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  1. documentclass[12pt]{report}
  2.  
  3. usepackage{amsmath}
  4. usepackage{physics}
  5. usepackage{parskip}
  6.  
  7. newcommand{set}[1]{left{#1right}}
  8.  
  9. begin{document}
  10.  
  11. We wish to rewrite this using
  12. begin{align*}
  13. int d^N psi exp[-frac{1}{2}psi_i A_{ij} psi_j + J_i psi_i] &= (2pi)^{N/2} (det A)^{-1/2} exp[frac{1}{2} J_i A^{-1}_{ij} J_j]\
  14. &=  (2pi)^{N/2}exp[-frac{1}{2}Tr(ln A)] exp[frac{1}{2} J_i A^{-1}_{ij} J_j]
  15. end{align*}
  16. or alternatively
  17. $$int d^N psi exp[-frac{1}{2}psi_i A_{ij} psi_j + J_i psi_i] =  (2pi)^{N/2}exp[-frac{1}{2}Tr(ln A)] exp[frac{1}{2} J_i A^{-1}_{ij} J_j]$$
  18. where we identify $J_i equiv sigma_i$ and $A_{ij}^{-1}/2 equiv K_{ij}^{-1}$ or $A_{ij} = K_{ij}/2$. This gives us
  19. $$Z = (2pi)^{-N/2} exp[frac{1}{2}Tr(ln(K/2))] sum_{set{sigma_i=pm 1}}int d^Npsi exp[-frac{1}{4}psi_i K_{ij} psi_j + sigma_i psi_i] expBig[h_i sigma_iBig]$$
  20. or
  21. $$Z = (2pi)^{-N/2} exp[frac{1}{2}Tr(ln(K/2))] int d^Npsi exp[-frac{1}{4}sum_{ij} psi_i K_{ij} psi_j] sum_{set{sigma_i=pm 1}}  prod_{i} expBig[(h_i +psi_i)sigma_iBig]$$
  22. performing the summation, we have
  23. $$Z = (2pi)^{-N/2} exp[frac{1}{2}Tr(ln(K/2))] int d^Npsi exp[-frac{1}{4}sum_{ij} psi_i K_{ij} psi_j] prod_{i}2, coshBig[(h_i +psi_i)sigma_iBig]$$
  24. After factoring out $N$ factors of $2$ from the product, the leading factor is simply an overall normalization which doesn't affect the physics, call it $mathcal{N}$, so we finally have the desired result
  25. $$Z = mathcal{N} int d^Npsi ,expleft{-left[frac{1}{4}sum_{ij} psi_i K_{ij} psi_j - sum_{i} ln[cosh(h_i +psi_i)]right]right}$$
  26. where
  27. $$mathcal{N} = left(frac{2}{pi}right)^{N/2} exp[frac{1}{2}Tr(ln(K/2))]$$
  28.  
  29. end{document}
  30.    
  31. documentclass[12pt]{report}
  32.  
  33. usepackage{amsmath}
  34. usepackage{physics}
  35. usepackage{parskip}
  36.  
  37. newcommand{set}[1]{left{#1right}}
  38.  
  39. begin{document}
  40.  
  41. We wish to rewrite this using
  42.  
  43. begin{align*}
  44. int d^N psi exp[-frac{1}{2}psi_i A_{ij} psi_j + J_i psi_i] &= (2pi)^{N/2} (det A)^{-1/2} exp[frac{1}{2} J_i A^{-1}_{ij} J_j]\
  45. &=  (2pi)^{N/2}exp[-frac{1}{2}Tr(ln A)] exp[frac{1}{2} J_i A^{-1}_{ij} J_j]
  46. end{align*}
  47.  
  48. or alternatively
  49.  
  50. $$int d^N psi exp[-frac{1}{2}psi_i A_{ij} psi_j + J_i psi_i] =  (2pi)^{N/2}exp[-frac{1}{2}Tr(ln A)] exp[frac{1}{2} J_i A^{-1}_{ij} J_j]$$
  51.  
  52. where we identify $J_i equiv sigma_i$ and $A_{ij}^{-1}/2 equiv K_{ij}^{-1}$ or $A_{ij} = K_{ij}/2$. This gives us
  53.  
  54. $$Z = (2pi)^{-N/2} exp[frac{1}{2}Tr(ln(K/2))] sum_{set{sigma_i=pm 1}}int d^Npsi exp[-frac{1}{4}psi_i K_{ij} psi_j + sigma_i psi_i] expBig[h_i sigma_iBig]$$
  55.  
  56. or
  57.  
  58. $$Z = (2pi)^{-N/2} exp[frac{1}{2}Tr(ln(K/2))] int d^Npsi exp[-frac{1}{4}sum_{ij} psi_i K_{ij} psi_j] sum_{set{sigma_i=pm 1}}  prod_{i} expBig[(h_i +psi_i)sigma_iBig]$$
  59.  
  60. performing the summation, we have
  61.  
  62. $$Z = (2pi)^{-N/2} exp[frac{1}{2}Tr(ln(K/2))] int d^Npsi exp[-frac{1}{4}sum_{ij} psi_i K_{ij} psi_j] prod_{i}2, coshBig[(h_i +psi_i)sigma_iBig]$$
  63.  
  64. After factoring out $N$ factors of $2$ from the product, the leading factor is simply an overall normalization which doesn't affect the physics, call it $mathcal{N}$, so we finally have the desired result
  65.  
  66. $$Z = mathcal{N} int d^Npsi ,expleft{-left[frac{1}{4}sum_{ij} psi_i K_{ij} psi_j - sum_{i} ln[cosh(h_i +psi_i)]right]right}$$
  67.  
  68. where
  69.  
  70. $$mathcal{N} = left(frac{2}{pi}right)^{N/2} exp[frac{1}{2}Tr(ln(K/2))]$$
  71.  
  72. end{document}
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