In Hall's book Quantum Theory for Mathematicians he gives a wonderful derivation of the Feynman path integral. His formal derivation is as follows.

Let $\psi \in L^2(\mathbb{R}^n)$ and $\hat{H}$ the Hamiltonian operator defined as $$-\frac{1}{\hbar} \hat{H} = \frac{\hbar}{2m} \Delta - \frac{1}{\hbar} V(X).$$

For simplicity I will be considering the free Schrodinger equation so $V(X) = 0$. Let $e^{it\hbar \Delta/(2mN)}$ be the solution operator/one-parameter subgroup generated by the operator $\hbar \Delta/(2mN)$. Thus the evolution of our state to a time $t$ in the future is given by $e^{it\hbar \Delta/(2mN)}\psi$.

The integral kernel associated to this solution operator is the fundamental solution to the free Schrodinger equation and thus the time evolution can also be represented as a convolution with this kernel. More explicitly: $$ e^{it\hbar \Delta/(2mN)} \psi(x_0) = \Big(\frac{mN}{it\hbar}\Big)^{n/2}\int_{\mathbb{R}^n} \exp{\Big(i\frac{mN}{2t\hbar} |x_1 - x_0|^2\Big)} \psi(x_1) dx_1.$$

Applying this operator $N$ times we get: $$ \Big(e^{it\hbar \Delta/(2mN)}\Big)^N \psi(x_0) = C \int_{\mathbb{R}^n} \exp{\Big(i\frac{mN}{2t\hbar} |x_1 - x_0|^2\Big)} \times \int_{\mathbb{R}^n} \exp{\Big(i\frac{mN}{2t\hbar} |x_2 - x_1|^2\Big)} \times \cdots \times \int_{\mathbb{R}^n} \exp{\Big(i\frac{mN}{2t\hbar} |x_N - x_{N-1}|^2\Big)} \times \psi(x_n) dx_N dx_{N-1} \cdots dx_1.$$

Upon rearranging the order of integration and taking a formal limit we obtain $$(e^{-it\hat{H}/\hbar} \psi)(x_0) = C \int_{\textrm{ paths with} \\ x(0) = x_0} \exp{\Big(\frac{i}{\hbar}\int_0^t \Big[\frac{m}{2} \Big|\frac{dx}{ds}\Big|^2\Big] ds \Big)} \psi(x(t)) \mathcal{D}x \tag{1}$$ where $\mathcal{D}x$ is formally an infinite dimensional Lebesgue measure.

Thus we have obtained an equivalent way to compute the time evolution of our state $\psi$, which is by computing a functional/Feynman path integral.

What is often of more interest than just the time evolution is the "transition amplitude", which is loosely defined as the probability for a particle that is at some position $A$ at time $t_0$ to be at position $B$ at time $t_1$. The difference is the measure in (1) can now be interpreted as a conditional probability measure where we also fix the endpoints of each path (i.e. we require $x(t_0) = A$ and $x(t_1) = B$.

I could not find a mathematically motivated derivation of this "transition amplitude" version of the path integral. The derivation does appear in many physics textbooks but as my background is in mathematics and not in physics I find many of these hard to follow.

Does anyone know where I may find a derivation of the "transition amplitude" version of the Feynman path integral similar to what Hall has done above or how I may modify his derivation to obtain it?

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    $\begingroup$ The trouble is that the limiting measure does not exist. I wonder if Hall mentions this.... $\endgroup$ Commented Apr 8 at 3:42
  • $\begingroup$ @MoisheKohan Sorry I should have been more clear in my original post. Hall does mention that the limiting measure does not exist and I was also aware of this fact. What I was looking for was a formal derivation for the Feynman path integral that calculates the probability for a particle to go from $A$ to $B$ in time $t$, i.e. a "transition amplitude". I was hoping someone could provide a derivation (even if it is not rigorous) that a mathematician could follow similar to what Hall has presented. $\endgroup$
    – CBBAM
    Commented Apr 8 at 5:22
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    $\begingroup$ Nakahara (I am sure you know that book) develops the Feynman path integral from transition amplitudes within the first 28 pages. Despite the division by $t$ that I don't understand it looks to me that Hall is doing the same, perhaps just not mentioning the word transition amplitude. $\endgroup$
    – Kurt G.
    Commented Apr 8 at 8:03
  • $\begingroup$ @KurtG. Thank you, this seems to be precisely what I was looking for. $\endgroup$
    – CBBAM
    Commented Apr 8 at 8:13
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    $\begingroup$ You are welcome. In this post you can find further mathematical references. It also contains the discussion that Mark Kac undoubtedly made the Euclidean path integral rigorous by introducing Wiener measure but Feynman objected to that for physical reasons. $\endgroup$
    – Kurt G.
    Commented Apr 8 at 8:23

1 Answer 1


Yes, Feynmans idea is really wonderful, but based on the illusory fact, that waves along "all" paths are interfering constructively along the classical path of stationary action; the rest of the "longer" paths does not carry a finite measure.

The rigorous part is called Euclidean Qantum Field theory, based on the mathematical sound case of imaginary time.

Then we deal with the Wiener processs as a limiting case sets of wild continuous paths, that represent points in a paths space with computable measures.

But at least in the wave case, by the Wick rotation of euclidean results, its possible to track analytically certain named terms in the Hamiltonian, as they show up in the solutions forms of elliptic PDE's.

But as everybody knows, wave and potential problems - despite the similarity of their characteristic quadratic operators, a form of presenting the metrics - in their complexified common Poincaré world of $\mathbb R^4$ have nearly nothing in common, because in general, they pose completly different problems of domains and their boundaries in atlasses of maps over manifolds.

While the euclidean case is smoothing - computational errors die out by the diffusion process, such that for $t \to \infty $ discetized approximations converge to solutions - in the discrete approximate solution to a wave equation, any error propagates time forward as a newly created partial wave.

  • $\begingroup$ Thank you for your answer. Perhaps I should have been more specific in my question, but what I was looking for was a derivation of the Feynman path integral used to calculate the probability of a particle propagating from a starting point $A$ to a final point $B$ in $t$ time. More specifically I was interested in how this affects Hall's derivation so that the Weiner paths now have a fixed endpoint. Even if such a derivation cannot be made rigorous, I was hoping for one that is written for mathematicians like Hall's derivation above. $\endgroup$
    – CBBAM
    Commented Apr 8 at 5:26
  • $\begingroup$ The Feynman-Kac idea is based on the Brownian bridge between two fixed points, extending Fermats principle to a greater class of paths with known measures but no definite path lengths, that could be tied to phase differences. But that is not so important, because quantum field theory is not based on the idea of following particles along its path. $\endgroup$
    – Roland F
    Commented Apr 17 at 14:39

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