Module 4: Deriving the Path Integral — The Full Construction
From Operators to Sums Over Paths: Every Step, No Hand-Waving
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This is the module where everything comes together. In Modules 1–3, we motivated the propagator, Feynman’s sum-over-paths idea, and computed the free-particle path integral almost by guessing the right procedure. Now we prove it. We will start from the standard operator formalism of quantum mechanics — the expression \(K = \langle x_b | e^{-i\hat{H}T/\hbar} | x_a \rangle\) — and, through a sequence of exact mathematical steps, derive Feynman’s path integral formula. No postulates, no hand-waving, just calculation.
By the end of this module, you will understand why the path integral works, not just that it works. You will also discover that the path integral naturally comes in two flavors — a phase-space version (more fundamental) and a configuration-space version (more commonly used) — and see exactly how the Lagrangian emerges from the Hamiltonian through the integration over momenta.
Textbook references: Shankar, Principles of Quantum Mechanics, Ch. 8 & 21. Feynman & Hibbs, Ch. 3. Kleinert, Ch. 2. Zinn-Justin, Ch. 2.
1 Starting Point: The Operator Expression
Let us begin with something exact and uncontroversial. In the operator formalism of quantum mechanics, the propagator (transition amplitude) for a particle to travel from position \(x_a\) at time \(t_a = 0\) to position \(x_b\) at time \(t_b = T\) is:
\[ \boxed{K(x_b, T; x_a, 0) = \langle x_b | \, e^{-i\hat{H}T/\hbar} \, | x_a \rangle} \tag{1}\]
This is exact. It follows directly from the time evolution of quantum states: if \(|\psi(0)\rangle\) is the state at time \(0\), then \(|\psi(T)\rangle = e^{-i\hat{H}T/\hbar} |\psi(0)\rangle\), and projecting onto \(\langle x_b|\) gives the wavefunction at position \(x_b\) and time \(T\).
There is nothing path-integral-ish about Equation 1 yet. The Hamiltonian \(\hat{H}\) is an operator acting on Hilbert space. Our goal for this entire module is a single, clear objective:
Convert the operator expression \(K = \langle x_b | e^{-i\hat{H}T/\hbar} | x_a \rangle\) into a sum over all paths weighted by \(e^{iS[\text{path}]/\hbar}\).
To do this, we need to somehow get rid of the operator \(\hat{H}\) and replace it with the classical Lagrangian \(L(x, \dot{x})\) — a function, not an operator. The key idea is breathtakingly simple: slice time into tiny pieces.
2 The Key Idea: Slicing Time
2.1 Breaking the exponential into factors
The time interval \([0, T]\) is divided into \(N\) equal segments, each of duration:
\[ \varepsilon = \frac{T}{N} \]
The intermediate times are \(t_j = j\varepsilon\) for \(j = 0, 1, 2, \ldots, N\), so \(t_0 = 0\) and \(t_N = T\).
Since the Hamiltonian \(\hat{H}\) is time-independent (we’ll assume this throughout), the time evolution operator factors:
\[ e^{-i\hat{H}T/\hbar} = \underbrace{e^{-i\hat{H}\varepsilon/\hbar} \cdot e^{-i\hat{H}\varepsilon/\hbar} \cdots e^{-i\hat{H}\varepsilon/\hbar}}_{N \text{ factors}} = \left( e^{-i\hat{H}\varepsilon/\hbar} \right)^N \tag{2}\]
This is exact — no approximation. We have simply used the rule \(e^{A+A+\cdots} = e^A \cdot e^A \cdots\) which holds because every factor is the same operator.
2.2 Inserting complete sets of position states
Now comes the crucial step. Between each pair of consecutive factors \(e^{-i\hat{H}\varepsilon/\hbar}\), we insert a complete set of position eigenstates. Recall the completeness relation (also called the resolution of the identity):
\[ \hat{1} = \int_{-\infty}^{+\infty} dx_j \, |x_j\rangle \langle x_j| \]
We insert \(N-1\) such resolutions — one between each pair of consecutive time evolution operators. The propagator becomes:
\[ K = \langle x_b | \, e^{-i\hat{H}\varepsilon/\hbar} \cdot \hat{1} \cdot e^{-i\hat{H}\varepsilon/\hbar} \cdot \hat{1} \cdots \hat{1} \cdot e^{-i\hat{H}\varepsilon/\hbar} \, | x_a \rangle \]
Writing this out with the integrals:
\[ K = \int dx_1 \, dx_2 \cdots dx_{N-1} \; \prod_{j=0}^{N-1} \langle x_{j+1} | \, e^{-i\hat{H}\varepsilon/\hbar} \, | x_j \rangle \tag{3}\]
where \(x_0 \equiv x_a\) and \(x_N \equiv x_b\) are the fixed endpoints.
What have we done? Each intermediate position \(x_j\) represents the particle’s position at time \(t_j\). By integrating over all possible intermediate positions \(x_1, x_2, \ldots, x_{N-1}\), we are already performing a sum over all piecewise-linear paths connecting \(x_a\) to \(x_b\)! Each “path” is a sequence of positions \((x_0, x_1, x_2, \ldots, x_N)\), and the amplitude for that path is the product of the short-time propagators \(\langle x_{j+1} | e^{-i\hat{H}\varepsilon/\hbar} | x_j \rangle\).
The problem has now been reduced to a simpler one: compute the short-time propagator \(\langle x_{j+1} | e^{-i\hat{H}\varepsilon/\hbar} | x_j \rangle\) for small \(\varepsilon\). This is where the Trotter product formula enters.
