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Mathematics course of the University of Cambridge

Past exam of the mathematics course of the University of Cambridge

2019

Paper 301

The vacuum has zero momentum and is invariant under translations. Hence
$$
e^{-i\mathbf P\cdot\mathbf y}|\mathbf q\rangle
=e^{-i\mathbf P\cdot\mathbf y}a_{\mathbf q}^\dagger
 e^{i\mathbf P\cdot\mathbf y}|0\rangle
=\boxed{e^{-i\mathbf q\cdot\mathbf y}|\mathbf q\rangle}.
$$
Thus a one-particle <momentum eigenstate> remains the same ray and acquires the translation phase appropriate to its momentum.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 301 / 1 / ii / d / Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 301 / 1 / ii / d / Solution

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Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 301 / 1 / ii / d / Solution

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