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The low-energy localized-well basis gives the tight-binding modelA Bloch state with coefficients $e^{ipna}$ is an eigenstate because the two neighbouring coefficients add to $2\cos (pa)e^{ipna}$. ThusHere $p$ is a Bloch wavenumber; physical momentum is $\hslash p$, and writing $p$ for physical momentum would require $pa/\hslash$. Spectral evolution of these eigenstates is exactly the Fourier integral in the preceding solution. The negative hopping lowers the symmetric $p=0$ state, and coherent quantum tunnelling spreads a formerly degenerate family of localized levels into a band of width $4\Delta ϵ$. Using the printed positive Euclidean prefactor instead would give on-site energy $-\hslash \omega /2$, showing its contradiction directly.

A homogeneous static saddle minimizes $f(\rho )=-\mu \rho +g{\rho }^2/2$ for $\rho =|\psi {|}^2\ge 0$. With $g>0$ and $\mu >0$,The mean-field Bose-Einstein condensate has a macroscopic coherent occupation of the uniform mode. Its continuous particle-number phase symmetry is $\psi \mapsto e^{i\chi }\psi$, the circle group $U(1)$ generated by particle number. Choosing one phase breaks it, and the saddle manifold is a circle. For $\mu \le 0$, the constrained minimum is the vacuum ${\rho }_0=0$, so the condensed saddle requires the stated positive-density regime.

Spontaneous symmetry breaking is understood through a phase-selected thermodynamic description: an exact finite-volume number eigenstate has vanishing field expectation. The resulting Goldstone boson is a gapless phase/sound mode, while phase gradients carry the superfluid velocity $\mathbf{v}=\nabla \varphi /m$. This is the saddle-point conclusion requested, not a proof of true condensate order in every dimension or temperature. Long-wavelength phase fluctuations can invalidate the assumed order, as quantified below.

For a normalized state and a self-adjoint operator $Q$ with finite second moment, the quantum uncertainty is defined by $\Delta Q=\Vert (Q-⟨Q⟩)\psi \Vert$. Since $⟨Q⟩$ is real,In an energy eigenstate of the quantum harmonic oscillator,Discarding the nonnegative squared means proves the stated lower bound. The Heisenberg uncertainty principle is $\Delta x\Delta p\ge \hslash /2$, following from $[x,p]=i\hslash$. Applying the arithmetic-geometric mean inequality to the two variance contributions givesHere $\omega >0$ is the oscillator frequency. The bound is the ground-state energy; equality requires zero means and the appropriate minimum-uncertainty width.

Since $U$ is a unitary operator, its eigenstates form an orthonormal basis. Expand $|\xi ⟩={\sum }_j{c}_j|v_j⟩$, with eigenphases ${\varphi }_j=y_j/2^n$. By linearity, the unmeasured quantum phase estimation output isThis is generally an entangled state, not a phase label attached to an unchanged pure system state. The distinct eigenvalues give distinct phase labels, so a computational basis measurement yields $y_j$ with Born rule probability $|c_j{|}^2$ and leaves $|v_j⟩$ in the system register. Before measurement, all relative phases remain coherent; that is essential for the following spectral transformation. If phases were degenerate, a measured label would instead select the corresponding eigenspace component.

The interaction picture removes the evolution generated by $H_0$: $|{\psi }_I(t)⟩=e^{i{H}_{0}t/\hslash }|{\psi }_S(t)⟩$. Differentiating the Schrödinger equation gives $i\hslash {\partial }_t|{\psi }_I⟩=\lambda V_I(t)|{\psi }_I⟩$, with $V_I=e^{i{H}_{0}t/\hslash }V(t)e^{-i{H}_{0}t/\hslash }$. The integral equation, iterated once, givesDistinct-energy eigenstates are orthogonal. Their transition amplitude is therefore $-i\lambda {\hslash }^{-1}{\int }_0^t⟨\varphi |V(s)|\chi ⟩e^{i(E^{\prime }-E)s/\hslash }ds+O({\lambda }^2)$. Squaring its modulus givesThe first-order amplitude produces a second-order probability.

