Module 6: Functional Integrals & Functional Calculus
The Mathematics of Infinite Dimensions
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To deeply understand path integrals, we need to transition from the mathematics of finite-dimensional spaces to the mathematics of infinite-dimensional spaces. In classical mechanics and standard quantum mechanics, we usually deal with functions: mathematical objects that map a number (like time \(t\)) to another number (like position \(x(t)\)).
However, in the path integral formalism, the central object is the action \(S[x]\), which maps an entire function (a path \(x(t)\)) to a single number. This requires us to generalize the rules of calculus. We must learn how to differentiate and Taylor-expand with respect to entire functions rather than simple variables. This machinery is known as functional calculus, and it is the universal language of modern Quantum Field Theory (QFT) and advanced statistical mechanics. Letâs build this toolkit from the ground up!
1 1. What is a Functional?
A standard function, \(f(x)\), takes a number \(x\) and returns another number \(f(x)\). A functional, denoted with square brackets as \(F[x(t)]\), takes an entire function \(x(t)\) and returns a single real or complex number.
Imagine a function as a simple machine: you drop a single coin (a number) into the slot, and it spits out a ticket (another number). A functional, on the other hand, is a much larger machine. You must feed it an entire strip of film (a full continuous trajectory \(x(t)\) over some time interval). Only after scanning the whole strip does the machine spit out a single ticket (a number).
1.1 Common Examples of Functionals
The Action Functional: The most famous functional in physics is the classical action: \[ S[x] = \int_{t_a}^{t_b} L(x(t), \dot{x}(t), t) \, dt \] You plug in a path \(x(t)\), evaluate the Lagrangian \(L\) at every instant, integrate over time, and get a single number \(S\).
Arc Length: The length of a curve \(y(x)\) between \(x_1\) and \(x_2\) in 2D space: \[ L[y] = \int_{x_1}^{x_2} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx \]
Local Evaluation Functional: A functional that simply evaluates the function at a specific point \(t_0\): \[ E_{t_0}[x] = x(t_0) \] This can be rewritten using the Dirac delta function, transforming it into an integral form: \[ E_{t_0}[x] = \int x(t) \delta(t - t_0) \, dt \]
2 2. Functional Derivatives
In ordinary calculus, the derivative \(df/dx\) tells us how the number \(f\) changes when we slightly vary the number \(x\). For a functional \(F[x]\), we want to know how the number \(F\) changes when we slightly vary the entire function \(x(t)\) by a small amount \(\delta x(t)\).
By analogy with the multivariable chain rule \(df = \sum_i (\partial f/\partial x_i) dx_i\), the variation of a functional is defined as: \[ \delta F = F[x + \delta x] - F[x] = \int \frac{\delta F[x]}{\delta x(t)} \delta x(t) \, dt \]
Here, \(\frac{\delta F[x]}{\delta x(t)}\) is the functional derivative. It tells us the sensitivity of \(F\) to a localized change in the function \(x\) at the specific time \(t\).
2.1 Formal Definition
To make this rigorous, we introduce a small parameter \(\epsilon\) and an arbitrary âtest functionâ \(\eta(t)\). We define the functional derivative via the first-order Taylor expansion in \(\epsilon\): \[ F[x + \epsilon \eta] = F[x] + \epsilon \int \frac{\delta F[x]}{\delta x(t)} \eta(t) \, dt + \mathcal{O}(\epsilon^2) \]
Letâs compute some explicit examples to build intuition.
2.2 Example A: The Evaluation Functional
Let \(F[x] = x(t')\). We want to find its functional derivative with respect to \(x(t)\). \[ F[x + \epsilon \eta] = x(t') + \epsilon \eta(t') \] Using the Dirac delta function, we can rewrite \(\eta(t')\) as an integral: \[ F[x + \epsilon \eta] = F[x] + \epsilon \int \delta(t - t') \eta(t) \, dt \] Comparing this with the formal definition, we immediately read off: \[ \boxed{ \frac{\delta x(t')}{\delta x(t)} = \delta(t - t') } \] This is the functional analog of the ordinary calculus rule \(\frac{\partial x_i}{\partial x_j} = \delta_{ij}\).
