Peeter Joot's (OLD) Blog.

Math, physics, perl, and programming obscurity.

Gaussian quadratic form integrals and multivariable approximation of exponential integrals

Posted by peeterjoot on January 26, 2014

[Click here for a PDF of this post with nicer formatting (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]

Motivation

In [1] eq. I.2.20 is the approximation

\begin{aligned}\int d\mathbf{q} e^{-f(\mathbf{q})/\hbar} \approx e^{-f(\mathbf{a})/\hbar} \sqrt{\frac{2 \pi \hbar}{\text{Det} f''(\mathbf{a})} },\end{aligned} \hspace{\stretch{1}}(1.1.1)

where [ f''(\mathbf{a}) ]_{ij} \equiv {\left.{{\partial^2 f/\partial q_i \partial q_j}}\right\vert}_{{\mathbf{q} = \mathbf{a}}}. Here \mathbf{a} is assumed to be an extremum of f. This follows from a generalization of the Gaussian integral result. Let’s derive both in detail.

Guts

First, to second order, let’s expand f(\mathbf{q}) around a min or max at \mathbf{q} = \mathbf{a}. The usual trick, presuming that one doesn’t remember the form of this generalized Taylor expansion, is to expand g(t) = f(\mathbf{a} + t \mathbf{q}) around t = 0, then evaluate at t = 1. We have

\begin{aligned}g'(t) &= \sum_i \frac{\partial {f(\mathbf{a} + t \mathbf{q})}}{\partial {(a_i + t q_i)}} \frac{d{{ (a_i + t q_i) }}}{dt} \\ &= \sum_i q_i \frac{\partial {f(\mathbf{a} + t \mathbf{q})}}{\partial {(a_i + t q_i)}} \\ &= \mathbf{q} \cdot \left(  {\left.{{\boldsymbol{\nabla}_\mathbf{q} f(\mathbf{q})}}\right\vert}_{{\mathbf{q} = \mathbf{a} + t \mathbf{q}}}  \right).\end{aligned} \hspace{\stretch{1}}(1.2.2)

The second derivative is

\begin{aligned}g''(t) = \sum_{i j} q_i q_j \frac{\partial {}}{\partial {(a_j + t q_j)}} \frac{\partial {f(\mathbf{a} + t \mathbf{q})}}{\partial {(a_i + t q_i)}},\end{aligned} \hspace{\stretch{1}}(1.2.3)

This gives

\begin{aligned}\begin{aligned}g'(0) &= \mathbf{q} \cdot \boldsymbol{\nabla}_\mathbf{q} f(\mathbf{q}) = \sum_i q_i \partial q_i f(\mathbf{q}) \\ g''(0) &= \left( \mathbf{q} \cdot \boldsymbol{\nabla}_\mathbf{q} \right)^2 f(\mathbf{q}) = \sum_{i j} q_i q_j \partial q_i \partial q_j f(\mathbf{q}).\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.2.4)

Putting these together, we have to second order in t is

\begin{aligned}f(\mathbf{a} + t \mathbf{q}) \approx f(\mathbf{a}) + \sum_i q_i \partial q_i f(\mathbf{q}) \frac{t^1}{1!}+ \sum_{i j} q_i q_j \partial q_i \partial q_j f(\mathbf{q}) \frac{t^2}{2!},\end{aligned} \hspace{\stretch{1}}(1.2.5)

or

\begin{aligned}f(\mathbf{a} + \mathbf{q}) \approx f(\mathbf{a}) + \sum_i q_i {\left.{{ \left(  \frac{\partial {f}}{\partial {q_i}} \right)}}\right\vert}_{\mathbf{a}}+ \frac{1}{{2}} \sum_{i j} q_i q_j {\left.{{\left(  \frac{\partial^2 f}{\partial q_i \partial q_j}  \right)}}\right\vert}_{\mathbf{a}}.\end{aligned} \hspace{\stretch{1}}(1.2.6)

