The path integral for some potential can be evaluated explicitly by discretion the space and performing $N$ Gaussian integrals then taking the limit as $N \to \infty$ for the case of a free particle. However, there is a simpler method which takes advantage of the completeness condition for a transition amplitude.

After splitting the action into the classical path plus a variation, $x = x_{cl} + \eta$ on it we have:

$$ \langle x_b, t_b|x_a, t_a\rangle = \int^{x_b}_{x_a}Dx e^{\frac{i}{h}S[x]} = e^{\frac{i}{h}S[x_{cl}]} \int^0_0 D\eta e^{\frac{i}{h} S[\eta]}$$

Now we can we must evaluate the prefactor, $F(T)$, given by the $\eta$ integral which does not depend on the endpoints.

$$F(T) = \int^0_0 D\eta e^{\frac{i}{h} S[\eta]} = \langle 0, t_b|0, t_a\rangle = \langle 0, T|0, 0\rangle $$ Inserting a complete set of states yields: $$ \langle 0, T|0, 0\rangle = \int^\infty_{-\infty} dx \langle 0, T|x, t\rangle\langle x, t|0, 0\rangle $$

I have a few questions about the use of this identity to evaluate the normalisation for the free particle and harmonic oscillator.

1. Free Particle

The free particle action is $$S[x] = \frac{m}{2}\frac{(x_b-x_a)^2}{t_b-t_a},$$ plugging this in gives:

$$ F(T) = \int_{-\infty}^\infty dx~ F(T)\exp(\frac{imx^2}{2h(T-t)} )F(t-0)\exp(\frac{imx^2}{2ht}) = \sqrt{\frac{2\pi ih(T-t)t}{mT}} F(T-t)F(t)\tag{2.27}$$ I am okay with the calculation up to this point. Now in my course notes it states that taking the limit for large times ($T \gg t$) gives: $$F(t) = \sqrt{\frac{m}{2\pi iht}} $$ which is the correct free particle normalisation. Firstly, why do we evaluate it in this limit? The last step seems true but only through "hand-wavy" arguments can this process be made more clear or rigorous somehow?

2. Harmonic Oscillator

The classical action for the harmonic oscillator is $$S[x_{cl}] = \frac{m\omega}{2\sin\omega T}[(x_a^2 + x_b^2)\cos\omega T -2x_ax_b]$$ Following a similar process as for the free particle gives: $$F(T) = \int_{-\infty}^\infty dx~F(T-t)\exp(\frac{im\omega x^2 \cos\omega (T-t)}{2h\sin\omega (T-t)}) F(t)\exp(\frac{im\omega x^2 \cos\omega t}{2h\sin\omega t}). $$ Performing the Gaussian integral yields: $$ F(T) = F(T-t)F(t) \sqrt{\frac{2\pi i h \sin\omega (T-t)\sin\omega t}{m\omega \sin\omega T}} \tag{2.35}$$ And finally taking in the limit $T >> t$ we have: $$F(t) = \sqrt{\frac{m\omega}{2\pi ih\sin\omega t}}$$ Again, for this case I cannot see why this final step is justified.

3. Explicit Calculation

I originally suspected that the previous calculations are only justifiable because they are in agreement with the explicit calculation by discretisation of the path (for a free particle): $$ \langle x_b, t_b | x_a, t_a \rangle = \lim_{N\to\infty} A_N (\prod^N_{n=1} \int^\infty_{-\infty} dx_n) \exp(\frac{i\epsilon m}{2h}\sum^{N+1}_{n=1}(\frac{x_n-x_{n-1}}{\epsilon}^2))$$ but in that case the normalisation factor for each infinitesimal step is assumed to take the following form: $$A_N = (\nu(\epsilon))^{N+1} , \nu(\epsilon) = \sqrt{\frac{m}{2\pi ih\epsilon}} $$

for which I can find no justification, other than agreement with the previous calculation.

Any attempt at an explanation will be greatly appreciated.


  1. Lecture Notes, Quantum Theory, The University of Edinburgh, School of Physics and Astronomy, September 2015. The course notes are available here: https://www2.ph.ed.ac.uk/~bjp/qt/qt.pdf ~p.13-15 for the relevant section.
  • $\begingroup$ Which course notes? Are they online? $\endgroup$
    – Qmechanic
    Apr 19, 2018 at 17:50
  • $\begingroup$ I´ve always assumed that only $\frac{F(T)}{F(t)}$ is physically meaningful anyways, since you only can apply the propagation for finite small time intervals. $\endgroup$
    – lurscher
    Apr 19, 2018 at 19:12
  • $\begingroup$ The course notes are available here www2.ph.ed.ac.uk/~bjp/qt/qt.pdf ~p13 for the relevant section. $\endgroup$
    – cyfirx
    Apr 19, 2018 at 22:19

1 Answer 1


OP seems to have a point that the argument presented in the lecture notes is not watertight. Let us argue as follows:

  1. Divide the $F$-functions with their sought-for formulas, and call the quotient $f$. Then eqs. (2.27) & (2.35) become on the form $$f(T)~=~f(T-t)f(t), \tag{A}$$ or equivalently, $$f(t+t^{\prime})~=~f(t)f(t^{\prime}). \tag{B}$$

  2. Let us additionally assume that $f$ is continuous, and not identically zero $f\not\equiv 0$. Then eq. (B) implies that $$ f(0)~=~1. \tag{C}$$

  3. Ignoring some mathematical technicalities, the functional eq. (B) implies that $f$ is an exponential function, i.e. there exists a constant $c$, so that $$ f(t)~=~e^{ct},\tag{D} $$ see e.g. my Phys.SE answer here.

  4. For small $t\lesssim\tau $ much smaller than some characteristic timescale$^1$ $\tau$, we can evaluate the Hamiltonian path integral directly (with no Feynman fudge factors!). The result is $$ F(t)~\simeq~ \sqrt{\frac{m}{2\pi ih t}} \quad\text{for} \quad t~\lesssim~ \tau, \tag{E}$$ see e.g. Section V of my Phys.SE answer here. Equivalently, $$ f(t)~\simeq~ 1 \quad\text{for} \quad t~\lesssim~ \tau, \tag{F}$$

  5. Comparing eqs. (D) & (F), we deduce that $$c ~\ll ~1/\tau .\tag{G}$$ Of course, the correct path integral normalization of the free particle and the harmonic oscillator can be directly calculated, and it is known that $c=0$ so that $$ f(t)~=~1. \tag{H}$$ And hence that the $F$-functions are given by their sought-for formulas.


$^1$ E.g. for the harmonic oscillator $\tau=\omega^{-1}$.


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