Past exam of the physics A course of the University of Cambridge / 2019 / Paper 1 / Section B / B9

Let $c=\sqrt{m\omega /(2\hslash )}$ and $d=1/\sqrt{2m\hslash \omega }$, so $a=cx+idp$ and $a^{†}=cx-idp$. The canonical commutation relation $[x,p]=i\hslash$ givesHence $\boxed{H=\hslash \omega (a^{†}a+1/2)}$. Similarly,Using the product rule for a commutator, $[AB,C]=A[B,C]+[A,C]B$, givesThus $a^{†}a$ is the number operator, while $a$ and $a^{†}$ are the annihilation operator and creation operator.

The commutators implyThus any nonzero resulting vector is the adjacent energy eigenstate. Since $a^{†}a|n⟩=n|n⟩$ and $a{a}^{†}=a^{†}a+1$, their squared norms areChoosing consistent phases for the eigenstates yieldsThe vanishing last expression terminates the lowering sequence at the ground state, rather than introducing a negative-energy oscillator state.

In the position representation $p=-i\hslash d/dx$, so $a{\psi }_0=0$ becomesIntegrating and using the Gaussian integral for wavefunction normalization givesThe creation operator gives $|1⟩=a^{†}|0⟩$, and thereforeIts norm is one since $\int x^2|{\psi }_0{|}^{2}dx=\hslash /(2m\omega )$. Moreover ${\psi }_0$ is even and ${\psi }_1$ is odd, so the overlap integrand ${\psi }_0^{\ast }{\psi }_1$ is odd and integrable. Integration over the whole real line gives $\boxed{⟨0|1⟩=0}$, establishing orthogonality.

Inverting the definitions of the creation operator and annihilation operator gives $x=\sqrt{\hslash /(2m\omega )}(a+a^{†})$. Their actions and the orthogonality of the oscillator energy eigenstates giveThus the electric dipole transition selection rule iswith the lower alternative absent for $n_1=0$. The field exchanges one oscillator quantum in first-order dipole coupling. For a long weak monochromatic drive, its frequency must also be resonant with the level separation, here $\omega$, to give a substantial transition probability. Higher-order multiphoton processes are beyond the stated single-position-matrix element approximation.