Past exam of the quantum theory course of the University of Oxford / 2015 / 2

The position operator $X$ and momentum operator $P$ are self-adjoint operators, so the adjoint operator isUse the canonical commutation relation with the order of factors preserved:Subtracting the first product from the second provesThese creation and annihilation operators are unscaled and carry the paper's phase convention; their commutator is not the unit-normalized oscillator commutator.

From $H=a_{+}{a}_{-}/(2m)+\hslash \omega /2$ and the commutator product rule,Consequently, if $H\psi =E\psi$,Thus the creation operator raises the energy eigenvalue by $\hslash \omega$, and the annihilation operator lowers it by $\hslash \omega$, provided the resulting vector is nonzero. The zero vector is not an eigenstate; in particular the ground state will be annihilated by $a_{-}$. The displayed operator identity holds even in that case.

The squared norm is obtained by moving the adjoint operator to the other side of the inner product:For an energy eigenstate, this isSince a squared norm is nonnegative, every nonzero energy eigenstate has $E\ge \hslash \omega /2$. Moreover $a_{-}\psi =0$ exactly when $E=\hslash \omega /2$.

Write $E_0=\hslash \omega /2$. Repeated application of the annihilation operator lowers the energy eigenvalue in steps of $\hslash \omega$. The lower bound from part (ii) forces this process to stop. To see why it also quantizes the starting energy, writeThe first $n$ lowering steps have positive squared norms. If $\delta >0$, the next step is also nonzero by part (ii), and its energy would be $E_0+(\delta -1)\hslash \omega <E_0$, contradicting the bound. Thus $\delta =0$, andExistence of the ground state can also be checked directly in the position representation. The equation $a_{-}{\psi }_0=0$ iswhose solution is proportional to $e^{-m\omega x^2/(2\hslash )}$. This Gaussian function is square-integrable, so it gives an actual ground state. Its normalization can be chosen using the Gaussian integral.

All the displayed levels occur: for a nonzero state at $E_n$,Thus repeated raising creates a nonzero energy eigenstate at every $E_n$.

Let $\psi$ have energy $E_n$ and put $\gamma =2m\hslash \omega$. The lowering norm formula gives, inductively,In particular $(a_{-}{)}^n\psi$ is a nonzero ground state. By the assumed nondegeneracy, it equals $c{\psi }_0$ for some $c\ne 0$.

It remains to show that raising this vector recovers $\psi$, rather than merely another state at the same energy. On an energy eigenstate at $E_n$, factorization gives $a_{+}{a}_{-}\psi =\gamma n\psi$. Iterating the factorization down the ladder yieldsApplying $(a_{+}{)}^n$ to $(a_{-}{)}^n\psi =c{\psi }_0$ therefore provesFor $n=0$ this is the original ground-state assumption. This proves nondegeneracy propagated by an oscillator ladder: every energy eigenspace is one-dimensional if the ground state is nondegenerate.

If $a_{-}\psi \ne 0$, it has energy $E-\hslash \omega$, distinct from the energy of $\psi$. Eigenstates of a self-adjoint operator with distinct eigenvalues are orthogonal. Explicitly,so $⟨\psi ,a_{-}\psi ⟩=0$. If $a_{-}\psi =0$, the same conclusion is immediate.

Dividing by $\Vert \psi {\Vert }^2$ and using the definition of $a_{-}$ givesBoth expectation values are real because $P$ and $X$ are self-adjoint operators. Its real and imaginary parts therefore vanish separately:An oscillator energy eigenstate is centered in both position and momentum, despite having nonzero quantum uncertainties.

Twice applying the creation operator or twice applying the annihilation operator changes the energy eigenvalue by $2\hslash \omega$, unless they give zero. The same orthogonality argument as in part (i) therefore givesAlso the factorization in part (a) givesNow $P=(a_{+}+a_{-})/2$ and $X=(a_{+}-a_{-})/(2im\omega )$. Squaring in the correct operator order yieldsBy part (i), the means are zero, so these are the variances. Taking positive square roots gives the requested standard deviations:These harmonic-oscillator number-state second moments also show equal mean kinetic energy and potential energy, each $E/2$. Their uncertainty product is $E/\omega =(n+\frac{1}{2})\hslash$, so the ground state saturates the Heisenberg uncertainty relation.