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Regular readers of this blog will know that I love my integrals. For those that share the same enjoyment, I recommend this channel that often covers integrals from the MIT integration bee. For example, see this case of writing an integral in terms of itself. Very cool.
In this post, I want to think about the m-dimensional Gaussian. It is a common integral in QFT. Here’s how we compute it when of the form,
[ z = int_{mathbb{R}^n} d^n xe^{-frac{1}{2} x^{T}Ax} ]
Two comments: $A$ is a real symmetric matrix, $A = A^T in mathbb{R}^{n times n} Leftrightarrow A_{ij} = A_{ji}$. X is a column vector,
[ x = begin{pmatrix}
x_1 \
x_2 \
. \
. \
. \
x_n \
end{pmatrix} ]
Since $A$ is real and symmetric, we make use of a result from spectral theorem. In particular, consider the following spectral theorem proposition:
[ A = A^T in mathbb{R}^{n times n} ]
This implies $A$ has eigenvalues $lambda_i in mathbb{R}$. It can also be diagonalised into a matrix $D = diag(lambda_1, lambda_2, … , lambda_n)$ by an orthogonal matrix $O$, such that
[ OAO^T = D = begin{pmatrix}
lambda & 0 & … & 0 \
0 & lambda_2 & … & 0 \
. & . & … & . \
. & . & … & . \
0 & 0 & … & lambda_n \
end{pmatrix} Leftrightarrow A = O^T DO ]
Here, it should be highlighted $O$ is an $n times m$ matrix. From these considerations,
[ implies z = int_{mathbb{R}^n} d^n xe^{-frac{1}{2} x^{T} Ax} ]
[ = int_{mathbb{R}^n} d^n xe^{-frac{1}{2} x^{T} O^T DO_x} ]
[ = int_{mathbb{R}^n} d^n xe^{-frac{1}{2} (O_x)^{T} D(O_x)} ]
At this point, we perform variable substitution. Notice,
[ y : = O_{x} ]
[ d^{n} y = det (frac{dy}{dx}) d^{n} x ]
[ implies d^{n} x = frac{1}{det (O)} d^{n} y ]
[ therefore z = int_{mathbb{R}^n} d^{n} y frac{1}{det (O)} e^{-frac{1}{2} y^{T} Dy} ]
[ = int_{mathbb{R}^n} d^{n} y frac{1}{det (O)} e^{-frac{1}{2} sum_{i=1}^{n} lambda_i y_{i}^2} ]
[ frac{1}{det (O)} prod_{i=1}^{n} int_{- infty}^{infty} dy_{i} e^{-frac{1}{2} lambda_i y_{i}^2 } ]
From the 1-dimensional case, we know: $e^{-frac{1}{2} lambda_i y_{i}^2} = sqrt{frac{2 pi}{ lambda_i}}$. So,
[ z = frac{1}{det(O)} prod_{i=1}^{n} sqrt{frac{2 pi}{ lambda_i}} ]
Now, recall that:
[ D = begin{pmatrix}
lambda & 0 & … & 0 \
0 & lambda_2 & … & 0 \
. & . & … & . \
. & . & … & . \
0 & 0 & … & lambda_n \
end{pmatrix} ]
From this, we can simplify our last result
[ z = frac{1}{det (O)} frac{(2 pi)^{frac{n}{2}}}{sqrt{det (D)}} ]
[ = frac{(2 pi)^{frac{n}{2}}}{sqrt{det (O^T DO)}} ]
[ therefore z = frac{(2 pi)^{frac{n}{2}}}{sqrt{det (A)}} ]
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