Past exam of the quantum theory course of the University of Oxford / 2014 / 3

The commutator of two linear operators is $\boxed{[A,B]=AB-BA}$. Write $T^{†}$ for the adjoint operator, using the inner product convention linear in its second argument. Since $A^{†}=A$ and $B^{†}=B$,Thus the commutator is anti-self-adjoint. With $C=-i[A,B]$, conjugation of the scalar givesAll products are understood on the stated common operator domain; for physical unbounded linear operators, the state must lie in the relevant product domains and have the finite moments used below.

A normalized quantum state satisfies $⟨\psi ,\psi ⟩=1$. Its expectation value is $\boxed{{\mathbb{E}}_{\psi }(A)=⟨\psi ,A\psi ⟩}$. The conjugate symmetry of the inner product and self-adjointness giveso the expectation value is real. If $A\varphi =\lambda \varphi$ with $\varphi \ne 0$, normalize $\varphi$ by dividing by its norm. ThenEvery eigenvalue of a self-adjoint operator is real.

Let $a={\mathbb{E}}_{\psi }(A)$, which is real. The dispersion, or quantum uncertainty, is the nonnegative standard deviationIndeed, self-adjointness and normalization giveA squared norm is nonnegative, so the square root is defined. It vanishes precisely when $(A-aI)\psi =0$, namely when $\psi$ is an eigenvector of $A$. Conversely, an eigenvector with eigenvalue $\lambda$ has $a=\lambda$ and hence zero dispersion. Thus zero dispersion is equivalent to being an eigenstate.

Set $u=A\psi$, $v=B\psi$ and $z=⟨u,v⟩$. The self-adjointness of $A,B$ impliesThe Cauchy-Schwarz inequality followed by $|z{|}^2\ge (\mathrm{Im}z{)}^2$ givesBoth inequalities must be equalities. The exact conditions are $A\psi$ and $B\psi$ are linearly dependent, and $\mathrm{Re}⟨A\psi ,B\psi ⟩=0$. Equivalently, equality holds if either vector is zero; when both are nonzero it holds precisely whenThe zero-vector cases matter: they saturate the inequality but cannot all be expressed by solving for $B\psi$ as a multiple of $A\psi$. The real-part condition can also be written ${\mathbb{E}}_{\psi }(AB+BA)=0$, using the anticommutator. This is the uncentered second-moment commutator inequality.

Centre the position operator and momentum operator by their real expectation values:Scalar multiples of the identity operator commute with every linear operator, so the canonical commutation relation givesThe second moments of the centred linear operators are the variances, ${\mathbb{E}}_{\psi }(A^2)={\Delta }_{\psi }(X{)}^2$ and ${\mathbb{E}}_{\psi }(B^2)={\Delta }_{\psi }(P{)}^2$. Applying the preceding inequality and taking nonnegative square roots proves the Heisenberg uncertainty relation:Subtracting the means is essential: the uncentred second moments alone do not equal the variances. The equality case of the Heisenberg uncertainty relation requires the centred vectors to be imaginary multiples, as in the preceding equality conditions.