Past exam of the quantum mechanics quantum theory and quantum computers course of the University of Oxford / 2007 / A

Differentiate the Lagrangian with respect to the generalized velocities:The Legendre transform cancels the vector-potential energy term, giving $H=p\cdot \dot{x}-L=m|\dot{x}{|}^2/2+V$. Eliminating velocity with $m\dot{x}=p-eA$ therefore yieldsFor the given gauge, set ${\pi }_1=p_1+eB{x}_2/2$, ${\pi }_2=p_2-eB{x}_1/2$, ${\pi }_3=p_3$. Hamilton's equations give ${\dot{x}}_j={\pi }_j/m$, ${\dot{p}}_1=(eB/2){\dot{x}}_2$, ${\dot{p}}_2=-(eB/2){\dot{x}}_1$, ${\dot{p}}_3=0$. Hence the conserved quantities areThe two transverse combinations are components of magnetic pseudomomentum. The printed conserved combinations omit the electric charge factor $e$. With the stated vector potential and general electric charge they need the factors displayed above; as printed they hold only if $e=1$ or if electric charge is already absorbed into the field convention.

The initial data give $K_1=K_2=p_3=0$. Thus $p_1=eB{x}_2/2$, $p_2=-eB{x}_1/2$ throughout, andSolving with the initial point givesConsequently $x_1^2+x_2^2=a^2$: the guiding centre is the origin and the signed cyclotron frequency fixes the rotation direction. The energy is the constant $e^2{B}^2{a}^2/(2m)$.

Inside the box, the time-independent Schrödinger equation is $-{\hslash }^2{\Psi }^{\prime \prime }/(2m)=E\Psi$. The zero endpoint values eliminate the cosine term and require $ka=n\pi$. There is no nonzero zero-energy solution with both endpoints zero. Normalization uses ${\int }_0^a{\sin }^2(n\pi x/a)dx=a/2$, soInsert the expansion into $i\hslash {\partial }_t\psi =H\psi$ and take its inner product with ${\Psi }_m$. Orthonormality gives $i\hslash {\dot{c}}_m=E_m{c}_m$ for each $m$, and conversely these equations make every expansion coefficient of the residual zero. Thus $c_n(t)=c_n(0)e^{-i{E}_{n}t/\hslash }$ in the usual Hilbert-space or strong-solution class appropriate to the expansion.

For the specified state,The modes are real, so the interference term is a cosine of the energy difference. Writing $s_n=\sin (n\pi x/a)$,Its integral remains one because the cross term integrates to zero. The probabilities of the two energy outcomes are fixed at $9/25$ and $16/25$, even though the position probability density oscillates.

For a decaying bound state take $\alpha >0$. Substituting $F=f{e}^{-\alpha r}$ gives $F^{\prime \prime }=e^{-\alpha r}(f^{\prime \prime }-2\alpha f^{\prime }+{\alpha }^{2}f)$. The term involving the energy cancels ${\alpha }^{2}f$, leaving the displayed equation withAt the lowest power of a series from the Frobenius method $f={\sum }_{k\ge 0}{a}_k{r}^{k+c}$, the indicial equation is $c(c-1)-\ell (\ell +1)=0$. Its roots are $\ell +1$ and $-\ell$. The latter makes $F\sim r^{-\ell }$ and the radial probability density integral $\int |F{|}^{2}dr$ diverge for positive integral $\ell$. Hence $\boxed{c=\ell +1}$.

Equating the remaining coefficients givesUnless the numerator vanishes, the regular confluent-hypergeometric series contains a large-radius growing branch with the exponential factor $e^{2\alpha r}$. Multiplication by the external $e^{-\alpha r}$ then leaves exponential growth, so it cannot be a normalizable bound state. Equivalently, the recurrence at large $k$ has $a_{k+1}/a_k\sim 2\alpha /k$, indicating that growing exponential scale; regularity at the origin has already fixed the combination of solutions, leaving termination as the way to remove it.

If the polynomial ends at degree index $N\ge 0$, its last recurrence numerator gives $A=2\alpha (N+\ell +1)$. With $n=N+\ell +1>\ell$,For the separate angle-independent exponential, normalization includes the full solid angle:Thus one can choose $\boxed{N=\sqrt{{\alpha }^3/\pi }}$, up to an overall phase. Angle independence corresponds to the $\ell =0$ extension, distinct from the positive-$\ell$ family used in the indicial discussion.

A pure state is a unit vector in a complex Hilbert space, with vectors differing by a global phase representing the same physical state. An observable is a self-adjoint operator, with measurement probabilities determined by its spectral projections. For vectors in the relevant operator domains,A squared norm vanishes exactly when its vector vanishes. Therefore zero dispersion is equivalent to $A\psi =E_{\psi }[A]\psi$.

In the position representation, $(X\psi )(x)=x\psi (x)$ and $(P\psi )(x)=-i\hslash {\psi }^{\prime }(x)$. On a common differentiable invariant domain,so $\boxed{[X,P]=i\hslash I}$.

For two observables set $a=\Vert A\psi {\Vert }^2$, $b=\Vert B\psi {\Vert }^2$, $c=E_{\psi }[i[A,B]]$. For real $t$,If $b>0$, nonpositivity of its discriminant yields $\boxed{4ab\ge c^2}$. Equality holds exactly when the minimum squared norm vanishes, namelyThe degenerate case must also be included: if $B\psi =0$, then $b=c=0$ and equality holds for every $A\psi$. Thus the complete criterion is that $B\psi =0$, or that the displayed real-$t$ relation holds. This includes $A\psi =0$ by choosing $t=0$ when $b>0$.

Apply the inequality to centred operators $A=P-E_{\psi }[P]I$, $B=X-E_{\psi }[X]I$. Their squared norms are the variances and their commutator is $[P,X]=-i\hslash I$, givingAn eigenvector of either operator would have its corresponding dispersion zero, contradicting this positive lower bound if the other dispersion were finite and the stated domain conditions held. In the usual $L^2(\mathbb{R})$ representation there are in fact no normalizable exact position or momentum eigenvectors; their generalized eigenvectors are not counterexamples. The equality condition above is the uncentered second-moment commutator inequality.