3 The Trotter Product Formula
3.1 The problem of non-commuting operators
For the standard non-relativistic Hamiltonian:
\[ \hat{H} = \hat{T} + \hat{V} = \frac{\hat{p}^2}{2m} + V(\hat{x}) \]
we would love to write:
\[ e^{-i(\hat{T}+\hat{V})\varepsilon/\hbar} \stackrel{?}{=} e^{-i\hat{T}\varepsilon/\hbar} \, e^{-i\hat{V}\varepsilon/\hbar} \]
If \(\hat{T}\) and \(\hat{V}\) were numbers, this would be trivially true. But they are operators, and \([\hat{T}, \hat{V}] \neq 0\) because \(\hat{T}\) depends on \(\hat{p}\) while \(\hat{V}\) depends on \(\hat{x}\), and \([\hat{x}, \hat{p}] = i\hbar\).
So the factorization above is not exact. But it becomes approximately correct for small \(\varepsilon\). Let us see exactly how.
3.2 Baker-Campbell-Hausdorff (BCH) formula
The Baker-Campbell-Hausdorff formula tells us how to combine two operator exponentials:
\[ e^{\hat{A}} \, e^{\hat{B}} = \exp\!\left( \hat{A} + \hat{B} + \frac{1}{2}[\hat{A}, \hat{B}] + \frac{1}{12}\bigl([\hat{A},[\hat{A},\hat{B}]] + [[\hat{A},\hat{B}],\hat{B}]\bigr) + \cdots \right) \tag{4}\]
Let us apply this with:
\[ \hat{A} = -\frac{i\hat{T}\varepsilon}{\hbar}, \qquad \hat{B} = -\frac{i\hat{V}\varepsilon}{\hbar} \]
Then:
\[ e^{\hat{A}} e^{\hat{B}} = \exp\!\left( -\frac{i(\hat{T}+\hat{V})\varepsilon}{\hbar} + \frac{1}{2}[\hat{A}, \hat{B}] + \cdots \right) \]
The commutator term is:
\[ \frac{1}{2}[\hat{A}, \hat{B}] = \frac{1}{2} \left(-\frac{i\varepsilon}{\hbar}\right)^2 [\hat{T}, \hat{V}] = -\frac{\varepsilon^2}{2\hbar^2} [\hat{T}, \hat{V}] \]
This is of order \(\varepsilon^2\). The next terms in BCH involve double commutators and are of order \(\varepsilon^3\), and so on.
3.3 The Trotter approximation
Therefore, we can write:
\[ \boxed{e^{-i(\hat{T}+\hat{V})\varepsilon/\hbar} = e^{-i\hat{T}\varepsilon/\hbar} \, e^{-i\hat{V}\varepsilon/\hbar} + \mathcal{O}(\varepsilon^2)} \tag{5}\]
This is the Trotter product formula (first-order version). The error in each time slice is \(\mathcal{O}(\varepsilon^2)\).
Now, we have \(N\) time slices, and \(\varepsilon = T/N\). The total accumulated error from all \(N\) slices is:
\[ \text{Total error} \sim N \times \mathcal{O}(\varepsilon^2) = N \times \mathcal{O}\!\left(\frac{T^2}{N^2}\right) = \mathcal{O}\!\left(\frac{T^2}{N}\right) \xrightarrow{N \to \infty} 0 \]
So in the limit \(N \to \infty\) (infinitely many infinitesimally thin time slices), the Trotter approximation becomes exact:
\[ e^{-i(\hat{T}+\hat{V})T/\hbar} = \lim_{N \to \infty} \left( e^{-i\hat{T}\varepsilon/\hbar} \, e^{-i\hat{V}\varepsilon/\hbar} \right)^N \tag{6}\]
This is the Trotter product theorem, and it is a rigorous mathematical result (proven by Trotter in 1959 for self-adjoint operators).
There is a more accurate version called the symmetric (or Strang) splitting: \[ e^{-i(\hat{T}+\hat{V})\varepsilon/\hbar} = e^{-i\hat{V}\varepsilon/2\hbar} \, e^{-i\hat{T}\varepsilon/\hbar} \, e^{-i\hat{V}\varepsilon/2\hbar} + \mathcal{O}(\varepsilon^3) \] This has error \(\mathcal{O}(\varepsilon^3)\) per slice instead of \(\mathcal{O}(\varepsilon^2)\). Both converge to the exact result as \(N \to \infty\), but the symmetric version converges faster. For the path integral derivation, either form works — we’ll use the simpler first-order form.
What happens if \(\hat{T}\) and \(\hat{V}\) commute, i.e., \([\hat{T}, \hat{V}] = 0\)? Does anything simplify in the Trotter formula?
If \([\hat{T}, \hat{V}] = 0\), then the BCH formula (Equation 4) truncates immediately: \(e^{\hat{A}} e^{\hat{B}} = e^{\hat{A}+\hat{B}}\) exactly. The Trotter formula becomes exact for any \(\varepsilon\) — no approximation needed!
The path integral would then trivially factorize into separate kinetic and potential contributions. The rich structure of the path integral arises precisely because \([\hat{x}, \hat{p}] = i\hbar \neq 0\). The non-commutativity of position and momentum is what makes the path integral non-trivial and interesting.
4 Matrix Elements: Building the Short-Time Propagator
Using the Trotter formula (Equation 5), the short-time propagator becomes:
\[ \langle x_{j+1} | \, e^{-i\hat{H}\varepsilon/\hbar} \, | x_j \rangle \approx \langle x_{j+1} | \, e^{-i\hat{T}\varepsilon/\hbar} \, e^{-i\hat{V}\varepsilon/\hbar} \, | x_j \rangle \]
We need to compute this matrix element. Let’s handle the two exponentials one at a time.
