Past exam of the quantum mechanics quantum theory and quantum computers course of the University of Oxford / 2009 / A

Inside the infinite well, the time-independent Schrödinger equation is $-{\hslash }^2{\Psi }^{\prime \prime }/(2m)=E\Psi$ with Dirichlet endpoints. Negative or zero energy gives no nonzero solution satisfying both endpoints. For $E>0$, the sine solution requires $ka=n\pi$. ThusThe normalization follows from ${\int }_0^a{\sin }^2(n\pi x/a)dx=a/2$. These functions form an orthonormal basis. Substitute the expansion into $i\hslash {\partial }_t\psi =H\psi$ and take the inner product with each basis vector: orthonormality gives $i\hslash {\dot{c}}_n=E_n{c}_n$, equivalentlyFor a general infinite expansion this is interpreted in the usual Hamiltonian domain or its unitary evolution; the particular state below has only two terms.

Put $\theta =\pi x/a$. The identity $\sin \theta (16{\cos }^2\theta -1)=4\sin 3\theta +3\sin \theta$ gives $c_1=3/5$, $c_3=4/5$, all others zero. Their squared magnitudes sum to one. HenceThe evolved state is $(3e^{-i{E}_{1}t/\hslash }{\Psi }_1+4e^{-i{E}_{3}t/\hslash }{\Psi }_3)/5$. Its position probability includes an interference term with frequency $\Omega =(E_3-E_1)/\hslash =4{\pi }^2\hslash /(m{a}^2)$. Specifically,The three required integrals are $\pi /8-1/4$, $\pi /8+1/12$ and $1/4$, respectively. ThereforeThe energy expectation is constant while the spatial probability oscillates, because relative energy phases affect interference but not the energy probabilities.

Let $u=rF$. Multiplication of the radial equation by $-2mr/{\hslash }^2$ gives $u^{\prime \prime }+[A/r-\lambda (\lambda +1)/r^2-{\alpha }^2]u=0$. With $u=e^{-\alpha r}f$ the constant terms cancel, leavingThe indicial equation is $c(c-1)=\lambda (\lambda +1)$, with roots $c=\lambda +1$ and $c=-\lambda$. The usual physical Coulomb radial origin condition selects $c=\lambda +1$: it imposes the regular reduced radial boundary condition $u(0)=0$. Normalizability alone at the origin is insufficient to make this selection when an irregular branch is locally square-integrable, notably at $\lambda =0$.

Write $f=r^{\lambda +1}{\sum }_{j\ge 0}{a}_j{r}^j$. Coefficient comparison givesA nonterminating regular series has the exponentially growing asymptotic branch, suggested by $a_{j+1}/a_j\sim 2\alpha /j$: it supplies an $e^{2\alpha r}$ component, which the prefactor $e^{-\alpha r}$ cannot normalize. Equivalently the regular confluent-hypergeometric solution grows unless its first parameter is a nonpositive integer. Thus a bound state requires the numerator to vanish at some $j=k\ge 0$, givingThis is series-termination quantization of the Coulomb radial equation. It requires $A>0$ for the assumed $\alpha >0$; at $A=0$ there are no such negative-energy Coulomb bound states.

For $k=0$, $F=N{r}^{\lambda }{e}^{-\alpha r}$. The angular integral cancels when computing $⟨1/r⟩$, while the radial volume element contributes $r^{2}dr$. ConsequentlyThe ratio follows by one integration by parts and remains valid for real $\lambda \ge 0$; no integer-only factorial identity is needed. Since now $\alpha =A/[2(\lambda +1)]$,For a classical particle with that negative total energy, nonnegative kinetic energy requires $E-V={\hslash }^2(A/r-{\alpha }^2)/(2m)\ge 0$, hence $\boxed{r\le A/{\alpha }^2}$. This is a necessary radial bound; extra angular momentum can impose a stronger restriction.

At $k=\lambda =0$, $\alpha =A/2$ and the classical boundary is $r=2/\alpha$. The normalized radial probability density is $4{\alpha }^3{r}^2{e}^{-2\alpha r}$. With $z=2\alpha r$, the exterior quantum probability isThe final integration gives $(4^2+2\cdot 4+2)e^{-4}/2$. This is the nonzero quantum tail beyond the classical turning region.

With the specified ladder-operator normalization, $A_{+}^{†}=A_{-}$ andTherefore $\Vert A_{-}\psi {\Vert }^2=2m(E-\hslash \omega /2)\Vert \psi {\Vert }^2\ge 0$. This proves $\boxed{E\ge \hslash \omega /2}$, with equality exactly when $A_{-}\psi =0$. In position representation this equation is $\hslash {\psi }^{\prime }+m\omega x\psi =0$, whose general solution isFor a unit vector $|C|=(m\omega /\pi \hslash {)}^{1/4}$; its phase is arbitrary. The commutators $[H,A_{\pm }]=\pm \hslash \omega A_{\pm }$ make $A_{+}$ raise and $A_{-}$ lower energies by $\hslash \omega$. Repeated raising of the Gaussian produces all levels $(n+1/2)\hslash \omega$. Conversely lowering any nonzero energy eigenvector must eventually terminate, since otherwise it would produce an energy below the lower bound. A terminating vector is annihilated by $A_{-}$ and hence has energy $\hslash \omega /2$. Thus no other eigenvalues occur.

Bilinearity of the two-dimensional commutators giveswhile all momentum-momentum and position-position commutators remain zero. Equal angles give a canonical position-momentum pair. Perpendicular angles give commuting components, which belong to independent coordinate directions.

The orthogonal coordinates $X_{+}=(X_1+X_2)/\sqrt{2}$, $X_{-}=(X_1-X_2)/\sqrt{2}$ preserve the kinetic sum, andHence $\boxed{{\omega }_{+}=\omega ,{\omega }_{-}=3\omega }$ and $H=H_{+}+H_{-}$. Every cross-mode position/momentum pair commutes, so $[H_{+},H_{-}]=0$. On the tensor-product oscillator basis,For each $N\ge 0$, allowed pairs have $n_{-}=0,1,\dots ,\lfloor N/3\rfloor$ and $n_{+}=N-3n_{-}$. ThusEach pair contributes one product eigenvector; counting them accounts for the energy-level degeneracy.