quantum mechanics - Normalization of the path integral

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 ?

Answer

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.

References:

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

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$^1$ 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.