4.1 The potential factor
The potential operator \(\hat{V} = V(\hat{x})\) is diagonal in the position basis:
\[ \hat{V} |x_j\rangle = V(x_j) |x_j\rangle \]
Therefore:
\[ e^{-i\hat{V}\varepsilon/\hbar} |x_j\rangle = e^{-iV(x_j)\varepsilon/\hbar} |x_j\rangle \]
The potential exponential acting on \(|x_j\rangle\) simply becomes a number (a c-number phase factor). We can pull it out:
\[ \langle x_{j+1} | \, e^{-i\hat{T}\varepsilon/\hbar} \, e^{-i\hat{V}\varepsilon/\hbar} \, | x_j \rangle = e^{-iV(x_j)\varepsilon/\hbar} \, \langle x_{j+1} | \, e^{-i\hat{T}\varepsilon/\hbar} \, | x_j \rangle \tag{7}\]
4.2 The kinetic factor: inserting momentum states
Now we need to compute \(\langle x_{j+1} | e^{-i\hat{T}\varepsilon/\hbar} | x_j \rangle\) where \(\hat{T} = \hat{p}^2/2m\). The key insight is that \(\hat{T}\) is diagonal in the momentum basis. So we insert a complete set of momentum eigenstates:
\[ \hat{1} = \int_{-\infty}^{+\infty} \frac{dp_j}{2\pi\hbar} \, |p_j\rangle \langle p_j| \]
(The factor \(1/(2\pi\hbar)\) appears because we normalize as \(\langle x | p \rangle = e^{ipx/\hbar}\).)
Inserting this identity:
\[ \langle x_{j+1} | \, e^{-i\hat{T}\varepsilon/\hbar} \, | x_j \rangle = \int \frac{dp_j}{2\pi\hbar} \; \langle x_{j+1} | p_j \rangle \; e^{-ip_j^2\varepsilon/2m\hbar} \; \langle p_j | x_j \rangle \]
Using \(\langle x | p \rangle = e^{ipx/\hbar}\) and \(\langle p | x \rangle = e^{-ipx/\hbar}\):
\[ \langle x_{j+1} | \, e^{-i\hat{T}\varepsilon/\hbar} \, | x_j \rangle = \int \frac{dp_j}{2\pi\hbar} \; e^{ip_j x_{j+1}/\hbar} \; e^{-ip_j^2\varepsilon/2m\hbar} \; e^{-ip_j x_j/\hbar} \]
Combining the exponents:
\[ \boxed{\langle x_{j+1} | \, e^{-i\hat{T}\varepsilon/\hbar} \, | x_j \rangle = \int \frac{dp_j}{2\pi\hbar} \; \exp\!\left\{ \frac{i}{\hbar}\left[ p_j(x_{j+1} - x_j) - \frac{p_j^2}{2m}\varepsilon \right] \right\}} \tag{8}\]
4.3 The complete short-time propagator
Combining the potential factor (Equation 7) and kinetic factor (Equation 8):
\[ \langle x_{j+1} | \, e^{-i\hat{H}\varepsilon/\hbar} \, | x_j \rangle = \int \frac{dp_j}{2\pi\hbar} \; \exp\!\left\{ \frac{i}{\hbar}\left[ p_j(x_{j+1} - x_j) - \left(\frac{p_j^2}{2m} + V(x_j)\right)\varepsilon \right] \right\} \tag{9}\]
Let us define the discrete velocity as \(\dot{x}_j \equiv (x_{j+1} - x_j)/\varepsilon\). Then \(x_{j+1} - x_j = \dot{x}_j \, \varepsilon\), and the exponent can be written suggestively as:
\[ \frac{i}{\hbar}\left[ p_j \, \dot{x}_j - H(x_j, p_j) \right]\varepsilon \]
where \(H(x, p) = p^2/2m + V(x)\) is the classical Hamiltonian. This is not a coincidence — it is the beginning of the connection between the quantum propagator and classical mechanics.
The exponent in the short-time propagator is \((i/\hbar)\) times \([p\dot{x} - H]\varepsilon\). If you recall your classical mechanics, the quantity \(p\dot{x} - H\) is precisely the integrand in the canonical action \(\int (p\dot{x} - H) \, dt\). The quantum path integral is directly building the classical action, one time slice at a time!
5 The Phase-Space Path Integral
5.1 Assembling the full propagator
Now we substitute the short-time propagator (Equation 9) back into the sliced expression (Equation 3):
\[ K = \int dx_1 \cdots dx_{N-1} \prod_{j=0}^{N-1} \left[ \int \frac{dp_j}{2\pi\hbar} \; \exp\!\left\{ \frac{i}{\hbar}\left[ p_j(x_{j+1}-x_j) - H(x_j, p_j)\,\varepsilon \right] \right\} \right] \]
We can pull all the \(x\)-integrals and \(p\)-integrals to the front, and combine all the exponentials (since the product of exponentials is the exponential of the sum):
\[ K = \int \left(\prod_{j=1}^{N-1} dx_j \right) \left(\prod_{j=0}^{N-1} \frac{dp_j}{2\pi\hbar} \right) \exp\!\left\{ \frac{i}{\hbar} \sum_{j=0}^{N-1} \left[ p_j(x_{j+1}-x_j) - H(x_j, p_j)\,\varepsilon \right] \right\} \tag{10}\]
Notice the different index ranges:
- There are \(N-1\) position integrals (\(x_1\) through \(x_{N-1}\)), because \(x_0 = x_a\) and \(x_N = x_b\) are fixed.
- There are \(N\) momentum integrals (\(p_0\) through \(p_{N-1}\)), one for each time slice. Unlike positions, momenta are not fixed at the endpoints!
