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# Untitled

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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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