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Mathematics course of the University of Cambridge

Past exam of the mathematics course of the University of Cambridge

2022

Paper 1

From the <number operator> commutator $[N,A]=-A$,
$$
N(A|n\rangle)=(n-1)A|n\rangle,
$$
so $A|n\rangle$ is proportional to $|n-1\rangle$. Its squared norm is
$$
\lVert A|n\rangle\rVert^2=\langle n|A^\dagger A|n\rangle=n.
$$
With the conventional phase choice,
$$
\boxed{A|n\rangle=\sqrt n\,|n-1\rangle}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / ii / Paper 1 / 34A / a / Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / ii / Paper 1 / 34A / a / Solution

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Past exam of the mathematics course of the University of Cambridge / 2022 / ii / Paper 1 / 34A / a / Solution

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