5.2 The continuum limit
In the limit \(N \to \infty\) (with \(\varepsilon = T/N \to 0\)), the sum in the exponent becomes an integral. With \(x_{j+1} - x_j = \dot{x}(t_j)\varepsilon\):
\[ \sum_{j=0}^{N-1} \left[ p_j \dot{x}_j - H(x_j, p_j) \right] \varepsilon \;\longrightarrow\; \int_0^T \left[ p(t)\,\dot{x}(t) - H\bigl(x(t), p(t)\bigr) \right] dt \]
The right-hand side is the canonical action functional:
\[ S_{\text{can}}[x, p] = \int_0^T \left[ p\,\dot{x} - H(x, p) \right] dt \]
And the products of integrals define the phase-space path integral measure:
\[ \int \mathcal{D}x \, \mathcal{D}p \equiv \lim_{N \to \infty} \int \left(\prod_{j=1}^{N-1} dx_j \right) \left(\prod_{j=0}^{N-1} \frac{dp_j}{2\pi\hbar} \right) \]
Putting it all together, we arrive at the phase-space path integral:
\[ \boxed{K(x_b, T; x_a, 0) = \int \mathcal{D}x \, \mathcal{D}p \; \exp\!\left\{ \frac{i}{\hbar} \int_0^T \left[ p\,\dot{x} - H(x, p) \right] dt \right\}} \tag{11}\]
with boundary conditions \(x(0) = x_a\), \(x(T) = x_b\), and \(p(t)\) unconstrained.
The formula Equation 11 works for any Hamiltonian \(H(x, p)\), not just the standard \(p^2/2m + V(x)\). This is the most general form of the path integral. The Lagrangian version (which we’ll derive next) requires the Hamiltonian to be quadratic in \(p\). Always remember: the phase-space form is the parent; the Lagrangian form is the child.
Is the phase-space path integral or the configuration-space (Lagrangian) path integral more fundamental? When does the Lagrangian version fail?
The phase-space version is more general and more fundamental — it works for any Hamiltonian \(H(x, p)\). The Lagrangian version requires performing the momentum integral, which is only a Gaussian when \(H\) is quadratic in \(p\) (i.e., \(H = p^2/2m + V(x)\) or similar).
For systems where \(H\) is not quadratic in \(p\) — such as relativistic particles where \(H = c\sqrt{p^2 + m^2c^2} + V(x)\), or systems on curved spaces where the kinetic term has \(x\)-dependent coefficients — the momentum integral cannot be done analytically, and one must either work in phase space or deal carefully with operator ordering ambiguities.
6 Integrating Out Momenta
For the standard non-relativistic Hamiltonian \(H(x, p) = p^2/2m + V(x)\), we can perform the momentum integrals explicitly. This is the step that converts the phase-space path integral into the more familiar Lagrangian form.
6.1 The Gaussian integral for one time slice
Let us focus on a single momentum integral from Equation 10. For the \(j\)-th time slice, we need:
\[ I_j = \int_{-\infty}^{+\infty} \frac{dp_j}{2\pi\hbar} \; \exp\!\left\{ \frac{i}{\hbar}\left[ p_j(x_{j+1}-x_j) - \frac{p_j^2}{2m}\varepsilon \right] \right\} \]
Let us define \(\Delta x_j \equiv x_{j+1} - x_j\) for brevity. The exponent is:
\[ \frac{i}{\hbar}\left[ p_j \, \Delta x_j - \frac{p_j^2\,\varepsilon}{2m} \right] = -\frac{i\varepsilon}{2m\hbar}\left[ p_j^2 - \frac{2m\,\Delta x_j}{\varepsilon} \, p_j \right] \]
6.2 Completing the square
We complete the square in \(p_j\):
\[ p_j^2 - \frac{2m\,\Delta x_j}{\varepsilon} \, p_j = \left(p_j - \frac{m\,\Delta x_j}{\varepsilon}\right)^2 - \frac{m^2(\Delta x_j)^2}{\varepsilon^2} \]
So the exponent becomes:
\[ -\frac{i\varepsilon}{2m\hbar}\left(p_j - \frac{m\,\Delta x_j}{\varepsilon}\right)^2 + \frac{i\varepsilon}{2m\hbar} \cdot \frac{m^2(\Delta x_j)^2}{\varepsilon^2} \]
The second term simplifies to:
\[ \frac{i}{\hbar} \cdot \frac{m(\Delta x_j)^2}{2\varepsilon} = \frac{i}{\hbar} \cdot \frac{m}{2}\left(\frac{\Delta x_j}{\varepsilon}\right)^2 \varepsilon \]
This is \((i/\hbar)\) times the discretized kinetic energy \(\frac{1}{2}m\dot{x}_j^2\) multiplied by \(\varepsilon\)!
6.3 Performing the Gaussian integral
The integral over \(p_j\) is now:
\[ I_j = \frac{1}{2\pi\hbar} \exp\!\left\{ \frac{im(\Delta x_j)^2}{2\hbar\varepsilon} \right\} \int_{-\infty}^{+\infty} dp_j \; \exp\!\left\{ -\frac{i\varepsilon}{2m\hbar}\left(p_j - \frac{m\,\Delta x_j}{\varepsilon}\right)^2 \right\} \]
Shifting the integration variable \(p_j \to p_j + m\,\Delta x_j/\varepsilon\) (translation invariance of the integral):
\[ I_j = \frac{1}{2\pi\hbar} \exp\!\left\{ \frac{im(\Delta x_j)^2}{2\hbar\varepsilon} \right\} \int_{-\infty}^{+\infty} dp_j \; \exp\!\left\{ -\frac{i\varepsilon}{2m\hbar} \, p_j^2 \right\} \]
This is a Fresnel-type Gaussian integral:
\[ \int_{-\infty}^{+\infty} dp \; e^{-\alpha p^2} = \sqrt{\frac{\pi}{\alpha}} \]
with \(\alpha = i\varepsilon/(2m\hbar)\). Using the analytic continuation of the Gaussian integral (see Module 7 for a careful treatment), we get:
\[ \int_{-\infty}^{+\infty} dp \; e^{-i\varepsilon p^2/2m\hbar} = \sqrt{\frac{2\pi m\hbar}{i\varepsilon}} = \sqrt{\frac{2\pi m\hbar \cdot (-i)}{\varepsilon}} = \sqrt{\frac{2\pi m \hbar}{i\varepsilon}} \]
Using \(1/i = -i = e^{-i\pi/2}\), we get \(\sqrt{2\pi m\hbar/(i\varepsilon)}\).
