When one defines the path integral propagator, there is the need to normalize the propagator (since it would give you a probability density). There are two formulas which are used.

1) Original (v1+v2): The first formula (which I can intuitively agree with) says that:

$$\tag{1} \int_{Dx_b}dx_b\left|K(x_bt_b|x_at_a)\right|^2=1$$

for all values of $x_a$ on fixed values of $t_a, t_b$ and where $Dx_b$ means the domain of $x_b$.

1') Update (v3+v4): I changed my mind (to get more into agreement with the Born-rules). The first formula (which I can intuitively agree with) says that:

$$\tag{1'} \left|\int_{Dx_b}dx_bK(x_bt_b|x_at_a)\right|^2=1$$

for all values of $x_a$ on fixed values of $t_a, t_b$ and where $Dx_b$ means the domain of $x_b$.

2) The second formula (which is actually also very intuïtive) says that:

$$\tag{2}\lim\limits_{t_b\rightarrow t_a}K(x_bt_b|x_a,t_a) = \delta(x_b-x_a).$$

Now these are usually treated as equivalent, but I can't directly see how this can be the case. Isn't the second formula less restrictive ?


2 Answers 2


I) Conceptually, OP's original eq. (1)

$$\int_{\mathbb{R}}\! \mathrm{d}x_f~ \left| K(x_f,t_f;x_i,t_i) \right|^2 ~\stackrel{?}{=}~1 \qquad(\leftarrow\text{Wrong!})\tag{1} $$

clashes (as OP independently realized) with the fundamental principle of the Feynman path integral that the amplitude

$$K( x_f ,t_f ; x_i ,t_i )~=~\sum_{\rm hist.}\ldots$$

is a sum of histories, while the probability

$$P( x_f ,t_f ; x_i ,t_i )~=~|K( x_f ,t_f ; x_i ,t_i )|^2~\neq~\sum_{\rm hist.}\ldots $$

is not a sum of histories.

Concretely, the failure of eq. (1) may also be seen as follows. If we assume that$^1$

$$ K( x_i ,t_i ; x_f ,t_f ) ~=~ \overline{K( x_f ,t_f ; x_i ,t_i ) }, \tag{A}$$

and the (semi)group property of Feynman propagators/kernels

$$ K(x_f,t_f;x_i,t_i) ~=~ \int_{\mathbb{R}}\!\mathrm{d}x_m ~ K(x_f,t_f;x_m,t_m) K(x_m,t_m;x_i,t_i),\tag{B}$$

then the lhs. of OP's original first eq. (1) with $(x_i,t_i)=(x_f,t_f)$ is not equal to $1$, but instead becomes infinite

$$\begin{align} K(x_f,t_f;x_i,t_i) ~=~&\delta(x_f-x_i)~=~\delta(0)~=~\infty, \cr x_i~=~&x_f,\qquad t_i~=~t_f, \end{align}\tag{C}$$

because of OP's second formula (2).

II) The infinite normalization result (C) can be intuitively understood as follows. Recall that the paths in the path integral satisfy Dirichlet boundary condition $x(t_i)=x_i$ and $x(t_f)=x_f$. In other words, the particle is localized in $x$-position space at initial and final times. On the other hand, a particle localized in $x$-position space corresponds to a delta function wave function $\Psi(x)=\delta(x-x_0)$, which is not normalizable, cf. e.g this and this Phys.SE posts.

III) Conceptually, OP's first eq. (1')

$$\left| \int_{\mathbb{R}}\! \mathrm{d}x_f~K(x_f,t_f;x_i,t_i) \right| ~\stackrel{?}{=}~1 \quad(\leftarrow\text{Turns out to be ultimately wrong!}) \tag{1'} $$

is the statement that a particle that is initially localized at a spacetime event $(x_i,t_i)$ must with probability 100% be within $x$-space $\mathbb{R}$ at a final time $t_f$, as our QM model does not allow creation or annihilation of particles. However, such notion of absolute probabilities of the Feynman kernel $K(x_f,t_f;x_i,t_i)$ cannot be maintained when a concept has to be converted into mathematical formulas, as discussed in detail in this Phys.SE post. In general, OP's first eq. (1') only holds for short times $\Delta t \ll \tau$, where $\tau$ is some characteristic time scale of the system.

