F}^L_n=\mathcal{F}^L_0 + i\mathcal {D} \mathcal{F}^L_0 + i\mathcal {D} \mathcal{F}^L_1 + \dots + i\mathcal {D} \mathcal{F}^L_{n-1} \ . \end{aligned}$$

(A.1)

If we assume that $n<L-1$, the second recurrence equation gives

$$\begin{aligned} \mathcal {D} \mathcal{F}^L_n=\left( {\begin{array}{c}n\\ 0\end{array}}\right) \mathcal {D} \mathcal{F}^L_0+\left( {\begin{array}{c}n\\ 1\end{array}}\right) \frac{i\mathcal {D}^2}{2!} \mathcal{F}^L_0+\dots =\sum _{j=0}^n{\left( {\begin{array}{c}n\\ j\end{array}}\right) \frac{i^j \mathcal {D}^{j+1}}{(j+1)!} \mathcal{F}^L_0} \ . \end{aligned}$$

(A.2)

Therefore we need to sum the series

$$\begin{aligned} \sum _{j=0}^{n-1}{i\mathcal {D} \mathcal{F}^L_j}=\sum _{j=0}^{n-1}{\sum _{k=0}^j{\left( {\begin{array}{c}j\\ k\end{array}}\right) \frac{i^{k+1} \mathcal {D}^{k+1}}{(k+1)!} \mathcal{F}^L_0}} \ . \end{aligned}$$

(A.3)

As a first step, we can commute the two sums as $\sum _{j=0}^{n-1}{\sum _{k=0}^j{}}=\sum _{k=1}^{n-1}{\sum _{j=k}^{n-1}{}}+\sum _{j=0}^{n-1}{\delta _{k,0}}$, because the index j only appears in the limit of the sum and in the binomial coefficient, so it is easier to perform first the sum over j. The second term is easy to evaluate because we only have to calculate $\sum _{j=0}^{n-1}{\left( {\begin{array}{c}j\\ 0\end{array}}\right) }=\left( {\begin{array}{c}n\\ 1\end{array}}\right)$. The sum over j of the first term can be computed using the properties of the binomial coefficients $\sum _{j=k}^{n-1}{\left( {\begin{array}{c}j\\ k\end{array}}\right) } = \left( {\begin{array}{c}n-1+1\\ k+1\end{array}}\right)$. The whole sum can thus be rewritten as

$$\begin{aligned} \mathcal{F}^L_n=\mathcal{F}^L_0+\sum _{k=1}^{n}{\left( {\begin{array}{c}n\\ k\end{array}}\right) \frac{i^{k} \mathcal {D}^{k}}{k!} \mathcal{F}^L_0} = \sum _{k=0}^{n}{\left( {\begin{array}{c}n\\ k\end{array}}\right) \frac{i^{k} \mathcal {D}^{k}}{k!} \mathcal{F}^L_0} \ . \end{aligned}$$

(A.4)

This equation is true as long as $n < L-1$. If we want to get an expression valid for $n\ge L-1$, we have to take into account derivatives of d of order greater than or equal to L, which can be performed independently of the calculation we have already done, because

$$\begin{aligned} \mathcal {D}^{L+\alpha -1} \mathcal{F}^L_{j+1}=\frac{i}{L+\alpha } \mathcal{F}^L_j \frac{\partial ^{L+\alpha } d}{\partial \lambda ^{L+\alpha }} + \cdots \ , \end{aligned}$$

where the dots stand for the part that we have already taken into account. Therefore, the d-contribution to $\mathcal {D}\mathcal{F}^L_M$ will be of the form

$$\begin{aligned} i\mathcal {D}\mathcal{F}^L_M&=\sum _{j=1}^{M-L+2}{\sum _{k=0}^{M+2-L-j}{\frac{i^{L+k-1}}{(L+k-1)!} \left( {\begin{array}{c}M-j\\ L+k-2\end{array}}\right) \mathcal {D}^{L+k-1} \mathcal{F}^L_j}} \\&=\sum _{j=1}^{M-L+2}{\sum _{k=0}^{M+2-L-j}{\frac{i^{L+k}}{(L+k)!} \left( {\begin{array}{c}M-j\\ L+k-2\end{array}}\right) \frac{\partial ^{L+k} d}{\partial \lambda ^{L+k}} \mathcal{F}^L_{j-1}}} \ , \end{aligned}$$

and the derivative of d can be calculated using Leibniz’s rule,

$$\begin{aligned} \left. \frac{\partial ^{L+k} d}{\partial \lambda ^{L+k}} \right| _\xi =\sum _{j=0}^{L+k}{\left( {\begin{array}{c}L+k\\ j\end{array}}\right) \frac{\partial ^{j} (\lambda -\xi )^L}{\partial \lambda ^{j}} \, \frac{\partial ^{L+k-j} (\lambda +\xi )^{-L}}{\partial \lambda ^{L+k-j}}} \ . \end{aligned}$$

Because we want to evaluate this expression at $\lambda =\xi$, the only non-zero contribution comes from the term with L derivatives acting on $(\lambda - \xi )^L$. This implies

$$\begin{aligned} \left. \frac{\partial ^{L+k} d}{\partial \lambda ^{L+k}} \right| _\xi =\left( {\begin{array}{c}L+k\\ L\end{array}}\right) \frac{\partial ^{L} (\lambda -\xi )^L}{\partial \lambda ^{L}} \, \frac{\partial ^{k} (\lambda +\xi )^{-L}}{\partial \lambda ^{k}}=\frac{(L+k)!}{k!} \, \frac{(L+k-1)!}{(L-1)!} \, \frac{(-1)^k}{i^{L+k}} \ . \end{aligned}$$

