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Past exam of the quantum physics course of the University of Oxford

2016

The <spherical harmonics> satisfy
$$
L^2Y_l^m=\hbar^2l(l+1)Y_l^m,\qquad L_zY_l^m=\hbar mY_l^m,
\quad l=0,1,\ldots,\quad -l\le m\le l.
$$
Commuting $L^2$ or $L_z$ past $L_\pm$ therefore gives
$$
\boxed{L^2(L_\pm Y_l^m)=\hbar^2l(l+1)(L_\pm Y_l^m),\qquad L_z(L_\pm Y_l^m)=\hbar(m\pm1)(L_\pm Y_l^m).}
$$
For a nonzero result these are the <eigenvalue equations> with unchanged $l$ and shifted <magnetic quantum number>. Taking the squared <norm>, using $L_\mp L_\pm=L^2-L_z^2\mp\hbar L_z$, fixes the conventional coefficient:
$$
L_\pm Y_l^m=\hbar\sqrt{l(l+1)-m(m\pm1)}\,Y_l^{m\pm1}.
$$
At the appropriate endpoint the coefficient is zero. The zero <vector> is not itself an <eigenstate>.

Solution

Past exam of the quantum physics course of the University of Oxford / 2016 / 8 / b / Solution

Past exam of the quantum physics course of the University of Oxford / 2016 / 8 / b / Solution

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Past exam of the quantum physics course of the University of Oxford / 2016 / 8 / b / Solution

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