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For a momentum eigenstate with eigenvalue $p$, the equation $-i\hslash {\psi }^{\prime }=p\psi$ has solution $\psi (x)=C{e}^{ipx/\hslash }$. The periodic boundary condition imposes $e^{2\pi ip/\hslash }=1$, so $p/\hslash \in \mathbb{Z}$. The wavefunction normalization gives $1=2\pi |C{|}^2$. Choosing a convenient global phase yieldsThese momentum eigenstates form an orthonormal basis, the Fourier basis for the particle on a circle:Here ${\delta }_{nk}$ is the Kronecker delta.

The identity ${\cos }^{2}x=\frac{1}{2}+\frac{1}{4}{e}^{2ix}+\frac{1}{4}{e}^{-2ix}$ gives the momentum eigenstate expansionUsing the orthonormal basis from part (a), the wavefunction normalization isHence $|A|=2/\sqrt{3\pi }$. The wavefunction normalization leaves a global phase free; the positive real choice used in the requested expansion givesThe two nonzero eigenvalues of the momentum operator have the same energy eigenvalue $E_{\star }=2{\hslash }^2/m$. Under unitary time evolution, their coefficients acquire the same dynamical phase. With $\Omega =E_{\star }/\hslash =2\hslash /m$,At $t=0$, $1+\cos (2x)=2{\cos }^{2}x$ recovers the initial wavefunction. Its probability density, useful below, is

The coefficient of the momentum eigenstate ${\psi }_2$ is $e^{-i\Omega t}/\sqrt{6}$. The Born rule therefore gives $\mathrm{Pr}(P=2\hslash )=1/6$. Conditional on this outcome, the projective measurement prepares the post-measurement state $e^{-i\Omega t}{\psi }_2$, equivalent to ${\psi }_2$ up to global phase.

Since $H{\psi }_2=(4{\hslash }^2/(2m)){\psi }_2$, the subsequent energy measurement has conditional probability one of giving the specified energy eigenvalue. The probability for the entire sequence is consequentlyAny intervening unitary time evolution under the same Hamiltonian operator preserves this energy eigenstate, up to global phase.

Using ${\sin }^{2}u=\frac{1}{2}-\frac{1}{4}{e}^{2iu}-\frac{1}{4}{e}^{-2iu}$, the energy eigenstate decomposition isOnly the energy eigenvalues zero and $E_{\star }=2{\hslash }^2/(m{R}^2)$ occur. The last two terms are distinct momentum eigenstates in the same energy eigenspace.

Let ${\Pi }_{\star }=|{\varphi }_2⟩⟨{\varphi }_2|+|{\varphi }_{-2}⟩⟨{\varphi }_{-2}|$ be the orthogonal projection for $E_{\star }$. The Lüders rule gives the normalized post-measurement stateThe factor $-e^{-i{E}_{\star }T/\hslash }$ is a global phase, so $\cos (2x/R)/\sqrt{\pi R}$ is an equivalent representative. The projective measurement preserves the quantum coherence between the two momentum eigenstates; it does not select one of them. This matters for the subsequent position probabilities.

A plane wave has constant amplitude and a phase linear in space and time. It is a momentum eigenstate and is not spatially normalizable.

A quantum time-reversal operator is an antiunitary operator reversing spatial momentum and angular momentum. For a time-reversal-invariant Hamiltonian operator, its conjugation of $i$ makes it intertwine forward and backward unitary time evolution without reversing the energy spectrum. A scalar momentum eigenstate may transform as $\hat{T}|\mathbf{p}⟩=e^{i\chi (\mathbf{p})}|-\mathbf{p}⟩$. With unit-modulus phases and a complete normalized basis, expansion of arbitrary states proves antiunitarity.

For a free particle on the line or a circle, $H_0=P^2/(2m)$ commutes with the momentum operator $P$. The commuting operator exponentials in the title therefore factorize. Since $P=-i\hslash {\partial }_x$, the factor $e^{isPt/\hslash }=e^{st{\partial }_x}$ translates a wavefunction to $f(x+st)$, as is verified on each momentum eigenstate. Thus the probability density translates with velocity $-s$ relative to the $H_0$ evolution. A real constant $C$ changes only the global phase. In particular a superposition in one degenerate energy eigenspace translates rigidly under the added term $-sP$.

For a quantum field with four-momentum operator $P^{\mu }$, spacetime dependence is implemented by the displayed conjugation. Taking a matrix element between momentum eigenstates gives $⟨q|\Phi (x)|p⟩=e^{i(q-p)\cdot x}⟨q|\Phi (0)|p⟩$. Differentiating gives $i{\partial }_t\Phi =[\Phi ,H]$ and $-i\nabla \Phi =[\Phi ,\mathbf{P}]$, connecting the Heisenberg equation of motion to spatial translation covariance.