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Physics / Chapter 40: All About Atoms / Section 40-6

The Pauli Exclusion Principle — Question 6

Section 40-6 · Physics pages 6682–6683

Question 6 Opposite measured spins do not by themselves identify a singlet Compare |s⟩= (| ↑↓⟩−| ↓↑⟩)/ √ 2 and |t0⟩= (| ↑↓⟩+ | ↓↑⟩)/ √ 2. Use | ± x⟩= (| ↑⟩± | ↓⟩)/ √ 2. electron 1 electron 2 z z Schematic; labels, not drawing lengths, specify the data. (a) Find their z-spin correlation ⟨S1zS2z⟩. (b) Expand each state in the x basis and compare ⟨S1xS2x⟩.
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Question 6 - Solution Opposite measured spins do not by themselves identify a singlet electron 1 electron 2 z z (a) Method and calculation S1zS2z| ↑↓⟩= −ℏ2| ↑↓⟩/4. ⟨S1zS2z⟩s = ⟨S1zS2z⟩t0 = −ℏ2/4 . (b) Interpretation and further calculation |s⟩= | −x, +x⟩−| + x, −x⟩ √ 2 , |t0⟩= | + x, +x⟩−| −x, −x⟩ √ 2 . ⟨S1xS2x⟩s = −ℏ2/4, ⟨S1xS2x⟩t0 = +ℏ2/4 . Physical check Both states always give opposite z results, but their relative signs create different correlations in another basis. The triplet requires a different spatial exchange symmetry.

Question 6

Opposite measured spins do not by themselves identify a singlet

Compare |s⟩=(|↑↓⟩−|↓↑⟩)/2|s\rangle=(|\uparrow\downarrow\rangle-|\downarrow\uparrow\rangle)/\sqrt2 and |t0⟩=(|↑↓⟩+|↓↑⟩)/2|t_0\rangle=(|\uparrow\downarrow\rangle+|\downarrow\uparrow\rangle)/\sqrt2. Use |±x⟩=(|↑⟩±|↓⟩)/2|\pm x\rangle=(|\uparrow\rangle\pm|\downarrow\rangle)/\sqrt2.

Diagram shown in the original SVG worksheet.

  1. Find their z-spin correlation ⟨S1zS2z⟩\langle S_{1z}S_{2z}\rangle.

  2. Expand each state in the x basis and compare ⟨S1xS2x⟩\langle S_{1x}S_{2x}\rangle.

Question 6 - Solution

Opposite measured spins do not by themselves identify a singlet

Diagram shown in the original SVG worksheet.

(a) Method and calculation

S1zS2z|↑↓⟩=−ℏ2|↑↓⟩/4.S_{1z}S_{2z}|\uparrow\downarrow\rangle=-\hbar^2|\uparrow\downarrow\rangle/4.⟨S1zS2z⟩s=⟨S1zS2z⟩t0=−ℏ2/4.\boxed{\langle S_{1z}S_{2z}\rangle_s=\langle S_{1z}S_{2z}\rangle_{t_0}=-\hbar^2/4}.

(b) Interpretation and further calculation

|s⟩=|−x,+x⟩−|+x,−x⟩2,|t0⟩=|+x,+x⟩−|−x,−x⟩2.|s\rangle=\frac{|-x,+x\rangle-|+x,-x\rangle}{\sqrt2},\quad |t_0\rangle=\frac{|+x,+x\rangle-|-x,-x\rangle}{\sqrt2}.⟨S1xS2x⟩s=−ℏ2/4,⟨S1xS2x⟩t0=+ℏ2/4.\boxed{\langle S_{1x}S_{2x}\rangle_s=-\hbar^2/4,\quad\langle S_{1x}S_{2x}\rangle_{t_0}=+\hbar^2/4}.

Physical check

Both states always give opposite z results, but their relative signs create different correlations in another basis. The triplet requires a different spatial exchange symmetry.

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