A Modern Approach to Quantum Mechanics · Problem 5.13

Wave Mechanics in One Dimension — John S. Townsend

Physics — Quantum Mechanics · positronium-annihilation-photon-polarization

Problem

The annihilation of positronium in its ground state produces two photons that travel back to back in the positronium rest frame along an axis taken to be the zz axis. The polarization state of the two-photon system is given by ∣ψ⟩=12∣R,R⟩−12∣L,L⟩|\psi\rangle = \frac{1}{\sqrt{2}}|R, R\rangle - \frac{1}{\sqrt{2}}|L, L\rangle (a) What is the probability that a measurement of the circular polarization state of the two photons will find them both right-handed? Both left-handed? (b) What is the probability that photon 1 will be found to be xx polarized and photon 2 will be found to be yy polarized, that is, that the system is in the state ∣x,y⟩|x, y\rangle? What is the probability that the system is in the state ∣y,x⟩|y, x\rangle? (c) Compare the probability for the two photons to be in the state ∣x,x⟩|x, x\rangle or in the state ∣y,y⟩|y, y\rangle with what you would obtain if the two-photon state were either ∣R,R⟩|R, R\rangle or ∣L,L⟩|L, L\rangle rather than the superposition ∣ψ⟩|\psi\rangle. Note: Since the photons are traveling back to back along the zz axis, if photon 1 is traveling in the positive zz direction, then photon 2 is traveling in the negative zz direction. Consequently, ∣R⟩2=12∣x⟩2−i2∣y⟩2and∣L⟩2=12∣x⟩2+i2∣y⟩2|R\rangle_2 = \frac{1}{\sqrt{2}}|x\rangle_2 - \frac{i}{\sqrt{2}}|y\rangle_2 \quad \text{and} \quad |L\rangle_2 = \frac{1}{\sqrt{2}}|x\rangle_2 + \frac{i}{\sqrt{2}}|y\rangle_2

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