Combinatorics and renormalization in quantum field theory by E. R. Caianiello

By E. R. Caianiello

This quantity relies upon paintings performed by means of the writer and his collaborators over a interval of roughly two decades and is devoted to a few chosen themes of quantum box conception, that have proved of accelerating value with the passing of time. There are 3 components: Combinatoric tools; Equations for Green's capabilities and Perturbative Expansions; Regularization, Renormalization, and Mass Equations. This paintings might be necessary to an individual drawn to studying or utilizing quantum box thought, many-body physics and, in addition, to many utilized mathematicians, since it introduces a few combinatoric and analytic instruments which tremendously simplify, and every now and then pass, remedies which more often than not take in lots of the bulk of the normal texts. it's a part of the Frontiers in Physics sequence, edited through David Pines.

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7 Consider the initial (t = 0) free-particle wave function ψ(x, 0) = N eik0 x e−α(x+x0 ) /2 + N e−ik0 x e−α(x−x0 ) /2 2 2 (a) Calculate the normalization constant N and the expectation values x 0 , x 2 0 . ˜ (b) Calculate the momentum wave function ψ(k, 0). (c) Write down the momentum probability density. Find the expectation values p , p 2 , as well as the uncertainty ( p)2 . Verify the Heisenberg uncertainty relation at time t = 0. (d) Find the evolved wave function ψ(x, t) at time t > 0. Write down the position probability density P(x, t).

B) Study the E < 0 case. Are there any bound-state solutions? Solution (a) Since the wave function has to vanish at the wall, the only choice in the region 0 < x < a is sin kx. Thus, the wave function is of the form ψ(x) = Csin kx, Ae−ikx + Beikx , 0a Continuity of the wave function at x = a gives C sin ka = Ae−ika + Beika However, the discontinuity of the derivative at that point gives −ik(Ae−ika − Beika ) − kC cos ka + gC sin ka = 0 or i Ae−ika − Beika = (k cos ka − g sin ka) C k 3 Simple potentials 51 Solving for B and A, we obtain C g g − e2ika i+ 2 2k 2k g g C − e−2ika −i + B= 2 2k 2k The ratio of the reflected to the incident amplitude is a pure phase: namely, we have A= B −i + g/2k − (g/2k)e−2ika = ≡ e2iδ A i + g/2k − (g/2k)e2ika so that tan δ ≡ −2k/g + sin 2ka 1 − cos 2ka At very high energies (k → ∞), tan δ ∼ k g sin2 ka =⇒ δ∼− π 2 At low energies (k → 0), tan δ ∼ 1 1 1− ka ga This gives either δ ∼ π/2 or δ ∼ −π/2, depending on how strong the delta-function potential is.

B) Calculate1 the expectation value of the energy at any time t > 0. e. u v0 . What is v0 ? Solution (a) The new eigenvalues and eigenfunctions of the modified system are En = h¯ 2 n 2 π 2 1 = En 2 32m L 4 1 x 1 ψ n (x) = √ einπ x/4L + (−1)n+1 e−inπ x/4L = √ ψn 2 2 2L 2 where n = 1, 2, . .. The wave function of the particle at times t > 0 will be ψ(x, t) = ∞ Cn e−i E n t/¯h ψ n (x) n=1 with coefficients Cn = 1 2L 1 ∗ d x ψ(x, 0) ψ n (x) = √ L −2L L d x cos −L You can make use of the series ∞ ν=0 (2ν + 1)2 π2 .

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