# Construction of the path integral

1. Apr 26, 2015

### aaaa202

I am reading about the construction of the path integral in which states that the propagator is given by:

<qf l exp(-iHt/ħ l qi > = ∫Dq exp(i/ħ∫dt L(q,q'))

and Dq is the integration measure given by limN→∞Πn=1N-1 dqn. Can someone help me understand this integration measure a bit better? I don't understand how you would carry out the integral and what this infinite product even means. Does it give us the integral over all possible paths in phase space?

2. Apr 26, 2015

### robphy

3. Apr 27, 2015

### Jilang

The infinite product arises since each possible path is broken down into an infinite number of steps. Each step had its own amplitude and to find the amplitude for one step followed by the next step followed by the next, the amplitudes are multiplied together in the same way as you would multiply probabilities together for consecutive events. Then all possible paths are summed up, an integral over real space.

4. Apr 27, 2015

### vanhees71

One should be aware that this is NOT the proper definition of the integral measure for the "Lagrangian path integral" over configuration-space trajectories. Unfortunately, although I tried very hard by studying the literature and also for myself, to give a rigorous definition in the continuum limit, I couldn't find a really satisfactory one. I'm not sure, whether there is a solution in the more mathematical literature, because this is hard to follow for a humble theoretical physicist.

There are two ways to deal with this problems in practice:

(a) Use the "Hamiltonian" path-integral over trajectories in phase-space and use a discrete time lattice (in QFT it's a space-time lattice, but that's a detail). There the integration measure is the naive one $\mathrm{d}^{N} x \mathrm{d}^{N+1} p/(2 \pi \hbar)^{N+1}$, where $N$ is the number of lattice sites. The integral wrt. $p$ is unrestricted, that for the position variables $x$ restricted by the boundary conditions $x(0)=q_i$, $x(t)=q_f$. For the free particle the lattice version gives the exact solution for the propagator immediately by taking both (!!!) Gaussian multi-dimensional integrals over $p$ and $x$. The calculation reveals that the $p$ integral alone, which you can do formally in many cases of non-relativistic quantum mechanics, because the Hamiltonian is quadratic in $p$ and the coefficients in front of the quadratic and linear components are $x$-independent (in Carstesian coordinates; there are complicated issues when using other than Cartesian coordinates), leads to an $x$-independent factor $\mathcal{N}'(t)$ which diverges in the continuum limit.

(b) You use the fact that this factor $\mathcal{N}'(t)$ is independent of the potential under the above mentioned constraints of the Hamiltonian, i.e., you can write the Lagrangian configuration path integral in the form
$$U(t;q_f,q_i)=\mathcal{N}'(t) \int_{(0,q_i)}^{(t,q_f)} \mathrm{D} x(t) \exp(\mathrm{i} S[x]/\hbar),$$
where you use the naive measure $\mathrm{D} x(t)$ over the paths in configuration space. Both the factor and this naive configuration-space path integral are ill-defined but the factor is independent of the potential and can thus determined in regularized form from the free-particle case, and the integral then should have a singularity such as to cancel the singularity of the factor. Using functional methods the formal result for the factor is
$$\mathcal{N}(t)=\sqrt{\frac{m}{2 \pi \mathrm{i} \hbar t}} \sqrt{\mathrm{det}\frac{m}{2 \pi \mathrm{i} \hbar}(-\mathrm{d}_t^2)}.$$
Of course, the determinant is ill-defined and must be regularized somehow (e.g., with the heat-kernel or $\zeta$-function regularization method).

The only other case, which can be simply solved analytically by either using the time-lattice version with the correct measure of the discretized path integral or the functional method is the harmonic oscillator, because there also the $x$-path integral is Gaussian. Then you can just leave the factor in the above given formal form and evaluate also the $x$-path integral leading to a functional determinant in the denominator in the same formal form. Then you can use the eigenvalues of the corresponding operators $-\mathrm{d}_t^2$ and $-\mathrm{d}_t^2-\omega^2$ (where $\omega$ is the frequency of the oscillator). The ratio of these two determinants leads to the correct finite result of the propagator of the harmonic oscillator.

In many calculations, particularly in the calculation of S-matrix elements in scattering theory (both in QM and QFT), the indefinite factor in the configuration-space path integral is irrelevant or cancels, which explains the usability of the path integral in practice. It develops its whole elegance when applied to the quantization of gauge theories, where the Faddeev-Popov formalism is much simpler than the later developed covariant operator-quantization methods by Kugo et al.

5. Apr 27, 2015

### stevendaryl

Staff Emeritus
You can get 95% of the way to the path integral just using a single fact about the green's function $g(x,t,x_0,t_0)$:

If $t_1$ is any time between $t_0$ and $t$, then

$g(x,t,x_0, t_0) = \int dx_1 g(x_1, t_1, x_0, t_0) g(x,t,x_1, t_1)$

Iterating this $N$ times gives:

$g(x,t,x_0, t_0) = \int dx_1 dx_2 dx_3 ... dx_N g(x_1, t_1, x_0, t_0) g(x_2, t_2, x_1, t_1) ... g(x, t, x_N, t_N)$

This is exactly true, no approximations or questionable limits. Then we can take the next step of saying, for $\delta t$ small, we can approximate

$g(x+\delta x, t+\delta t, x, t) \approx A(\delta t) e^{\frac{i}{\hbar} L(x, \frac{\delta x}{\delta t}, t) \delta t}$, where $L(x,\dot{x},t)$ is the classical Lagrangian and where $A(\delta t)$ is chosen to normalize $g$ correctly. With this approximation, we can rewrite the above:

$g(x,t,x_0, t_0) \approx \int dx_1 A(\delta t_1) dx_2 A(\delta t_2) dx_3 ... dx_N A(\delta t_N) e^{\frac{i}{\hbar} \sum_j L(x_j, v_j, t_j) \delta t_j}$ where $\delta t_j = t_{j} - t_{j-1}$, and with $v_j = \frac{x_j - x_{j-1}}{\delta t_j}$

The next step is to appoximate the discrete sum by an integral:

$\sum_j L(x_j, v_j, t_j) \delta t_j \approx \int dt L(x,\frac{dx}{dt}, t)$

where $x(t)$ is chosen to be a function that smoothly extrapolates $x(t_j) = x_j$.

So we have a finite approximation:

$g(x,t,x_0, t_0) \approx \int dx_1 A(\delta t_1) dx_2 A(\delta t_2) dx_3 ... dx_N A(\delta t_N) e^{\frac{i}{\hbar} \int dt L(x,\frac{dx}{dt}, t)}$

This is pretty much noncontroversial. The only issue is whether there is some sense in which the right-hand side can be said to "coverge" to something in the limit as $N \rightarrow \infty$

6. Apr 28, 2015

### vanhees71

This is a great approach too. It's not clear to me, how you come so quickly to the configuration-space relation, but I guess that can be worked out somehow.

As I said in my previous posting, to my knowledge, there seems not to be a clear and unambiguous definition of the configuration-space path-integral measure. What I summarized, is the best I could come up with, and it oviously only works in Cartesian coordinates.

Already the introduction of spherical coordinates and the path integral for the hydrogen atom is pretty tough, although it was solved by Hagen Kleinert in collaboration with Feynman. Kleinert has written "the bible" about path integrals. It's available from World Scientific for a really fair prize, and I can only recommend it, if you want to really dive deeply into all sorts of path-integral applications, including even "econophysics":

https://www.amazon.com/Integrals-Qu...10733&sr=8-1&keywords=Kleinert+Path+Integrals

Last edited by a moderator: May 7, 2017