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Short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford

Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford

2018

Choose the overall <phase> so that $A$ is positive. <Normalization> requires
$$
1=|A|^2\int_{-a}^a(a^2-x^2)^2dx
=|A|^2\frac{16a^5}{15}.
$$
Therefore
$$
\boxed{A=\frac{\sqrt{15}}{4a^{5/2}}.}
$$
An arbitrary constant <phase> multiplying this answer has no effect on any expectation value.

Solution

Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2018 / S3 / 2 / a / Solution

Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2018 / S3 / 2 / a / Solution

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Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2018 / S3 / 2 / a / Solution

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