Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

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 ?

share|improve this question

2 Answers 2

up vote 5 down vote accepted

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

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

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$^1$

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

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

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

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

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

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) Ideologically, OP's first eq. (1')

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

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 ideology 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

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

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

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

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

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

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

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

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.

share|improve this answer
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 ? –  Dominique Oct 2 '13 at 11:55
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. –  Dominique Oct 3 '13 at 17:03
I found the error in my first formula, now the two equations should be correct, sorry for my mistake :s –  Dominique Oct 3 '13 at 22:02
OK. I plan an update. –  Qmechanic Oct 3 '13 at 23:41
@Nick the propagator cannot in general be normalized in this way, it yields relative probability distributions. Feynman's 'Path Integrals on quantum Mechanics' discusses this divergence in an exercise in the 3rd chapter, do refer to it for greater clarity. In fact Dirac states in his book that in non-normalizable amplitudes like this we can only hope to get relative probabilities and not absolute probabilities. –  dj_mummy Oct 4 '13 at 4:27

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:


share|improve this answer
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 ? –  Dominique Oct 2 '13 at 8:57
@Nick I misunderstood your question. My original answer was for another question. I have changed my answer. I apologize for any confusion I might have caused. I added a reference. It is very nice, please do read it if you can. –  dj_mummy Oct 2 '13 at 14:53
@Nick I think you got confused in the first place because you wrote the first equation incorrectly. I think the you overlooked the arguments in the feynman propagator, x takes on indiscrete values. –  dj_mummy Oct 2 '13 at 15:03
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. –  Dominique Oct 3 '13 at 17:04
@Nick Schrodinger's equation is essentially gives a scheme for the time evolution of the system, the path integral approach to time slices and exponential terms come from the relationship between the hamiltonian and time. For more details look at chapter 4 in Feynman and Hibbs. Feynman himself derives the normalization constant from schrodinger's equation for the non-relativistic particle. –  dj_mummy Oct 4 '13 at 4:09

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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