If we substitute that expression, we obtain

$$\begin{aligned} i\mathcal {D}\mathcal{F}^L_M =\sum _{j=1}^{M-L+2}{\sum _{k=0}^{M+2-L-j}{\left( {\begin{array}{c}M-j\\ L+k-2\end{array}}\right) \frac{(-1)^k (L+k-1)!}{(L-1)! k!} \mathcal{F}^L_{j-1}}} \ . \end{aligned}$$

After some algebra, the sum over k takes the form

$$\begin{aligned}&\sum _{k=0}^{m+2}{(-1)^k \left( {\begin{array}{c}L+m\\ L+k-2\end{array}}\right) \left( {\begin{array}{c}L+k-1\\ k\end{array}}\right) }=\frac{(L+m)!}{(L-1)!} \sum _{k=0}^{m+2}{\frac{(L+k-1)!}{(L+k-2)!} \frac{(-1)^k}{(m-k+2)! k!}} \\&=\frac{(L+m)!}{(L-1)! (m+2)!} \sum _{k=0}^{m+2}{\left[ (-1)^k (L-1) \left( {\begin{array}{c}m+2\\ k\end{array}}\right) +(-1)^k k\left( {\begin{array}{c}m+2\\ k\end{array}}\right) \right] } \ , \end{aligned}$$

where $m=M-L-j$. Properties of the binomial coefficients say that the first sum is zero (unless there is a single term, that is, if $m+2=0$) and the second sum is also zero (except if there are two terms, so that $m+2=1$). Then, the total contribution of these terms will amount to

$$\begin{aligned}&\sum _{M=L-1}^{n-1}{i\mathcal {D}\mathcal{F}^L_M}=\sum _{M=L-1}^{n-1}{\sum _{j=1}^{M-L+2}}{\frac{(M-j)!}{(L-1)! (M-L-j+2)!}} \nonumber \\&\times \big [ (L-1)\delta _{M-L-j+2,0} - (M-L-j+2) \delta _{M-L-j+2,1}) \big ] \mathcal{F}^L_{j-1} \ , \end{aligned}$$

(A.5)

which telescopes, so that

$$\begin{aligned} \sum _{M=L-1}^{n-1}{i\mathcal {D}\mathcal{F}^L_M} = \mathcal{F}^L_{n-L} \ . \end{aligned}$$

(A.6)

Therefore, the most general form of the correlation function $\mathcal{F}^L_n$ is

$$\begin{aligned} \mathcal{F}^L_n=\sum _{k=0}^{n}{\left( {\begin{array}{c}n\\ k\end{array}}\right) \frac{i^{k} \mathcal {D}^{k}}{k!} \mathcal{F}^L_0}+\theta (n-L) \mathcal{F}^L_{n-L} \ . \end{aligned}$$

(A.7)

where $\theta (x)$ is the Heaviside step function, with $\theta (x)=0$ if $x<0$ and $\theta (x)=1$ if $x \ge 0$.

B Extension to SL(2) and SU(1|1) Sectors

Along most of this article, we have considered the SU(2) spin 1/2 homogeneous Heisenberg spin chain. The whole analysis that we have presented relies on the commutation relations of the elements of the monodromy matrix, (2.19), (2.20) and (2.21), the explicit form of the eigenvalues $a(\lambda )$ and $d(\lambda )$, and the S-matrix (and thus the Bethe equations). In this appendix, we want to give a taste of how the computations change for spin chains with symmetries SL(2) and SU(1|1).

1.1 B.1 SL(2) Sector

The case of the SL(2) spin chain seems at first sight rather similar to the SU(2) chain, because the commutation relationships between $(A+D)$ and B are the same in both cases. However, the eigenvalues a and d are exchanged,

$$\begin{aligned} a_{SL(2)}(\lambda )=d_{SU(2)}(\lambda ) \ , \quad d_{SL(2)}(\lambda )=a_{SU(2)}(\lambda )=1 \ . \end{aligned}$$

(B.1)

Fortunately, this does not prevent us from repeating the analysis that we have presented in Section 4.3 to obtain for instance a set of recurrence equations for correlation functions with two spin operators, like (4.39) and (4.52). The derivation is just the same as in that section, but keeping terms in a rather than terms in d. The final result is

$$\begin{aligned} \mathcal{F}^{L,(-1)}_{n+1}(\alpha )&\!=\! \big [1\!+\!a(\xi \!+\!\alpha )\!-\!i\left. \frac{\partial ^{} a}{\partial \lambda ^{}} \right| _{\xi +\alpha } \big ] \mathcal{F}^{L,(-1)}_n (\alpha ) + i\big [ a(\xi +\alpha )-1 \big ] \mathcal {D} \mathcal{F}^{L,(-1)}_n (\alpha ) \ , \nonumber \\ \mathcal {D}^m \mathcal{F}^{L,(-1)}_{n+1} (\alpha )&=\left[ 1+a(\xi +\alpha ) \right] \mathcal {D}^m \mathcal{F}^{L,(-1)}_{n} (\alpha ) - \frac{i}{m+1} \frac{\partial ^{m+1} a}{\partial \alpha ^{m+1}} \mathcal{F}^{L,(-1)}_{n} (\alpha ) \nonumber \\&+\frac{i}{m+1} \big [ a(\xi +\alpha ) -1 \big ] \mathcal {D}^{m+1} \mathcal{F}^{L,(-1)}_{n} (\alpha ) \ , \\ \mathcal {D}^m \mathcal{F}^{L,(-1)}_{0} (\alpha )&=\frac{\partial ^{m} \mathcal{F}^{L,(-1)}_{0} (\alpha )}{\partial \alpha ^{m}} \ , \nonumber \end{aligned}$$

