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

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.