How does path integral formulation explain bound states?

It seems to me that the intuitive explanation of path integrals in quantum mechanics describes scattering processes only.

You have a particle going from A to B and you compute the probability amplitude to go from A to B.

But in a bound state, such as an infinite potential well or an hydrogen atom, it doesn't seem natural to consider a particle going from one place to other. Specially if the state is stationary.

How is the path integral useful in bound states?

• The path integral formulation is mathematically rigorous in "imaginary time" (see the Feynman-Kac formula). That means it is suited to study spectral properties, such as eigenvalues (and thus bound states), maybe more than to study the "real time" dynamics. See for example this paper, as an application of the path integral to the ground state energy and energy crossing of the Rabi model. – yuggib Oct 30 '14 at 10:06
• @yuggib Thanks, but my question was more oriented to an intuitive explanation. – jinawee Oct 30 '14 at 10:21
• There are loads of articles in Googlespace on path integral treatment of the simple harmonic oscillator. This might be a good start. – John Rennie Oct 30 '14 at 10:31
• What if I posed the question thus: Consider a QFT of a proton field and an electron field. Initial state: Free electron and free proton --> Final state Hydrogen atom in the 1s state. How would one calculate this transition amplitude? – Siva Oct 30 '14 at 13:50
• Once I can compute that, I could use the optical theorem to calculate the one-loop correction from such bound states to the Moller-like scattering of the electron and proton. Going that way, I think there might be a path (forgive the pun) to incorporate the effects of bound states into the formalism. – Siva Oct 30 '14 at 13:59

There is no conceptual problem to compute a probability amplitude for a system with bound-states in this formalism. For example, if the system is initially in a state $\psi(x)$ localized around $A=0$, the probability amplitude $a(B)$ to find the particle in $B$ after a time $t$ is $$a_t(B)\propto \int dx' K_t(B|x')\psi(x')$$ where $K$ is given by the path integral $$K_t(x|x')=\int_{x(0)=x'}^{x(t)=x} Dx(\tau) \,e^{i S[x(\tau)]}$$ with $S$ the classical action (of the Hydrogen atom problem). If $B$ is very far from $A$, and if $\psi$ is overlap manly on the bound-states (localized around A), this amplitude will be very small, as expected. If $\psi$ overlap mainly on the scattering states, the probability might be much closer to one.