2.3 Example B: A Quadratic Functional
Let \(F[x] = \int x(t')^2 \, dt'\). \[ F[x + \epsilon \eta] = \int \left( x(t') + \epsilon \eta(t') \right)^2 \, dt' \] \[ F[x + \epsilon \eta] = \int \left( x(t')^2 + 2\epsilon x(t')\eta(t') + \epsilon^2 \eta(t')^2 \right) \, dt' \] Dropping the \(\mathcal{O}(\epsilon^2)\) term, we have: \[ F[x + \epsilon \eta] \approx F[x] + \epsilon \int \left[ 2 x(t') \right] \eta(t') \, dt' \] Changing the dummy variable \(t'\) to \(t\) in the integral, we identify: \[ \boxed{ \frac{\delta}{\delta x(t)} \int x(t')^2 \, dt' = 2x(t) } \]
2.4 Example C: Derivative of the Kinetic Energy (Integration by Parts!)
Let \(F[x] = \int_{t_a}^{t_b} \frac{1}{2} m \dot{x}(t')^2 \, dt'\). \[ F[x + \epsilon \eta] = \int_{t_a}^{t_b} \frac{1}{2} m \left( \dot{x}(t') + \epsilon \dot{\eta}(t') \right)^2 \, dt' \] \[ F[x + \epsilon \eta] \approx F[x] + \epsilon \int_{t_a}^{t_b} m \dot{x}(t') \dot{\eta}(t') \, dt' \] Notice that the variation involves \(\dot{\eta}(t')\), but our definition of the functional derivative requires the integral to be of the form \(\int (\cdots) \eta(t') \, dt'\). We must move the derivative off of \(\eta(t')\) and onto \(x(t')\). We do this using integration by parts: \[ \int_{t_a}^{t_b} m \dot{x}(t') \dot{\eta}(t') \, dt' = \left[ m \dot{x}(t') \eta(t') \right]_{t_a}^{t_b} - \int_{t_a}^{t_b} m \ddot{x}(t') \eta(t') \, dt' \] Assuming the variation \(\eta(t')\) vanishes at the boundaries \(t_a\) and \(t_b\) (as is standard in the calculus of variations since the endpoints of the path are fixed), the boundary term drops out. We are left with: \[ F[x + \epsilon \eta] \approx F[x] + \epsilon \int_{t_a}^{t_b} \left[ -m \ddot{x}(t') \right] \eta(t') \, dt' \] Thus, swapping the dummy variable \(t'\) for \(t\): \[ \boxed{ \frac{\delta}{\delta x(t)} \int \frac{1}{2} m \dot{x}(t')^2 \, dt' = -m \ddot{x}(t) } \]
Whenever the functional contains time derivatives of \(x(t)\), computing the functional derivative will almost always require integration by parts to isolate the arbitrary variation \(\eta(t)\). Always pay attention to the boundary terms!
3 3. Key Identities
The rules of functional calculus beautifully mirror those of ordinary calculus.
Linearity: \[ \frac{\delta}{\delta x(t)} \left( a F[x] + b G[x] \right) = a \frac{\delta F[x]}{\delta x(t)} + b \frac{\delta G[x]}{\delta x(t)} \]
Product Rule (Leibniz Rule): \[ \frac{\delta}{\delta x(t)} \left( F[x] G[x] \right) = \frac{\delta F[x]}{\delta x(t)} G[x] + F[x] \frac{\delta G[x]}{\delta x(t)} \]
Chain Rule: If \(F\) depends on \(x\) through another function \(y(t)\), where \(y\) is a local function of \(x\): \[ \frac{\delta F[y(x)]}{\delta x(t)} = \int \frac{\delta F[y]}{\delta y(t')} \frac{\delta y(t')}{\delta x(t)} \, dt' \] Notice that the sum from the multivariable chain rule \(\sum_i\) becomes an integral \(\int dt'\) in the functional case.
Higher-order functional derivatives are symmetric, just like partial derivatives. That is, \(\frac{\delta^2 F}{\delta x(t_1) \delta x(t_2)} = \frac{\delta^2 F}{\delta x(t_2) \delta x(t_1)}\), assuming the functional is sufficiently smooth.
