In the context of quantum mechanics via path integrals the normalization of the propagator as

$$\left | \int d x K(x,t;x_0,t_0) \right |^2 ~=~ 1$$

is incorrect. But why?

It gives the correct pre-exponential factor for the free particle and the incorrect one for the harmonic oscillator.

It seems to me that the propagator should describe the probability for the particle beginning at $(x_0,t_0)$ to arrive at $(x,t)$. So, the probability for the particle starting at some particular space-point to go somewhere (any x) must be 100%. The result that

$$\left | \int d x K(x,t;x_0,t_0) \right |^2 ~=~ 1$$

isn't satisfied seems to attack this ideas.. What is wrong with my though?

Other thing... There is some normalization procedure to obtain the pre-exponential term using the phase? (avoiding functional determinants)

**Comment 1 : ** Now I found the question Normalization of the path integral which is similar. But seems to say that the normalization procedure is correct! I am not finding this to the harmonic oscillator. So, I don't have an answer yet.


Just found by chance the correct normalization condition that allows the propagator to be interpreted as a probability amplitude. It reads

\begin{equation} \int dx_0' \int d x_1 K^\star(x_1,t;x_0',t_0) K(x_1,t;x_0,t_0) = 1, \quad \forall x_0 \end{equation}

Unlike other normalization prescriptions, this one gives the correct normalization factor when one considers the harmonic oscilador, for example.

For completeness, I leave here the prove. (Extracted from Path Integral for the Hydrogen Atom, by Anders Svensson, 2016)

\begin{align*} 1 &= \int dx_0' \delta (x_0'-x_0) = \int dx_0' \langle x_0' | x_0 \rangle = \int dx_0' \langle x_0' | \hat{U}^\dagger(t,t_0) \hat{U}(t,t_0)| x_0 \rangle\\ &= \int dx_0' \int dx_1 \langle x_0' | \hat{U}^\dagger(t,t_0)|x_1\rangle\langle x_1 | \hat{U}(t,t_0)| x_0 \rangle\\ &= \int dx_0' \int dx_1 \langle x_1 | \hat{U}(t,t_0)|x_0'\rangle^\star \langle x_1 | \hat{U}(t,t_0)| x_0 \rangle\\ &= \int dx_0' \int dx_1 K^\star(x_1,t;x_0',t_0) K(x_1,t;x_0,t_0) \end{align*}


1 Answer 1


I) OP is right, ideologically speaking. Ideologically, OP's first eq.

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

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 ideology has to be converted into mathematical formulas. E.g. for the harmonic oscillator, one has

$$\begin{align} \left| \int_{\mathbb{R}}\!\mathrm{d}x_f ~ K(x_f,t_f;x_i,t_i)\right| ~=~&\frac{1}{\sqrt{\cos\omega \Delta t}}, \cr \Delta t ~:=~&t_f-t_i,\end{align} \tag{A}$$

which only becomes unity for $\omega \Delta t \to 0$. The problem can ultimately be traced to the fact that there is no normalizable uniform probability distribution on the real axis $\mathbb{R}$, i.e. the $x$-position space. 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.

II) Let us review how normalization appears in the Feynman path integral from first principles. The main tool to determine the Feynman propagator/kernel/amplitude $K(x_b,t_b;x_a,t_a)$ is the (semi)group property

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

III) Equivalently, if we identify

$$ K(x_f,t_f;x_i,t_i)~=~\langle x_f,t_f \mid x_i,t_i \rangle \tag{C}$$

with an overlap of instantaneous$^1$ position eigenstates in the Heisenberg picture, then eq. (B) follows from the (first of) the completeness relations

$$ \int \!\mathrm{d}x ~|x,t \rangle \langle x,t |~=~{\bf 1}, \qquad \text{and} \qquad \int \!\mathrm{d}p~ |p,t \rangle \langle p,t |~=~{\bf 1}.\tag{D}$$

These instantaneous position and momentum eigenstates have overlap$^2$

$$ \langle p,t \mid x,t \rangle~=~\frac{1}{\sqrt{2\pi\hbar}}\exp\left[\frac{px}{i\hbar}\right].\tag{E}$$