Putting it all together:
\[ I_j = \frac{1}{2\pi\hbar} \sqrt{\frac{2\pi m \hbar}{i\varepsilon}} \; \exp\!\left\{ \frac{im(\Delta x_j)^2}{2\hbar\varepsilon} \right\} \]
Simplifying:
\[ \boxed{I_j = \left(\frac{m}{2\pi i\hbar\varepsilon}\right)^{1/2} \exp\!\left\{ \frac{i}{\hbar} \frac{m}{2} \left(\frac{x_{j+1}-x_j}{\varepsilon}\right)^2 \varepsilon \right\}} \tag{12}\]
6.4 Combining with the potential
Including the potential factor \(e^{-iV(x_j)\varepsilon/\hbar}\) from Equation 7, each time slice contributes:
\[ \left(\frac{m}{2\pi i\hbar\varepsilon}\right)^{1/2} \exp\!\left\{ \frac{i}{\hbar}\left[ \frac{m}{2}\left(\frac{x_{j+1}-x_j}{\varepsilon}\right)^2 - V(x_j) \right]\varepsilon \right\} \]
The quantity in square brackets is precisely the discrete Lagrangian:
\[ L_j = \frac{m}{2}\left(\frac{x_{j+1}-x_j}{\varepsilon}\right)^2 - V(x_j) = \frac{1}{2}m\dot{x}_j^2 - V(x_j) \]
This is the Lagrangian \(L = T - V\) evaluated on the \(j\)-th time slice!
Where did the Lagrangian come from? We started with the Hamiltonian \(H = p^2/2m + V(x)\). When we integrated out the momentum \(p\), the Gaussian integral performed a Legendre transform automatically. The momentum integral found the stationary point \(p^* = m\dot{x}\) (the classical relation!), and substituting back gave \(p^*\dot{x} - H(x, p^*) = m\dot{x}^2 - (m\dot{x}^2/2 + V) = m\dot{x}^2/2 - V = L(x, \dot{x})\). The path integral naturally performs the Legendre transform from the Hamiltonian to the Lagrangian.
7 The Lagrangian Path Integral: Feynman’s Formula
7.1 Assembling the full result
Multiplying together all \(N\) time slices and integrating over all intermediate positions:
\[ K(x_b, T; x_a, 0) = \lim_{N \to \infty} \left(\frac{m}{2\pi i\hbar\varepsilon}\right)^{N/2} \int dx_1 \cdots dx_{N-1} \; \exp\!\left\{ \frac{i}{\hbar} \sum_{j=0}^{N-1} L_j\,\varepsilon \right\} \tag{13}\]
In the continuum limit, the sum becomes the action:
\[ \sum_{j=0}^{N-1} L_j \, \varepsilon \;\longrightarrow\; \int_0^T L\bigl(x(t), \dot{x}(t)\bigr) \, dt = S[x(t)] \]
The products of integrals with the normalization factors define the path integral measure:
\[ \int \mathcal{D}x \equiv \lim_{N \to \infty} \left(\frac{m}{2\pi i\hbar\varepsilon}\right)^{N/2} \int dx_1 \cdots dx_{N-1} \]
And we arrive at Feynman’s path integral formula:
\[ \boxed{K(x_b, T; x_a, 0) = \int \mathcal{D}x(t) \; \exp\!\left\{ \frac{i}{\hbar} \int_0^T L\bigl(x, \dot{x}\bigr) \, dt \right\} = \int \mathcal{D}x \; e^{iS[x]/\hbar}} \tag{14}\]
with boundary conditions \(x(0) = x_a\), \(x(T) = x_b\).
🎉 This is it. This is Feynman’s formula. Let us pause and appreciate what we have accomplished:
- We started from the exact operator expression \(K = \langle x_b | e^{-i\hat{H}T/\hbar} | x_a \rangle\).
- We sliced time into \(N\) pieces and inserted complete sets of position states.
- We used the Trotter formula to split \(e^{-i\hat{H}\varepsilon/\hbar}\) into kinetic and potential factors.
- We inserted momentum states to compute the kinetic matrix element.
- We performed the Gaussian momentum integral, which gave us the Lagrangian.
- We took \(N \to \infty\) and obtained the continuum path integral.
No postulates were needed. The path integral is a consequence of the standard operator formalism, not an alternative to it.
Why do we get the Lagrangian \(L = T - V\) in the final formula, and not the Hamiltonian \(H = T + V\)? After all, we started with the Hamiltonian!
Integrating out momenta performs a Legendre transform! Here’s the chain of logic:
- The phase-space path integral has the integrand \(\exp\{(i/\hbar)\int[p\dot{x} - H(x,p)]\,dt\}\).
- For \(H = p^2/2m + V(x)\), the \(p\)-integral is Gaussian with stationary point at \(p^* = m\dot{x}\).
- Substituting \(p^* = m\dot{x}\) into \(p\dot{x} - H\): \[ p^*\dot{x} - H(x, p^*) = m\dot{x}^2 - \left(\frac{m\dot{x}^2}{2} + V\right) = \frac{m\dot{x}^2}{2} - V = L(x, \dot{x}) \]
The Legendre transform \(L = p\dot{x} - H\) is built into the structure of the phase-space path integral. The momentum integral simply evaluates this transform by performing the stationary-phase approximation (which, for a Gaussian, is exact).
8 What “Sum Over All Paths” Really Means
8.1 The path integral measure
Let us be honest about the mathematical status of the expression \(\int \mathcal{D}x \; e^{iS[x]/\hbar}\).