4 4. Euler-Lagrange from Functionals
We are now equipped to derive the Euler-Lagrange equations using our compact functional derivative notation. The principle of least action states that the classical path \(x_{cl}(t)\) is an extremum of the action \(S[x]\). In functional language, this means the functional derivative of \(S[x]\) evaluated at \(x_{cl}\) must vanish exactly: \[ \left. \frac{\delta S[x]}{\delta x(t)} \right|_{x = x_{cl}} = 0 \]
Letâs derive this for a general action \(S[x] = \int_{t_a}^{t_b} L(x(t'), \dot{x}(t')) \, dt'\). We vary \(x\) by \(\epsilon \eta(t')\): \[ S[x + \epsilon \eta] = \int_{t_a}^{t_b} L(x(t') + \epsilon \eta(t'), \dot{x}(t') + \epsilon \dot{\eta}(t')) \, dt' \] Taylor expanding the Lagrangian to first order in \(\epsilon\): \[ S[x + \epsilon \eta] \approx \int_{t_a}^{t_b} \left[ L(x, \dot{x}) + \epsilon \frac{\partial L}{\partial x}\eta(t') + \epsilon \frac{\partial L}{\partial \dot{x}}\dot{\eta}(t') \right] dt' \] \[ = S[x] + \epsilon \int_{t_a}^{t_b} \left[ \frac{\partial L}{\partial x}\eta(t') + \frac{\partial L}{\partial \dot{x}}\dot{\eta}(t') \right] dt' \] To isolate \(\eta(t')\), we integrate the second term by parts: \[ \int_{t_a}^{t_b} \frac{\partial L}{\partial \dot{x}}\dot{\eta}(t') \, dt' = \left[ \frac{\partial L}{\partial \dot{x}}\eta(t') \right]_{t_a}^{t_b} - \int_{t_a}^{t_b} \frac{d}{dt'}\left( \frac{\partial L}{\partial \dot{x}} \right) \eta(t') \, dt' \] Since the endpoints are fixed, \(\eta(t_a) = \eta(t_b) = 0\), the boundary term vanishes. Substituting this back: \[ S[x + \epsilon \eta] \approx S[x] + \epsilon \int_{t_a}^{t_b} \left[ \frac{\partial L}{\partial x} - \frac{d}{dt'}\left( \frac{\partial L}{\partial \dot{x}} \right) \right] \eta(t') \, dt' \] Comparing this to the formal definition \(S[x + \epsilon \eta] = S[x] + \epsilon \int \frac{\delta S}{\delta x(t')} \eta(t') \, dt'\), we identify the functional derivative: \[ \boxed{ \frac{\delta S[x]}{\delta x(t)} = \frac{\partial L}{\partial x} - \frac{d}{dt}\left( \frac{\partial L}{\partial \dot{x}} \right) } \] Setting this functional derivative to zero precisely yields the celebrated Euler-Lagrange equation.
4.1 đ Worked Example 1: Action for a particle in a potential
Problem: Compute \(\frac{\delta S}{\delta x(t)}\) for the action \(S = \int \left[ \frac{1}{2}m\dot{x}^2 - V(x) \right] dt\), and show that setting it to zero gives Newtonâs Second Law.
Solution: We can use the general formula we just derived, or compute it directly from scratch. Letâs do it directly to build muscle memory. Vary \(x \to x + \epsilon \eta\): \[ S[x + \epsilon \eta] = \int \left[ \frac{1}{2}m(\dot{x} + \epsilon \dot{\eta})^2 - V(x + \epsilon \eta) \right] dt \] Expand to first order in \(\epsilon\): \[ \approx \int \left[ \frac{1}{2}m\dot{x}^2 + \epsilon m\dot{x}\dot{\eta} - V(x) - \epsilon V'(x)\eta \right] dt \] \[ = S[x] + \epsilon \int \left[ m\dot{x}\dot{\eta} - V'(x)\eta \right] dt \] Integrate the \(m\dot{x}\dot{\eta}\) term by parts: \[ \int m\dot{x}\dot{\eta} \, dt = \left[ m\dot{x}\eta \right] - \int m\ddot{x}\eta \, dt = - \int m\ddot{x}\eta \, dt \] Substitute back: \[ S[x + \epsilon \eta] \approx S[x] + \epsilon \int \left[ -m\ddot{x} - V'(x) \right] \eta(t) \, dt \] We extract the functional derivative: \[ \frac{\delta S}{\delta x(t)} = -m\ddot{x} - V'(x) \] Setting this to zero yields \(-m\ddot{x} - V'(x) = 0 \implies m\ddot{x} = -V'(x)\), which is precisely \(F = ma\).
5 5. Functional Taylor Expansion
Just as a function \(f(x)\) can be expanded around a point \(x_0\), a functional \(F[x]\) can be expanded around a specific function \(x_0(t)\). Letâs write the variation as \(y(t)\), so we are evaluating \(F[x_0 + y]\).