IV) OP's first eq. (1) is equivalent to the statement that

$$ \left| \langle p_f=0,t_f \mid x_i,t_i \rangle \right| ~\stackrel{?}{=}~\frac{1}{\sqrt{2\pi\hbar}},\qquad(\leftarrow\text{ Ultimately wrong!}) \tag{F}$$

due to the identification (C) and

$$ \langle p_f,t_f \mid x_i,t_i \rangle ~\stackrel{(D)+(E)}{=}~\int_{\mathbb{R}}\!\frac{\mathrm{d}x_f}{\sqrt{2\pi\hbar}}\exp\left[\frac{p_fx_f}{i\hbar}\right] \langle x_f,t_f \mid x_i,t_i \rangle. \tag{G}$$

Eq. (F) is violated for e.g. the harmonic oscillator, where one has

$$ \left| \langle p_f,t_f \mid x_i,t_i \rangle \right| ~=~\frac{1}{\sqrt{2\pi\hbar\cos\omega \Delta t}}. \tag{H}$$

V) For sufficiently short times $\Delta t\ll \tau$, one derives from the Hamiltonian formulation (without introducing arbitrary normalization/fudge factors!) that

$$\begin{align} &\langle x_f,t_f \mid x_i,t_i\rangle\cr ~\stackrel{(D)}{=}~&\int_{\mathbb{R}} \!\mathrm{d}p~ \langle x_f,t_f \mid p,\bar{t} \rangle \langle p,\bar{t} \mid x_i,t_i\rangle \cr ~=~&\int_{\mathbb{R}} \!\mathrm{d}p~\langle x_f,\bar{t} \mid \exp\left[-\frac{i\Delta t}{2\hbar}\hat{H}\right]\mid p,\bar{t} \rangle \cr &\qquad \langle p,\bar{t} \mid \exp\left[-\frac{i\Delta t}{2\hbar}\hat{H}\right]\mid x_i,\bar{t}\rangle\cr ~\approx~&\int_{\mathbb{R}} \!\mathrm{d}p~ \langle x_f,\bar{t} \mid p,\bar{t} \rangle \langle p,\bar{t} \mid x_i,\bar{t}\rangle \exp\left[-\frac{i\Delta t}{\hbar} H(\bar{x},p) \right]\cr ~\stackrel{(E)}{=}~& \int_{\mathbb{R}} \!\frac{\mathrm{d}p}{2\pi\hbar} \exp\left[\frac{i}{\hbar}\left(p\Delta x -\left(\frac{p^2}{2m} + V(\bar{x})\right)\Delta t\right) \right]\cr ~=~& \sqrt{\frac{A}{\pi}} \exp\left[-A(\Delta x)^2-\frac{i}{\hbar}V(\bar{x})\Delta t\right], \qquad A~:=~\frac{m}{2 i\hbar} \frac{1}{\Delta t},\cr ~=~&\sqrt{\frac{m}{2\pi i\hbar} \frac{1}{\Delta t}} \exp\left[ \frac{i}{\hbar}\left(\frac{m}{2}\frac{(\Delta x)^2}{\Delta t}-V(\bar{x})\Delta t\right)\right],\end{align}\tag{I}$$


$$ \begin{align} \Delta t~ :=~&t_f-t_i, \quad \bar{t}~ :=~ \frac{t_f+t_i}{2}, \cr \Delta x~ :=~&x_f-x_i, \quad \bar{x}~ :=~ \frac{x_f+x_i}{2} .\end{align}\tag{J}$$

The oscillatory Gaussian integral (I) over momentum $p$ was performed by introducing the pertinent $\Delta t\to\Delta t-i\epsilon$ prescription. Eq. (I) implies that

$$ K(x_f,t_f;x_i,t_i) ~\longrightarrow~\delta(\Delta x) \quad \text{for} \quad \Delta t \to 0^{+}, \tag{K}$$

which in turn implies OP's first eq. (1) in the short time limit $\Delta t \to 0^{+}$. More generally, Eq. (I) implies OP's first eq. (1) for $\Delta t\ll \tau$.