For the isotropic quantum harmonic oscillator, parity makes $\int x_i{x}_3|\chi {|}^2{d}^{3}x=0$ for $i=1,2$. Rotational symmetry and the given radial integral give $\int x_3^2|\chi {|}^2{d}^{3}x=\hslash /(2m\omega )$. Thus $⟨{\varphi }_i|{\hat{x}}_3|\chi ⟩={\delta }_{i3}\sqrt{\hslash /(2m\omega )}$, and the energy gap is $\hslash \omega$. The time integral is ${\int }_0^{\infty }{e}^{(-\mu +i\omega )s}ds=(\mu -i\omega {)}^{-1}$. ConsequentlyThe perturbation picks the excitation polarized along the third coordinate, rather than an arbitrary member of the degenerate first excited level.

The Hamiltonian is the sum of three commuting one-dimensional harmonic-oscillator Hamiltonians. Its product eigenstates arewith energiesEquivalently, the level with $N=n_x+n_y+n_z$ has energy $E_N=\hslash \omega (N+3/2)$, as in the three-dimensional isotropic harmonic oscillator.

Up to normalization, the ground-state wavefunction isIt is radial, so $\nabla {\Psi }_{000}$ is parallel to the position vector. Since the orbital angular momentum is $L=-i\hslash r\times \nabla$, every component of $L$ annihilates this state. Thereforeand the ground state has $\ell =m=0$.

The first excited level has $N=1$ and is spanned byThe statesatisfies $L_z{\Phi }_0=0$. Applying the ladder operators givesHence convenient $m=\pm 1$ eigenstates areup to normalization and irrelevant overall phases.

Finally, the isotropic Hamiltonian is rotationally invariant. The rotational invariance of a central-potential Hamiltonian givesThese commuting self-adjoint operators can be simultaneously diagonalized within each energy eigenspace, which is why joint eigenstates of $H,L^2,L_z$ must exist.

Inside the infinite square well, the time-independent Schrödinger equation has Dirichlet boundary conditions at $0$ and $a$. Its normalized energy eigenstates and energy eigenvalues are

A $Z$ measurement is a quantum measurement in the computational basis. To measure $X$, apply a Hadamard gate, measure $Z$, and apply another Hadamard gate to the measured qubit. Since $HXH=Z$, this gives the $X$ outcome and leaves the qubit in the corresponding $X$ eigenstate.

An $n$-qubit stabilizer state is the unique simultaneous $+1$ eigenstate of an abelian group $G$ of Pauli operators having $2^n$ elements and not containing $-I$. The group $G$ is the state's stabilizer group; equivalently, it is generated by $n$ independent commuting Hermitian Pauli operators.

Apply exact quantum phase estimation to $U(t)$ with the supplied eigenstate $|\Psi ⟩$. Sincethe promise that the phase has an $N$-bit representation makes an $N$-qubit control register recover $\varphi$ exactly. Multiplying by $-2\pi$ modulo $2\pi$ gives $\lambda t\mathrm{mod}2\pi$. Equivalently, phase estimation may be run on $U(t{)}^{†}$, whose eigenphase is $\lambda t/(2\pi )$ modulo one.

Write the local Hamiltonian as $H={\sum }_{j=1}^m{h}_j$. Since its terms commute, their matrix exponentials factor exactly:Each factor acts on at most two qubits and can be compiled over a fixed universal quantum gate set to operator norm error at most $\epsilon /m$. The telescoping bound for products of operators then bounds the total error by the sum of the factor errors, at most $\epsilon$. Because $m$ is polynomial in $n$ and the Solovay--Kitaev theorem gives gate count polynomial in $\log (m/\epsilon )$ for each fixed-dimensional factor, this is an efficient commuting local Hamiltonian simulation. Finally, the eigenvalue equation $H|\Psi ⟩=\lambda |\Psi ⟩$ impliesso $|\Psi ⟩$ remains an eigenstate and its eigenvalue is $e^{-i\lambda t}$.

Since ${\hat{x}}_i=P_{1,i}$ and $I-{\hat{x}}_i=P_{0,i}$, the diagonal Hamiltonian iswith the identity on every unlisted qubit. Each computational-basis state is an eigenstate, and its eigenvalue is minus the cost of the corresponding cut. Part i supplies cost four, while the assumed bound $|{\lambda }_{\min }|\le 4$ rules out a lower energy. Thus the ground-state energy is $-4$. For the assignment $(x_1,x_2,x_3,x_4,x_5)=(0,1,0,0,1)$, one ground state isIts bitwise complement $|10110⟩$ is another ground state, as are the basis states corresponding to the other maximum cuts.

