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The input is the Bell state $\rho =|{\Phi }^{+}⟩⟨{\Phi }^{+}|$, where $|{\Phi }^{+}⟩=(|00⟩+|11⟩)/\sqrt{2}$. Acting on the first qubit reverses the signs of all three terms containing its Pauli matrices, so in computational basis order $00,01,10,11$ the output isIts eigenvalues are $\boxed{-1/2,1/2,1/2,1/2}$: in particular, its expectation value in $|{\Phi }^{+}⟩$ is $-1/2$. It therefore fails to be positive semidefinite, although its trace remains one. This single entangled state proves that the universal NOT gate is positive but not completely positive, so it cannot be a deterministic physical quantum channel on an arbitrary input.

Define the Hadamard basis states $|\pm ⟩=(|0⟩\pm |1⟩)/\sqrt{2}$. The initial two Hadamard gates prepare $|+⟩|+⟩$. For one bracket, the two control branches giveUsing the Pauli X gate $X=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$, unitary conjugation by the Hadamard gate gives $H{U}_{0}H=-X$, so $G=-X{U}_f$. Keeping every sign, the target states along the control-one branch, in the exact gate order, areThe last column also includes the final target Hadamard gate, which lies outside the bracket. Writing $|{\chi }_f⟩=HG|+⟩$, applying the final top Hadamard gate gives the exact normalized two-qubit quantum stateThus the complete outputs, in top-then-bottom computational basis order, areFor the first qubit, the Pauli matrix ${\sigma }_z=Z$ has expectation value $P(0)-P(1)$ by the Born rule. The two $k=1$ cases each have $P(0)=P(1)=1/2$, while the $k=0$ and $k=2$ cases are deterministic. Equivalently, this Hadamard test measuresThis explains the approximate quantum counting interpretation: repeated preparations estimate the expectation value, and hence $k$. A single top-qubit computational-basis measurement need not identify $k$.

For a normalized pure state, expand $|\psi ⟩={\sum }_m{c}_m|m⟩$ in an orthonormal basis of eigenvectors of the measured Hermitian operator. The Born rule assigns outcome $m$ the probability $p_m=|c_m{|}^2$. Under the ideal projective measurement update, a nondegenerate outcome leaves $|m⟩$, up to a global phase. The sum of the outcome probabilities is one, and the expectation value is ${\sum }_{m}m|c_m{|}^2$; a single trial reports one eigenvalue, not that average.

If $M|m⟩=m|m⟩$, thensince $M=M^{†}$. A nonzero eigenvector has positive squared norm, so $m$ is real. For two eigenvectors,Thus $(m-n)⟨n|m⟩=0$: distinct eigenvalues have orthogonal eigenvectors.

The number operator is $\hat{N}=a^{†}a$. The number state expansion gives $a|\alpha ⟩=\alpha |\alpha ⟩$ by shifting the summation index; taking the adjoint gives $⟨\alpha |a^{†}={\alpha }^{\ast }⟨\alpha |$. Thus the expectation value isFor the second moment, use $a{a}^{†}=a^{†}a+I$:Its expectation value in the coherent state is $⟨{\hat{N}}^2⟩=|\alpha {|}^4+|\alpha {|}^2$. ThereforeThe variance equals the mean, exactly as for the Poisson distribution found in b. For $\mu >0$ the fractional standard deviation is $1/\sqrt{\mu }$, the usual photon shot noise scaling. For the vacuum state, both the mean photon number and its variance vanish; this does not imply zero optical quadrature fluctuations. At $\mu =0$ the fractional fluctuation is undefined.

For the normalized coherent state, $⟨a⟩=\alpha$ and $⟨a^{†}⟩={\alpha }^{\ast }$. The dimensionless electric field and magnetic field optical quadratures consequently have expectation valuesThe physical electric field and magnetic field strengths have the corresponding proportionality factors and mode functions. In free evolution replace $\alpha$ by $\alpha e^{-i\omega t}$ to recover the classical oscillation.

Expanding the squared operators cancels the $a^2$ and $(a^{†}{)}^2$ terms:On the other hand, $(\alpha +{\alpha }^{\ast }{)}^2+[-i(\alpha -{\alpha }^{\ast }){]}^2=4|\alpha {|}^2$. HenceThe difference is the sum of the variances, not an extra mean field. Indeed $\mathrm{Var}(\mathcal{E})=\mathrm{Var}(\mathcal{B})=1$, independent of $\alpha$. The commutator $[\mathcal{E},\mathcal{B}]=2iI$ requires ${\sigma }_{\mathcal{E}}{\sigma }_{\mathcal{B}}\ge 1$, and the coherent state attains equality. The residual two units are the vacuum fluctuations of the two optical quadratures.