(B.2)

where the $(-1)$ superindex reminds that now we are calculating the correlation function in a SL(2) spin chain. Finally, as in the SU(2) sector,

$$\begin{aligned} \mathcal{F}^{L,(-1)}_n =\sum _{k=0}^{n}{\left( {\begin{array}{c}n\\ k\end{array}}\right) \frac{(-i)^{k} \mathcal {D}^{k}}{k!} \mathcal{F}^{L,(-1)}_0}+\theta (n-L) \mathcal{F}^{L,(-1)}_{n-L} \ . \end{aligned}$$

(B.3)

1.2 B.2 SU(1|1) Sector

The case of a SU(1|1) spin chain is slightly more complex to handle because of the grading of the algebra. However, the evaluation of correlation functions turns out to be simpler than in the SU(2) and SL(2) sectors. The commutation relations are given by [47],

$$\begin{aligned} B(\mu ) B(\lambda )&=-\frac{\mu -\lambda +i}{\mu -\lambda -i} B(\lambda ) B(\mu ) \ , \nonumber \\ A (\mu ) B(\lambda )&=\left( 1+\frac{i}{\mu -\lambda } \right) B(\lambda ) A(\mu ) +\frac{i}{\mu -\lambda } B(\mu ) A (\lambda ) \ , \nonumber \\ D (\mu ) B(\lambda )&=\left( 1+\frac{i}{\mu -\lambda } \right) B(\lambda ) D(\mu ) +\frac{i}{\mu -\lambda } B(\mu ) D (\lambda ) \ , \nonumber \\ C(\lambda ) A (\mu )&=\left( 1+\frac{i}{\lambda -\mu } \right) A(\mu ) C(\lambda ) -\frac{i}{\lambda -\mu } A (\lambda ) C(\mu ) \ , \nonumber \\ C(\lambda ) D (\mu )&=\left( 1+\frac{i}{\lambda -\mu } \right) D(\mu ) C(\lambda ) -\frac{i}{\lambda -\mu } D (\lambda ) C(\mu ) \ . \end{aligned}$$

(B.4)

These commutation relations present some differences with respect to their SU(2) counterparts. The most important one is that they have the same form both for A and D. Another important difference is that the transfer matrix has to be graded and thus new we have $T(\lambda )=A(\lambda )-D(\lambda )$ instead of $A(\lambda )+D(\lambda )$. On the contrary, the form of the functions a and d does not change.

We can now follow Section 4.3 and proceed to find the commutation relations between the transfer matrix and the C operators in the limit where the rapidities are equal. We obtain

$$\begin{aligned}&\lim _{\beta \rightarrow \alpha } C(\alpha ) (A-D)(\beta )=\lim _{\beta \rightarrow \alpha } \frac{\alpha -\beta +i}{\alpha -\beta } (A-D)(\beta ) C(\alpha )-\frac{i}{\alpha -\beta } (A-D)(\alpha ) C(\beta ) \nonumber \\&=(A-D)(\alpha ) C(\alpha ) +i \left[ (A-D)(\alpha ) \left. \frac{\partial ^{} C(\lambda )}{\partial \lambda ^{}} \right| _{\lambda =\alpha } - \left. \frac{\partial ^{} (A-D)(\lambda )}{\partial \lambda ^{}} \right| _{\lambda =\alpha } C(\alpha ) \right] \ . \end{aligned}$$

(B.5)

We can also calculate the derivatives and thus the recurrence relations become

$$\begin{aligned} \mathcal{F}^{L,(0)}_{n+1} (\alpha )&=(1-d+i\partial d) \mathcal{F}^{L,(0)}_n (\alpha ) +i (1-d) \mathcal{D} \mathcal{F}^{L,(0)}_n (\alpha ) \nonumber \ , \\ \mathcal{D}^m \mathcal{F}^{L,(0)}_{n+1} (\alpha )&= (1-d) \mathcal{D}^m \mathcal{F}^{L,(0)}_n (\alpha ) \nonumber \\&+\frac{i}{m+1} \left[ (1-d) \mathcal{D}^{m+1} \mathcal{F}^{L,(0)}_n (\alpha ) +\frac{\partial ^{m+1} d}{\partial \alpha ^{m+1}} \mathcal{F}^{L,(0)}_n (\alpha ) \right] \nonumber \ , \\ \mathcal{D}^m \mathcal{F}^{L,(0)}_0 (\alpha )&=\frac{\partial ^{m} \mathcal{F}^{L,(0)}_0 (\alpha )}{\partial \alpha ^{m}} \ , \end{aligned}$$

(B.6)

where the (0) superindex states that now we are calculating a correlation function in the case of an SU(1|1) spin chain, and where, as usual $d=d(\xi +\alpha )$ and $\partial d=\left. \frac{\partial ^{} d(\lambda )}{\partial \lambda ^{}} \right| _{\lambda =\xi +\alpha }$. These recurrence equations can again be written in terms of some starting condition corresponding to $n=0$, using

$$\begin{aligned} \mathcal{F}^{L,(0)}_n=\sum _{k=0}^{n}{\left( {\begin{array}{c}n\\ k\end{array}}\right) \frac{i^{k} \mathcal {D}^{k}}{k!} \mathcal{F}^{L,(0)}_0} \ , \end{aligned}$$

(B.7)

provided we keep $n<L-1$.

C Correlation Functions Involving States with Three Magnons

Computing correlation functions using the ABA becomes a challenge when the number of magnons increases. In this appendix, we give a taste of how the complexity increases for the case of correlations functions involving states with three magnons.

