Past exam of the mathematics course of the University of Cambridge / 2006 / ii / Paper 1 / 32D (Principles of Quantum Mechanics)

Generalized position eigenstates satisfy $⟨x|x^{\prime }⟩=\delta (x-x^{\prime })$ and $\int |x⟩⟨x|dx=I$. Define the position-space wavefunction $\psi (x)=⟨x|\psi ⟩$. Completeness then gives $⟨\psi |\psi ⟩=\int |\psi (x){|}^{2}dx$. Position acts by multiplication, $⟨x|\hat{x}|\psi ⟩=x\psi (x)$. Using $⟨x|p⟩=(2\pi \hslash {)}^{-1/2}{e}^{ipx/\hslash }$ and momentum completeness, differentiating the Fourier integral gives $⟨x|\hat{p}|\psi ⟩=-i\hslash {\psi }^{\prime }(x)$ on the momentum operator's domain.

For the unit-mass, unit-frequency oscillator defineSince $[H,a]=-\hslash a$, a simultaneous energy eigenstate and $a$-eigenstate would satisfy $[H,a]|{\psi }_{\alpha }⟩=0=-\hslash \alpha |{\psi }_{\alpha }⟩$, forcing $\alpha =0$. For $\alpha =0$, the energy is $\hslash /2$, the smallest possible because $⟨a^{†}a⟩=\Vert a\psi {\Vert }^2\ge 0$.

The position-space equation is $(x+\hslash {\partial }_x){\psi }_{\alpha }=\sqrt{2\hslash }\alpha {\psi }_{\alpha }$. Solving and normalizing, with $\alpha =u+iv$, gives up to a constant phaseIts squared modulus is a normalized Gaussian, so every complex $\alpha$ gives a normalizable coherent state.

For a creation eigenstate, $(x-\hslash {\partial }_x)\psi =\sqrt{2\hslash }\beta \psi$ instead gives $\psi =C\exp [x^2/(2\hslash )-\sqrt{2/\hslash }\beta x]$. Every nonzero such function grows quadratically exponentially at infinity and fails square integrability. Hence there are no nonzero normalizable eigenstates of $a^{†}$. This last symbol is $a^{†}$ in the original PDF, although the converted TeX drops its dagger.