Unveiling the Mysteries of Functional Derivatives: A Tale of Quantum Field Theory for the Gifted Amateur (Exercise 1.2)

Table of Contents

## Introduction

Once upon a time in the fantastical realm of Quantum Field Theory, three functional giants roamed the land: $H[f]$ , $I[f]$ , and $J[f]$ . These functionals were peculiar entities, creatures that didn’t just evaluate a number at a point but instead swallowed entire functions and churned out a single value. Today, dear reader, we embark on an odyssey to dissect the inner workings of these giants and understand how they respond to infinitesimal changes. In the language of mathematics, we seek the “functional derivatives” of these giants.

## The Case of $H[f]$

The functional $H[f]$ is defined as:

$H[f] = \int G(x, y) f(y) \, dy$

We want to find $\frac{\delta H[f]}{\delta f(z)}$ following the formal definition:

$\frac{\delta H}{\delta f(z)} = \lim_{\epsilon \rightarrow 0} \frac{H\left[f(y) + \epsilon \delta(z - y)\right] - H[f(y)]}{\epsilon}$

First, let’s find $H\left[f(y) + \epsilon \delta(z - y)\right]$ :

$H\left[f(y) + \epsilon \delta(z - y)\right] = \int G(x, y) \left[ f(y) + \epsilon \delta(z - y) \right] \, dy$

Upon expanding, this becomes:

$= \int G(x, y) f(y) \, dy + \epsilon \int G(x, y) \delta(z - y) \, dy$

$= H[f] + \epsilon G(x, z)$

Therefore, the functional derivative becomes:

$\frac{\delta H}{\delta f(z)} = \lim_{\epsilon \rightarrow 0} \frac{H[f] + \epsilon G(x, z) - H[f]}{\epsilon}$

$= G(x, z)$

## The Mystery of $I[f]$

### The First Functional Derivative of $I[f^3]$

The functional $I[f]$ under consideration is $I[f^3] = \int_{-1}^{1} f(x)^3 \, dx$ .

Using the formal definition of functional derivative:

$\frac{\delta I[f^3]}{\delta f(x_0)} = \lim_{\epsilon \rightarrow 0} \frac{I\left[f(x)^3 + \epsilon \delta(x - x_0)^3\right] - I\left[f(x)^3\right]}{\epsilon}$

The integral with the small change $\epsilon \delta(x - x_0)$ becomes:

$I\left[f(x)^3 + \epsilon \delta(x - x_0)\right] = \int_{-1}^{1} \left[ f(x) + \epsilon \delta(x - x_0) \right]^3 \, dx$

$= \int_{-1}^{1} f(x)^3 \, dx + 3 \epsilon f(x_0)^2$

Using this in the functional derivative:

$\frac{\delta I[f^3]}{\delta f(x_0)} = \lim_{\epsilon \rightarrow 0} \frac{I[f^3] + 3 \epsilon f(x_0)^2 - I[f^3]}{\epsilon}$

$= 3 f(x_0)^2$

### The Second Functional Derivative of $I[f^3]$

For the second functional derivative $\frac{\delta^2 I[f^3]}{\delta f(x_0) \delta f(x_1)}$ , we take the derivative of the first functional derivative with respect to $f(x_1)$ :

$\frac{\delta^2 I[f^3]}{\delta f(x_0) \delta f(x_1)} = \frac{\delta}{\delta f(x_1)} (3 f(x_0)^2)$

$= 6 f(x_0) \delta(x_1 - x_0)$

## The Enigma of $J[f]$

The functional $J[f]$ is given by:

$J[f] = \int \left( \frac{\partial f}{\partial y} \right)^2 \, dy$

We want to find $\frac{\delta J[f]}{\delta f(x)}$ using the formal definition:

$\frac{\delta J[f]}{\delta f(x)} = \lim_{\epsilon \rightarrow 0} \frac{J\left[ f(y) + \epsilon \delta(x - y) \right] - J[f(y)]}{\epsilon}$

First, consider the term inside $J[f] + \delta J[f]$ :

$J\left[ f(y) + \epsilon \delta(x - y) \right] = \int \left( \frac{\partial}{\partial y} \left[ f(y) + \epsilon \delta(x - y) \right] \right)^2 \, dy$

$= \int \left( \frac{\partial f}{\partial y} \right)^2 dy + 2 \epsilon \frac{\partial f}{\partial y} \frac{\partial}{\partial y} \delta(x - y) + \epsilon^2 \left( \frac{\partial}{\partial y} \delta(x - y) \right)^2 \, dy$

The last term involves $\epsilon^2$ and will vanish in the limit $\epsilon \rightarrow 0$ .

Now, the middle term after integration will look like:

$2 \epsilon \int \frac{\partial f}{\partial y} \frac{\partial}{\partial y} \delta(x - y) \, dy$

$= -2 \epsilon \frac{\partial^2 f}{\partial y^2} \Bigg|_{y=x}$

The negative sign appears due to integration by parts. Thus, $\delta J[f] = -2 \epsilon \frac{\partial^2 f}{\partial y^2} \Bigg|_{y=x}$ .

Finally, using this in the formal definition:

$\frac{\delta J[f]}{\delta f(x)} = \lim_{\epsilon \rightarrow 0} \frac{-2 \epsilon \frac{\partial^2 f}{\partial y^2} \Bigg|_{y=x}}{\epsilon}$

$= -2 \frac{\partial^2 f}{\partial y^2} \Bigg|_{y=x}$

And there we have it! The functional derivative $\frac{\delta J[f]}{\delta f(x)} = -2 \frac{\partial^2 f}{\partial y^2} \Bigg|_{y=x}$ is found in a rigorous manner following the formal definition. This reveals that $J[f]$ is sensitive to the curvature of the function $f(x)$ at the point $x$ .

## Conclusion

Thus, we’ve rigorously computed the functional derivatives for $H[f]$ , $I[f]$ , and $J[f]$ based on the formal definitions. The enigmas are solved, and the giants’ secrets are laid bare, all through the power of rigorous mathematical formulation.

—

And so, dear reader, we’ve unmasked the mysteries of our three functional giants. May this tale serve as a guide on your own journey through the wondrous world of Quantum Field Theory.

Happy questing!

## Introduction

## The Case of

## The Mystery of

### The First Functional Derivative of

### The Second Functional Derivative of

## The Enigma of