Why Path Integrals Are Subtle

A path-integral symbol can represent an operator limit, a probability expectation, a distributional kernel, or an asymptotic expansion. These are different mathematical objects, with different convergence requirements. The central issue is identifying which object an expression constructs and which operations its construction permits. The standard physical derivation from propagators belongs to Propagators to Path Integrals.

Helpful background. Tracking Topologies distinguishes strong, weak, and norm convergence. Continuous-Basis Notation explains why coordinate kernels need not be ordinary functions.

A continuum symbol contains several separate claims

For a self-adjoint Hamiltonian $H$, the spectral theorem already constructs $U(t)=e^{-itH/\hbar}$. A formal expression

$$\langle x|U(t)|y\rangle =\int_{q(0)=y}^{q(t)=x}e^{iS[q]/\hbar}\,\mathcal Dq$$

then proposes a representation of an existing operator. It does not automatically define a measure $\mathcal Dq$, a pointwise kernel, or a convergent integral over classical trajectories.

A time-sliced construction begins instead with a finite product $U_N(t)$ of known operators. Inserting coordinate resolutions may express this product by finitely many integrals. Three questions remain distinct: do those integrals have a specified meaning, do the products converge, and does their limit equal the intended $U(t)$?

Self-adjointness of the target Hamiltonian answers none of these representation questions by itself. Conversely, the absence of an ordinary path probability measure does not prevent a valid operator limit.

Strong convergence controls packets, not arbitrary kernel manipulations

The usual unbounded-operator Trotter Product Formula gives strong convergence under its hypotheses:

$$\|(U_N(t)-U(t))\psi\|\longrightarrow0 \quad\text{for every }\psi\in\mathcal H.$$

It follows that transition amplitudes between fixed Hilbert vectors converge, because

$$|\langle\phi,(U_N-U)\psi\rangle| \leq\|\phi\|\,\|(U_N-U)\psi\|.$$

Position kets are not Hilbert vectors, so this estimate does not prove convergence of $\langle x|U_N|y\rangle$ at each pair of points. Kernel convergence may require a distributional formulation, stronger estimates, or restrictions away from singular times. Those are additional results.

Nor does strong convergence permit exchanging a limit with an infinite trace. On $\ell^2$, let $P_n$ project onto the $n$th basis vector. For every $\psi$, $\|P_n\psi\|=|\psi_n|\to0$, but $\operatorname{Tr}P_n=1$ for every $n$. A path-integral expression for a partition function therefore needs trace-class control beyond convergence on individual packets.

Euclidean probability changes the construction

For a suitable real potential and a semibounded Schrödinger Hamiltonian, the Feynman–Kac Formula realizes the heat semigroup through Brownian expectation. The kinetic term is encoded in a Gaussian process. It is not evaluated as the classical kinetic action of a typical differentiable trajectory.

This distinction matters because Brownian paths almost surely lack the finite kinetic regularity of classical paths. Wiener Measure versus Feynman Measure gives the quadratic-variation calculation. A positive potential weight is then an ordinary measurable random variable when the potential hypotheses make its time integral meaningful.

Returning to real time is a further operation. Spectral analytic continuation can give a strong boundary value of the heat semigroup, but it does not convert Wiener measure into a complex probability law by replacing time with an imaginary number.

Exactness and asymptotic accuracy are different conclusions

For quadratic actions, Gaussian integration can recover the exact propagator with an appropriate phase and boundary prescription. For a general smooth action, stationary phase usually gives an asymptotic expansion in a small parameter. Its finite-dimensional remainder depends on derivatives, Hessian bounds, support, and the integration dimension.

A proof for each fixed time-slice number does not supply estimates uniform as that number grows. Interchanging the continuum limit and a semiclassical expansion is therefore a separate problem. Stationary Phase and Semiclassical Rigor isolates this issue without repeating the standard saddle formula.

This is a useful way to read a successful formal calculation: determine which conclusion has been established. An exact answer for a solvable case, an asymptotic coefficient, and a convergent construction answer different mathematical questions.

Exercise: weak limits can lose norm

On $L^2(0,2\pi)$, let $(U_n\psi)(x)=e^{inx}\psi(x)$. Show that $U_n$ converges weakly to zero but cannot converge strongly to zero. Explain the relevance to oscillatory approximations.

Solution

Each $U_n$ is unitary. For fixed $\phi,\psi\in L^2$, the product $\overline\phi\psi$ lies in $L^1$, and the Riemann–Lebesgue lemma gives $\langle\phi,U_n\psi\rangle\to0$. Thus the weak operator limit is zero. Nevertheless $\|U_n\psi\|=\|\psi\|$, so strong convergence to zero fails for every nonzero $\psi$. Oscillatory cancellation of fixed matrix elements does not establish convergence of the evolved state in norm, much less preservation of unitary dynamics by the limit.

References

• Reed, Michael, and Barry Simon, Methods of Modern Mathematical Physics I: Functional Analysis, revised and enlarged ed., Academic Press, 1980, Chapter VIII — operator convergence and unitary product formulas.
• Simon, Barry, Functional Integration and Quantum Physics, 2nd ed., AMS Chelsea Publishing, 2005, doi:10.1090/chel/351 — probability constructions and operator meanings of functional integrals.