The Functional Taylor Expansion is given by: \[ F[x_0 + y] = F[x_0] + \int dt \left. \frac{\delta F}{\delta x(t)} \right|_{x_0} y(t) + \frac{1}{2!} \int dt \int dt' \left. \frac{\delta^2 F}{\delta x(t) \delta x(t')} \right|_{x_0} y(t) y(t') + \dots \]
This is profoundly useful when \(F[x]\) is the action \(S[x]\) and we expand around the classical path \(x_{cl}(t)\): \[ S[x_{cl} + y] = S[x_{cl}] + \int dt \left. \frac{\delta S}{\delta x(t)} \right|_{x_{cl}} y(t) + \frac{1}{2!} \int dt \int dt' \left. \frac{\delta^2 S}{\delta x(t) \delta x(t')} \right|_{x_{cl}} y(t) y(t') + \dots \]
Because \(x_{cl}\) satisfies the Euler-Lagrange equations, the first functional derivative is zero! Therefore, the linear term in \(y(t)\) vanishes entirely: \[ \left. \frac{\delta S}{\delta x(t)} \right|_{x_{cl}} = 0 \]
We are left with: \[ S[x_{cl} + y] = S[x_{cl}] + S_{fluct}[y] + \mathcal{O}(y^3) \] where the quadratic term represents the action of the quantum fluctuations \(y(t)\) around the classical path. This provides the rigorous mathematical justification for the semiclassical decomposition we used in Module 5 for the Harmonic Oscillator!
What is the functional derivative of the action with respect to \(x(t)\)? What does it equal when evaluated exactly on the classical path?
The functional derivative \(\frac{\delta S}{\delta x(t)}\) gives the Euler-Lagrange expression: \(\frac{\partial L}{\partial x} - \frac{d}{dt}(\frac{\partial L}{\partial \dot{x}})\). When evaluated on the classical path \(x_{cl}\), this functional derivative is identically zero, which simply means that the action is stationary at the classical path.
6 6. Generating Functionals & Source Terms
In ordinary probability theory, to compute the moments \(\langle x \rangle, \langle x^2 \rangle, \dots\) of a probability distribution \(P(x)\), we introduce a moment-generating function: \[ Z(J) = \int e^{J x} P(x) \, dx \] By taking derivatives with respect to \(J\) and setting \(J=0\), we easily pull down factors of \(x\): \[ \langle x^n \rangle = \left. \frac{d^n Z(J)}{dJ^n} \right|_{J=0} \]
In the path integral formulation of quantum mechanics and QFT, we construct a generating functional, \(Z[J]\). We modify the standard path integral by adding a âsource termâ \(J(t)x(t)\) to the action. Here, \(J(t)\) is an arbitrary external driving force.
\[ \boxed{ Z[J] = \int \mathcal{D}x \, \exp\left\{ \frac{i}{\hbar} S[x] + i \int J(t) x(t) \, dt \right\} } \]
Textbook reference: Shankar, Ch. 8. Also widely covered in Zee, QFT in a Nutshell, Ch. I.3.
This \(Z[J]\) is the master object of Quantum Field Theory. If you know \(Z[J]\), you mathematically know absolutely everything there is to know about the quantum system!
Why add \(J(t)\)? Think of \(J(t)\) as an artificial external hand poking and prodding the quantum system at different times \(t\). By observing how the systemâs âvacuumâ (the sum over paths) reacts to this prodding (by taking functional derivatives with respect to \(J\)), we can deduce the internal dynamics and correlations of the particle itself. Once we have our answers, we set \(J=0\) to turn the prodding off.
7 7. Correlation Functions from Z[J]
How do we use \(Z[J]\)? Suppose we want to compute the expectation value of the particleâs position at time \(t_1\), known as a 1-point correlation function \(\langle x(t_1) \rangle\).