For second-qubit inputs $|0⟩$ and $|1⟩$, c gives control expectation values $\mathrm{Re}{U}_{00}$ and $\mathrm{Re}{U}_{11}$. Preparing the maximally mixed state $I/2$ is equivalent to choosing these inputs independently with equal probability and discarding the choice. Linearity of the density operator evolution and quantum measurement probabilities therefore givesHenceThe factor two is essential: the Hadamard test directly estimates the normalized trace. More generally, a target density operator $\rho$ gives $⟨Z⟩=\mathrm{Re}\mathrm{Tr}(\rho U)$. This argument never requires knowledge of the eigenvectors of $U$.

The initial wavefunction has unit norm, since ${\int }_0^a(2/a){\sin }^2(\pi x/a)dx=1$. On its nonzero half, $-{\hslash }^2{\psi }^{\prime \prime }/(2m)=E_{\ast }\psi$ with $E_{\ast }={\hslash }^2{\pi }^2/(2m{a}^2)$. A small domain qualification is important: the derivative jumps at zero, so the state is not in the operator domain of the Dirichlet Hamiltonian operator. Its finite energy expectation value is nevertheless defined by the quadratic form of a positive quantum Hamiltonian. The energy form of a half-well sine state givesThis is also the rigorous weak interpretation of the printed $\int \psi H\psi$. Indeed the distributional second derivative is ${\psi }^{\prime \prime }=-(\pi /a{)}^2\psi +\sqrt{2/a}(\pi /a){\delta }_0$ inside the well. The Dirac delta term pairs to zero because $\psi (0)=0$. One should not instead conclude that this state is an energy eigenstate from its equation on just the positive half.

Its coefficients in the full-well energy eigenstates areFor even $n$, sine orthogonality gives $a_2=-1/\sqrt{2}$ and all other even coefficients zero. For $n=2p+1$, set $u=x/a$ and use $\sin [n\pi (u+1)/2]=(-1{)}^p\cos (n\pi u/2)$. Product-to-sum integration then givesSince these coefficients are $O(n^{-2})$, ${\sum }_n{E}_n|a_n{|}^2$ converges, confirming that the state has finite energy. With $E_n=n^2{E}_1$, the spectral expression isThe kinetic-energy quadratic form equals this spectral sum: expanding the weak derivative in the corresponding orthonormal cosine modes and using Parseval's identity gives $\int |{\psi }^{\prime }{|}^2={\sum }_n{k}_n^2|a_n{|}^2$. Equating it to the directly integrated value $4E_1$ provesIn contrast, ${\sum }_n{E}_n^2|a_n{|}^2$ diverges, consistently showing why $H\psi$ is not an ordinary square-integrable vector even though the energy expectation is finite.

For $H=p^2/(2m)+V(x)$ and a time-independent operator $O$, differentiate its expectation value and use the Schrödinger equation and its adjoint. Moving $H$ across the inner product by integration by parts givesWork with wavefunctions in the displayed operators' domains; the vanishing at infinity removes the associated boundary terms. This is the Ehrenfest theorem derived from time evolution.

The canonical commutation relation follows directly from differentiating $x\psi$: $[p,x]\psi =-i\hslash \psi$. Thus $[p^2,x]=p[p,x]+[p,x]p=-2i\hslash p$, while $[V(x),x]=0$. It follows thatFor the momentum equation, apply both products to a test wavefunction:The kinetic term commutes with $p$, so $[H,p]=i\hslash V^{\prime }$. ThereforeThese are the classical equations from Ehrenfest theorem for position and momentum expectations; the force expectation need not equal the force evaluated at the expected position.

Write $B=b+ic$ with $b>0$. Normalization of the Gaussian wave packet gives $|A{|}^2\int e^{-2b{x}^2}dx=1$, hence $|A{|}^2=\sqrt{2b/\pi }$. The supplied Gaussian integral ratio then givesThe momentum operator obeys $p\Psi =2i\hslash Bx\Psi$, and direct second differentiation gives $p^2\Psi =(2{\hslash }^{2}B-4{\hslash }^2{B}^2{x}^2)\Psi$. Taking the expectation value and substituting the first result givessince $2b-B=\bar{B}$. These are the second moments of a complex Gaussian wave packet.