We can put the terms up to second order in a nice tidy matrix forms

\begin{aligned}\mathbf{b} = {\left.{{\left(  \boldsymbol{\nabla}_\mathbf{q} f \right)}}\right\vert}_{\mathbf{a}}\end{aligned} \hspace{\stretch{1}}(1.0.7a)

\begin{aligned}A = {\begin{bmatrix}{\left.{{ \left(  \frac{\partial^2 f}{\partial q_i \partial q_j} \right)}}\right\vert}_{\mathbf{a}}\end{bmatrix}}_{i j}.\end{aligned} \hspace{\stretch{1}}(1.0.7b)

Note that eq. 1.0.7b is a real symmetric matrix, and can thus be reduced to diagonal form by an orthonormal transformation. Putting the pieces together, we have

\begin{aligned}f(\mathbf{a} + \mathbf{q}) \approx f(\mathbf{a}) + \mathbf{q}^\text{T} \mathbf{b} + \frac{1}{{2}} \mathbf{q}^\text{T} A \mathbf{q}.\end{aligned} \hspace{\stretch{1}}(1.0.8)

Integrating this, we have

\begin{aligned}\int dq_1 dq_2 \cdots dq_N \exp\left( -\left(  f(\mathbf{a}) + \mathbf{q}^\text{T} \mathbf{b} + \frac{1}{{2}} \mathbf{q}^\text{T} A \mathbf{q}  \right)\right)=e^{-f(\mathbf{a})}\int dq_1 dq_2 \cdots dq_N \exp\left(  -\mathbf{q}^\text{T} \mathbf{b} - \frac{1}{{2}} \mathbf{q}^\text{T} A \mathbf{q}  \right).\end{aligned} \hspace{\stretch{1}}(1.0.9)

Employing an orthonormal change of variables to diagonalizae the matrix

\begin{aligned}A = O^\text{T} D O,\end{aligned} \hspace{\stretch{1}}(1.0.10)

and \mathbf{r} = O \mathbf{q}, or r_i = O_{ik} q_k, the volume element after transformation is

\begin{aligned}dr_1 dr_2 \cdots dr_N &= \frac{\partial(r_1, r_2, \cdots, r_N)}{\partial(q_1, q_2, \cdots, q_N)}dq_1 dq_2 \cdots dq_N \\ &= \begin{vmatrix}O_{11} & O_{12} & \cdots & O_{1N} \\ O_{21} & O_{22} & \cdots & O_{2N} \\ \dot{v}s & \dot{v}s & \dot{v}s & \dot{v}s \\  O_{N1} & O_{N2} & \cdots & O_{NN} \\ \end{vmatrix}dq_1 dq_2 \cdots dq_N \\ &= (\text{Det} O)dq_1 dq_2 \cdots dq_N \\ &= dq_1 dq_2 \cdots dq_N \end{aligned} \hspace{\stretch{1}}(1.0.10)

Our integral is

\begin{aligned}e^{-f(\mathbf{a})}\int dq_1 dq_2 \cdots dq_N \exp\left(  -\mathbf{q}^\text{T} \mathbf{b} - \frac{1}{{2}} \mathbf{q}^\text{T} A \mathbf{q}  \right) &= e^{-f(\mathbf{a})}\int dr_1 dr_2 \cdots dr_N \exp\left(  -\mathbf{q}^\text{T} O^\text{T} O \mathbf{b} - \frac{1}{{2}} \mathbf{q}^\text{T} O^\text{T} D O \mathbf{q}  \right) \\ &= e^{-f(\mathbf{a})}\int dr_1 dr_2 \cdots dr_N \exp\left(  -\mathbf{r}^\text{T} (O \mathbf{b}) - \frac{1}{{2}} \mathbf{r}^\text{T} D \mathbf{r}  \right) \\ &= e^{-f(\mathbf{a})}\int dr_1 e^{ -\frac{1}{{2}} r_1^2 \lambda_1 - r_1 (O \mathbf{b})_1 }\int dr_2 e^{ -\frac{1}{{2}} r_2^2 \lambda_2 - r_2 (O \mathbf{b})_2 }\cdots \int dr_N e^{ -\frac{1}{{2}} r_N^2 \lambda_N - r_N (O \mathbf{b})_N }.\end{aligned} \hspace{\stretch{1}}(1.0.10)

