By Hagen Kleinert, Verena-Schulte Frohlinde

ISBN-10: 9810246587

ISBN-13: 9789810246587

ISBN-10: 9810246595

ISBN-13: 9789810246594

ISBN-10: 981279994X

ISBN-13: 9789812799944

Established upon lecture notes for a direction taught by way of Kleinert, this monograph explains intimately how you can practice perturbation expansions in quantum box conception to excessive orders. The authors additionally describe how you can extract the serious homes of the speculation from the ensuing divergent strength sequence. Kleinert teaches physics on the Freie U. in Berlin and Schulte-Frohlinde is a traveling scientist at Harvard. Annotation c. ebook information, Inc., Portland, OR (booknews.com)

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7 Consider a one-dimensional delta function potential V (x) = gδ(x) and the scattering of particles of energy E > 0 at it. Without loss of generality assume that the particles are incident from the left. (a) Applying the appropriate continuity conditions at x = 0, find the wave function ψ E (x) and write it in the form ψ E (x) = eikx + Feik|x| where k ≡ 2m E/¯h 2 . 46 Problems and Solutions in Quantum Mechanics (b) Compute the probability current density and demonstrate that it is everywhere continuous.

B) Determine the matrix U defined by the relation √ √ qC U11 U12 kA √ = √ U21 U22 qD kB Show that U is a unitary matrix. (c) Write down the probability current conservation and show that it is directly related to the unitarity of the matrix U. Solution (a) Continuity at the point x = 0 gives A+B =C+D ik(A − B) = iq(C − D) These relations are equivalent to √ 1 − q/k √ 2 q/k √ √ qC =− qD+ kA 1 + q/k 1 + q/k √ √ 2 q/k √ 1 − q/k √ kB= qD+ kA 1 + q/k 1 + q/k 42 Problems and Solutions in Quantum Mechanics (b) From the last pair of relations we can conclude immediately that √ 2 q/k U11 = U22 = 1 + q/k 1 − q/k U12 = −U21 = − 1 + q/k Unitarity requires that |U11 |2 + |U12 |2 = |U22 |2 + |U21 |2 = 1 ∗ ∗ + U12U22 =0 U11U21 These relations are easily seen to be satisfied and so U is unitary.

These values are thus n 2h¯ 2 π 2 8ma 2 for n = 1, 2, . . The physical reason for the overall vanishing of reflection is destructive interference between the waves reflected at x = −a and those reflected at x = a; it is analogous to the phenomenon occurring in optics. It is immediately obvious from the expression for the reflected amplitude, E n = −V0 + B = e−2ika (k + q)(1 − e−4iqa ) (k − q) − e−4iqa (k + q)2 /(k − q) that for 2qa = nπ, B = 0. e. odd and even functions. Thus, we have 54 Problems and Solutions in Quantum Mechanics an even solution for C = 0, D=A and an odd solution for B = 0, D = −A The continuity conditions for the even solutions are A = Beaκ cos qa, tan qa = κ/q and those for the odd solutions are A = −Ceaκ sin qa, tan qa = −q/κ The energy eigenvalue conditions can be expressed in a more transparent form if we introduce ξ ≡ qa, Then we have κa = β 2 ≡ 2mV0 a 2 /¯h 2 β 2 − ξ 2 and the conditions are written as tan ξ = β2 − 1, ξ2 ξ tan ξ = − β2 − ξ2 for the even and the odd solutions respectively.

### Critical Properties of o4 Theories by Hagen Kleinert, Verena-Schulte Frohlinde

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