The path integral measure \(\mathcal{D}x\) is defined only through the limiting procedure:
\[ \int \mathcal{D}x(\cdot) \equiv \lim_{N \to \infty} \left(\frac{m}{2\pi i\hbar\varepsilon}\right)^{N/2} \int_{-\infty}^{+\infty} dx_1 \int_{-\infty}^{+\infty} dx_2 \cdots \int_{-\infty}^{+\infty} dx_{N-1} \tag{15}\]
Several important points:
The normalization factor matters. The prefactor \((m/2\pi i\hbar\varepsilon)^{N/2}\) diverges as \(N \to \infty\) (since \(\varepsilon \to 0\)). This divergence is necessary — it cancels against the fact that the integrals \(\int dx_1 \cdots dx_{N-1}\) produce a vanishing result in the same limit. The ratio is finite and well-defined.
Infinitely many integrations. We are integrating over a function \(x(t)\), which is a point in an infinite-dimensional space. This is fundamentally different from ordinary integration.
The Minkowski path integral is formal. The oscillatory integrand \(e^{iS/\hbar}\) does not converge absolutely — every path contributes with the same modulus \(|e^{iS/\hbar}| = 1\). There is no rigorous measure on path space for the Minkowski signature path integral. It is defined only as the limit of finite-dimensional integrals.
The Euclidean version is rigorous. After Wick rotation \(t \to -i\tau\) (Module 9), the oscillatory \(e^{iS/\hbar}\) becomes a damping factor \(e^{-S_E/\hbar}\), and the path integral becomes a well-defined integral with respect to Wiener measure — the measure on Brownian motion paths. This is the mathematical foundation of the path integral, due to Kac.
The paths that dominate the Wiener measure are continuous but nowhere differentiable — they are fractal-like, with fractal dimension 2. This means \(\dot{x}(t)\) does not exist for a typical path! The “velocity” \(\dot{x}_j = (x_{j+1}-x_j)/\varepsilon\) diverges as \(\varepsilon \to 0\) for generic paths. The kinetic energy \(\frac{1}{2}m\dot{x}^2\) is infinite on a typical path. Nevertheless, the action (which involves the integral \(\int \frac{1}{2}m\dot{x}^2\,dt\)) is well-defined in a distributional sense because the divergent kinetic energy is compensated by the measure factor \((m/2\pi i\hbar\varepsilon)^{N/2}\). One should always think of the path integral as a limit of finite-dimensional integrals, never as a pre-existing infinite-dimensional integral.
8.2 Why this doesn’t bother physicists (much)
Despite the lack of mathematical rigor in the Minkowski case, the path integral:
- Reproduces the correct Schrödinger equation (we proved this in Module 3 for the free particle).
- Gives correct propagators for all solvable systems (free particle, harmonic oscillator, etc.).
- Provides the most efficient framework for quantum field theory, where operator methods become unwieldy.
- Has a rigorous Euclidean version, which suffices for most calculations after Wick rotation.
Physicists use the path integral as a computational tool, always keeping in mind that it is defined through the time-slicing limit. The formal continuum expression \(\int \mathcal{D}x \, e^{iS/\hbar}\) is a shorthand for this limit.
9 Worked Example 1: Full Derivation for \(H = p^2/2m + V(x)\)
Let us put all the steps together in one place, with no interruptions, for the standard non-relativistic particle.
Given: \(\hat{H} = \frac{\hat{p}^2}{2m} + V(\hat{x})\)
Goal: Derive \(K = \int \mathcal{D}x \; e^{iS[x]/\hbar}\) where \(S = \int L \, dt\) with \(L = \frac{1}{2}m\dot{x}^2 - V(x)\).
Step 1: Time slicing.
\[ K = \langle x_b | e^{-i\hat{H}T/\hbar} | x_a \rangle = \langle x_b | \left(e^{-i\hat{H}\varepsilon/\hbar}\right)^N | x_a \rangle, \quad \varepsilon = T/N \]
Step 2: Insert \(N-1\) complete sets of position states.
\[ K = \int dx_1 \cdots dx_{N-1} \prod_{j=0}^{N-1} \langle x_{j+1} | e^{-i\hat{H}\varepsilon/\hbar} | x_j \rangle \]
Step 3: Trotter split each factor.
\[ \langle x_{j+1} | e^{-i\hat{H}\varepsilon/\hbar} | x_j \rangle \approx \langle x_{j+1} | e^{-i\hat{T}\varepsilon/\hbar} e^{-i\hat{V}\varepsilon/\hbar} | x_j \rangle \]
Step 4: Evaluate the potential factor.
\[ e^{-i\hat{V}\varepsilon/\hbar} |x_j\rangle = e^{-iV(x_j)\varepsilon/\hbar} |x_j\rangle \]
Step 5: Insert momentum states for the kinetic factor.
\[ \langle x_{j+1} | e^{-i\hat{T}\varepsilon/\hbar} | x_j \rangle = \int \frac{dp_j}{2\pi\hbar} \; e^{ip_j(x_{j+1}-x_j)/\hbar} \; e^{-ip_j^2\varepsilon/2m\hbar} \]
Step 6: Combine into the phase-space path integral.
\[ K = \int \prod_{j=1}^{N-1} dx_j \prod_{j=0}^{N-1} \frac{dp_j}{2\pi\hbar} \; \exp\!\left\{ \frac{i}{\hbar}\sum_{j=0}^{N-1} \left[ p_j \frac{x_{j+1}-x_j}{\varepsilon} - \frac{p_j^2}{2m} - V(x_j) \right]\varepsilon \right\} \]
Step 7: Perform the Gaussian momentum integrals. For each \(j\):
\[ \int \frac{dp_j}{2\pi\hbar} \exp\!\left\{ \frac{i\varepsilon}{\hbar}\left[ p_j \dot{x}_j - \frac{p_j^2}{2m} \right] \right\} = \left(\frac{m}{2\pi i\hbar\varepsilon}\right)^{1/2} \exp\!\left\{ \frac{i\varepsilon}{\hbar} \frac{m\dot{x}_j^2}{2} \right\} \]
Step 8: Assemble the Lagrangian path integral.
