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Write ${\gamma }^{\pm }=-i{\beta }^{\pm }$, $\overline{\beta }=({\beta }^{+}+{\beta }^{-})/2$ and $\chi =({\beta }^{+}-{\beta }^{-})z/2$. Then $\gamma z=-i\overline{\beta }z$ and $\delta z=-i\chi$. With $\cosh (-i\chi )=\cos \chi$ and $\sinh (-i\chi )=-i\sin \chi$, the corrected matrix becomesThe scalar is a common propagation wave phase and the real rotation matrix rotates the input linear polarization through $\chi$. There is no attenuation and no generated ellipticity: the field map is a unitary matrix. The two circular eigenmodes accumulate different wave phases, giving circular birefringence and optical activity.

The correct unfactored matrix remains finite where $\cos \chi =0$. In contrast, the printed $\mathrm{sech}$ form would give $e^{i\overline{\beta }z}{\sec }^2\chi$ times the rotation matrix, including unphysical divergent gain in this stipulated lossless case. This provides a direct physical check on the correction in part (d).

Choose the first polarizer's pass axis as ${\mathbf{e}}_1$ and write the second as $\mathbf{a}=({\mathbf{e}}_1+{\mathbf{e}}_2)/\sqrt{2}$. The positive signed rotation takes ${\mathbf{e}}_1$ toward $\mathbf{a}$ by the short $\pi /4$ angle. Let $B_{\parallel }$ be signed according to $\theta ={\rm Y}L{B}_{\parallel }$. The Jones matrices of ideal polarizers are their orthogonal projection matrices,Removing the unimportant common propagation phase, the lossless glass block has the rotation matrixThe elements act from right to left, soWith the common optical phase retained, multiply this by $e^{i\bar{\varphi }}$. Extra scalar attenuation would also multiply the result, but none is supplied. A single Jones vector does not represent unpolarized light; the device's Jones matrix is nevertheless well defined for each incident component and can be applied to their statistical mixture.

Put $v=P_2-P_1$ and $w=Q_2-Q_1$. They are nonzero vectors with the same Euclidean norm. Choose the rotation matrix $R$ taking the direction $v/\Vert v\Vert$ to $w/\Vert w\Vert$; equal lengths then give $Rv=w$. DefineThis is a translation by $Q_1-P_1$ followed by a rotation about $Q_1$, so it is an isometry. Direct substitution gives $t(P_1)=Q_1$ and $t(P_2)=Q_1+w=Q_2$.

Put $v=P_2-P_1$ and $w=Q_2-Q_1$. They are nonzero vectors with the same Euclidean norm. Choose the rotation matrix $R$ taking the direction $v/\Vert v\Vert$ to $w/\Vert w\Vert$; equal lengths then give $Rv=w$. DefineThis is a translation by $Q_1-P_1$ followed by a rotation about $Q_1$, so it is an isometry. Direct substitution gives $t(P_1)=Q_1$ and $t(P_2)=Q_1+w=Q_2$.

A mass eigenstate $|{\nu }_i⟩$ has a definite neutrino mass $m_i$ and acquires a definite propagation phase at fixed momentum. An interaction eigenstate is the neutrino flavor selected by a weak charged current: ${\nu }_e$ is produced or detected together with an electron, and ${\nu }_{\mu }$ together with a muon. The two bases are related by a unitary matrix, here the real rotation matrix $U$.

Inverting the supplied transformation gives $|{\nu }_e⟩=c|{\nu }_1⟩+s|{\nu }_2⟩$ and $|{\nu }_{\mu }⟩=-s|{\nu }_1⟩+c|{\nu }_2⟩$, where $c=\cos \theta$, $s=\sin \theta$. If $\theta =0$, these bases coincide: each produced neutrino flavor propagates as one mass eigenstate, and there is no neutrino oscillation, even if the two masses differ. Conversely, equal masses also prevent vacuum neutrino oscillation, even when a nonzero basis-rotation angle is written down.

