# A generalized Path integral ?

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I'm trying to rederive quantum mechanics from feynman's idea of integrating over all possible histories of quantum systems. Following this line of thought , I have found that the path integral can be considered as just matrix multiplication with Einstein's summation convention in the continuum limit. Every random walk in the discretized integral can be considered as a choice of certain values for the dummy indices. If we generalized this situation by considering arbitrary contractions with higher rank tensors instead of just matrix multiplication and passing to the continuum limit , can this lead to any new form of path integral or is it just a path integral with more integration variables ? What about considering contractions using spinors ?

edited Mar 23

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Quantum mechanics bases on states in the Hilbert space; these can be seen like vectors (representable in an orthonormal basis system), where a scalar product can be defined. If one multiplies two different vectors $a_i,b_j$, where the indices $i,j$ run over the complete Hilbert space basis by the scalar product

$<a_i,b_j> = c_{ij}$

one obtains a matrix $c_{ij}$. To obtain higher rank tensors you have to define first a new space of states, e.g. the states of the $L^p$ space with $p>2$. Moreover you have to define e.g. a generalized scalar product

$<a_i,b_j,c_k,...> = d_{ijk ...}$

with orthogonality conditions. Suppose you can generalize the spectral theorem as follows:

$H = \sum_{\lambda \in \Gamma} H_{\gamma} e_\gamma$

where $H$ is an operator, $H_\lambda$ the eigenvalues of it, $\Gamma$ the generalized spectrum and $e_\gamma$ a generalized projector on the eigenstate basis. You will compute now products like $e_\gamma(t) e_\lambda(t+dt)$ with time coordinate $t$. Such products you have to define and in general, these are some higher rank tensors.

At the end you will get a path integral very similar to the ordinary one, but with integrals over the generalized state basis (due to insertion of the $\sum_{\lambda \in \Gamma}$). However, if such a spectral theorem can not be applied, the path integral cannot be defined rigorously.

answered Mar 24 by (40 points)

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