Expand $A(r)=A(r_0)+A^{\prime }(r_0)s+O(s^2)$ about $qA(r_0)=p_z$. Then $p_z-qA\simeq -q{A}^{\prime }(r_0)s$ and conservation of energy givesHere $\Omega =q{A}^{\prime }(r_0)/m$ can be signed, but the physical oscillation frequency and amplitude are nonnegative. The small-energy condition ensures that this local expansion samples only a narrow range about $r_0$, provided the field varies smoothly there.

The initial condition makes $p_{\theta }=0$ at all times. Accessible radii obey $|p_z-qA(r)|\le \sqrt{2mH}$. For $q>0$ and $p_z>\sqrt{2mH}$, monotonicity of $A$ gives exactly the two stated turning equations and confinement to the interval between their unique roots.

The squared condition alone omits a sign qualification. More generally both positive roots require $p_z/q>0$, and are the sorted roots $A(r)=(p_z\pm \sqrt{2mH})/q$. For example $q=-1$, $A=r$, $p_z=-3$, $2mH=1$ confines the orbit to $2\le r\le 4$, but the source's named $r_1,r_2$ equations reverse those labels. This is a convention issue, not a failure of confinement.

The Legendre transform cancels the velocity-linear magnetic term. Expressing velocities in terms of momenta givesThe Hamiltonian is time independent and hence conserved. It equals kinetic energy because a static magnetic field does no work; its dependence on $A$ is through the mechanical momentum combination.

Velocity differentiation gives $\boxed{p_r=m\dot{r},p_{\theta }=m{r}^2\dot{\theta },p_z=m\dot{z}+qA(r)}$. The coordinates $\theta$ and $z$ are cyclic, so $p_{\theta },p_z$ are conserved. The radial momentum generally changes. In particular the conserved axial canonical momentum differs from the axial mechanical momentum by $qA$.

In cylindrical coordinates the orthogonal velocity components are $\dot{r}$, $r\dot{\theta }$, $\dot{z}$. Their squared magnitudes add, and $\dot{\mathbf{r}}\cdot \mathbf{A}=\dot{z}A(r)$. Thus $\boxed{\mathcal{L}=m({\dot{r}}^2+r^2{\dot{\theta }}^2+{\dot{z}}^2)/2+q\dot{z}A(r)}$. Taking the curl gives $B_{\theta }=-A^{\prime }(r)$, consistent with the prescribed field.

A quarter-wave plate needs a specified axis angle. If its fast axis makes angle $\eta$ to horizontal, choose its Jones matrix $Q=R(\eta )\mathrm{diag}(1,i)R(-\eta )$. ThenAfter the preceding successful lower-path detection the total state is $(aH+bV)|{\downarrow }_1⟩$. The next crystal sends it toFor axes at $45^{\circ }$, the state before splitting is circular up to phase, $a=(1-i)/2$, $b=(1+i)/2$, and the probability is $1/2$. Detection in ${\downarrow }_2$ always leaves vertical polarization. If no earlier lower-path measurement was actually performed, retain the untouched upper amplitude: the final state is $(|H,{\uparrow }_1⟩+a|H,{\uparrow }_2⟩+b|V,{\downarrow }_2⟩)/\sqrt{2}$, and the last probability is $|b{|}^2/2$, or $1/4$ at $45^{\circ }$. At axis angle zero the conditional probability is instead one, a concrete demonstration that an unspecified orientation does not determine a unique number.

Orthogonal path projectors give an upper-path probability $1/2$. A successful non-destructive lower-path detection projects onto $|V⟩|{\downarrow }_1⟩$ and its conditional polarization is therefore pure vertical. Retaining the detection record removes the upper amplitude for that run; it would be incorrect to continue evolving the original coherent two-path state as though the detection had not occurred.

Choose the upper route for horizontal polarization, the lower route for vertical polarization, and absorb fixed optical phases into the path kets. The normalized output isThis is entanglement between two degrees of freedom of one photon. Interchanging the path convention interchanges the labels but not any probabilities.

For a fixed propagation direction and spatial mode, a photon has two independent transverse polarization amplitudes. Orthogonal horizontal and vertical states span its two-dimensional Hilbert space: $\alpha |H⟩+\beta |V⟩$, with $|\alpha {|}^2+|\beta {|}^2=1$. Polarization analyzers measure bases in this space, and phase plates implement transformations of its relative amplitudes. Additional spatial or frequency modes enlarge the space and must not be silently discarded.

A sudden wall displacement leaves the initial wavefunction unchanged: it is $\sqrt{2/L}\sin (\pi x/L)$ on $[0,L]$ and zero on $(L,2L]$. It is not an eigenfunction of the new Hamiltonian, so there is no single certain new energy. Its energy expectation immediately afterward remains $E_1^{\mathrm{old}}$, as seen from ${\hslash }^2\int |{\psi }^{\prime }{|}^2/(2m)$; the measurement outcomes are $E_n^{\mathrm{new}}=n^2{E}_1^{\mathrm{old}}/4$.

The new ground state is $L^{-1/2}\sin (\pi x/(2L))$. Its overlap is $4\sqrt{2}/(3\pi )$, soThe discontinuity of the derivative at the former wall does not invalidate the finite quadratic-form expectation, although it matters when applying the differential Hamiltonian pointwise. Adiabatic expansion would instead keep the state in the ground level and reduce its energy by four.

Inside the well, ${\psi }^{\prime \prime }+k^2\psi =0$. Vanishing at zero removes the cosine term; vanishing at $L$ requires $k_{n}L=n\pi$. The nonzero states have positive integer $n$ because $n=0$ gives the zero function and negative $n$ only repeats a state up to sign. Normalization follows from ${\int }_0^L{\sin }^2(n\pi x/L)dx=L/2$. ThusThe wavefunction vanishes outside the infinite barriers, so there is no additional contribution to its norm.

