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

2023

The phases of the three <basis vectors> can be chosen so that $A$ and $B$ are nonnegative real numbers. Matrix multiplication then gives
$$
[J_+,J_-]=\operatorname{diag}(A^2,B^2-A^2,-B^2).
$$
The <angular momentum commutation relations> require this to equal $2\hbar J_3=2\hbar^2\operatorname{diag}(1,0,-1)$. Equating diagonal entries yields
$$
A^2=2\hbar^2,\qquad B^2-A^2=0,\qquad B^2=2\hbar^2.
$$
Hence
$$
\boxed{A=B=\sqrt2\,\hbar,\qquad
J_+=\sqrt2\,\hbar\begin{pmatrix}0&1&0\\0&0&1\\0&0&0\end{pmatrix},\qquad
J_-=\sqrt2\,\hbar\begin{pmatrix}0&0&0\\1&0&0\\0&1&0\end{pmatrix}.}
$$
These are also the $j=1$ special case of the normalized <spin ladder operator> coefficients $\hbar\sqrt{j(j+1)-\mu(\mu\pm1)}$.

Solution

Past exam of the quantum theory course of the University of Oxford / 2023 / 3 / b / ii / Solution

Past exam of the quantum theory course of the University of Oxford / 2023 / 3 / b / ii / Solution

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