Write $\bar{\chi }=({\chi }_1+{\chi }_2)/2$ and $\delta \chi ={\chi }_2-{\chi }_1$. Since the supplied rotation matrix is real and orthogonal, $U^{-1}=U^T$. Direct matrix multiplication gives the flavor-basis HamiltonianThe common term $\bar{\chi }I$ produces only an overall quantum phase. The off-diagonal elements $H_{e\mu }^{\prime }=H_{\mu e}^{\prime }=(\delta \chi /2)\sin 2\theta$ couple the two interaction eigenstates in the Schrödinger equation. Flavor change requires both $\delta \chi \ne 0$ and $\sin 2\theta \ne 0$. For ultrarelativistic neutrinos in vacuum, $\delta \chi \simeq \Delta m^2/(2E)$, so a nonzero squared-mass difference is needed. Observable neutrino oscillation also requires coherent production and propagation of the two components and a resolvable nonzero relative phase; special propagation times can return the initial flavor even when these Hamiltonian conditions hold.

A useful distinction for the next part is that $H_{ee}^{\prime }$ is an energy, not a time-dependent survival probability amplitude. The latter is the $ee$ element of $S(t)=\exp (-i{H}^{\prime }t)$.

The characteristic polynomial of the rotation matrix isIts roots are $\cos U\pm i\sin U$, soFor example $(1,i{)}^T$ is an eigenvector for $e^{iU}$ and $(1,-i{)}^T$ for $e^{-iU}$. At $U=0$ or $\pi$ the two eigenvalues coincide; these events have probability zero under the uniform distribution.

For an active clockwise rotation of column vectors in the usual oriented $xy$ plane, takeDirect matrix multiplication givesThus the rotation matrix is an orthogonal matrix, preserving the Euclidean inner product, with determinant $+1$.

For a passive one-dimensional Lorentz boost, use rapidity $\eta$, $\beta =\tanh \eta$:This is not $I$ unless $\eta =0$. Hence a nontrivial Lorentz boost is not an orthogonal matrix for the Euclidean inner product. Instead it preserves the Minkowski metric:by ${\cosh }^2\eta -{\sinh }^2\eta =1$. Thus it is pseudo-orthogonal. The zero-rapidity identity is the explicit exception to the assertion that a Lorentz boost is not orthogonal.

Translate the vertex centroid to the origin, rotate, and translate back:This follows the supplied rotation matrix, so positive $\theta$ rotates clockwise in the usual Cartesian axes. The opposite signs give the conventional counterclockwise matrix. Since the vertex centroid of the centered vertices is zero and a linear map commutes with their average, the final centroid remains c.

Rotate the specified line onto the horizontal axis, stretch horizontally by two while leaving the orthogonal component unchanged, then rotate back. With $\theta =\pi /3$ and counterclockwise rotation matrix $R(\theta )$, the requested product isThe rightmost rotation acts first. In particular, $R(\theta )=\left(\begin{array}{cc}1/2& -\sqrt{3}/2\\ \sqrt{3}/2& 1/2\end{array}\right)$ and $R(-\theta )=R(\theta {)}^T$. For the unit direction $u=(1/2,\sqrt{3}/2{)}^T$, the same transformation is $T=I+u{u}^T$. It sends $u$ to $2u$ and fixes every orthogonal vector, verifying that the product is a directional stretch rather than a shear mapping or scaling in every direction. The constructed second panel in the geometric figure in Question 5 illustrates this separation.

A planar rotation about the origin is the linear map represented by the rotation matrix $R_{\theta }=\left(\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right)$. It preserves inner products, lengths, oriented angles and area, with positive angles measured anticlockwise. Its inverse is $R_{-\theta }$ and its determinant is one.

Differentiate the planar rotation matrix at angle zero:The even and odd powers in the matrix exponential therefore giveThis exact formula gives the whole one-parameter subgroup. Taking $X=J$ means unit angular velocity: at $t=\pi /2$, $\exp (tJ)=J$ sends $(1,0{)}^T$ to $(0,1{)}^T$, a counterclockwise quarter-turn. At $t=\pi$, it equals $-I$; at $t=2\pi$, it returns to $I$. Choosing $X=2J$ reaches the quarter-turn at $t=\pi /4$. Changing the generator changes the speed, while preserving the circular path.

For a moving vector $v(t)=R_{t}v(0)$, the ordinary differential equation $v^{\prime }(t)=Jv(t)$ says its velocity is perpendicular to its position. The beautiful compression is that one constant matrix specifies every angle of this endless family of rotations.

