4
$\begingroup$

In canonical quantization, the particles arise as quantized excitations on the vacuum $|0\rangle$. For example, a one-particle state with four momentum $p=(E,\textbf{p})$ is given by $$|p\rangle\sim a^\dagger_{p}|0\rangle.$$ Is it possible to arrive at the particle picture in the path-integral formulation of quantum field theory (QFT)? What is the notion of a particle in path-integral way of doing QFT?

$\endgroup$
6
$\begingroup$

Path integrals compute transition amplitudes between quantum states.

Disclaimer.

I don't want to go into issues of rigged Hilbert spaces and mathematical correctness of definitions here. This answer gives the general rule of thumb for the path integral formulation.

Notation.

In this answer round brackets denote ordinary function arguments (like $\Psi(\vec{x})$), while square brackets denote functional arguments (like $S[\phi(x)]$), which are themselves functions (elements of infinite-dimensional spaces).

As a simple toy model, consider quantum mechanics.

Particle state can be described by a wavefunction $\Psi(\vec{x})$. Moreover, transition amplitudes are characterized by initial and final wavefunctions:

$$ \Psi_I(\vec{x}), \Psi_F(\vec{x}) $$

at times $0$ and $\tau$ respectively. The transition amplitude is given by

$$ \left< \Psi_F \right| \hat{U} \left| \Psi_I \right> = \int d^3 x_I \Psi_I (\vec{x}_I) \int d^3 x_F \Psi_F^{*}(\vec{x}_F) \cdot U(\vec{x}_I, \vec{x}_F), $$

where $U(\vec{x}_I, \vec{x}_F)$ are the matrix elements of the evolution operator. These are given by path integrals:

$$ U(\vec{x}_I, \vec{x}_F) = \intop_{\vec{x}(0)=\vec{x}_I}^{\vec{x}(\tau)=\vec{x}_F} {\cal D}x(t) \, e^{i S[\vec{x}(t)]}. $$

The boundary conditions are essential, because they determine the matrix element that is to be evaluated.

Generalization to QFT.

In QFT, states live on the infinite-past and infinite-future hyperplanes. We can loosely associate functionals of form

$$ \Psi_I[\phi(\vec{x})], \Psi_F[\phi(\vec{x})] $$

to them. Note that $\Psi$ depends on values of the field at the (initial and final) 3d hyperplanes.

The transition amplitude is given by the path integral

$$ \left< \Psi_F \right| \hat{U} \left| \Psi_I \right> = \int {\cal D}\phi_I \Psi_I[\phi_I] \int {\cal D}\phi_F \Psi_F^{*}[\phi_F] \cdot U[\phi_I(\vec{x}), \phi_F(\vec{x})], $$ $$ U[\phi_I(\vec{x}), \phi_F(\vec{x})] = \intop_{\phi(t_I, \vec{x}) = \phi_I(\vec{x})}^{\phi(t_F, \vec{x}) = \phi_F(\vec{x})} {\cal D}\phi e^{i S[\phi]}, $$

where the integral $\int {\cal D}\phi$ is over field configurations in between of two boundaries. It depends on the chosen boundary field configurations $\phi_I$ and $\phi_F$, and thus can't be factored out.

General rule.

States and Hilbert spaces are associated to boundaries. Path integrals are over the bulk region and compute transition amplitudes for the given pair of boundary states.

Transition amplitudes between particle states.

In QFT, it is logical to choose the Fock basis (and to label functionals associated to it with asymptotic particle states). Thus, path integrals give transition amplitudes between particle states.

The Fock basis is the same as in canonical quantization. It spans the space of rapidly decreasing functionals $\Psi[\phi(\vec{x})]$, and there are actually expressions for the $\Psi$ functional associated to elements of the Fock space. These are given by Hermitian polynomials times decreasing exponentials, just like for the simple Harmonic oscillator.

Conclusion

Path integrals are a tool for computation of quantum dynamics, i.e. transition amplitude between quantum states. They don't replace the canonical formalism of Hilbert spaces and self-adjoint operators. They instead complement it by providing a covariant way of deriving transition amplitudes. You still have to do the Hilbert space quantization, and there will be elementary particles as before.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.