We now have products of terms that are of the regular Gaussian form. One such integral is

\begin{aligned}\int e^{-a x^2/2 + J x} &= \int \exp\left(-\frac{1}{{2}} \left(\left(  \sqrt{a} x - J/\sqrt{a}  \right)^2- \left(  J/\sqrt{a}  \right)^2\right)\right) \\ &= e^{J^2/2a} \sqrt{2 \pi \int_0^\infty r dr e^{-a r^2/2}}\end{aligned} \hspace{\stretch{1}}(1.0.10)

This is just

\begin{aligned}\int e^{-a x^2/2 + J x}= e^{J^2/2a} \sqrt{ \frac{2 \pi}{a} }.\end{aligned} \hspace{\stretch{1}}(1.0.10)

Applying this to the integral of interest, writing m_i = (O \mathbf{b})_i

\begin{aligned}\begin{aligned}e^{-f(\mathbf{a})}\int &dq_1 dq_2 \cdots dq_N \exp\left(  -\mathbf{q}^\text{T} \mathbf{b} - \frac{1}{{2}} \mathbf{q}^\text{T} A \mathbf{q}  \right) \\ &=e^{-f(\mathbf{a})}e^{-m_1^2/2\lambda_1} \sqrt{ \frac{2 \pi}{\lambda_1}}e^{-m_2^2/2\lambda_2} \sqrt{ \frac{2 \pi}{\lambda_2}}\cdots e^{-m_N^2/2\lambda_N} \sqrt{ \frac{2 \pi}{\lambda_N}} \\ &=e^{-f(\mathbf{a})}\sqrt{\frac{2 \pi}{\text{Det} A}}\exp\left(-\frac{1}{{2}}\left(  -m_1^2/\lambda_1 -m_2^2/\lambda_2 \cdots -m_N^2/\lambda_N  \right) \right).\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.10)

This last exponential argument can be put into matrix form

\begin{aligned}-m_1^2/\lambda_1-m_2^2/\lambda_2\cdots -m_N^2/\lambda_N &= (O \mathbf{b})^\text{T} D^{-1} O \mathbf{b} \\ &= \mathbf{b}^\text{T} O^\text{T} D^{-1} O \mathbf{b} \\ &= \mathbf{b}^\text{T} A^{-1} \mathbf{b},\end{aligned} \hspace{\stretch{1}}(1.0.10)

Finally, referring back to eq. 1.0.7, we have

\begin{aligned}\int d\mathbf{q} e^{-f(\mathbf{q})} \approx e^{-f(\mathbf{a})}\sqrt{\frac{2 \pi}{\text{Det} A}}e^{-\mathbf{b}^\text{T} A^{-1} \mathbf{b}/2}.\end{aligned} \hspace{\stretch{1}}(1.0.10)

Observe that we can recover eq. 1.1.1 by noting that \mathbf{b} = 0 for that system was assumed (i.e. \mathbf{a} was an extremum point), and by noting that the determinant scales with 1/\hbar since it just contains the second partials.

An afterword on notational sugar:

We didn’t need it, but it seems worth noting that we can write the Taylor expansion of eq. 1.0.8 in operator form as

\begin{aligned}f(\mathbf{a} + \mathbf{q}) = \sum_{k = 0}^\infty \frac{1}{{k!}} {\left.{{ \left( \mathbf{q} \cdot \boldsymbol{\nabla}_{\mathbf{q}'} \right)^k f(\mathbf{q}') }}\right\vert}_{{\mathbf{q}' = \mathbf{a}}}={\left.{{ e^{\mathbf{q} \cdot \boldsymbol{\nabla}_{\mathbf{q}'}} f(\mathbf{q}') }}\right\vert}_{{\mathbf{q}' = \mathbf{a}}}.\end{aligned} \hspace{\stretch{1}}(1.0.18)

References

[1] A. Zee. Quantum field theory in a nutshell. Universities Press, 2005.

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