# I Field eigenstates and path integral formulation

1. Jul 6, 2016

### Alhaurin

Recently I have reviewed by reference books to get a better understanding of the fundamentals of QFT and there is one thing I still do not understand. In the QFT derivation of the path integral formula, it seems that every book and online resource makes the assumption that for the field operator f(x,t)_op there are "simultaenous eigenstates" |g(x,t)> such that f(x,t)_op |g(x,t)> = g(x,t) |g(x,t)> for every x and t. The same goes for the conjugate operator pi(x,t)_op. For sake of simplicity we can focus in the case of a real scalar field that obeys the Klein-Gordon equation.

I see that by making that assumption it is straightforward to move from the path integral formulation in traditional quantum mechanics (with p and q operators) to the one used in QFT, leading to equations similar in form. However I have an issue with the existence of these eigenstates, they seem to come out of nowhere so to me they look like some sort of "input" to the theory, like an assumption and not a direct consequence of the formalism of QFT via "Action in terms of the Lagrangian -> Klein Gordon equation -> f(x,t)_op in terms of a(k)_op and a'(k)_op -> conmutation relations for the operators".

Can somebody explain to me or give me a reference where I can see the mathematic reasoning behind these |g(x,t)> states, or otherwise just confirm that they are indeed a necessary input to the theory.

2. Jul 7, 2016

### vanhees71

Put the field operators on a spatial grid (time in the Heisenberg picture can be kept as a continuous parameter for that purpose). For simplicity let's take a neutral scalar boson, i.e., take Hermitean field operators. Then the canonical equal-time commutator relations read
$$[\hat{\phi}(t,\vec{x}),\hat{\phi}(t,\vec{x}')]=[\hat{\Pi}(t,\vec{x}),\hat{\Pi}(t,\vec{x}')]=0, \quad [\hat{\phi}(t,\vec{x}),\hat{\Pi}(t,\vec{x}')] =\mathrm{i} \delta_{\vec{x},\vec{x}'}.$$
Then you just have the Heisenberg algebra in a space of large dimension with the degrees of freedom labeled by $\vec{x}$ on the grid. By pure algebra you get the exponential functions as generalized eigenfunctions of the fields and their conjugate momenta, in complete analogy to the corresponding plane-wave solutions in non-relativistic "first-quantization" formulation in terms of position and momentum observables of a many-body system.

The continuum limit is, of course, very subtle (and mathematically still not rigorously well defined anyway!), and usually one uses hand-waving arguments. Note that neither the definition of the path integral (not even for the Euclidean QFT!) nor the operator formalism is rigorously well defined for interacting QFTs in (1+3) dimensions! So you must live with some handwaving unfortunately.

3. Jul 7, 2016

### Alhaurin

Thank you very much for your answer, it is clear to me now that some handwaving is indeed necessary for the path integral in QFT.

With regards to your first paragraph I understand that since operators $\phi$ and $\Pi$ evaluated at $(x,t)$ behave just like operators $\hat{x}$ and $\hat{p}$ of QM, then if there exist eigenstates of $\hat{x}$ and $\hat{p}$ with any possible real eigenvalue (in space that would be delta functions for $\hat{x}$ and plane waves for $\hat{p}$) then necessarily there have to be abstract eigenstates for $\phi$ and $\Pi$ with the same range of eigenvalues?

4. Jul 7, 2016

### A. Neumaier

The path integral is completely well-defined only on a finite lattice approximation of the 3-dimensional space (and discrete or continuous time). In the continuum one works with analogies derived from the finite case.

Note that simultaneous eigenstates exist only at fixed times, and the eigenstates are (as $p,q$ for a free particle) generalized, i.e., unnormalizable. Moreover, in the continuum limit everything is distribution-valued, i.e., needs additional smearing in space (and probably also in time).

5. Jul 8, 2016

### vanhees71

Yes, in the lattice version (see the previous posting by A. Neumeier) from the canonical equal-time commutation relations the eigenvalues and eigenvectors follow. It's just the Heisenberg algebra, and thus gives the analogous results as for $x$ and $p$ in non-relativistic quantum mechancis.

The "hand-waving continuum" limit then makes integrals out of the sums, and the multidimensional usual integrals become (also only formally and not strictly defined) functional integrals. For real calculations you have to regularize them somehow again. There are various techniques like the Schwinger proper-time method to define Gaussian functional integrals and perturbation theory (i.e., the saddle-point approximation, aka $\hbar$ expansion), finally leading to renormalized perturbation theory.