For the solution in part (b), the trigonometric identitygives, writing $D=\cosh (2t)+\cos (2\varphi )$,Thus normalizability requires $\sin (2\varphi )>0$, and the exact expectation values areSince $\cosh (2t)\sim e^{2t}/2$,Their difference is the constant $-\hslash \cot (2\varphi )$, so the mean energy $⟨H⟩=-\hslash \cot (2\varphi )/2$ stays constant despite exponential spreading in both observables.

The expectation value of an observable is the mean outcome of repeated measurements on identically prepared systems in the normalized state. For a self-adjoint $Q$ it is real and equals $⟨\psi ,Q\psi ⟩$. Assume the state belongs to the domains needed for the second moments and the commutator below, so all these quantities exist.

For real $\lambda$, positivity of the squared norm givesThe commutator expectation is purely imaginary, so this is a real quadratic in $\lambda$. If $⟨P^2⟩>0$, its minimum is nonnegative exactly when its discriminant is nonpositive. If $⟨P^2⟩=0$, nonnegativity for all real $\lambda$ forces the linear coefficient to vanish, yielding the same conclusion. ThusReplace the observables by their centred versions $Q-⟨Q⟩I$, $P-⟨P⟩I$. Their commutator is unchanged, and their second moments are the variances $(\Delta Q{)}^2,(\Delta P{)}^2$. Taking square roots gives the Robertson uncertainty principleFor the quantum harmonic oscillator, $p=-i\hslash {\partial }_x$ gives $[x,p]\psi =i\hslash \psi$. The uncentred inequality above consequently gives $⟨x^2⟩⟨p^2⟩\ge {\hslash }^2/4$. The arithmetic-geometric mean inequality now yieldsThis is a bound for every normalized state of finite oscillator energy, not just an eigenstate. It is attained by the centred Gaussian ground state, for which $⟨p^2⟩=m^2{\omega }^2⟨x^2⟩$ and the uncertainty product equals $\hslash /2$.

For a state in the domain of the operator, its expectation value isFor a normalized state the denominator is one. Use an orthonormal energy eigenstate basis, choosing orthogonal vectors within degenerate eigenspaces. Inserting an expansion into the time-dependent Schrödinger equation gives $i\hslash {\dot{a}}_n(t)=E_n{a}_n(t)$. ConsequentlyThe phases preserve normalization. Projection onto a specified basis eigenstate gives probability $\boxed{|a_p{|}^2}$, independent of time. If the measurement distinguishes only the energy and $E_p$ is degenerate, sum $|a_n{|}^2$ over all $n$ with $E_n=E_p$.

Using orthonormality to eliminate all cross terms,This is time independent whenever the expectation is defined. The time-independent Hamiltonian changes only relative phases, not the energy probabilities.

Use target signature $(-,+,\dots ,+)$ and a closed-string spatial coordinate $0\le \sigma \le 2\pi$. Write $Y$ for the compact coordinate and $X^A$ for the noncompact coordinates. In conformal gauge, the wave equation factorizes, so every coordinate is a sum of a function of $\tau +\sigma$ and a function of $\tau -\sigma$. The closed-string boundary conditions areThus the derivatives of each chiral part are periodic and have integer Fourier modes; only their worldsheet zero modes can produce the winding displacement. Integrating these Fourier series gives the closed-string mode expansionThe oscillator reality conditions are ${\alpha }_m^{†}={\alpha }_{-m}$ and similarly for the tilded modes. The coefficient of $\tau$ follows by integrating the canonical momentum density $\dot{Y}/(2\pi {\alpha }^{\prime })$. Single-valued wavefunctions of the center coordinate $y\sim y+2\pi R$ have momentum $n/R$, with $n\in \mathbb{Z}$. Equivalently, the compact chiral worldsheet zero modes are ${\alpha }^{\prime }{p}_L(\tau +\sigma )/2$ and ${\alpha }^{\prime }{p}_R(\tau -\sigma )/2$, whereHere tildes denote the left-moving sector. This convention will also fix the sign in compact-circle closed-string level matching.