There are four non-vanishing correlation functions whose most populated state contains three magnons. The first one is just the scalar product $\left\langle \lambda _1 ,\lambda _2 ,\lambda _3 | \mu _1 ,\mu _2 ,\mu _3 \right\rangle$ and can be directly calculated using the Gaudin formula (2.38). The second one is the form factor of a single spin operator, $\left\langle \lambda _1 ,\lambda _2 \left| \sigma ^+_k \right| \mu _1 ,\mu _2 ,\mu _3 \right\rangle$, which can be evaluated in a straightforward extension of the computation of $\left\langle \lambda \left| \sigma ^+_k \right| \mu _1 \mu _2 \right\rangle$ in Section 4.2. The third kind of correlation function involving three magnons is $\left\langle \lambda \left| \sigma ^+_k \sigma ^+_l \right| \mu _1 \mu _2 \mu _3 \right\rangle$, and the fourth one is $\left\langle 0 \left| \sigma ^+_k \sigma ^+_l \sigma ^+_m \right| \mu _1 \mu _2 \mu _3 \right\rangle$. These last two types of correlators are the ones that we will consider in this appendix. We will start studying the third one (the one that involves two spin operators), and along the computation we will find that it involves correlation functions of the form $\left\langle 0 \left| \sigma ^+_k \sigma ^+_{k+1} \sigma ^+_{k+n+2} \right| \mu _1 \mu _2 \mu _3 \right\rangle$, which are a particular case of the fourth type of correlator.

We will start by writing the problem in terms of operators of the monodromy matrix using (2.32),

$$\begin{aligned} \langle \lambda |\sigma ^+_k \sigma ^+_l |\mu _1 \mu _2 \mu _3\rangle = \left\langle 0 \left| C(\lambda )(A+D)^{k-1} (\xi ) C(\xi ) (A+D)^{n} (\xi ) C(\xi ) (A+D)^{L-l} (\xi ) \right| \mu _1 \mu _2 \mu _3 \right\rangle \ , \end{aligned}$$

(C.1)

where, as before, $n=L+l-k-1$. The factor $(A+D)^{k-1}$ acts on $C(\lambda )$ to give $e^{-ip_\lambda (k-1)}$, and the factor $(A+D)^{L-l}$ acts on the three-magnon state to give $e^{-i(p_1+p_2+p_3)\cdot (L-l)}=e^{i(p_1+p_2+p_3)l}$, where we have used the periodicity condition for the Bethe roots. Therefore, our main problem will be to find the correlation function

$$\begin{aligned} \mathcal{H}^L_{n}(\alpha )=\left\langle 0 \left| C(\lambda ) C(\xi +\alpha ) (A+D)^n (\xi ) C(\xi ) B(\mu _1) B(\mu _2) B(\mu _3) \right| 0 \right\rangle \ . \end{aligned}$$

(C.2)

Following the procedure that we have constructed along the article, this can be done by relating $\mathcal{H}^L_{n+1}(\alpha )$ to $\mathcal{H}^L_{n}(\alpha )$. In order to do this, let us start by introducing

$$\begin{aligned} \mathcal{H}^L_{n+1} (\lambda ,\alpha , \delta )=\lim _{\beta \rightarrow \alpha } \left\langle 0 \left| C(\lambda ) C(\xi +\alpha ) (A+D)(\xi +\beta ) \mathcal{O} (\delta ) \right| 0 \right\rangle \ . \end{aligned}$$

(C.3)

Now we just need to apply the commutation relations (2.20) two times in each step, which gives

$$\begin{aligned}&\mathcal{H}^L_{n+1} (\lambda ,\alpha , \delta ) = \lim _{\beta \rightarrow \alpha } \{ [ 1+d(\xi +\beta ) ] \mathcal{H}^L_{n} (\lambda ,\alpha , \delta ) \nonumber \\&-\frac{i}{\lambda -\xi -\beta } \left[ (d(\xi +\beta )-1) \mathcal{H}^L_{n} (\lambda ,\alpha , \delta )-(d(\lambda )-1) \mathcal{H}^L_{n} (\xi +\beta ,\alpha , \delta ) \right] \nonumber \\&-\frac{i}{\alpha -\beta } \left[ (d(\xi +\beta ) -1) \mathcal{H}^L_{n} (\lambda ,\alpha , \delta ) -(d(\xi +\alpha )-1) \mathcal{H}^L_{n} (\lambda ,\beta , \delta ) \right] \nonumber \\&+\frac{i}{\alpha -\beta } \ \frac{i}{\lambda -\xi -\beta } \left[ (d( \xi +\beta )+1) \mathcal{H}^L_{n} (\lambda ,\alpha , \delta ) -(d(\lambda )+1) \mathcal{H}^L_{n} (\xi +\beta ,\alpha , \delta ) \right] \nonumber \\&-\frac{i}{\alpha -\beta } \ \frac{i}{\lambda -\xi -\alpha } \left[ (d( \xi +\alpha )+1) \mathcal{H}^L_{n} (\lambda ,\beta , \delta ) -(d(\lambda )+1) \mathcal{H}^L_{n} (\xi +\beta ,\alpha , \delta ) \right] \} \ . \end{aligned}$$

(C.4)

Taking the limit and applying the Bethe equation for the rapidity $\lambda$ we obtain

$$\begin{aligned}&\mathcal{H}^L_{n+1} (\lambda ,\alpha , \delta )=\Big ( 1+d+i\partial d+\frac{\partial d-i(d-1)}{\lambda -\xi -\alpha }+\frac{d+1}{(\lambda -\xi -\alpha )^2} \Big ) \mathcal{H}^L_{n} (\lambda ,\alpha , \delta ) \nonumber \\&+\Big [ i(1-d)-\frac{d+1}{\lambda -\xi -\alpha } \Big ] \frac{\partial ^{} \mathcal{H}^L_{n} (\lambda ,\alpha , \delta )}{\partial \alpha ^{}} -\frac{2}{(\lambda -\xi -\alpha )^2} \mathcal{H}^L_{n} (\xi +\alpha ,\alpha , \delta ) \ , \end{aligned}$$