Letâs take the functional derivative of \(Z[J]\) with respect to the source \(J(t_1)\): \[ \frac{\delta Z[J]}{\delta J(t_1)} = \int \mathcal{D}x \, \frac{\delta}{\delta J(t_1)} \exp\left\{ \frac{i}{\hbar} S[x] + i \int J(t) x(t) \, dt \right\} \] Using the chain rule and the fact that \(\frac{\delta}{\delta J(t_1)} \int J(t)x(t)dt = x(t_1)\): \[ \frac{\delta Z[J]}{\delta J(t_1)} = \int \mathcal{D}x \, \left( i x(t_1) \right) \exp\left\{ \frac{i}{\hbar} S[x] + i \int J(t) x(t) \, dt \right\} \] If we multiply by \(-i\) and then set the artificial source \(J=0\), the exponential just becomes the standard amplitude \(e^{iS/\hbar}\): \[ \left. -i \frac{\delta Z[J]}{\delta J(t_1)} \right|_{J=0} = \int \mathcal{D}x \, x(t_1) e^{\frac{i}{\hbar}S[x]} \] To get the normalized expectation value, we must divide by \(Z[0] = \int \mathcal{D}x \, e^{iS/\hbar}\): \[ \boxed{ \langle x(t_1) \rangle = \frac{1}{Z[0]} \left. \left( -i \frac{\delta}{\delta J(t_1)} \right) Z[J] \right|_{J=0} } \]
We can extend this to find the expectation value of the product of positions at two different times, the 2-point correlation function (or Greenâs function): \[ \langle x(t_1) x(t_2) \rangle = \frac{\int \mathcal{D}x \, x(t_1) x(t_2) e^{\frac{i}{\hbar}S[x]}}{\int \mathcal{D}x \, e^{\frac{i}{\hbar}S[x]}} \] This is generated by taking two functional derivatives: \[ \boxed{ \langle x(t_1) x(t_2) \rangle = \frac{1}{Z[0]} \left. \left( -i \frac{\delta}{\delta J(t_1)} \right) \left( -i \frac{\delta}{\delta J(t_2)} \right) Z[J] \right|_{J=0} } \]
In general, an \(n\)-point correlation function is given by taking \(n\) functional derivatives of \(Z[J]\). The physical meaning of \(\langle x(t_1) x(t_2) \rangle\) is the probability amplitude for a particle to propagate from \(x(t_2)\) to \(x(t_1)\); it characterizes how fluctuations at time \(t_2\) influence the particle at time \(t_1\).
Why is \(Z[J]\) called a âgenerating functionalâ? What exactly does it generate?
\(Z[J]\) is called a generating functional because it generates all possible correlation functions (expectation values of products like \(x(t_1)x(t_2)\dots x(t_n)\)) of the quantum theory. By repeatedly applying the functional derivative \(\delta/\delta J(t)\) and then setting the source \(J\) to zero, we âpull downâ variables into the path integral, generating the entire family of \(n\)-point correlation functions, exactly like a moment generating function generates statistical moments.
7.1 đ Worked Example 2: The Free Particle Correlator
Problem: Compute \(Z[J]\) for a free particle where the action is \(S_0 = \int \frac{1}{2}m\dot{x}^2 dt\) plus a source term \(\int Jx dt\). Then use it to find \(\langle x(t_1)x(t_2) \rangle\).
Solution: We will use units where \(\hbar = 1\) for simplicity. The total exponent is: \[ iS_{tot} = i \int \left[ \frac{1}{2}m\dot{x}^2 + J(t)x(t) \right] dt \] Integrating the kinetic term by parts (assuming boundary terms vanish as usual): \[ \int \frac{1}{2}m\dot{x}^2 dt = -\int \frac{1}{2} x(t) m\frac{d^2}{dt^2} x(t) dt \] Letâs define the differential operator \(O = -m \frac{d^2}{dt^2}\). The exponent is a quadratic form: \[ iS_{tot} = \frac{i}{2} \int x(t) O x(t) dt + i \int J(t)x(t) dt \] By âcompleting the squareâ for functional integrals (which we will rigorously prove step-by-step in Module 7), we can shift variables \(x \to x + x_0\) to perform the integral. The result for the generating functional is an exponential of a quadratic in \(J\): \[ Z[J] = Z[0] \exp\left\{ -\frac{i}{2} \int \int J(t) G(t, t') J(t') \, dt \, dt' \right\} \] Here, \(G(t, t')\) is the Greenâs function, which is the inverse of our differential operator \(O\): \[ -m \frac{d^2}{dt^2} G(t, t') = \delta(t - t') \]
Now, letâs extract the 2-point correlator! \[ \langle x(t_1) x(t_2) \rangle = \frac{1}{Z[0]} \left. (-i)^2 \frac{\delta^2 Z[J]}{\delta J(t_1) \delta J(t_2)} \right|_{J=0} \] Let \(W[J] = -\frac{i}{2} \int \int J(t) G(t, t') J(t') dt dt'\) so that \(Z[J] = Z[0] e^{W[J]}\). First derivative (using the chain rule and the symmetric property \(G(t,t') = G(t',t)\)): \[ \frac{\delta Z}{\delta J(t_2)} = Z[0] e^{W[J]} \frac{\delta W}{\delta J(t_2)} = Z[J] \left( -i \int G(t_2, t') J(t') dt' \right) \] Second derivative with respect to \(J(t_1)\): \[ \frac{\delta^2 Z}{\delta J(t_1) \delta J(t_2)} = \frac{\delta Z}{\delta J(t_1)} \left( -i \int G(t_2, t') J(t') dt' \right) + Z[J] \left( -i G(t_2, t_1) \right) \] Now, evaluate at \(J=0\). The first term vanishes entirely because the integral over \(J\) is zero. We are left with just the second term: \[ \left. \frac{\delta^2 Z}{\delta J(t_1) \delta J(t_2)} \right|_{J=0} = Z[0] ( -i G(t_2, t_1) ) \] Plug this into the correlator formula: \[ \langle x(t_1) x(t_2) \rangle = \frac{1}{Z[0]} (-1) Z[0] (-i G(t_2, t_1)) = i G(t_1, t_2) \] This beautiful result shows that the 2-point correlation function of a free particle is exactly proportional to its Greenâs function!
