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The Born rule and probability current in one dimension giveMultiply the Time-dependent Schrödinger equation by ${\Psi }^{\ast }$, subtract its complex conjugate multiplied by $\Psi$, and rearrange to obtain the continuity equation ${\rho }_t+j_x=0$.

For the stated superposition, the interference terms in $j$ are real and cancel from its imaginary part. Henceindependent of $x$ and $t$.

Each of the four valuesoccurs for three of the twelve inputs. The corresponding basis states in the first register are orthogonal, so the Born rule gives

The angular momentum commutation relations implyThus $S_{\pm }|s,\sigma ⟩$ has the same total-spin eigenvalue and magnetic quantum number $\sigma \pm 1$. Its squared norm follows fromChoosing the conventional positive phases gives the spin ladder operator formula

For $s=1$, in the ordered basis $|1,1⟩,|1,0⟩,|1,-1⟩$,Since $S_x=(S_{+}+S_{-})/2$,

The incident state is the spin coherent state obtained by rotating $|1,1⟩$ through $\theta$ about the $y$ axis. Its amplitudes in the $S_z$ basis areup to an irrelevant phase convention. The Born rule therefore gives

Inside the infinite square well, the stationary Schrödinger equation for unit mass isIts normalized eigenstates and energies areThe general normalized time-dependent solution is

Immediately after removing the barrier, the normalized wavefunction isIts Fourier sine series coefficients in the full-well basis areThey areThe Born rule therefore gives

A stationary state has the formFor a time-independent quantum observable $Q$, the global phases cancel in the expectation value:It is therefore independent of time.

For the infinite square well,The possible energy measurement results arewith respective Born rule probabilities $|c_1{|}^2$ and $|c_2{|}^2$. Normalization gives $|c_1{|}^2+|c_2{|}^2=1$.

At time $t$,The required matrix elements areConsequentlywhere the angular frequency isFinally, $2|c_1{c}_2|\le |c_1{|}^2+|c_2{|}^2=1$, so

The spectral projector for outcome $+1$ isFor $|1⟩=(0,1{)}^T$, the Born rule gives

The Born rule says that the probability of a projective-measurement outcome is the squared norm of the state projected onto the corresponding eigenspace. The basis vectors in the state whose second qubit is one areThey are mutually orthogonal, so

The Born rule says that each probability is the squared norm of the corresponding ancilla component. Since the states are normalized,Consequently

Before the change, the normalized ground state of the infinite square well isFor an allowed post-quench energy $E\in (0,U_0)$ satisfying part (i), define the unnormalized eigenfunctionIts normalization factor isThe Born rule therefore givesFor a value of $E$ that is not an eigenvalue, this probability is zero. The sudden change leaves the wavefunction fixed, while the energy eigenbasis changes.

The states ${\chi }_1$ and ${\chi }_2$ are orthonormal energy eigenstates with hydrogenic energiesThe Born rule therefore gives

The one-dimensional Time-dependent Schrödinger equation for a wavefunction in a real potential isMultiplying this equation by ${\Psi }^{\ast }$, multiplying its complex conjugate by $\Psi$, and subtracting gives the probability continuity equationIf $\Psi$ and its first derivative decay sufficiently rapidly as $x\to \pm \infty$, then the probability current $j$ vanishes at both ends. Integrating the continuity equation and using the fundamental theorem of calculus yieldsThis conservation of quantum probability is required by the Born rule: once a normalizable wavefunction has total probability one, its time evolution must preserve that normalization.

For the stated Gaussian wave packet, put $\beta =\hslash t/m$. Sinceits squared modulus isThe Gaussian integral then giveswhich is independent of time, as required.

The Born rule depends only on transition probabilitiesbetween rays. A physical transformation must preserve these probabilities. By Wigner theorem, such a ray transformation is induced by a unitary or antiunitary operator. For ordinary continuously connected transformations one chooses the unitary branch, so

The states $U(g_1)U(g_2)|\psi ⟩$ and $U(g_1{g}_2)|\psi ⟩$ represent the same ray. They may therefore differ by a state-independent quantum phase:Thus quantum transformations form a projective unitary representation.

If $g$ is a symmetry of the Hamiltonian operator $H$, it also obeysequivalently $[U(g),H]=0$. For a differentiable one-parameter subgroup, writewith a self-adjoint generator $Q$. Differentiating the symmetry equation at $s=0$ yieldsFor a time-independent $Q$, the Heisenberg picture equation isHence the generator is a conserved observable. Conversely, a self-adjoint conserved $Q$ commutes with $H$ and its unitary group generates symmetries. This is the conserved generator of a continuous quantum symmetry.

The specified initial state is normalized. The requested energy is $E_2$, and its probability amplitude isThus the Born rule givesUnitary evolution changes each energy coefficient only by a phase, so this probability remains $1/2$ at both $T$ and $2T$. If an energy measurement returns $E_2$, the state collapses to the nondegenerate eigenstate ${\chi }_2$; a subsequent energy measurement therefore returns $E_2$ with probability

Forthe phase of the $n$th stationary state isReflection in the midpoint givessoThe minus sign is a global quantum phase and therefore does not change the physical state. The relative even--odd phases first acquire this reflection pattern at $T$. Applying it twice givesIf $f$ vanishes on the right half of the well, then $\psi (x,T)$ vanishes on the left half. The Born rule consequently gives zero probability of finding the particle in $0\le x\le a/2$ at time $T$.