(C.5)

where, as before, $d=d(\xi +\alpha )$ and $\partial d=\left. \frac{\partial ^{} d}{\partial \lambda ^{}} \right| _{\xi +\alpha }$. The next step of the calculation is a little bit more involved than in the previous cases because, according to (C.5), information about both functions $\mathcal{H}^L_{n} (\lambda ,\alpha , \delta )$ and $\mathcal{H}^L_{n+1} (\xi +\alpha ,\alpha , \delta )$ is now needed. This will make the computation slightly more difficult, but still manageable. In the expressions below, we will define $\mathcal{H}^L_{n+1} (\alpha +\xi ,\alpha , \delta )=\hat{\mathcal{H}}^L_{n+1} (\alpha ,\alpha , \delta )$ for convenience. This function $\hat{\mathcal{H}}$ has a nice interpretation because

$$\begin{aligned} \left\langle 0 \left| \sigma ^+_k \sigma ^+_{k+1} \sigma ^+_{k+n+2} \right| \mu _1 \mu _2 \mu _3 \right\rangle&=\left\langle 0 \left| C(\xi ) C(\xi ) (A+D)^{n} (\xi ) C(\xi ) (A+D)^{L-n-k-2} (\xi ) \right| \mu _1 \mu _2 \mu _3 \right\rangle \nonumber \\&=\hat{\mathcal{H}}^L_{n} e^{i(p_1+p_2+p_3)(n+k+2)} \ . \end{aligned}$$

(C.6)

Our starting point is thus to find the recursive equation for $\hat{\mathcal{H}}$. This can be obtained setting $\lambda =\gamma +\xi$ in expression (C.4) and taking the limit $\gamma \rightarrow \alpha$,

$$\begin{aligned}&\hat{\mathcal{H}}^L_{n+1} (\alpha ,\alpha , \delta )=\lim _{\beta \rightarrow \alpha }\frac{1}{\beta -\alpha } \Big [ (1+d) \left. \frac{\partial ^{} \hat{\mathcal{H}}^L_{n} (\lambda ,\alpha , \delta )}{\partial \lambda ^{}} \right| _{\lambda =\alpha } \!\! - (1+d) \left. \frac{\partial ^{} \hat{\mathcal{H}}^L_{n} (\alpha ,\lambda , \delta )}{\partial \lambda ^{}} \right| _{\lambda =\alpha } \Big ] \nonumber \\&+ \big (1+d+2i\partial d-\frac{1}{2} \partial ^2 d \big ) \, \hat{\mathcal{H}}^L_{n} (\alpha ,\alpha , \delta ) + \big [ 2i(1-d) +\partial d \big ] \left. \frac{\partial ^{} \hat{\mathcal{H}}^L_{n} (\lambda ,\alpha , \delta )}{\partial \lambda ^{}} \right| _{\lambda =\alpha } \nonumber \\&+ \frac{1+d}{2} \left. \frac{\partial ^{2} \hat{\mathcal{H}}^L_{n} (\lambda ,\alpha , \delta )}{\partial \lambda ^{2}} \right| _{\lambda =\alpha }-(1+d) \left. \frac{\partial ^2 \hat{\mathcal{H}}^L_{n} (\lambda _1 ,\lambda _2, \delta )}{\partial \lambda _1 \partial \lambda _2} \right| _{\begin{array}{c} \lambda _{1} = \alpha \\ \lambda _{2} = \alpha \end{array}} \ . \end{aligned}$$

(C.7)

Note that, although the first term in this expression seems divergent, it vanishes because of the commutator of two C operators vanishes, which makes the two derivatives equal. However, this way of calculating recursively $\hat{\mathcal{H}} (\alpha , \alpha , \delta )$ is going to create more problem than it solves, as it will imply calculating the recurrence equation of the derivative of $\hat{\mathcal{H}} (\lambda ,\mu )$ with respect to either the first or the second argument. Therefore, we are going to present the recursion relation of $\hat{\mathcal{H}} (\beta , \alpha , \delta )$ but without taking the limit $\beta \rightarrow \alpha$. To obtain this recurrence relation, we only need to substitute $\lambda =\xi +\beta$ in (C.5), but without imposing $d(\lambda )=1$,

$$\begin{aligned}&\hat{\mathcal{H}}^L_{n+1} (\beta ,\alpha , \delta )=\Big ( 1+d+i\partial d+\frac{\partial d-i(d-1)}{\beta -\alpha }+\frac{d+1}{(\beta -\alpha )^2} \Big ) \hat{\mathcal{H}}^L_{n} (\beta ,\alpha , \delta ) \nonumber \\&+\Big [ i(1-d)-\frac{d+1}{\beta -\alpha } \Big ] \frac{\partial ^{} \hat{\mathcal{H}}^L_{n} (\beta ,\alpha , \delta )}{\partial \alpha ^{}} +\left[ \frac{i(d'-1)}{\beta -\alpha } -\frac{d'+1}{(\beta -\alpha )^2} \right] \lim _{\gamma \rightarrow \alpha } \hat{\mathcal{H}}^L_{n} (\gamma ,\alpha , \delta ) \ , \end{aligned}$$

(C.8)

where $d'=d(\xi +\beta )$. Note that if we take $\beta \rightarrow \alpha$ (C.8) gives (C.7). Now in the recurrence relation we need to include $\lim _{\gamma \rightarrow \alpha } \hat{\mathcal{H}}^L_{n} (\gamma ,\alpha , \delta )$, but this quantity is obviously known once we know $\hat{\mathcal{H}}^L_{n} (\beta ,\alpha , \delta )$.