The parity of a wavefunction is its eigenvalue under spatial reflection: $\psi (-x)=+\psi (x)$ for even parity and $\psi (-x)=-\psi (x)$ for odd parity. If $V(-x)=V(x)$, the Hamiltonian commutes with reflection, so its energy eigenspaces admit a parity eigenbasis. In particular, every nondegenerate bound-state energy eigenfunction has definite parity; within a degenerate scattering eigenspace one must choose the even and odd linear combinations.

Write $c=\cos (\theta /2)$ and $s=\sin (\theta /2)$. ThenThus $|\psi ⟩$ is unchanged. It is the spin-one-half singlet state, so joint rotations leave it invariant. Equivalently, direct use of the Pauli matrices showsIt is therefore an eigenstate of every Cartesian component of the combined spin, with eigenvalue zero.

Using the states from part (a),The Born rule therefore givesAfter the equal-sign outcomes the state is respectively $|\uparrow \uparrow ⟩$ or $|\downarrow \downarrow ⟩$. These are triplet states, hence eigenstates of ${\mathbf{S}}^2$ with eigenvalueThe opposite-sign product states are each a nontrivial linear combination of the singlet and the $m=0$ triplet, so they are not eigenstates of ${\mathbf{S}}^2$.

Insert the stated expansion into the Time-dependent Schrödinger equationWhen the derivative acts on each free phase, its energy factor is cancelled by the action of $H_0$ on the corresponding eigenstate. The remaining terms obeyDivision by $i\hslash$ gives

The known active neutrinos are too light and too fast to supply the dominant cold dark matter. Their relativistic decoupling leaves substantial free streaming, which erases rather than seeds sufficiently small-scale structure; their allowed mass density is also inadequate for the dominant cold component. Changing a light active neutrino from Dirac to Majorana does not remove those problems. What can change the picture is an enlarged neutral-fermion sector containing sterile states with different masses, production and decay rates. A sufficiently long-lived sterile neutrino in a Majorana state can be a dark-matter candidate; Majorana character alone supplies neither stability nor the correct abundance.

For the spinor calculation, put $a=p_z/(E+m)$ and define $A=N(1-a)/2$, $B=N(1+a)/2$. Applying the given chiral projectors givesConsequentlywhere the last equality uses $E^2-p_z^2=m^2$. Same-chirality Lorentz scalar products instead vanish: for example ${\overline{U}}_{1L}{U}_{1L}=0$. The nonzero overlaps demonstrate the Lorentz scalar structures coupling left and right chirality. For $U$ to $U$ this is a particle-number-preserving Dirac coupling; for $U$ to the charge-conjugate $V$ sector it can describe a lepton-number-violating Majorana coupling.

There is a normalization subtlety: both displayed overlaps equal the on-shell mass parameter used to define the spinors. A chiral scalar spinor overlap verifies a coupling structure without fixing an independent Lagrangian coefficient $M_L$. Repeating the on-shell construction with mass parameter $M_L$ would instead give the corresponding overlap $M_L$; this is a different input, rather than a newly derived parameter. The independent coefficients of a Dirac mass term and a Majorana mass term are defined byThe Majorana terms connect a field with its charge conjugate and violate the corresponding continuous lepton phase symmetry by two units. With only unbroken Standard Model gauge symmetry, a left-handed Majorana mass also requires electroweak symmetry breaking or an appropriate higher-dimensional operator; it is not an arbitrary gauge-invariant bare mass for a lepton doublet.

Use an all-left-handed Weyl basis ${\mathbf{n}}_L=({\nu }_L,{\nu }_R^c{)}^T$. For real parameters the one-family Dirac-Majorana mass matrix isWriting $({\nu }_{1L},{\nu }_{2L}{)}^T=\left(\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right)({\nu }_L,{\nu }_R^c{)}^T$ gives an orthogonal diagonalization, with $\tan 2\theta =2m/(M_R-M_L)$ up to the rotation-sign convention. Massive-particle chirality and helicity are not identical; the familiar left/right helicity language becomes an approximation in the relativistic regime.

