# How to path-integrate over the half-line?

Consider the path-integral over a scalar field $\varphi$: $$Z=\int_{\mathcal S}\ \mathrm e^{iS[\varphi]}\mathrm d\varphi$$ where $\mathcal S$ is some function space (say, Schwartz or its dual).

How can we implement the condition $\varphi\ge0$? what effect does this restriction have, both at the perturbative level and at the non-perturbative level?

Note that this is not merely an "out of curiosity" question. The situation above occurs in practice (in its simplest manifestation, in the Higgs mechanism where we decompose $H(x)=\rho(x)\ \mathrm e^{i\sigma(x)}$; here the integral over $\rho$ is only over the half-line $\rho\ge0$).

• Could you not stick a functional delta function or Heaviside step function in the integral. I have seen something similar done for path integrals on manifolds in Statistical physics. – Quantum spaghettification Mar 27 '18 at 18:58
• @Quantumspaghettification a delta function would select $\varphi\equiv0$, so I don't think that would work. A Heaviside, on the other hand, does select $\varphi\ge0$, so that's good. I'm not sure how to turn $\Theta(\varphi)$ into something manageable though. In any case, it's food for thought, so thanks! – AccidentalFourierTransform Mar 27 '18 at 19:12
• Again going from what I know from statistical physics - you could write the Heaviside in terms of its integral representation which I assume can be generalized to the functional case. – Quantum spaghettification Mar 27 '18 at 19:20
• In perturbation theory, one usually works in the case where $\langle\rho\rangle\neq 0$ (say, in the ordered/non-vanishing vev phase), and expands $\rho=\langle\rho\rangle+\delta \rho$. Then the integral over $\delta \rho$ is not restricted (which is usually fine). – Adam Mar 27 '18 at 20:28
• Other ideas are to write $\rho$ as, say, $\rho= \phi^2$ or $\rho=e^{\phi}$, where $\phi\in\mathbb{R}$. – Qmechanic Mar 28 '18 at 11:49

For a free particle on the half line, the requirement that the Hamiltonian:

$$H = \frac{p^2}{2m}$$

$$(\psi ,H \phi) = (H\psi , \phi) = \phi(0) \frac{d\psi}{dx}(0)-\psi(0) \frac{d\phi}{dx}(0)$$

(this condition means that there is no probability current crossing $x=0$), leads to a family of self adjoint extensions characterized by the boundary conditions:

$$\psi(0) = \gamma \frac{d\psi}{dx}(0)$$

The propagator of the general solution of the Schrodinger equation with this boundary condition is given by:

$$G(x_1, x_2) = G_0(x_2-x_1)+G_0(x_2+x_1) - 2 \gamma \int_0^{\infty}d\lambda e^{-\gamma \lambda} G_0(x_2+x_1+\lambda)$$

($G_0$ is the free unconstrained propagator; Please, see Gamboa)

showed that this propagator can be obtained from a path integral quantization of the free action with a delta potential at the origin for paths extending from $-\infty$ to $+\infty$.

$$L = \frac{m \dot{x}^2}{2} + \gamma \delta(x)$$

However, this is not the unique way to quantize the motion on the half line:

Isham , defined a quantization scheme based on the canonical transformation:

$$(x, p) \rightarrow (e^x, e^{-x}p)$$

Another quantization scheme is based on the quatization of $\mathbb{R}^+$ as a quotient space:Please see Tanimura

$$\mathbb{R}^+ = \mathbb{R}^2/SO(2),$$

This method leads to another family of Hamiltonians:

I am not aware of any path integral formulation of the last two methods. Clearly, the last method leads to inequivalent quantizations with respect to the first one. I am quite confident that both methods can be merged to a unified method including both families of quatizations.