Choose a noncompact light-cone pair and impose light-cone gauge in string theory $X^{+}=x^{+}+{\alpha }^{\prime }{p}^{+}\tau$, with $p^{+}\ne 0$. The Virasoro constraints $({\partial }_{\pm }X{)}^2=0$ then solve for ${\partial }_{\pm }{X}^{-}$ in terms of the transverse coordinates. For example, with ${\partial }_{\pm }={\partial }_{\tau }\pm {\partial }_{\sigma }$, one has ${\partial }_{\pm }{X}^{-}=({\partial }_{\pm }{X}^i{)}^2/(2{\alpha }^{\prime }{p}^{+})$. Thus there are two independent transverse oscillator Fock spaces, including $Y$ among the transverse coordinates. Quantization gives $[{\alpha }_m^i,{\alpha }_k^j]=m{\delta }^{ij}{\delta }_{m+k,0}$, and the same relation for the tilded oscillators, with the two sectors commuting. Lorentz consistency gives the critical dimension of the bosonic string $D=26$ and string intercept one in each sector. Accordinglycount 24 transverse oscillator species. The two zero-mode constraints give the lower-dimensional bosonic string mass spectrumTaking their difference and average yieldsEvery choice of transverse string oscillators and integer charges satisfying this closed-string level matching is a physical string state in light-cone gauge. The eliminated longitudinal oscillators do not supply additional polarizations.

To classify all massless states, positivity of the two compact terms implies $N+\tilde{N}\le 2$. When the sum is two, masslessness forces $n=w=0$, and closed-string level matching then forces $N=\tilde{N}=1$. The states ${\alpha }_{-1}^i{\tilde{\alpha }}_{-1}^j|0⟩$ have $24^2=576$ polarizations. In 25 noncompact dimensions, their decomposition gives a graviton, a Kalb–Ramond field, a dilaton, two Abelian gauge vectors from the metric and two-form with one compact index, and the radius scalar field. Their physical polarization counts areThe first two numbers are the symmetric traceless second-rank tensor and antisymmetric second-rank tensor of the massless little group $SO(23)$.

When $N+\tilde{N}=1$, closed-string level matching requires $|nw|=1$. The compact contribution obeysand equality occurs precisely at $R=\sqrt{{\alpha }^{\prime }}$. The charges $n=w=\pm 1$ have $(N,\tilde{N})=(1,0)$; the charges $n=-w=\pm 1$ have $(N,\tilde{N})=(0,1)$. Each charge has 24 polarizations supplied by its single excited sector. At a general radius these have

Finally, $N=\tilde{N}=0$ requires $nw=0$. The exceptional massless ground states of a bosonic circle occur atEach charge is a scalar field. These special radii must be included if “general radius” is to mean a complete classification, rather than only a generic radius. At a generic radius the massless spectrum is the neutral 576-state spectrum above; at exceptional radii one adds exactly the states just listed. The neutral ground state remains tachyonic.

The mapsends $p_L$ to $p_L$ and $p_R$ to $-p_R$. Reversing all right-moving compact string oscillators, ${\alpha }_m^Y\mapsto -{\alpha }_m^Y$, preserves their commutators and oscillator number; the left-moving oscillators are unchanged. Both the mass formula and compact-circle closed-string level matching are therefore unchanged, with a one-to-one map of the complete oscillator basis. This proves T-duality of the full spectrum, including winding, massive and tachyonic states.

At the self-dual circle the four single-oscillator charge families give $4\cdot 24=96$ new massless states. Each family decomposes into a 25-dimensional gauge vector with 23 polarizations and one scalar field. There are also four massless ground states, with charges $(\pm 2,0)$ and $(0,\pm 2)$. Thus the massless spectrum at the bosonic self-dual circle hasThe gauge interpretation follows directly from the chiral compact-boson operators. With $⟨Y_L(z)Y_L(0)⟩=-{\alpha }^{\prime }\log z/2$, the exponential $e^{ik{Y}_L}$ has conformal weight ${\alpha }^{\prime }{k}^2/4$. At the self-dual radius,have weight one and satisfy $J^3(z)J^3(0)\sim 1/(2z^2)$, $J^3(z)J^{\pm }(0)\sim \pm J^{\pm }(0)/z$, and $J^{+}(z)J^{-}(0)\sim z^{-2}+2J^3(0)/z$. These are the level-one $SU(2)$ current relations; the right sector supplies another copy. Multiplying these currents by a noncompact oscillator produces six gauge vectors in total. The nine scalar string vertex operators $J_L^a{J}_R^b$ form the $(3,3)$ representation, accounting for the radius scalar field and eight new scalar fields. The remaining neutral fields are the graviton, two-form and dilaton.

