Paper 1, Section II, D

Principles of Quantum Mechanics | Part II, 2011

Two individual angular momentum states j1,m1,j2,m2\left|j_{1}, m_{1}\right\rangle,\left|j_{2}, m_{2}\right\rangle, acted on by J(1)\mathbf{J}^{(1)} and J(2)\mathbf{J}^{(2)} respectively, can be combined to form a combined state J,M|J, M\rangle. What is the combined angular momentum operator J\mathbf{J} in terms of J(1)\mathbf{J}^{(1)} and J(2)\mathbf{J}^{(2)} ? [Units in which =1\hbar=1 are to be used throughout.]

Defining raising and lowering operators J±(i)J_{\pm}^{(i)}, where i{1,2}i \in\{1,2\}, find an expression for J2\mathbf{J}^{2} in terms of J(i)2,J±(i)\mathbf{J}^{(i)^{2}}, J_{\pm}^{(i)} and J3(i)J_{3}^{(i)}. Show that this implies

[J2,J3]=0\left[\mathbf{J}^{2}, J_{3}\right]=0

Write down the state with J=j1+j2J=j_{1}+j_{2} and with J3J_{3} eigenvalue M=j1j2M=-j_{1}-j_{2} in terms of the individual angular momentum states. From this starting point, calculate the combined state with eigenvalues J=j1+j21J=j_{1}+j_{2}-1 and M=j1j2+1M=-j_{1}-j_{2}+1 in terms of the individual angular momentum states.

If j1=3j_{1}=3 and j2=1j_{2}=1 and the combined system is in the state 3,3|3,-3\rangle, what is the probability of measuring the J3(i)J_{3}^{(i)} eigenvalues of individual angular momentum states to be 3-3 and 0 , respectively?

[You may assume without proof that standard angular momentum states j,m|j, m\rangle are joint eigenstates of J2\mathbf{J}^{2} and J3J_{3}, obeying

J±j,m=(jm)(j±m+1)j,m+1,J_{\pm}|j, m\rangle=\sqrt{(j \mp m)(j \pm m+1)}|j, m+1\rangle,

and that

[J±,J3]=±J±]\left.\left[J_{\pm}, J_{3}\right]=\pm J_{\pm} \cdot\right]

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