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magic_gaussian_integral [2015/04/15 13:42] nikolaj |
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| ===== Magic Gaussian integral ===== | ===== Magic Gaussian integral ===== | ||
| ==== Partial function ==== | ==== Partial function ==== | ||
| - | | @#55CCEE: context | @#55CCEE: $ m\in\mathbb N $ | @#55CCEE: | | ||
| - | | @#55CCEE: | @#55CCEE: $ A $ .... self-adjoint $n\times n$ square matrix over $\mathbb C$ | @#55CCEE: | | ||
| - | | @#55CCEE: | @#55CCEE: $ A_\varepsilon $ ... $n\times n$ square matrix over $\mathbb C$ | @#55CCEE: $ \lim_{\varepsilon\to 0}A_\varepsilon = A $ | | ||
| - | | @#FFBB00: definiendum | @#FFBB00: $Z: (\mathbb C^m\to \mathbb C) \to \mathbb C^m \to \mathbb C$ | @#FFBB00: | | ||
| - | | @#FFBB00: | @#FFBB00: $Z_f(J):=\lim_{\varepsilon\to 0}\int_{-\infty}^\infty\,f(\phi)\,\mathrm e^{\frac{1}{2} \left\langle\phi\left|\,A_\varepsilon\,\right|\phi\right\rangle +i\,\left\langle\phi\left|\right.J\right\rangle}\prod_{i=1}^m \mathrm d\phi_i $ | @#FFBB00: | | ||
| ==== Discussion ==== | ==== Discussion ==== | ||
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| which is obtained, for example, with the basic trick of switching to polar coordinates. | which is obtained, for example, with the basic trick of switching to polar coordinates. | ||
| - | Remark: Having found the normalization of ${\mathrm e}^{-\tfrac{1}{2}a\,\phi^2}$ lets us define the function ${\mathrm{erf}}$ with $\lim_{d\to\infty}{\mathrm{erf}}(d)=1$. Then we can show that $\frac{1}{2}\pi^\frac{1}{2}{\mathrm{erf}}(d) = \sum_{k=0}^\infty\frac{1}{k!}(-1)^k\frac{d^{2k+1}}{2k+1}$ and with $\int_0^d(-\phi^2)^k{\mathrm d}\phi = (-1)^k\frac{d^{2k+1}}{2k+1}$ we can evaluate the initial integral on $[-d,d]$, not just all of $\mathbb R$. | + | (Side remark: Having found the normalization of ${\mathrm e}^{-\tfrac{1}{2}a\,\phi^2}$ lets us define the function ${\mathrm{erf}}$ with $\lim_{d\to\infty}{\mathrm{erf}}(d)=1$. Then we can show that $\frac{1}{2}\pi^\frac{1}{2}{\mathrm{erf}}(d) = \sum_{k=0}^\infty\frac{1}{k!}(-1)^k\frac{d^{2k+1}}{2k+1}$ and with $\int_0^d(-\phi^2)^k{\mathrm d}\phi = (-1)^k\frac{d^{2k+1}}{2k+1}$ we can evaluate the initial integral on $[-d,d]$, not just all of $\mathbb R$.) |
| Adding a imaginary linear term in the exponent results in the statement that the Fourier transform of a Gaussian curve is again a Gaussian curve | Adding a imaginary linear term in the exponent results in the statement that the Fourier transform of a Gaussian curve is again a Gaussian curve | ||
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| === Theorems === | === Theorems === | ||
| - | Let's now discuss the more general integral. Notice that via diagonalization of the matrix and knowledge of the first integral above, we get | + | | @#55CCEE: $ A_\varepsilon $ ... $n\times n$ square matrix over $\mathbb C$ | @#55CCEE: $ \lim_{\varepsilon\to 0}A_\varepsilon = A $ | |
| + | | @#FFBB00: $Z_f(J):=\lim_{\varepsilon\to 0}\int_{-\infty}^\infty\,f(\phi)\,\mathrm e^{\frac{1}{2} \left\langle\phi\left|\,A_\varepsilon\,\right|\phi\right\rangle +i\,\left\langle\phi\left|\right.J\right\rangle}\prod_{i=1}^m \mathrm d\phi_i $ | @#FFBB00: | | ||
| + | |||
| + | Here $\langle u|v\rangle$ denotes the inner product in ${\mathbb C}^n$ as vector space. | ||
| + | |||
| + | 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)^{1/2}(\det A)^{-1/2} $ ^ | ||
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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) = f\left(-i\frac{\partial}{\partial J}\right)\, Z_1(J) $ ^ | + | ^ $ 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 }$ ^ |
| - | i.e. | ||
| - | |||
| - | ^ $ \frac{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 } $ ^ | ||
| 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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| === Related === | === Related === | ||
| [[Retarded propagator . time-independent Hamiltonian]] | [[Retarded propagator . time-independent Hamiltonian]] | ||
| + | |||