The characteristic polynomial isIts roots areFor Majorana fermions the nonnegative physical masses, as explained by physical Majorana mass and signed eigenvalue, are $|{\lambda }_{\pm }|$ after a possible field rephasing; a negative real eigenvalue is not a negative physical rest mass. When $M_R\gg |m|,|M_L|$,The light state is predominantly ${\nu }_L-(m/M_R){\nu }_R^c$, while the heavy state is predominantly ${\nu }_R^c+(m/M_R){\nu }_L$. Each mass eigenstate combines its Weyl component and charge conjugate to form a Majorana field.

For $M_R=10^{12}\mathrm{GeV}$, the heavy mass is approximately $10^{12}\mathrm{GeV}$. In the type-I limit $M_L\simeq 0$,The seesaw light-mass estimate needs a Dirac scale: no unique numerical light mass follows because the Dirac mass and $M_L$ were not supplied. As an explicitly conditional illustration, choosing $m=100\mathrm{GeV}$ and $M_L=0$ gives $10\mathrm{eV}$; this is an assumed Dirac scale, not additional exam data or a prediction from $M_R$ alone. A heavy seesaw state is also not automatically stable enough to be dark matter.

The conventional minimal Pati-Salam model gauge group isThe group printed in this paper is different: it contains two $SU(4)$ factors and only one $SU(2)$. This is a discrepancy in the original PDF, not just in the transcription. The standard-model discussion below uses the conventional intended Pati-Salam model model. A literal theory with the printed product would require additional fermion representations and symmetry-breaking data that are not specified.

The Pati-Salam model unifies the three quark color charges with lepton number as a fourth color charge. A family forms $(4,2,1)$ left-handed and $(4,1,2)$ right-handed multiplets, including a right-handed neutrino. The extended color charge group contains quark-lepton gauge bosons, while the right-handed weak factor supplies right-handed charged currents after symmetry breaking. The Pati-Salam charge embedding isThe original paper explicitly develops the smaller local subgroup $SU(2{)}_L\times SU(2{)}_R\times SU(4{)}_C$, as well as a larger underlying global structure; neither distinction licenses silently replacing the printed group.^[ref] The product group is a quark-lepton unification scheme rather than a simple gauge group with one automatically common coupling. Breaking the right-handed and $B-L$ symmetries can generate a large $M_R$ and light/heavy Majorana neutrinos through the seesaw. The gauge group alone does not uniquely fix their masses or guarantee a dark-matter lifetime.

The most important laboratory test of Majorana character is neutrinoless double beta decay,with no outgoing neutrinos. Under light-neutrino exchange, the amplitude involvesThe coherent squared mixing elements allow phase cancellations. A null result constrains this combination within the assumed mechanism, rather than proving that all neutrinos are Dirac.

The observable is a peak in summed electron energy at the decay Q value, unlike the continuous two-neutrino spectrum. Enriched source material, large exposure, good energy resolution, underground operation and radiopure construction suppress backgrounds. germanium arrays use the source as a semiconductor detector; xenon time-projection chambers combine ionization, light and event-position information; bolometers measure deposited energy through a temperature rise. The GERDA collaboration's Phase-I study is a concrete example of an enriched-germanium source-detector search with background suppression.^[ref] energy calibration, resolution tails, radon and material backgrounds, cosmic-ray vetoes, signal-like event topology and sideband fits all need independent checks before a Q-value excess is convincing. Collider same-sign dilepton processes can test heavier Majorana states, but require a specified production and decay mechanism. Lepton-number violation, absolute-mass measurements and oscillation data provide complementary information; oscillations alone do not settle Majorana character.

The angular momentum commutation relations are $[L_x,L_y]=i\hslash L_z$ and cyclic permutations. The product rule yields $[L_z,L_x^2]=i\hslash (L_y{L}_x+L_x{L}_y)$ and $[L_z,L_y^2]=-i\hslash (L_x{L}_y+L_y{L}_x)$, so $[L_z,L^2]=0$. Simultaneous eigenstates have eigenvalues ${\hslash }^{2}l(l+1)$ and $\hslash m$; orbital angular momentum has integer $l\ge 0$ and $m=-l,\dots ,l$.

By cyclic symmetry, $L^2$ commutes with every component and hence every squared component. Therefore $[H,L^2]=0$: an energy eigenbasis can be chosen with definite $l$. For unequal transverse coefficients, $L_z$ need not commute with $H$, so definite $m$ is not generally available.