Since $P|\psi ⟩={\lambda }_P|\psi ⟩$, $P^2=I$, and $P$ anticommutes with $Q$,so the expectation value of $Q$ is zero. The two spectral projectors of the Hermitian Pauli operator $Q$ are $(I\pm Q)/2$, and hence the Born rule gives

Let $q=({\mathbf{q}}_1,\dots ,{\mathbf{q}}_N)$ denote the particle positions. In the Ghirardi--Rimini--Weber model, the wavefunction obeys the ordinary Schrödinger equationbetween random collapses. Each particle $i$ has an independent Poisson process of collapse times with rate $\lambda$. At such a time the state jumps according towhere the random centre has probability density $\Vert L_i(\mathbf{X})\psi {\Vert }^2$. The normalization is arranged so that $\int d^{3}X{L}_i(\mathbf{X}{)}^2=I$.

The original GRW scales are approximatelyAn isolated microscopic particle is therefore exceedingly unlikely to collapse during a laboratory experiment. A macroscopic pointer containing about $N\sim 10^{23}$ relevant particles has total collapse rate $N\lambda \sim 10^7{\mathrm{s}}^{-1}$ and collapse time about $10^{-7}\mathrm{s}$. After a measurement interaction correlates different microscopic outcomes with pointer positions separated by much more than $r_C$, one constituent's localization suppresses all incompatible pointer branches. This GRW amplification mechanism produces one definite macroscopic outcome with probabilities given by the Born rule, while leaving ordinary microscopic unitary time evolution almost unchanged.

For one spatial coordinate, average over the random centre of one collapse. The resulting density operator has position-space kernelIts diagonal is unchanged, so $⟨x⟩$ and $⟨x^2⟩$ are unchanged. The first derivative of the Gaussian factor vanishes at $x=x^{\prime }$, so $⟨p⟩$ is also unchanged. Its second derivative does not vanish, and the position representation of the momentum operator givesIn three dimensions the increase in total ${\mathbf{p}}^2$ is $3{\hslash }^2/(2r_C^2)$, corresponding to kinetic-energy increase $3{\hslash }^2/(4m{r}_C^2)$ per collapse. These are ensemble statements: conditioning on one specified collapse centre can shift the position moments. Repetition at rate $\lambda$ predicts GRW spontaneous heating, so precision searches for anomalous bulk heating, spontaneous radiation, momentum diffusion, and loss of matter-wave interference test the model.

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 ${\mathbf{S}}^2$.

Let a projective measurement have mutually orthogonal projections $P_i$ satisfying ${\sum }_i{P}_i=I$. On a density operator $\rho$, the Born rule and the Lüders rule giveconditional on outcome $i$. On the first part of a bipartite quantum system, replace $P_i$ by $P_i\otimes I_2$.

The subsystems are isolated when there is no interaction term coupling them. Their Hamiltonian operator has the formso their subsequent unitary time evolution factorizes as $U_1\otimes U_2$.

The reduced density matrix of subsystem 1 iswhere the partial trace is characterized byfor every local observable $M_1$. Thus ${\rho }_1$ contains exactly the statistics accessible by measurements on subsystem 1.

Suppose a projective measurement $\{Q_j\}$ is performed on subsystem 2 and its outcome is not communicated. The resulting nonselective state isFor every $M_1$, the cyclic property of the trace and ${\sum }_j{Q}_j^2=I_2$ giveHence ${\rho }_1^{\prime }={\rho }_1$. No local projective measurement on subsystem 1 can reveal whether the remote unreported measurement occurred. This is quantum no-signalling, which prevents a choice made at a spacelike separation from transmitting information faster than light and makes the measurement formalism compatible with relativistic causality.

The proposed nondisturbing device would violate no information without disturbance. In an ordinary quantum instrument, let $M_{i\alpha }$ be the Kraus operators associated with classical output $i$. If every pure state $|\psi ⟩$ remains unchanged even conditional on the displayed outcome, every nonzero $M_{i\alpha }|\psi ⟩$ must be parallel to $|\psi ⟩$. A linear operator for which every vector is an eigenvector is a scalar multiple of the identity, so $M_{i\alpha }=c_{i\alpha }I$. Its output probabilityis independent of the state. It cannot equal $⟨\psi |P_i|\psi ⟩$ for arbitrary projectors. Equivalently, repeated nondisturbing samples would permit quantum state tomography of one specimen and then quantum cloning, contradicting ordinary quantum theory.

Such devices would not necessarily enable superluminal signalling. One consistent operational extension could make every sequence of outputs depend only on the local reduced density matrix and local settings. Since an unreported remote measurement leaves that matrix unchanged, all local device statistics would remain unchanged too. Other extensions could add nonlocal outcome-dependent rules and permit signalling, but that behavior is additional to the device specification rather than forced by it.

Let the apparatus begin in a ready state $|A_0⟩$. An ideal unitary measurement interaction is defined on the relevant subspace byThus an initial $|\varphi ⟩={\sum }_i{a}_i|e_i⟩$ evolves to the entangled stateand orthogonality of the pointer states gives

Before the interaction, the Born rule givesAfterward,The missing cross term is the lost interference between the $i$ and $j$ branches. Entanglement with orthogonal pointer records therefore explains quantum decoherence and the appearance of a classical mixture to the subsystem. It does not solve the quantum measurement problem: unitary evolution alone does not explain why one definite pointer value is observed.

One consistent five-postulate formulation is the following; different textbooks group the same content differently.

The position-dependent probability density is $|\psi {|}^2$, and its flux is the probability current. These distinguish the wave amplitude from the measurable particle flow in the following scattering problem.