VI) Note that the short time probability

$$\begin{align} P(x_f,t_f;x_i,t_i)~=~&|K(x_f,t_f;x_i,t_i)|^2~\stackrel{(I)}{\approx}~\frac{m}{2\pi \hbar} \frac{1}{\Delta t} , \cr \Delta t~\ll~& \tau, \end{align} \tag{L}$$

is independent of initial and final positions, $x_i$ and $x_f$, respectively. For fixed initial position $x_i$, the formula (L) can be interpreted as a uniform and unnormalizable probability distribution in the final position $x_f\in\mathbb{R}$. This reflects the fact that the instantaneous eigenstate $|x_i,t_i \rangle$ is not normalizable in the first place, and ultimately dooms the notion of absolute probabilities.

VII) For finite times $\Delta t$ not small, the interaction term $V$ becomes important. In the general case, the functional determinant typically needs to be regularized by introducing a cut-off and counterterms. But regularization is not the (only) source of violation of OP's first eq. (1), or equivalently, eq. (F). Rather it is a generic feature that the $px$ matrix elements of an unitary evolution operator

$$ \frac{\langle p,t \mid \exp\left[-\frac{i\Delta t}{\hbar}\hat{H}\right] \mid x,t\rangle}{\langle p,t \mid x,t\rangle} \tag{M} $$

is not just a phase factor away from the short time approximation $\Delta t\ll \tau$.

VIII) Example: Consider the Hermitian Hamiltonian

$$ \hat{H}~:= \frac{\omega}{2}(\hat{p}\hat{x}+\hat{x}\hat{p}) ~=~ \omega(\hat{p}\hat{x}+\frac{i\hbar}{2}). \tag{N}$$


$$ \begin{align}\frac{\langle p,t \mid \exp\left[-\frac{i\Delta t}{\hbar}\hat{H}\right] \mid x,t\rangle}{\langle p,t \mid x,t\rangle} ~=~&1 - \omega\Delta t\left(\frac{1}{2}-i\frac{px}{\hbar} \right)\cr ~+~&\frac{(\omega\Delta t)^2}{2}\left(\frac{1}{4}-2i\frac{px}{\hbar} - \left(\frac{px}{\hbar} \right)^2\right)\cr ~+~&{\cal O}\left((\omega\Delta t)^3\right),\end{align}\tag{O} $$

which is not a phase factor if $\omega\Delta t\neq 0$. To see this more clearly, take for simplicity $px=0$.


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

  2. J.J. Sakurai, Modern Quantum Mechanics, 1994, Section 2.5.


$^1$ Instantaneous eigenstates are often introduced in textbooks of quantum mechanics to derive the path-integral formalism from the operator formalism in the simplest cases, see e.g. Ref. 2. Note that the instantaneous eigenstates $\mid x,t \rangle $ and $\mid p,t \rangle $ are time-independent states (as they should be in the Heisenberg picture).

$^2$ Here we assume that possible additional phase factors in the $px$ overlap (E) have been removed via appropriate redefinitions, cf. this Phys.SE answer.

  • 1
    $\begingroup$ @Qmechanic: In your opinion, does this simply mean that a particle can never be in a position eigenstate (because, though they provide a basis, such states are just not normalizable)? If so, would you say the reason for this is the fact, that a (massive) particle has some spacial extent (because there are no problems if e.g. we consider the initial state to be a Gaussian wave packet)? $\endgroup$
    – André
    Dec 19, 2013 at 11:37
  • $\begingroup$ One interesting feature of approximation I is that $\langle x_f,t_f \mid x_i,t_i\rangle$ is independent of $x_f,x_i$. For the harmonic oscillator and the free particle, this is moreover true for all times. Is this a generic feature, or are there Hamiltonians for which $|\langle x_f,t_f \mid x_i,t_i\rangle|$ depends on $x_f,x_i$? $\endgroup$ Jul 22, 2018 at 20:06
  • $\begingroup$ @Semiclassical: It is undoubtedly a special (=non-generic) feature. $\endgroup$
    – Qmechanic
    Jul 24, 2018 at 15:31

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