Expanding the ladder operators gives $L_{+}^2+L_{-}^2=2(L_x^2-L_y^2)$ and $L^2-L_z^2=L_x^2+L_y^2$. Comparing coefficients givesFor $l=1,m=0$, both squared ladder operators vanish, since two raising or lowering steps would leave the multiplet. Thus $E_0=2\beta {\hslash }^2={\hslash }^2(a_x+a_y)$. In the remaining basis $\{|1,1⟩,|1,-1⟩\}$ the Hamiltonian isIts symmetric and antisymmetric combinations have energies ${\hslash }^2(a_x+a_z)$ and ${\hslash }^2(a_y+a_z)$, respectively. ConsequentlyThe energies are related by permutation of the principal axes, not generally equal. Two coincide when the corresponding coefficients coincide; all three become $2a{\hslash }^2$ in the isotropic limit.

Stationary states. Inside the infinite square well, the time-independent Schrödinger equation has sinusoidal solutions. The boundary conditions $\varphi (0)=\varphi (a)=0$ select $k_{n}a=n\pi$; normalization then givesEach wavefunction vanishes outside the well. There is no nonzero $n=0$ state.

Immediate energy measurement after expansion. For the one-sided sudden expansion of an infinite square well, take the original wall at $x=0$ as fixed, so the expanded well occupies $0<x<2a$. In the sudden approximation, the initial wavefunction remains $\sqrt{2/a}\sin (\pi x/a)$ on $0<x<a$, extended by zero elsewhere. The new ground state is ${\chi }_1=a^{-1/2}\sin (\pi x/(2a))$. Its probability amplitude isThe Born rule therefore givesThe motion of the walls matters: a symmetric expansion about the old midpoint would instead give $64/(9{\pi }^2)$. The stated width alone does not explicitly identify the final endpoints. The first result uses the natural continuation of the specified left boundary; if both boundaries are moved, the final location must be supplied. For a new well $(b,b+2a)$ containing $(0,a)$, the overlap is $c_1(b)=(4\sqrt{2}/(3\pi ))[\cos (\pi b/(2a))-\sin (\pi b/(2a))]$.

Evolution of one energy eigenstate. For a time-independent Hamiltonian operator, write $|\psi (t)⟩=f(t)|{\varphi }_n⟩$. Substitution into the Time-dependent Schrödinger equation gives $i\hslash \dot{f}=E_{n}f$, soOnly a global phase changes, leaving all measurement probabilities unchanged.

Delayed energy measurement. Expand the unchanged initial state in the complete new energy eigenbasis,For the fixed-left-wall geometry, direct overlap integration givesorthogonality isolates the new ground-state amplitude as $c_1{e}^{-i{E}_1^{\prime }t/\hslash }$. Its phase cancels on squaring, so the ground-state probability remains $32/(9{\pi }^2)$ for every delay, provided the expanded well remains fixed and no intervening measurement or environment changes the state. The position probability density generally evolves because different eigenstates acquire different relative phases.

Factorization. Put $\alpha =\sqrt{m\omega /(2\hslash )}$ and use the canonical commutation relation $[x,p]=i\hslash$:It follows thatThe $-1/2$ arises from quantum operator ordering; treating $x$ and $p$ as commuting would lose the zero-point energy.

commutators. Direct use of $[x,p]=i\hslash$ gives $[a,a^{†}]=1$. The product rule for commutators then yields $[a^{†}a,a]=-a$ and $[a^{†}a,a^{†}]=a^{†}$, hence

Raising the energy. If $H|\varphi ⟩=E|\varphi ⟩$, thenThus the creation operator produces an energy eigenstate with energy $E+\hslash \omega$. It is nonzero: for a normalized physical oscillator eigenstate, $\Vert a^{†}\varphi {\Vert }^2=⟨a{a}^{†}⟩=E/(\hslash \omega )+1/2>0$. Normalize by this norm if a normalized state is wanted.

Lowest wavefunctions and orthogonality. In the position representation in quantum mechanics,The annihilation condition therefore gives ${\psi }_0^{\prime }=-(m\omega /\hslash )x{\psi }_0$. Applying $a^{†}$ to its solution givesTheir inner product is proportional to ${\int }_{-\infty }^{\infty }x{e}^{-m\omega x^2/\hslash }dx=0$, since the integrand is odd. These are the even and odd lowest wavefunctions shown in A2, with energies $\hslash \omega /2$ and $3\hslash \omega /2$.