IV) Example. Finally, let us consider the example of a non-relativistic free particle in 1D. The Feynman propagator then reads

$$\begin{align} K( x_f ,t_f ; x_i ,t_i ) ~=~& \sqrt{\frac{A}{\pi}} e^{-A(\Delta x)^2}\cr ~=~& \sqrt{\frac{m}{2\pi i\hbar} \frac{1}{\Delta t}} \exp\left[ \frac{im}{2\hbar}\frac{(\Delta x)^2}{\Delta t}\right],\cr A~:=~&\frac{m}{2 i\hbar} \frac{1}{\Delta t} , \cr \Delta x~:=~&x_f-x_i, \cr \Delta t~:=~&t_f-t_i ~\neq ~0. \end{align}\tag{D}$$

[It is an instructive exercise to show that formula (D) satisfies eqs. (A-C) and OP's second formula (2).] The Gaussian integral over $x_f$ is one

$$ \int_{\mathbb{R}}\!\mathrm{d}x_f ~ K(x_f,t_f;x_i,t_i)~=~1,\tag{E} $$

which shows that OP's first eq. (1') actually holds for a free particle. The integrand

$$ \begin{align} |K(x_f,t_f;x_i,t_i)|^2~=~& \frac{|A|}{\pi}~=~ \frac{m}{2\pi \hbar}\frac{1}{|\Delta t|}, \cr \Delta t ~\neq ~&0,\end{align}\tag{F}$$

on the lhs. of OP's original first eq. (1) is independent of the midpoint $x_m$. Hence the integral over $x_m$ (i.e. lhs. of OP's first eq. (1)) becomes infinite

$$\begin{align} \int_{\mathbb{R}}\!\mathrm{d}x_f ~ |K(x_f,t_f;x_i,t_i)|^2~=~& \frac{m}{2\pi \hbar}\frac{1}{|\Delta t|} \int_{\mathbb{R}}\!\mathrm{d}x_f ~=~\infty, \cr \Delta t ~\neq ~&0,\end{align}\tag{G} $$

in agreement with what we found in eq. (C) in section I.


  1. R.P. Feynman and A.R. Hibbs, Quantum Mechanics and Path Integrals, 1965.


$^1$ Note that Ref. 1 defines $K(x_f,t_f;x_i,t_i)=0$ if $t_i>t_f$, see Ref. 1 between eq. (4-27) and eq. (4-28). Here we assume property (A) instead.

  • $\begingroup$ Yes of course ! Excuse me, I dare to generalise to quick, indeed according to the second equation this should hold ! So i should also demand that as $t_a\rightarrow t_b$ that $x_a \neq x_b$ because my delta-function would indeed blow up ! Or is this reasoning not correct ? $\endgroup$
    – Nick
    Commented Oct 2, 2013 at 11:55
  • $\begingroup$ I adjusted my equations, I dont know wether they seem correct now. But I believe the first equation demands that the propagator is normalized for fixed values of time that are not equal and that the second equation demands that for fixed values of space the limit holds. $\endgroup$
    – Nick
    Commented Oct 3, 2013 at 17:03
  • $\begingroup$ @Qmechanic, again thanks for the great update. The exact formulas can of course also be derived in the books of Kleinert and Schulman. Their derivation was less of my concern actually (altough very educational of course). For me it was more about the question wether both are needed. Of course the analog with quantum mechanics shows this in a beautiful way! I would have maybe hoped for arguments that don't use anything of quantum mechanics, but this will do! :) $\endgroup$
    – Nick
    Commented Oct 6, 2013 at 22:04
  • $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic
    Commented Oct 15, 2013 at 17:58
  • $\begingroup$ @Cosmas Zachos: Thanks! $\endgroup$
    – Qmechanic
    Commented Jul 5, 2021 at 20:58

