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magic_gaussian_integral [2017/07/01 12:20]
nikolaj
magic_gaussian_integral [2019/09/27 00:19] (current)
nikolaj
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 Notice that via diagonalization of the matrix and knowledge of basic Gaussian integral above, we get Notice that via diagonalization of the matrix and knowledge of basic Gaussian integral above, we get
  
-^ $Z_1(0):​=\int_{-\infty}^\infty \mathrm e^{-\tfrac{1}{2}\left\langle\phi\left|\,​A\,​\right|\phi\right\rangle} \prod_{i=1}^m \mathrm d\phi_i = (2\pi)^{1/2}(\det A)^{-1/2} $ ^+^ $Z_1(0):​=\int_{-\infty}^\infty \mathrm e^{-\tfrac{1}{2}\left\langle\phi\left|\,​A\,​\right|\phi\right\rangle} \prod_{i=1}^m \mathrm d\phi_i = (2\pi)^{m/2}(\det A)^{-1/2} $ ^
  
 Taking care of the vector $J$, we can obtain ​ Taking care of the vector $J$, we can obtain ​
  
-^ $Z_1(J):​=\int_{-\infty}^\infty \mathrm e^{-\tfrac{1}{2}\left\langle\phi\left|\,​A\,​\right|\phi\right\rangle +i\,​\left\langle\phi\left|\right.J\right\rangle}\prod_{i=1}^m\mathrm d\phi_i = Z_1(0)\cdot\mathrm e^{-\frac{1}{2}\left\langle J\left|\,​A^{-1}\,​\right|J\right\rangle }$ ^+^ $Z_1(J):​=\int_{-\infty}^\infty \mathrm e^{-\tfrac{1}{2}\left\langle\phi\left|\,​A\,​\right|\phi\right\rangle +i\,​\left\langle\phi\left|\right.J\right\rangle}\prod_{i=1}^m\mathrm d\phi_i = Z_1(0)\cdot\mathrm e^{-\frac{1}{2}\left\langle J\left|\,​A^{-1}\,​\right|J\right\rangle } $ ^
  
 Now from a physical perspective,​ it's actually better to write this as $\propto\mathrm e^{-\frac{1}{2}\left\langle A^{-1}J\left|\,​A\,​\right|A^{-1}J\right\rangle }$. In the path integral treatment of the diffusion equation, the propagator involves quantity $\phi\equiv p,J\equiv q$ and roughly speaking $A\propto \delta t,\ A^{-1}\propto\tfrac{\mathrm d}{\mathrm dt}$. The above integral plays a role in passing from the Hamiltonian perspective to the Lagrangian one: $\left\langle\phi\left|\,​A\,​\right|\phi\right\rangle\propto p^2\Delta t$ to a sort of conjugate $\left\langle A^{-1}J\left|\,​A\,​\right|A^{-1}J\right\rangle\propto {\dot q}^2\Delta t$. Now from a physical perspective,​ it's actually better to write this as $\propto\mathrm e^{-\frac{1}{2}\left\langle A^{-1}J\left|\,​A\,​\right|A^{-1}J\right\rangle }$. In the path integral treatment of the diffusion equation, the propagator involves quantity $\phi\equiv p,J\equiv q$ and roughly speaking $A\propto \delta t,\ A^{-1}\propto\tfrac{\mathrm d}{\mathrm dt}$. The above integral plays a role in passing from the Hamiltonian perspective to the Lagrangian one: $\left\langle\phi\left|\,​A\,​\right|\phi\right\rangle\propto p^2\Delta t$ to a sort of conjugate $\left\langle A^{-1}J\left|\,​A\,​\right|A^{-1}J\right\rangle\propto {\dot q}^2\Delta t$.
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 Lastly, notice that $-i\frac{\partial}{\partial J_i}e^{i\,​\left\langle\phi\left|\right.J\right\rangle}=\phi_i\,​ \mathrm e^{i\,​\left\langle\phi\left|\right.J\right\rangle}$ and therefore Lastly, notice that $-i\frac{\partial}{\partial J_i}e^{i\,​\left\langle\phi\left|\right.J\right\rangle}=\phi_i\,​ \mathrm e^{i\,​\left\langle\phi\left|\right.J\right\rangle}$ and therefore
  
-^ $ Z_f(J) = Z_1(0)\ f\left(-i\frac{\partial}{\partial J}\right)\,​\mathrm e^{-\frac{1}{2}\left\langle J\left|\,​A^{-1/​2}\,​\right|J\right\rangle }$ ^+^ $ Z_f(J) = (2\pi)^{m/​2}(\det A)^{-1/2}\cdot f\left(-i\frac{\partial}{\partial J}\right)\,​\mathrm e^{-\frac{1}{2}\left\langle J\left|\,​A^{-1/​2}\,​\right|J\right\rangle }$ ^
  
 +$ \int_{\mathbb R}\,​f(\phi)\,​\mathrm e^{\frac{1}{2} \left\langle\phi\left|\,​A\,​\right|\phi\right\rangle +i\,​\left\langle\phi\left|\right.J\right\rangle} \prod_{i=1}^m ​ \mathrm d\phi_i = (2\pi)^{m/​2}(\det A)^{-1/​2}\cdot f\left(-i\frac{\partial}{\partial J}\right)\,​\mathrm e^{-\frac{1}{2}\left\langle J\left|\,​A^{-1/​2}\,​\right|J\right\rangle }$
  
 We are interested in that expression as the solution of the integral, because in quantum field theory, the path integral is often an infinite dimensional variant it. There the exponent in the defining integral is the action functional, the operator $A$ involves a hard to invert differential operator (the inverse being strongly related to the response function/​green function) and $f$ encodes the type of process and the interaction. The terms from the expansion of $f$ are encoded by Feynman diagrams. ​ We are interested in that expression as the solution of the integral, because in quantum field theory, the path integral is often an infinite dimensional variant it. There the exponent in the defining integral is the action functional, the operator $A$ involves a hard to invert differential operator (the inverse being strongly related to the response function/​green function) and $f$ encodes the type of process and the interaction. The terms from the expansion of $f$ are encoded by Feynman diagrams. ​
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