Past exam of the mathematics course of the University of Cambridge / 2017 / ib / Paper 3 / 16B (Quantum Mechanics)

The orbital angular momentum operators areIn particular ${\hat{L}}_3=-i\hslash (x_1{\partial }_2-x_2{\partial }_1)$. For smooth compactly supported functions $\varphi ,\psi$, integration by parts produces no boundary term, and ${\partial }_2{x}_1={\partial }_1{x}_2=0$. HenceThis proves Hermiticity on the common test-function domain, and the same calculation applies on suitable decaying domains. As an unbounded operator, it is not defined on every $L^2$ function: the usual self-adjoint operator realization is the generator of unitary rotations about the third axis. Thus the integration statement includes its necessary domain/boundary assumptions.

Take a normalized state for which the indicated variances and operator pairings exist. Set $A={\hat{L}}_3-⟨{\hat{L}}_3⟩$ and $B={\hat{x}}_1-⟨{\hat{x}}_1⟩$. For real $\lambda$, positivity of the squared norm givesThe canonical commutation relations yield $[{\hat{L}}_3,{\hat{x}}_1]=i\hslash {\hat{x}}_2$. Thus the quadratic polynomial in $\lambda$ is nonnegative everywhere only if its discriminant is nonpositive. This gives the Robertson uncertainty principleIf either variance is zero, the same polynomial argument forces the commutator expectation to vanish, so that edge case is included. Centering the operators is essential to obtain variances rather than uncentered second moments.

Use the radial exponential $e^{-\alpha r}$ shown in the original PDF; the TeX's $e^{-\alpha x}$ is a transcription defect. Every orbital angular momentum operator annihilates a radial function, so it acts only on $P=x_1+x_2+2x_3$. The angular momentum of a radial function times a linear polynomial is determined by its degree-one angular factor: direct application of the rotation derivatives givesEquivalently its restriction to the sphere is a degree-one spherical harmonic. ThusFurthermore ${\hat{L}}_3\psi =-i\hslash K(x_1-x_2)e^{-\alpha r}$. Its expected value is proportional to the integral of $(x_1+x_2+2x_3)(x_1-x_2)e^{-2\alpha r}$. The cross terms are odd in a coordinate, and the integrals of $x_1^2$ and $x_2^2$ agree by rotational symmetry. HenceThe state need not be an eigenstate of ${\hat{L}}_3$. If normalization is desired, $\int P^2{e}^{-2\alpha r}{d}^{3}x=6\pi /{\alpha }^5$, so $K={\alpha }^{5/2}/\sqrt{6\pi }$; the zero expectation holds after normalization for any nonzero initial $K$.