Read Online or Download Critical Properties of o4 Theories PDF

Best quantum physics books

• Theoretical Atomic Spectroscopy
• Methodes en theorie des champs =: Methods in field theory
• Quantum Reality
• Quantenmechanik fuer Fortgeschrittene
• Arrival times in quantum mechanics
• The Quantum Theory of Measurement

Additional info for Critical Properties of o4 Theories

Example text

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.

Download PDF sample