We also need a recurrence equation for the derivatives. For the case of $\mathcal{H}_{n}$ we have

$$\begin{aligned}&\mathcal{D}^n \mathcal{H}^L_{m+1} (\lambda ,\alpha , \delta )=(1+d) \mathcal{D}^n \mathcal{H}^L_{m} -\frac{i}{\lambda -\xi -\alpha } \left[ (d-1) \mathcal{D}^n \mathcal{H}^L_{m} \right] \nonumber \\&+\frac{i}{n+1} (\partial ^{n+1} d) \mathcal{H}^L_{m}+(1-d)\frac{i}{n+1} \mathcal{D}^{n+1} \mathcal{H}^L_{m} \nonumber \\&+\sum _{k=0}^n \sum _{l=0}^{k+1}{\frac{n!}{(n-k)! (k+1-l)!} \frac{1}{(\lambda -\xi -\alpha )^{l+1}} \left[ \partial ^{k+1-l} (d+1) \mathcal{D}^{n-k} \mathcal{H}^L_{m} \right. } \nonumber \\&\left. -(d(\lambda )+1) \mathcal{D}^{n-k}_1 \mathcal{D}^{k+1-l}_2 \hat{\mathcal{H}}^L_{m} \right] \nonumber \\&-\sum _{k=0}^n{\sum _{l=0}^{n-k}{ \frac{n!}{(k+1)! (n-l-k)!} \frac{1}{(\lambda -\xi -\alpha )^{l+1}} \left[ \partial ^{n-k-l}(d+1) \mathcal{D}^{k+1} \mathcal{H}^L_{m} \right. }} \nonumber \\&\left. -(d(\lambda ) +1) \mathcal{D}^{n-k-l}_1 \mathcal{D}^{k+1}_2 \hat{\mathcal{H}}^L_{m} \right] \ . \end{aligned}$$

(C.9)

The last two sums cancel themselves except for the terms with $k=n$. Therefore

$$\begin{aligned}&\mathcal{D}^n \mathcal{H}^L_{m+1} (\lambda ,\alpha , \delta )=(1+d) \mathcal{D}^n \mathcal{H}^L_{m} -\frac{i}{\lambda -\xi -\alpha } (d-1) \mathcal{D}^n \mathcal{H}^L_{m} \nonumber \\&+\frac{i}{n+1} (\partial ^{n+1} d) \mathcal{H}^L_{m}+(1-d)\frac{i}{n+1} \mathcal{D}^{n+1} \mathcal{H}^L_{m} \nonumber \\&+\sum _{l=0}^{n+1}{\frac{n!}{(n+1-l)!} \frac{1}{(\lambda -\xi -\alpha )^{l+1}} \left[ \partial ^{n+1-l} (d+1) \mathcal{H}^L_{m} - 2 \, \mathcal{D}^{n+1-l} \hat{\mathcal{H}}^L_{m} \right] } \nonumber \\&-\frac{1}{n+1} \frac{1}{(\lambda -\xi -\alpha )} \left[ (d+1) \mathcal{D}^{n+1} \mathcal{H}^L_{m} - 2 \, \mathcal{D}^{n+1} \hat{\mathcal{H}}^L_{m} \right] \ , \end{aligned}$$

(C.10)

where we have used that $d(\lambda )=1$. In a similar way, we can obtain an expression for the derivatives of $\hat{\mathcal{H}}$,

$$\begin{aligned}&\mathcal{D}^n \hat{\mathcal{H}}^L_{m+1} (\beta ,\alpha , \delta )=(1+d) \mathcal{D}^n \hat{\mathcal{H}}^L_{m} -\frac{i}{\beta -\alpha } (d-1) \mathcal{D}^n \hat{\mathcal{H}}^L_{m} \nonumber \\&+\frac{i}{n+1} (\partial ^{n+1} d) \hat{\mathcal{H}}^L_{m}+(1-d)\frac{i}{n+1} \mathcal{D}^{n+1} \hat{\mathcal{H}}^L_{m} +\frac{i}{\beta -\alpha } (d' -1) \lim _{\gamma \rightarrow \alpha } \frac{\partial ^{n} \hat{\mathcal{H}}^L_{n} (\gamma ,\alpha , \delta )}{\partial \alpha ^{n}} \nonumber \\&+\sum _{l=0}^{n+1}{\frac{n!}{(n+1-l)!} \frac{1}{(\beta -\alpha )^{l+1}} \left[ \partial ^{n+1-l} (d+1) \hat{\mathcal{H}}^L_{m} - (d'+1)\, \lim _{\gamma \rightarrow \alpha } \frac{\partial ^{n+1-l} \hat{\mathcal{H}}^L_{m} (\gamma ,\alpha ,\delta )}{\partial \alpha ^{n+1-l}} \right] } \nonumber \\&-\frac{1}{n+1} \frac{1}{(\beta -\alpha )} \left[ (d+1) \mathcal{D}^{n+1} \hat{\mathcal{H}}^L_{m} - (d'+1) \, \lim _{\gamma \rightarrow \alpha } \frac{\partial ^{n+1} \hat{\mathcal{H}}^L_{m} (\gamma ,\alpha ,\delta )}{\partial \alpha ^{n+1}} \right] \ . \end{aligned}$$

(C.11)

