Magnetic Resonance — Question 5

Section 40-5 · Physics pages 6660–6661

Question 5 Thermal Zeeman polarization is governed by a hyperbolic tangent Two equally degenerate spin levels have energies ∓∆/2. Their thermal populations are Nl, Nu. Define p = (Nl −Nu)/(Nl + Nu). −∆/2 +∆/2 ∆E Schematic; labels, not drawing lengths, specify the data. (a) Derive p as a function of the gap and temperature. (b) Find the high-temperature limit, then evaluate the exact p when ∆/(kBT) = ln 3.
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Question 5 - Solution Thermal Zeeman polarization is governed by a hyperbolic tangent −∆/2 +∆/2 ∆E (a) Method and calculation Nu/Nl = e−∆/(kBT), p = 1 −e−∆/(kBT) 1 + e−∆/(kBT) . p = tanh[∆/(2kBT)] . (b) Interpretation and further calculation ∆≪kBT : p ≃∆/(2kBT) . ∆/(kBT) = ln 3 ⇒p = (1 −1/3)/(1 + 1/3) = 1/2 . Physical check The approximation applies only when the level gap is small relative to thermal energy. At positive temperature the lower level is more populated, so equilibrium does not give inversion.

Question 5

Thermal Zeeman polarization is governed by a hyperbolic tangent

Two equally degenerate spin levels have energies ∓Δ/2\mp\Delta/2. Their thermal populations are Nl,NuN_l,N_u. Define p=(Nl−Nu)/(Nl+Nu)p=(N_l-N_u)/(N_l+N_u).

Diagram shown in the original SVG worksheet.

  1. Derive p as a function of the gap and temperature.

  2. Find the high-temperature limit, then evaluate the exact p when Δ/(kBT)=ln⁡3\Delta/(k_BT)=\ln3.

Question 5 - Solution

Thermal Zeeman polarization is governed by a hyperbolic tangent

Diagram shown in the original SVG worksheet.

(a) Method and calculation

Nu/Nl=e−Δ/(kBT),p=1−e−Δ/(kBT)1+e−Δ/(kBT).N_u/N_l=e^{-\Delta/(k_BT)},\quad p=\frac{1-e^{-\Delta/(k_BT)}}{1+e^{-\Delta/(k_BT)}}.p=tanh⁡[Δ/(2kBT)].\boxed{p=\tanh[\Delta/(2k_BT)]}.

(b) Interpretation and further calculation

Δ≪kBT:p≃Δ/(2kBT).\Delta\ll k_BT:\quad\boxed{p\simeq\Delta/(2k_BT)}.Δ/(kBT)=ln⁡3⇒p=(1−1/3)/(1+1/3)=1/2.\Delta/(k_BT)=\ln3\Rightarrow\boxed{p=(1-1/3)/(1+1/3)=1/2}.

Physical check

The approximation applies only when the level gap is small relative to thermal energy. At positive temperature the lower level is more populated, so equilibrium does not give inversion.