\[ K = \lim_{N \to \infty} \left(\frac{m}{2\pi i\hbar\varepsilon}\right)^{N/2} \int dx_1 \cdots dx_{N-1} \; \exp\!\left\{ \frac{i}{\hbar}\sum_{j=0}^{N-1}\left[\frac{m\dot{x}_j^2}{2} - V(x_j)\right]\varepsilon \right\} \]
Step 9: Take the continuum limit.
\[ \boxed{K(x_b, T; x_a, 0) = \int \mathcal{D}x \; \exp\!\left\{ \frac{i}{\hbar}\int_0^T \left[\frac{1}{2}m\dot{x}^2 - V(x)\right] dt \right\} = \int \mathcal{D}x \; e^{iS[x]/\hbar}} \]
This is Feynman’s result. ∎
10 Worked Example 2: Velocity-Dependent Potentials and Ordering Ambiguities
10.1 The problem: charged particle in a magnetic field
Consider a charged particle in an electromagnetic field. The Hamiltonian is:
\[ \hat{H} = \frac{1}{2m}\left(\hat{p} - \frac{e}{c}\hat{A}(\hat{x})\right)^2 + e\Phi(\hat{x}) \]
where \(A(x)\) is the vector potential and \(\Phi(x)\) is the scalar potential. Expanding:
\[ \hat{H} = \frac{\hat{p}^2}{2m} - \frac{e}{2mc}\left(\hat{p}\cdot\hat{A}(\hat{x}) + \hat{A}(\hat{x})\cdot\hat{p}\right) + \frac{e^2}{2mc^2}\hat{A}(\hat{x})^2 + e\Phi(\hat{x}) \]
The cross term \(\hat{p}\cdot\hat{A}(\hat{x}) + \hat{A}(\hat{x})\cdot\hat{p}\) involves products of non-commuting operators \(\hat{p}\) and \(\hat{A}(\hat{x})\). The ordering of these operators matters because \([\hat{x}, \hat{p}] = i\hbar \neq 0\).
10.2 The ordering ambiguity
In the phase-space path integral, the Hamiltonian appears as a classical function \(H(x, p)\). But which classical function? Consider the term \(\hat{p}\,\hat{A}(\hat{x})\):
- If we put \(\hat{V}\) to the right in the Trotter splitting (\(e^{-i\hat{T}\varepsilon/\hbar} e^{-i\hat{V}\varepsilon/\hbar}\)), the potential is evaluated at \(x_j\).
- If we put \(\hat{V}\) to the left, it would be evaluated at \(x_{j+1}\).
- The symmetric Trotter splitting gives evaluation at the midpoint \((x_j + x_{j+1})/2\).
These choices differ by terms of order \(\varepsilon\), which might contribute finite corrections in the limit \(N \to \infty\). For the standard Hamiltonian \(H = p^2/2m + V(x)\), these corrections are absent because \(V\) depends only on \(x\). But for velocity-dependent potentials (where \(H\) involves mixed \(x\)-\(p\) terms), the ordering ambiguity becomes physical.
10.3 The midpoint prescription
The correct prescription, which reproduces the Schrödinger equation, is the Weyl-ordered or midpoint rule:
\[ \langle x_{j+1} | e^{-i\hat{H}\varepsilon/\hbar} | x_j \rangle \approx \int \frac{dp_j}{2\pi\hbar} \exp\!\left\{ \frac{i}{\hbar}\left[ p_j(x_{j+1}-x_j) - H\!\left(\frac{x_j+x_{j+1}}{2}, p_j\right)\varepsilon \right]\right\} \]
where the Hamiltonian is evaluated at the midpoint \(\bar{x}_j = (x_j + x_{j+1})/2\).
For the charged particle, this gives the correct Lagrangian:
\[ L = \frac{1}{2}m\dot{x}^2 + \frac{e}{c}\dot{x} \cdot A(x) - e\Phi(x) \]
The cross term \(\dot{x} \cdot A\) emerges correctly only with the midpoint prescription. Other prescriptions (endpoint evaluations) give incorrect results that differ by terms proportional to \(\hbar\).
The path integral is not independent of the operator formalism — it is derived from it. Whenever operator ordering matters (which it does for velocity-dependent potentials, curved spaces, etc.), the correct path integral must be inferred from the operator ordering of the quantum Hamiltonian. The path integral does not resolve ordering ambiguities; it inherits them. This is a subtle but important point that many textbooks gloss over.
11 Comparison of Formulations
Let us summarize the three equivalent ways of expressing the propagator:
| Feature | Operator Formalism | Phase-Space Path Integral | Configuration-Space (Lagrangian) Path Integral |
|---|---|---|---|
| Formula | \(K = \langle x_b \| e^{-i\hat{H}T/\hbar} \| x_a \rangle\) | \(K = \int \mathcal{D}x\,\mathcal{D}p \; e^{(i/\hbar)\int[p\dot{x}-H]\,dt}\) | \(K = \int \mathcal{D}x \; e^{(i/\hbar)\int L\,dt}\) |
| Variables | Operators \(\hat{x}, \hat{p}\) | Classical functions \(x(t), p(t)\) | Classical function \(x(t)\) only |
| Integrand | Hamiltonian operator \(\hat{H}\) | Classical Hamiltonian \(H(x,p)\) | Classical Lagrangian \(L(x,\dot{x})\) |
| Boundary conditions | \(x_a, x_b\) fixed | \(x_a, x_b\) fixed; \(p\) free | \(x_a, x_b\) fixed |
| Valid for | Any \(\hat{H}\) | Any \(H(x,p)\) | \(H\) quadratic in \(p\) only |
| Operator ordering | Explicit | Encoded in discretization | Encoded in discretization |
| Mathematical rigor | Fully rigorous | Defined via time-slicing limit | Defined via time-slicing limit |
| Key advantage | Exact, non-perturbative | Most general path integral form | Elegant; connects to classical mechanics |
Textbook reference: Kleinert, Path Integrals in QM, Statistics, Polymer Physics, and Financial Markets, Ch. 2 for a thorough comparison.