At this point the problem is, at least formally, solved. We have found the recursion relation for $\hat{\mathcal{H}}$ and its derivatives, with $\left\langle 0 \left| C(\xi +\beta ) C(\xi +\alpha ) (\xi ) C(\xi ) \right| \mu _1 \mu _2 \mu _3 \right\rangle =\hat{\mathcal{H}}^L_{n} (\beta , \alpha )$ as the initial condition. These functions can then be substituted in the recursion relation for $\mathcal H$ and thus we can obtain the desired correlation function. However, we are not going to present the general form for the correlation function $\mathcal{H}^L_n$ as a function of $\mathcal{H}^L_0$, $\hat{\mathcal{H}}^L_0$ and their derivatives, because it becomes rather lengthy. This is because when we substitute the expression for the derivatives, the recursion relations turn out to depend on all the $\mathcal{H}_i$ with $0\le i\le n$, even after we take the limit $\alpha \rightarrow 0$. Instead, we can present the case of correlation functions with n small, to exhibit the nested procedure to write the result in terms of the initial functions $\mathcal{H}^L_0$ and $\hat{\mathcal{H}}^L_0$. In particular, we are going to consider the first three functions, with $n=1$, $n=2$ and $n=3$. Thus, we can safely assume that $n<L-1$ so that all the d and $\partial ^k d$ factors can be set to zero in the limit $\alpha \rightarrow 0$. The first of these correlation functions is given by

$$\begin{aligned} \mathcal{H}^L_{1} = \big ( 1+ i c(\lambda ) + c(\lambda )^2 \big ) \, \mathcal{H}^L_0 + \big ( i-c(\lambda ) \big ) \left. \frac{\partial ^{} \mathcal{H}^L_0 (\lambda ,\alpha )}{\partial \alpha ^{}} \right| _{\alpha =0} - 2 c(\lambda )^2 \, \hat{\mathcal{H}}^L_0 \ , \end{aligned}$$

(C.12)

where, for convenience, we have defined $c(\lambda ) = 1/(\lambda -\xi )$. For simplicity, if no arguments of these functions are given, $\mathcal{H}^L (\lambda , 0)$ and $\hat{\mathcal{H}}^L (0, 0)$ must be understood. The last step of the computation reduces to calculating some initial conditions, which now are

$$\begin{aligned} \mathcal{H}^L_0 (\lambda , \alpha )&=\left\langle 0 \left| C(\lambda ) C(\xi +\alpha ) C(\xi ) B(\mu _1) B(\mu _2) B(\mu _3) \right| 0 \right\rangle \ , \end{aligned}$$

(C.13)

$$\begin{aligned} \hat{\mathcal{H}}^L_0 (\alpha , \beta )&= \mathcal{H}^L_0 (\xi +\alpha , \beta ) \ . \end{aligned}$$

(C.14)

These functions can be easily computed using (2.37). However, we are not going to present the explicit expression for these scalar products because of its length and because we want to show the way to solve the recurrence relation rather than obtaining the explicit value of the correlation function.

The functional dependence of $\mathcal{H}_1$ on $\mathcal{H}_0$ is repeated for a given value of n and the lower correlator. That is, in the limit $\alpha \rightarrow 0$ the recurrence relation for $\mathcal{H}^L_{n+1}$ is given by

$$\begin{aligned} \mathcal{H}^L_{n+1} = \big ( 1+ i c(\lambda ) + c(\lambda )^2 \big ) \mathcal{H}^L_n + \big ( i-c(\lambda ) \big ) \mathcal{D} \mathcal{H}^L_n - 2 c(\lambda )^2 \hat{\mathcal{H}}_n \, \ . \end{aligned}$$

(C.15)

Therefore, for the second correlation function, we have

$$\begin{aligned} \mathcal{H}^L_{2} = \big ( 1+ i c(\lambda ) + c(\lambda )^2 \big ) \mathcal{H}^L_1 + \big ( i-c(\lambda ) \big ) \mathcal{D} \mathcal{H}^L_1 - 2 c(\lambda )^2 \hat{\mathcal{H}}_1 \ . \end{aligned}$$

(C.16)

As we already know $\mathcal{H}^L_1$, it only remains to find the other two functions entering (C.16). This can be done using the equations that we have obtained along this appendix. We get

$$\begin{aligned} \mathcal{D} \mathcal{H}^L_{1}&=c(\lambda )^3 \mathcal{H}^L_0+(1+ic(\lambda )) \left. \frac{\partial ^{} \mathcal{H}^L_0 (\lambda ,\alpha )}{\partial \alpha ^{}} \right| _{\alpha =0}+\frac{i(1+ic(\lambda ))}{2} \left. \frac{\partial ^{2} \mathcal{H}^L_0 (\lambda ,\alpha )}{\partial \alpha ^{2}} \right| _{\alpha =0} \nonumber \\&-2c(\lambda )^3 \hat{\mathcal{H}}^L_0 -2c(\lambda )^2 \left. \frac{\partial ^{} \hat{\mathcal{H}}^L_0 (0 ,\alpha )}{\partial \alpha ^{}} \right| _{\alpha =0} \ , \end{aligned}$$

(C.17)

$$\begin{aligned} \hat{\mathcal{H}}^L_1 (\beta , 0)&= \Big ( 1+\frac{i}{\beta }+\frac{1}{\beta ^2} \Big ) \hat{\mathcal{H}}^L_0 (\beta , 0) +\Big [ i-\frac{1}{\beta } \Big ] \left. \frac{\partial ^{} \hat{\mathcal{H}}^L_0 (\beta ,\alpha )}{\partial \alpha ^{}} \right| _{\alpha =0}-\left[ \frac{i}{\beta } +\frac{1}{\beta ^2} \right] \hat{\mathcal{H}}^L_{0} \ , \end{aligned}$$

(C.18)

$$\begin{aligned} \hat{\mathcal{H}}^L_1&= \hat{\mathcal{H}}^L_{0} +2i \left. \frac{\partial ^{} \hat{\mathcal{H}}^L_0 (0 ,\alpha )}{\partial \alpha ^{}} \right| _{\alpha =0} +\frac{1}{2} \left. \frac{\partial ^{2} \hat{\mathcal{H}}^L_0 (0 ,\alpha )}{\partial \alpha ^{2}} \right| _{\alpha =0}-\left. \frac{ \partial ^2 \hat{\mathcal{H}}^L_0 (\alpha ,\beta )}{\partial \alpha \partial \beta } \right| _{\begin{array}{c} \alpha =0 \\ \beta =0 \end{array}} \ , \end{aligned}$$

(C.19)

which reduce again to some dependence on the initial conditions we have described before.