A radius deformation as a Higgs mechanism is the expectation value of the Cartan–Cartan scalar string vertex operator $J_L^3{J}_R^3\propto \partial Y_L\bar{\partial }{Y}_R$. Its stabilizer is $U(1{)}_L\times U(1{)}_R$, soThe four compact single-oscillator scalar field polarizations become longitudinal polarizations of the four massive vectors, exactly as in the Higgs mechanism. The other four extra scalar fields have masses $4/R^2-4/{\alpha }^{\prime }$ or $4R^2/{\alpha }^{\prime 2}-4/{\alpha }^{\prime }$ and need not remain massless; one pair becomes tachyonic on either side of the self-dual point. Together with the original bosonic tachyon, this means that the gauge-symmetry interpretation is formal and does not assert a stable bosonic-string vacuum.

Take $0\le \sigma \le \pi$ and split the coordinates into $a=0,\dots ,p$ along the branes and $I=p+1,\dots ,25$ transverse to them. In conformal gauge the wave equation permits a separated mode $e^{-i\omega \tau }f(\sigma )$ with $f^{\prime \prime }+{\omega }^{2}f=0$. For Neumann boundary conditions at both ends, $f^{\prime }(0)=f^{\prime }(\pi )=0$ selects $\cos m\sigma$ and integer $m$. The zero-frequency solution is independent of $\sigma$, with a constant position and a term linear in time. Therefore the open-string mode expansion isThe canonical momentum density integrated over this interval is $k^a$.

For Dirichlet boundary conditions, subtract the static interpolation between fixed endpoint positions $y_0^I$ and $y_{\pi }^I$. The remaining field vanishes at both ends, selecting $\sin m\sigma$ with integer $m$ and excluding a momentum worldsheet zero mode. The expansion isHere ${\alpha }_m^{I†}={\alpha }_{-m}^I$. There is no factor $i$ in the sine series with this reality convention: changing $m$ to $-m$ reverses both the sine and the denominator. These are the integer-mode expansion with parallel D-brane endpoints; they are not the half-integer modes of a coordinate with a Neumann condition at one endpoint and a Dirichlet condition at the other.

The endpoints can move on $(p+1)$-dimensional worldvolumes but are fixed in the transverse directions. Thus these are open strings attached to D-branes, or stretched between parallel D-branes at $y_0$ and $y_{\pi }$. The tangential momenta describe propagation on the brane; the transverse positions describe its embedding. Let $L=|y_{\pi }-y_0|$. The linear spatial worldsheet zero mode contributes $L^2/(4{\pi }^2{\alpha }^{\prime })$ to $L_0$. Quantization with string intercept one givesThe first term is also the square of the classical stretched-string energy $TL$, since the string tension is $T=1/(2\pi {\alpha }^{\prime })$.

For two coincident D3-branes, label the ends of an oriented open string by $i,j\in \{1,2\}$. The four endpoint combinations give Chan-Paton factors ${\lambda }_{ij}$, which form the full algebra of $2\times 2$ matrices. Under a change of brane basis these transform as $\lambda \mapsto U\lambda U^{-1}$, the adjoint action of $U(2)$. At $L=0$, the first oscillator level is massless. Its tangential polarization supplies a four-dimensional vector $A_a$, while the 22 transverse polarizations supply real adjoint scalar fields ${\Phi }^I$. This is the bosonic D3-brane low-energy field content.

The endpoint matrices multiply in their boundary order in disk amplitudes. The difference between the two orders in the three-vector interaction gives the matrix commutator; the kinematic factor is the one-derivative Yang-Mills theory cubic vertex. Factorization and the vector gauge redundancy then supply the quartic interaction with the same commutator. Equivalently, the leading low-energy action is the dimensional reduction of 26-dimensional Yang-Mills theory, withFor Hermitian scalar fields the last displayed term corresponds to a nonnegative potential, because their commutators are anti-Hermitian. Higher-derivative terms are suppressed at energies well below $1/\sqrt{{\alpha }^{\prime }}$. Ignoring the tachyon, the gauge sector is four-dimensional $U(2)$ Yang-Mills theory, accompanied by 22 adjoint scalar fields. It is not pure Yang-Mills theory, nor the supersymmetric six-scalar D3-brane theory of a ten-dimensional superstring.