12 Summary of Key Results
| Step | Formula | Equation |
|---|---|---|
| Starting point | \(K = \langle x_b \| e^{-i\hat{H}T/\hbar} \| x_a \rangle\) | Equation 1 |
| Time slicing | \(e^{-i\hat{H}T/\hbar} = (e^{-i\hat{H}\varepsilon/\hbar})^N\) | Equation 2 |
| Trotter formula | \(e^{-i(\hat{T}+\hat{V})\varepsilon/\hbar} = e^{-i\hat{T}\varepsilon/\hbar}e^{-i\hat{V}\varepsilon/\hbar} + \mathcal{O}(\varepsilon^2)\) | Equation 5 |
| Single-slice kinetic element | \(\int \frac{dp}{2\pi\hbar} e^{ip\Delta x/\hbar - ip^2\varepsilon/2m\hbar}\) | Equation 8 |
| Momentum integral result | \((m/2\pi i\hbar\varepsilon)^{1/2} \, e^{im(\Delta x)^2/2\hbar\varepsilon}\) | Equation 12 |
| Phase-space path integral | \(K = \int \mathcal{D}x\,\mathcal{D}p \; e^{(i/\hbar)\int[p\dot{x}-H]\,dt}\) | Equation 11 |
| Feynman’s formula | \(K = \int \mathcal{D}x \; e^{iS[x]/\hbar}\) | Equation 14 |
| Path integral measure | \(\int \mathcal{D}x = \lim_{N\to\infty} (m/2\pi i\hbar\varepsilon)^{N/2}\int dx_1\cdots dx_{N-1}\) | Equation 15 |
Before moving on, make sure you can:
- Start from \(K = \langle x_b | e^{-i\hat{H}T/\hbar} | x_a \rangle\) and explain each step leading to the path integral.
- State the Trotter product formula and explain why the error vanishes as \(N \to \infty\).
- Derive the short-time propagator Equation 9 by inserting momentum states.
- Perform the Gaussian momentum integral to go from the phase-space to the Lagrangian path integral.
- Explain why the Lagrangian (not the Hamiltonian) appears in Feynman’s formula.
- State when the configuration-space path integral fails and the phase-space version is needed.
- Explain the midpoint prescription and why operator ordering matters.
13 Practice Problems
Problem 1 (Warm-up): Verify the BCH formula to second order. Expand \(e^{\hat{A}}e^{\hat{B}}\) and \(e^{\hat{A}+\hat{B}+\frac{1}{2}[\hat{A},\hat{B}]}\) each to second order in the operators, and show they agree to \(\mathcal{O}(\hat{A}^2, \hat{A}\hat{B}, \hat{B}^2)\). (Hint: \(e^{\hat{A}} = 1 + \hat{A} + \hat{A}^2/2 + \cdots\))
Problem 2 (Trotter error): Consider the symmetric Trotter splitting \(e^{-i\hat{V}\varepsilon/2\hbar}\,e^{-i\hat{T}\varepsilon/\hbar}\,e^{-i\hat{V}\varepsilon/2\hbar}\). Use the BCH formula to show that the error is \(\mathcal{O}(\varepsilon^3)\) instead of \(\mathcal{O}(\varepsilon^2)\). Specifically, show that the \(\mathcal{O}(\varepsilon^2)\) commutator terms cancel.
Problem 3 (Gaussian integral practice): Compute the integral
\[ I = \int_{-\infty}^{+\infty} \frac{dp}{2\pi\hbar} \exp\!\left\{ \frac{i}{\hbar}\left[ p \cdot \Delta x - \frac{p^2}{2m}\varepsilon \right] \right\} \]
by completing the square, and verify that you obtain \((m/2\pi i\hbar\varepsilon)^{1/2}\exp\{im(\Delta x)^2/2\hbar\varepsilon\}\).
Problem 4 (Free particle check): Starting from the Lagrangian path integral Equation 13 with \(V = 0\), verify that for \(N = 2\) (one intermediate integration), you recover the free-particle propagator \(K = (m/2\pi i\hbar T)^{1/2}\exp\{im(x_b - x_a)^2/2\hbar T\}\). (Hint: You need to do one Gaussian integral over \(x_1\), using the result from Module 3.)
Problem 5 (Phase-space for the harmonic oscillator): Write down the discrete phase-space path integral for the harmonic oscillator, \(H = p^2/2m + \frac{1}{2}m\omega^2 x^2\). Perform the momentum integrals to obtain the Lagrangian form. What is the discrete Lagrangian \(L_j\) that appears?
Problem 6 (Ordering ambiguity — challenging): Consider a one-dimensional system with the Hamiltonian \(\hat{H} = f(\hat{x})\hat{p}^2\), where \(f(x)\) is some function of position (this arises, for example, for motion on a curved space). Show that the Trotter splitting \(e^{-i\hat{H}\varepsilon/\hbar} \approx e^{-if(\hat{x})\hat{p}^2\varepsilon/\hbar}\) is ambiguous: the matrix element \(\langle x_{j+1}| e^{-if(\hat{x})\hat{p}^2\varepsilon/\hbar} |x_j\rangle\) depends on whether \(f(\hat{x})\) acts to the left or the right. Write down the matrix element using the midpoint prescription, \(f \to f(\bar{x}_j)\) where \(\bar{x}_j = (x_j + x_{j+1})/2\), and show that the resulting Lagrangian includes an extra quantum correction term proportional to \(\hbar^2\).