Past exam of the physics A course of the University of Cambridge / 2012 / Paper 1 / Section B

The canonical commutation relations are $[x,p_x]=[y,p_y]=i\hslash$, with $[x,y]=[p_x,p_y]=[x,p_y]=[y,p_x]=0$. Operators from different Cartesian modes commute, so $(p_x{p}_y{)}^{†}=p_y{p}_x=p_x{p}_y$ and $(xy{)}^{†}=yx=xy$. Thus $J_1$ is Hermitian and represents an observable on the usual oscillator domain.

Introduce annihilation operators $a_x=\sqrt{m\omega /(2\hslash )}x+i{p}_x/\sqrt{2m\hslash \omega }$ and similarly $a_y$. Set $N_x=a_x^{†}{a}_x$, $N_y=a_y^{†}{a}_y$ and $N=N_x+N_y$. The Schwinger boson representation givesFor $A=a_x^{†}{a}_y$ and $B=a_y^{†}{a}_x$, $AB=N_x(N_y+1)$ and $BA=N_y(N_x+1)$. ThereforeThe two-mode oscillator Casimir identity isSince $[N,A]=[N,B]=0$, $[H,J_1]=0$, and $J_1$ is conserved. However, $[J_1,J_2]=i\hslash J_3$ is not the zero operator, so there is no general joint eigenbasis. The exception is important: a common eigenstate of two angular-momentum components must satisfy $J_3\psi =0$, then the other commutators force $J_1\psi =J_2\psi =0$. Thus only the $j=0$ sector admits a common state; here that is the oscillator ground state.

Using $J^2={\hslash }^{2}j(j+1)$, with $j=0,1/2,1,\dots$, and selecting the positive energy branch givesEquivalently $N=2j=0,1,\dots$, so $E_N=\hslash \omega (N+1)$ and the degeneracy is $N+1$, corresponding to $N_x=0,\dots ,N$.

For observables with finite variances, the Heisenberg uncertainty principle gives $\Delta A\Delta B\ge |⟨[A,B]⟩|/2$; in particular $\Delta x\Delta p\ge \hslash /2$. It concerns the spreads of measurements in a prepared state, rather than only disturbance by a measuring device. Identical integer-spin particles have symmetric full states and Bose-Einstein statistics; half-integer-spin particles have antisymmetric full states and Fermi-Dirac statistics. The latter imply Pauli exclusion from one complete single-particle state, including spin.

Inside an infinite square well, solutions satisfying both zero boundary conditions arewith zero wavefunction outside. For the ground state, direct integration givesThe infinite-well ground-state uncertainty product follows:The ground state is not a minimum-uncertainty Gaussian. Here momentum variance is the usual kinetic-momentum second moment of the confined state; the hard-wall momentum operator has domain subtleties that do not alter this kinetic-energy result.

For spin degeneracy in filling a one-dimensional well, let $E_1={\pi }^2{\hslash }^2/(2m{L}^2)$ and assume the spin levels are degenerate and available. Four spin-zero bosons all occupy the lowest orbital, giving $4E_1$. Four spin-$1/2$ fermions fill two spin states in each of $n=1,2$, giving $10E_1$. Spin-$3/2$ fermions have four spin projections, so all four can occupy $n=1$ in different complete states, giving $4E_1$. These fillings do not impose full spin polarization.

matrix multiplication gives ${\sigma }_i{\sigma }_j={\delta }_{ij}I+i{ϵ}_{ijk}{\sigma }_k$, hence $[{\sigma }_i,{\sigma }_j]=2i{ϵ}_{ijk}{\sigma }_k$. For physical neutron spin, use $\mathbf{S}=(\hslash /2)\mathbf{\sigma }$, so $[S_i,S_j]=i\hslash {ϵ}_{ijk}{S}_k$, $S^2=3{\hslash }^2/4$, and measurement along any axis gives $\pm \hslash /2$.

For the specified field, $H=(\alpha B\hslash /2){\sigma }_y$. The Time-dependent Schrödinger equation is $i\hslash \dot{\psi }=H\psi$ with $\psi (0)=(1,0{)}^T$. The uniform-field spin precession follows from ${\sigma }_y^2=I$:

The sign of the precession is contained in the given $\alpha$. The normalized $x$-axis eigenstates are $(1,1{)}^T/\sqrt{2}$ and $(1,-1{)}^T/\sqrt{2}$.

The measurement direction is $\mathbf{n}=(0,\sin \theta ,\cos \theta )$, as confirmed by the original diagram. Write $c=\cos (\theta /2)$ and $s=\sin (\theta /2)$. The spinors ${\chi }_{+}=(c,is{)}^T$ and ${\chi }_{-}=(is,c{)}^T$ have unit norm and inner product $ics-isc=0$. Applying $\mathbf{n}\cdot \mathbf{\sigma }$ gives eigenvalues $+1$ and $-1$. At $\theta =0$ they are the usual $z$ basis; at $\pi /2$ they are the $y$ basis up to a phase; at $\pi$ the $z$ up/down roles are exchanged, with a factor $i$; and at $2\pi$ both return with a minus sign. A spinor changes sign under a full rotation, while its isolated-state measurement probabilities are unchanged.

For spin measurement in the yz plane, projection givesFor $\theta =0$ these are ${\cos }^2\varphi ,{\sin }^2\varphi$; for $\pi /2$ both are $1/2$ at all times; for $\pi$ they are exchanged; and for $2\pi$ they are the same as at zero angle. The constant $y$ probabilities also follow from conservation of $S_y$.

For the two inline operator labels, (a) $L_z$ has eigenvalue $m_{\ell }\hslash$, and (b) $L^2$ has eigenvalue $\ell (\ell +1){\hslash }^2$, with $\ell =0,1,\dots$ and $m_{\ell }=-\ell ,\dots ,\ell$. The subscript distinguishes the magnetic quantum number from the electron mass.

Put $b=a_0(\ell +1)$ and $\kappa =e^2/(4\pi {ϵ}_0)$. For the proposed radial shape,The angular-momentum term cancels the inverse-square part. The inverse-radius coefficient vanishes if ${\hslash }^2(\ell +1)/(mb)=\kappa$. ThereforeThis nodeless Coulomb radial eigenstate has principal quantum number $n=\ell +1$, not the complete set of hydrogen radial states. Since the spherical harmonic is normalized, radial normalization requiresup to an arbitrary phase.

For the hydrogen ground-state nuclear-volume probability, use $R=2a_0^{-3/2}{e}^{-r/a_0}$. With $u=r_p/a_0\ll 1$,No numerical proton radius is supplied or required. A sharp proton boundary is an approximation to a charge distribution, and the point-Coulomb potential changes inside an extended proton. Finite proton mass, relativistic and electromagnetic corrections also limit the ideal wavefunction's precision.

Classically the kinetic energy must satisfy $E-V(r)\ge 0$. For $E=-\kappa /(2a_0)$ this demands $r\le 2a_0$. The hydrogen ground-state probability beyond the classical turning point isA substantial probability lies beyond the classical turning radius; the bound-state wavefunction decays there rather than vanishing.