For a stationary nonrelativistic state, $\boxed{[-{\hslash }^2{\nabla }^2/(2m)+V(\mathbf{r})]\psi =E\psi }$. The wavefunction $\psi$ is a probability amplitude, $m$ the particle mass and $E$ its energy eigenvalue; the potential energy $V$ is measured in the same energy units. In one dimension the Laplacian becomes $d^2/d{x}^2$.

Separate center-of-mass motion and replace $m$ by the reduced mass $\mu =mM/(m+M)$ in the relative-coordinate formulas. Then $r_n$ is the proton-electron separation, energy levels and the Rydberg constant scale by $\mu /m$, and transition wavelengths increase by $m/\mu$. The electron's own orbital radius about the center of mass is $M{r}_n/(M+m)$; the proton radius is $m{r}_n/(M+m)$. The relative speed retains $\alpha c/n$, while the electron's individual speed acquires $M/(M+m)$. Replacing the mass in the radius alone but leaving energies unchanged would be inconsistent.

The emitted photon has energy $E_3-E_2=\mathrm{Ry}(1/4-1/9)=5\mathrm{Ry}/36$. ThereforeThis is the Balmer-alpha transition in this approximation. Absorption in the reverse direction has the same photon energy; fine structure and radiative corrections are beyond the model.

Substitute the quantized radius in $mvr=n\hslash$ to get $v_n=e^2/(4\pi {ϵ}_0\hslash n)$. Coulomb potential energy is $-e^2/(4\pi {ϵ}_0{r}_n)$ and the stipulated virial relation makes the total one half of this value. HenceThe negative sign means a bound state relative to a separated stationary proton and electron.

The Bohr model assumes Coulomb attraction and classical circular motion between transitions; selected stationary orbits do not radiate; their angular momentum is $mvr=n\hslash$; and a transition exchanges a photon of energy $h\nu =E_i-E_f$. The first two dynamical equations giveHere $n=1,2,\dots$, $m$ is the electron mass under the fixed-proton approximation, $e$ its charge magnitude, and ${ϵ}_0$ the vacuum permittivity. The stationary-orbit postulate deliberately suspends the classical radiation prediction; this is a historical model, not a consistent deduction from classical electrodynamics.

Homogeneity makes the inertial transformation linear. Align origins at $t=t^{\prime }=0$, let the second origin have velocity $v$, and impose isotropy and reciprocity. Then $x^{\prime }=\gamma (v)(x-vt)$, $y^{\prime }=y$, $z^{\prime }=z$. If the same limiting signal speed $c$ has trajectories $x=\pm ct$ in both systems, substituting these two trajectories fixes $t^{\prime }=\gamma (v)(t-vx/c^2)$. Requiring the inverse to have the same form with $-v$ fixes ${\gamma }^2(1-v^2/c^2)=1$. Continuity at rest selects the positive square root:Light enters this derivation through the empirical/in-principle operational identification of the invariant speed and Einstein synchronization. Group, homogeneity and isotropy arguments can instead derive a family with a universal parameter $K$, $t^{\prime }=\gamma (t-Kvx)$ and $\gamma =(1-K{v}^2{)}^{-1/2}$. They do not by themselves determine whether $K=0$ or a positive finite value. Empirical light propagation or other relativistic dynamics identifies $K=1/c^2$ for our world. Consequently a derivation need not assume light as its first premise, but it must acquire the physical invariant-speed structure somewhere rather than extract it from linearity alone.^[ref]

A coordinate transformation considered solely as a relabeling of events has no independent observable outcome: changing labels changes neither a detector record nor a clock comparison. But the Lorentz transformations also express substantive relations between operationally realized inertial coordinates. With clocks and rulers specified, their velocity-dependent time dilation, length relations and velocity composition make testable claims about how material systems and electromagnetic signals behave.

The distinction matters for synchronization. One-way coordinate light speed depends on how distant clocks are synchronized; a conventional alteration changes the displayed transformation law while leaving two-way timing observations unchanged. It would therefore be incorrect to regard every coefficient in an unspecified coordinate formula as a separately measurable physical fact. Testable content comes from the transformation together with the protocol and dynamical behavior of the systems implementing the coordinates.

Tests can compare clock proper times, decay lifetimes or resonance behavior across relative motion and orientation. Agreement supports Lorentz-invariant kinematics and the associated matter dynamics over the range tested; it does not prove exact invariance at every possible scale. A rival theory with an undetectable preferred frame can reproduce the same observations by changing its explanatory commitments. Empirical equivalence of that rival does not make the observable clock relations untestable.

Thus the answer is qualified: transformations as passive mathematical changes are not experiments, while Lorentzian relations among specified measuring procedures have substantial empirical consequences. Separating these roles avoids both treating coordinate convention as a measured force and declaring the entire relativistic structure immune to evidence.

If economic activity is adapted to the old climate, additional warming generally pushes temperatures, water availability and extremes further from that adapted range. With convex aggregate damages, marginal annual damage per degree $D^{\prime }$ therefore tends to rise as the world warms. This is an argued expectation under the stated adaptation baseline, not a theorem: adaptation, wealth changes, regional benefits and thresholds alter both the level and curvature of damages. A constant $D^{\prime }$ in the preceding integration is a simplifying local approximation, and may understate later losses when convexity dominates.

For constant marginal annual damage $D^{\prime }$ and positive discount rate $r$, use the Laplace transform of the preceding equations:The approximation uses $r\ll b_1$ but does not require choosing an order between $r$ and $k$. At zero discount rate the integral diverges in this model because the pulse leaves permanent positive warming and damages remain constant per degree.