PROBLEM FOR PHYSICS LOVERS ❤❤❤❤
#QUANTUM MECHANICS
Answers
Answer:
Answer:
one-dimensional potential well is given in the form of a delta function at x = 0, V(x) = Cδ(x), C < 0. A stream of non-relativistic particles of mass m and energy E approaches the origin from one side.
(a) Derive an expression for the reflectance R(E).
(b) Can you express R(E) in terms of sin2(δ), where δ is the phase sift of the transmitted wave.
Solution:
Concepts:
This is a "square potential" problem. We solve HΦ(x) = EΦ(x) in regions where V(x) is constant and apply boundary conditions.
Reasoning:
V(x) = 0 everywhere except at x = 0.
Details of the calculation:
(a) Φ1(x) = A1 exp(ikx) + A1'exp(-ikx) for x < 0. k2 = 2mE/ħ2.
Φ2(x) = A2 exp(ikx) for x > 0.
Φ is continuous at x = 0. Φ1(0) = Φ2(0). A1 + A1' = A2.
∂2Φ(x)/∂x2 + (2m(E - V(x))/ħ2)Φ(x) = 0.
Let us evaluate this equation at x = ε and at x = -ε and write down a difference equation.
∂Φε(x1 + ε)/∂x + ∂Φε(x1 - ε)/∂x = (2m/ħ2)∫x1-εx1+ε (Cδ(x) - E) Φ(x) dx
= (2mC/ħ2)Φ(0)
If V does not remain finite at the step, then ∂Φ/∂x has a finite discontinuity at the step.
iA1k - iA1'k = iA2k - A22mC/ħ2, A1 - A1' = A2 - A22mC/(ikħ2) = [1 - 2mC/(ikħ2)]A2
Eliminate A1':
2A1 = [2 - 2mC/(ikħ2)]A2, A1 = [1 + imC/(kħ2)]A2,
T(E) = (k|A2|2)/(k|A1|2) = ħ4k2/(ħ4k2 + m2C2) = E/(E + mC2/(2ħ2))
R(E) = 1 - T(E) = [mC2/(2ħ2)]/(E + mC2/(2ħ2)) = m2C2/(ħ4k2 + m2C2)
(b) A2 = |A2|exp(iδ).
sinδ = [mC/(kħ2)]/[1 + m2C2/(k2ħ4)]½.
sin2δ = [m2C2/(k2ħ4)]/[1 + m2C2/(k2ħ4)] = R(E).
Problem:
Answer:
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