Your first formula is incorrect. This distribution cannot normalized. We can only get relative probability distributions from the absolute square of the kernel. It has a normalization factor, but this is a different factor, this factor relates to the definition of the path integral. Refer to section 4.1 in Feynman's 'Path Integrals in Quantum Mechanics' to understand how this factor is obtained. We know$${\lmoustache_{Dx_b}}{K(x_ct_c|x_bt_b)}{dx_b}{K(x_bt_b|x_at_a)}={K(x_ct_c|x_at_a)}$$ where $t_c>t_b>t_a$

In your second formula $t_b>t_a$, so the limit is a left-hand limit.

Applying the limit of $t_c\rightarrow t_a$ to the second integral we obtain(which should have been your first formula) $${\lmoustache_{Dx_b}}{K(x_ct_c|x_bt_b)}{dx_b}{K(x_bt_b|x_at_a)}={\delta}{(x_c-x_a)}$$

Thus we can show, in the limit $t_c\rightarrow t_a$ $${K(x_ct_c|x_at_a)}={\delta}{(x_c-x_a)}$$

The absolute value feynman propagators multiplied into $dx_c$ will give you a relative probability and not an exact probability. This is why the integral in your equation should diverge. If the observable $x$ took on a set of finite values ${x_1,....,x_N}$, then we would replace the integral with a simple sum and you would get in the same limit:


  • $\begingroup$ So actually both demands are needed ? The first one tells us something about the global normalisation, the second one tells us something about *the phase of the propagator and how it should behave when $t_b\rightarrow t_a$, and of course the centering since it's describing particles and we want them to evolve in a continuous way. Is this more right then ? $\endgroup$
    – Nick
    Commented Oct 2, 2013 at 8:57
  • $\begingroup$ Yes but I'm currently trying to follow a path in which we define the path integral and quantum mechanics without using Schrödinger of any quantum knowledge. $\endgroup$
    – Nick
    Commented Oct 3, 2013 at 17:04
  • $\begingroup$ Your line of reasoning indeed makes perfect sense to me if I were to go from the known results in quantum mechanics to a path-integral formulation. Now i adjusted my queston yesterday to make the first formula more correct (I had posted in wrong). What the text that I'm reading is trying to do is to formulate quantum mechanics in such a way that it is in the same line as classical mechanics (principle of "least" action) WITHOUT prior knowledge of quantum mechanics. We start with the 2-slit experiment en impose Born's rule. That rule states that we should sum all paths (end of part 1) $\endgroup$
    – Nick
    Commented Oct 4, 2013 at 9:38
  • $\begingroup$ (part 2) this gives us the idea of the path-integral (indeed this is what Feynman says as summing over all of the paths). Now I can hear your next question: "Okay, so how do you determine the 'weight' of a path". Well together with the born-rules we simply state that the weight (or phase of whatever) should have the form $exp(-\frac{i}{\hbar}S)$, where the i is to indeed make it a phase (so that we can get the interference-effects that are present in the double-slit experiment), $\hbar$ is a very small number which demands that (according to the principle of least action) (end of part2) $\endgroup$
    – Nick
    Commented Oct 4, 2013 at 9:43
  • $\begingroup$ (part 3), we should have $\delta S = 0$, now since $\hbar\neq 0$, the demand with this phase becomes $\delta s \approx 0$. And in the classical limit, where $S >> \hbar$ this actually gives $\delta S = 0$. So we have defined a phase which makes a perfect transition. I know you can derive this from quantum mechanics with the time-evolution operator, but that's one of the things that are NOT used. But to finish it all, I really like your comment since it has given me insight in the link between orthogonality of the wave function and my second formula! I will check my Feynman as well. thanks! $\endgroup$
    – Nick
    Commented Oct 4, 2013 at 9:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.