8 8. Dictionary: Ordinary Calculus vs Functional Calculus
To consolidate your learning, keep this translation dictionary handy as we move deeper into the course:
| Concept | Ordinary Calculus (Finite \(N\)) | Functional Calculus (Infinite Dim) |
|---|---|---|
| Index / Coordinate | Discrete index \(i\) | Continuous time \(t\) |
| Variable | Vector \(\vec{x} = \{x_1, x_2, \dots, x_N\}\) | Function \(x(t)\) |
| Function(al) | Function \(f(\vec{x})\) | Functional \(F[x(t)]\) |
| Summation | Sum \(\sum_i (\dots)\) | Integral \(\int dt (\dots)\) |
| Identity Matrix | Kronecker delta \(\delta_{ij}\) | Dirac delta \(\delta(t - t')\) |
| Derivative | Partial derivative \(\frac{\partial}{\partial x_i}\) | Functional derivative \(\frac{\delta}{\delta x(t)}\) |
| Chain Rule | \(\frac{\partial f}{\partial y_i} = \sum_j \frac{\partial f}{\partial x_j} \frac{\partial x_j}{\partial y_i}\) | \(\frac{\delta F}{\delta y(t)} = \int dt' \frac{\delta F}{\delta x(t')} \frac{\delta x(t')}{\delta y(t)}\) |
| Taylor Series | \(f(\vec{x}_0 + \vec{y}) = f(\vec{x}_0) + \sum_i \frac{\partial f}{\partial x_i} y_i + \dots\) | \(F[x_0 + y] = F[x_0] + \int dt \frac{\delta F}{\delta x(t)} y(t) + \dots\) |
Review Checklist - Can you explain the difference between a function and a functional in your own words? - Do you understand why integration by parts is essential when computing functional derivatives of Lagrangians with \(\dot{x}\)? - Can you write down the general formula for extracting an expectation value \(\langle x(t) \rangle\) from the generating functional \(Z[J]\)?
9 Practice Problems
- Easy: Compute the functional derivative of \(F[x] = \int_{0}^{T} x(t)^3 \, dt\) with respect to \(x(t')\).
- Easy: For the functional \(G[x] = \int x(t) y(t) \, dt\) where \(y(t)\) is a fixed function, what is \(\frac{\delta G}{\delta x(t')}\)?
- Medium: Let \(F[x] = \exp\left( \int x(t)^2 dt \right)\). Compute \(\frac{\delta F}{\delta x(t')}\) using the functional chain rule.
- Medium: The length of a string hanging between two posts is given by the functional \(L[y] = \int \sqrt{1 + y'(x)^2} dx\). Compute \(\frac{\delta L}{\delta y(x)}\). Set it to zero to find the differential equation for the curve of minimum length (which you should recognize as a straight line if there are no other forces).
- Hard: Given the Lagrangian for a particle in an electromagnetic field, \(L = \frac{1}{2}m\dot{x}^2 + \frac{e}{c} \dot{x} A(x) - e \phi(x)\), use functional differentiation of the action \(\frac{\delta S}{\delta x(t)} = 0\) to derive the Lorentz force law. (Assume 1D for simplicity, but treat \(A\) and \(\phi\) as functions of \(x\)).
- Hard: Using the expression for \(Z[J]\) from Worked Example 2, perform the functional derivatives to find the explicit form of the 4-point correlation function \(\langle x(t_1) x(t_2) x(t_3) x(t_4) \rangle\) in terms of the Greenâs function \(G(t, t')\). (This is your first glimpse of Wickâs Theorem!).