Under an orthogonal coordinate transformation $R$, a Cartesian pseudotensor has the usual tensor transformation law multiplied by $\det R$. Thus proper rotation matrices act as on ordinary tensors, while an orientation-reversing change contributes an extra minus sign. Interpreted as a rank-three pseudotensor, the Levi-Civita symbol has the same numerical components in every orthonormal frame, because its ordinary three-index transformation already contributes a factor $\det R$.

An axial vector transforms as ${\mathbf{a}}^{\prime }=(\det Q)Q\mathbf{a}$ under an orthogonal matrix $Q$. It agrees with a polar vector under a proper rotation matrix and has the additional determinant sign under an orientation-reversing transformation. Angular momentum and angular velocity are axial vectors.

For every real angle $\theta$,It is the equation of the unit circle written in terms of its sine and cosine coordinates. For an acute angle, dividing the Pythagorean theorem by the square of the hypotenuse gives the same relation in a right triangle. The circle interpretation explains why it remains true outside that first-quadrant triangle.

For example, if $\sin \theta =3/5$, the identity gives ${\cos }^2\theta =16/25$, so $\cos \theta$ is either $4/5$ or $-4/5$. The sign needs the angle's location: it is positive in the first quadrant and negative in the second. Thus the identity determines the magnitude of the missing coordinate without determining the direction.

The relation is useful for eliminating one squared trigonometric function, checking that a coordinate pair has unit length, and checking that the columns of a rotation matrix are unit vectors. It also holds for complex arguments under the complex exponential definitions, although then the squares are algebraic squares rather than squared moduli.

Let $J$ be the plane rotation matrix through $\pi /2$, generating a cyclic group of order four. A quadratic form $a{x}^2+b{y}^2$ has its two coefficients exchanged by a quarter-turn and unchanged by a half-turn. Thus the pullback action on this two-dimensional space gives the displayed group representation: $D(J^k)=D(J{)}^k$, with $D(J{)}^2=I$. It is not a faithful representation, because its kernel of a group homomorphism is $\{I,J^2\}$. Its two invariant coefficient lines correspond to $x^2+y^2$ and $x^2-y^2$, on which a quarter-turn acts by $+1$ and $-1$, respectively. Restricting to this diagonal space is valid for these signed coordinate permutations; a general linear transformation can introduce an $xy$ term.

A real rotation matrix is a unitary matrix because it is an orthogonal matrix. It is a Hermitian matrix precisely when it is symmetric, equivalently when its square is the identity. Its eigenvalues are then $\pm 1$. In three dimensions a proper rotation is therefore Hermitian only for the identity or a half-turn. A half-turn about a unit vector $n$ has matrix $2n{n}^T-I$: it fixes $n$ and reverses the perpendicular plane. For every real rotation $R$, the difference $R-R^T$ is anti-Hermitian; it is also Hermitian only in the zero case.

For a constant orthogonal matrix $R$, the Laplacian is unchanged by the coordinate transformation $\xi =Rx$. If $u$ is twice continuously differentiable, the chain rule givesbecause ${\sum }_i{R}_{ji}{R}_{ki}={\delta }_{jk}$ by orthogonality of the rows. Here ${\delta }_{jk}$ is the Kronecker delta. In two dimensions, the rotation matrix $\left(\begin{array}{cc}a& -b\\ b& a\end{array}\right)$ with $a^2+b^2=1$ makes the mixed partial derivatives cancel explicitly. The same argument works for reflections and translations, so the Laplacian depends on the Euclidean geometry rather than the orientation or origin of the coordinates.

Under rotation by a rotation matrix, a symmetric second-rank tensor transforms by conjugation, so its traces of powers are invariant. For a three-dimensional traceless second-rank tensor, the Cayley-Hamilton theorem implies $\mathrm{Tr}{Q}^4=(\mathrm{Tr}{Q}^2{)}^2/2$. In two dimensions its eigenvalues are $s,-s$, so $\mathrm{Tr}{Q}^3=0$. These identities organize the Landau-de Gennes free energy.

An object has rotational symmetry when a nontrivial rotation matrix leaves it unchanged.