Separate the two branes and use the D-brane scalar-position normalizationA Yang-Mills gauge transformation preserves this expectation value only when it is diagonal if $y_1\ne y_2$. Thus the unbroken group is $U(1)\times U(1)$. The scalar field kinetic term supplies a mass for the off-diagonal vectors, sinceSumming over $I$ gives $m_{12}^2=L^2/(4{\pi }^2{\alpha }^{\prime 2})$, exactly the mass of the stretched $N=1$ string. The two diagonal gauge vectors remain massless. This identifies brane separation with an adjoint Higgs mechanism, rather than an explicit deletion of the off-diagonal strings.

For the ground state of an open string joining the two different branes, $N=0$, so the bosonic stretched-string tachyon threshold isAt equality it is massless, and above it the stretched ground state has positive mass squared. The same-brane ground strings still have $M_0^2=-1/{\alpha }^{\prime }$; this calculation removes only the stretched-string instability, not the original instability of the bosonic branes.

For a normalized wavefunction in the required operator domains and a self-adjoint operator $A$, its expectation value and standard deviation areEquivalently the squared uncertainty is $\Vert (A-⟨A{⟩}_{\psi })\psi {\Vert }^2$, which is visibly nonnegative. Since the expectation value of a self-adjoint operator is real, expansion of the square and $\int |\psi {|}^2=1$ giveFor a state not normalized to one, divide each defining expectation integral by $\int |\psi {|}^2$ before using this formula.

For the quantum harmonic oscillator, split the two expectation values into their variances and squared means:The last two terms are nonnegative, soApply the arithmetic-geometric mean inequality to the two remaining nonnegative terms, then the Heisenberg uncertainty principle:The argument applies to any normalized state with finite oscillator energy; the stated stationary state is a particular case. Equality requires zero position and momentum means, minimum quantum uncertainty, and balanced kinetic and potential contributions, as in the oscillator ground state.

For a time-independent Quantum Hamiltonian, a stationary state is an energy eigenstate. The Schrödinger equation separates its time dependence asIts probability density $|\psi (x,t){|}^2=|\chi (x){|}^2$ is time independent. Linearity of the Schrödinger equation givesPut $A_{jk}={\int }_{-\infty }^{\infty }{\chi }_j^{\ast }\hat{A}{\chi }_{k}dx$. The operator for a time-independent observable is a Hermitian operator, so $A_{21}=\overline{A_{12}}$. Substituting the evolved wavefunction into the expectation value givesand therefore, for a normalized state,The normalization is automatic for orthogonal ${\chi }_1,{\chi }_2$, in particular when $E_1\ne E_2$: self-adjointness of $\hat{H}$ gives $(E_1-E_2)⟨{\chi }_1,{\chi }_2⟩=0$. If the energies coincide, normalization of the individual eigenstates alone does not ensure normalization of their displayed sum; its squared norm is $1+\mathrm{Re}⟨{\chi }_1,{\chi }_2⟩$. For a nonzero sum not having unit norm, the physical expectation value is the displayed numerator divided by this squared norm. With orthonormal energy states, the stated formula holds in both the distinct and degenerate cases.

For a normalized state $|\psi (t_F;f)⟩$ driven by the admissible control $f$, maximizeHere $A$ is the specified Hermitian operator, $t_F$ the final time, and the dependence on $f$ is through the Schrödinger equation. The objective is real and is exactly the desired expectation value, so increasing it increases the requested observable. A control-resource penalty may be included when the admissible set does not already constrain field strength or fluence.

The deterministic law cannot be transplanted unchanged into measurement-based quantum feedback. Evaluating its state-dependent scalar requires knowledge of the current quantum state. In fact $C=\mathrm{Tr}(i[H_1,{\rho }_d]\rho )$ is the expectation value of a Hermitian observable, but that expectation cannot be obtained exactly from a single system without measurement backaction. Tomography uses many independently prepared copies, not a nondisturbing readout of one evolving state.

Continuous weak measurements can supply a noisy record and a conditional state estimate, but then the state follows stochastic measurement dynamics with additional disturbance and decoherence, rather than the closed unitary equation used in part (a). A new stochastic stability analysis is required. One may instead simulate the corrected Lyapunov quantum control from known initial data and apply its waveform as open-loop control, or learn a pulse across repeated experiments as in Question 3. Those procedures do not constitute real-time state feedback on a single quantum trajectory. The separate source error in part (a) remains: even ideal knowledge of the state would not make its printed overlap law monotone.