An identical computation can be done for $\mathcal{H}^L_{3}$,

$$\begin{aligned} \mathcal{H}^L_{3} = \big ( 1+ i c(\lambda ) + c(\lambda )^2 \big ) \mathcal{H}^L_2 + \big ( i-c(\lambda ) \big ) \mathcal{D} \mathcal{H}^L_2 - 2 c(\lambda )^2 \hat{\mathcal{H}}_2 \ . \end{aligned}$$

(C.20)

Now, besides $\mathcal{H}^L_{2}$, that has been calculated just before, we need

$$\begin{aligned} \mathcal{D} \mathcal{H}^L_{2}&=c(\lambda )^3 \mathcal{H}^L_1+(1+ic(\lambda )) \mathcal{D} \mathcal{H}^L_{1}+\frac{i(1+ic(\lambda ))}{2} \mathcal{D}^ 2 \mathcal{H}^L_{1} \nonumber \\&-2c(\lambda )^2 \Big ( c(\lambda ) \hat{\mathcal{H}}^L_1 +\mathcal{D} \hat{\mathcal{H}}^L_{1} \Big ) \ , \end{aligned}$$

(C.21)

$$\begin{aligned} \mathcal{D}^2 \mathcal{H}^L_{1}&=2 c(\lambda )^4 \mathcal{H}^L_0+(1+ic(\lambda )) \left. \frac{\partial ^{2} \mathcal{H}^L_0 (\lambda ,\alpha )}{\partial \alpha ^{2}} \right| _{\alpha =0}+\frac{i(1+ic(\lambda ))}{3} \left. \frac{\partial ^{3} \mathcal{H}^L_0 (\lambda ,\alpha )}{\partial \alpha ^{3}} \right| _{\alpha =0} \nonumber \\&-4c(\lambda )^4 \hat{\mathcal{H}}^L_0 -4 c(\lambda )^3 \left. \frac{\partial ^{} \hat{\mathcal{H}}^L_0 (0 ,\alpha )}{\partial \alpha ^{}} \right| _{\alpha =0} -2c(\lambda )^2 \left. \frac{\partial ^{2} \hat{\mathcal{H}}^L_0 (0 ,\alpha )}{\partial \alpha ^{2}} \right| _{\alpha =0} \ ,\end{aligned}$$

(C.22)

$$\begin{aligned} \mathcal{D} \hat{\mathcal{H}}^L_{1}&=\left. \frac{\partial ^{} \hat{\mathcal{H}}^L_0 (0 ,\alpha )}{\partial \alpha ^{}} \right| _{\alpha =0} + \frac{i}{2}\left. \frac{\partial ^{2} \hat{\mathcal{H}}^L_0 (0 ,\alpha )}{\partial \alpha ^{2}} \right| _{\alpha =0}+i\left. \frac{ \partial ^2 \hat{\mathcal{H}}^L_0 (\alpha ,\beta )}{\partial \alpha \partial \beta } \right| _{\begin{array}{c} \alpha =0 \\ \beta =0 \end{array}} \nonumber \\&+ \frac{1}{3!} \left. \frac{\partial ^{3} \hat{\mathcal{H}}^L_0 (0 ,\alpha )}{\partial \alpha ^{3}} \right| _{\alpha =0} - \frac{1}{2} \left. \frac{ \partial ^3 \hat{\mathcal{H}}^L_0 (\alpha ,\beta )}{\partial \alpha \partial \beta ^ 2} \right| _{\begin{array}{c} \alpha =0 \\ \beta =0 \end{array}} \ , \end{aligned}$$

(C.23)

$$\begin{aligned} \hat{\mathcal{H}}^L_{2}&=\hat{\mathcal{H}}^L_{0} +4i \left. \frac{\partial ^{} \hat{\mathcal{H}}^L_0 (0 ,\alpha )}{\partial \alpha ^{}} \right| _{\alpha =0} -4 \left. \frac{ \partial ^2 \hat{\mathcal{H}}^L_0 (\alpha ,\beta )}{\partial \alpha \partial \beta } \right| _{\begin{array}{c} \alpha =0 \\ \beta =0 \end{array}}+\frac{i}{2}\left. \frac{\partial ^{3} \hat{\mathcal{H}}^L_0 (0 ,\alpha )}{\partial \alpha ^{3}} \right| _{\alpha =0} \nonumber \\&- \frac{3i}{2} \left. \frac{ \partial ^3 \hat{\mathcal{H}}^L_0 (\alpha ,\beta )}{\partial \alpha \partial \beta ^2} \right| _{\begin{array}{c} \alpha =0 \\ \beta =0 \end{array}} -\frac{1}{3!} \left. \frac{ \partial ^4 \hat{\mathcal{H}}^L_0 (\alpha ,\beta )}{\partial \alpha \partial \beta ^3} \right| _{\begin{array}{c} \alpha =0 \\ \beta =0 \end{array}} +\frac{1}{2!^2} \left. \frac{ \partial ^4 \hat{\mathcal{H}}^L_0 (\alpha ,\beta )}{\partial \alpha ^2 \partial \beta ^2} \right| _{\begin{array}{c} \alpha =0 \\ \beta =0 \end{array}} \ . \end{aligned}$$

(C.24)

The cases with higher values of n can be obtained along similar lines.

To conclude our analysis, we will brief comment on the calculation of correlation functions $\left\langle 0 \left| \sigma ^+_k \sigma ^+_l \sigma ^+_m \right| \{ \mu \} \right\rangle$, with general values of k, l and m. In this case, the value of n in $\hat{\mathcal{H}}^L_n$ will be proportional to the separation of l and m. But it still remains to separate k from l. This last step can be solved using the tools from Section 4.3.3, because the problem in both cases is the same.