An introduction into the Feynman path integral by Grosche C.

By Grosche C.

During this lecture a brief advent is given into the speculation of the Feynman course crucial in quantum mechanics. the final formula in Riemann areas should be given in line with the Weyl- ordering prescription, respectively product ordering prescription, within the quantum Hamiltonian. additionally, the speculation of space-time ameliorations and separation of variables could be defined. As user-friendly examples I speak about the standard harmonic oscillator, the radial harmonic oscillator, and the Coulomb strength.

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25), the angular dependence comes out wrong. 29) which is also vanishing for r ′ , r ′′ → 0, but not in the right manner. The points r ′ = r ′′ = 0 are essential singularities, where as the correct vanishing is powerlike. 22) seems very suggestive, but it is quit useless in explicit calculations. e. 30) cannot be calculated. The integrals turn out to be repeated integrals over errorfunctions which are not tractable. Also the method of Arthurs [4] fails. In this method one assumes that the integrations in the limit ǫ → 0 are effectively from −∞ to +∞.

58) N→∞ where KlN (T ) is defined by the iterated integrals. Furthermore we have set α = m/ǫ¯ h 2 2 (j) and β(j) = α[1 − ǫ mω (t )/2]. 59) γ| < π/4 and α, β > 0. 60) is valid for ν > −1 and ℜ(α) > 0. 60) we obtain for KlN (T ): KlN (T ) = α i N 2 exp i β ′2 2 (r + r ′ ) 2 ∞ 0 ∞ r(1) dr(1) · · · r(N−1) dr(N−1) 0 2 2 2 × exp i(β(1) r(1) + β(2) r(2) + · · · + β(N−1) r(N−1) ) × Il+ D−2 (− i αr(0) r(1) ) . . 61) Important Examples where the coefficients αN , pN and qN are given by N−1 αN = α k=1 pN qN α = − 2                       α 2γk N−1 k=1 α2k 4γk α2 α = − 2 4γN−1                      k α1 = α, αk+1 = α γ1 = β1 , α 2γk j=1 γk+1 = βk+1 − (k ≥ 1) α2 .

63) α yk we obtain γk+1 = βk+1 − α2 4γk ⇐⇒ yk+1 − 2yk + yk−1 + ωk2 yk = 0. 64) In the limit N → ∞ this gives a differential equation for y y¨ + ω 2 (t)y = 0 with solution y = η(t′ + t) ≡ η(t). Since on one hand side y1 ≃ y0 + ǫy˙ 0 → y0 2 y1 ≃ y0 γ1 → 2y0 , α (ǫ → 0), (ǫ → 0), and on the other y˙ 0 = 1 2 y1 − y0 = γ1 − 1 y0 → 1, ǫ ǫ α (ǫ → 1), we have the boundary conditions η(0) = 0, η(0) ˙ = 1. 2 The Radial Harmonic Oscillator is satisfied up to second order in ǫ. 64) α η[(k + 2)ǫ] + O(ǫ3 ) α yk+2 = .

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