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A chi-squared test compares squared deviations with the size expected from the assumed random errors. For independent observations with a normal distribution $y_i$ with known standard deviations ${\sigma }_i$ and predictions ${\mu }_i(\theta )$, useFor a correct fixed model, the standardized independent residuals are unit-variance normal distributions, so $Q$ has a chi-squared distribution with $\nu =N$. Fitting $p$ independent linear parameters removes $p$ residual directions, giving $\nu =N-p$. This distribution is exact for a full-rank linear model with known errors with a normal distribution; it is an approximation for many nonlinear fits. With known correlated errors with a multivariate normal distribution, replace the sum by $Q={\mathbf{r}}^{\mathsf{T}}{\Sigma }^{-1}\mathbf{r}$, where $\Sigma$ is the covariance matrix and its rank and fitted constraints determine the available statistical degrees of freedom.

Under the null hypothesis, $\mathbb{E}Q=\nu$ and $\mathrm{Var}Q=2\nu$. The reduced statistic $Q/\nu$ should therefore be near one, with standard deviation $\sqrt{2/\nu }$, rather than equal to one exactly. An upper-tail p-value isCompare it with a significance threshold chosen in advance. A small value indicates tension with the model together with its error assumptions; it can arise from a wrong mean model, underestimated uncertainties or neglected correlations. An unusually small statistic can suggest overestimated uncertainties, correlations or overfitting. A p-value is not the probability that the model is true, and failure to reject is not proof of the model.

For counts in categories, the Pearson form is $Q={\sum }_{j=1}^k(O_j-E_j{)}^2/E_j$. Its chi-squared distribution is an asymptotic approximation requiring adequate expected counts. For a fixed-total multinomial sample there is one normalization constraint, so the usual statistical degrees of freedom are $k-1-p$ after fitting $p$ other parameters. Independent Poisson bins instead start with $k$ statistical degrees of freedom before parameters are fitted. Sparse counts require an appropriate count likelihood or a calibrated exact or simulation-based test; plugging tiny expected counts into the asymptotic formula is unreliable. The error model, binning, fitted parameters and chosen test must therefore be reported with the result.

Digital acquisition converts a continuous voltage signal into timed numerical samples. A practical chain is a suitably buffered and amplified sensor signal, an analogue anti-aliasing filter, a sampling or sample-and-hold stage, an analog-to-digital converter, and a clocked data-storage system. Choose the gain so that the interesting variation uses the available input range without clipping. Finite input impedance can load a sensor, so an appropriate buffer or operational amplifier may be needed.

Sampling makes time discrete. If the conditioned signal is confined below frequency $f_{\max }$, choose a sampling rate $f_s>2f_{\max }$, allowing a transition band for the physical filter. The Nyquist–Shannon sampling theorem permits ideal reconstruction of a suitably bandlimited function from exact samples. Without filtering, components separated by integer multiples of $f_s$ have identical samples; in particular real sinusoids at $f$ and $f_s-f$ can be indistinguishable. This aliasing cannot generally be undone by later digital filtering. Oversampling can ease filter design, but does not replace the requirement to suppress unwanted high-frequency content before sampling.

Amplitude quantization is separate from sampling. A uniform $b$-bit converter covering voltage span $V_{\mathrm{span}}$ has step $q\simeq V_{\mathrm{span}}/2^b$. Rounding an in-range input gives the uniform-quantization error bound $|ϵ|\le q/2$. If the rounding errors can additionally be modeled as uniform and uncorrelated, their standard deviation is $q/\sqrt{12}$. This statistical estimate is not a guarantee for a constant input or a coherent periodic signal. More bits increase resolution, but accuracy also depends on voltage-reference calibration, offset, gain error, nonlinearity and electronic noise. Out-of-range signals clip and invalidate the rounding bound.

Clock quality matters too. Timing jitter $\delta t$ gives a voltage error approximately $(dV/dt)\delta t$; for a sinusoid its scale grows with $2\pi f{V}_p{\sigma }_t$. A clock calibration error biases measured frequencies and time intervals. Simultaneous multichannel signals need known relative timing, and the sampling and analogue settling times must be appropriate to the signal bandwidth.

For $n_c$ channels the raw recording rate is approximately $n_c{f}_{s}b$ bits per second, before overhead, so storage and sustained transfer speed must prevent missing data. Store units, channel identities, timestamps, sampling rate, input ranges and calibration information along with the readings. Preserve the original samples when applying digital filters or other processing. Averaging can improve precision for independent fluctuating errors, but cannot restore aliased frequencies, clipped peaks or a fixed rounding bias. Digital recording makes reproducible analysis and transmission possible when its sampling, amplitude and timing limitations are all controlled.

The standard deviations defining the position and momentum uncertainties satisfy $(\Delta x{)}^2=⟨x^2⟩-⟨x{⟩}^2$ and $(\Delta p{)}^2=⟨p^2⟩-⟨p{⟩}^2$. Part (a) makes both means zero. Using the second moments from part (b),Multiplication givesThe Heisenberg uncertainty relation requires only $\Delta x\Delta p\ge \hslash /2$. Among the oscillator energy eigenstates, equality occurs solely for the ground state $n=0$; the excited states have larger uncertainty products.

Assume independent approximately Gaussian log-rate errors with known common standard deviation $s=0.15$. The weighted least squares objective isSince all weights agree, the normal equations for linear least squares reduce toFor these data, $\bar{x}=1.5\mathrm{ps}$, $\bar{y}=4.1596533$, $S_{xx}=4.375{\mathrm{ps}}^2$ and $S_{xy}=-10.3782409\mathrm{ps}$. Hence $\hat{m}=-2.3721693{\mathrm{ps}}^{-1}$ and $\hat{c}=7.7179073$.

The covariance matrix of the fitted slope and intercept follows from the inverse weighted design matrix product, or directly by writing both estimates as linear combinations of the independent $y_i$:This givesThe parameter conversion and first-order propagation of uncertainty areConsequently,These are one-standard deviation errors. The estimates are correlated, since $\mathrm{Cov}(\hat{m},\hat{c})=-0.0077143{\mathrm{ps}}^{-1}$. The minimum is ${\chi }_{\min }^2=4.6537$ for four residual degrees of freedom. Because $s$ is supplied as a known measurement uncertainty, it should not be rescaled by the observed reduced ${\chi }^2$. This is a log-linear exponential-decay fit.

For an independent count with a Poisson distribution, collected for exposure $T_i$ at rate $R_i$, the expected count is $N_i=R_i{T}_i$ and its standard deviation is $\sqrt{N_i}$. The fractional rate uncertainty is consequentlyA target of $0.15$ requires $N_i\simeq (0.15{)}^{-2}=44.44$ counts at each proper-time setting. The exposure time for a Poisson rate at fixed fractional precision is therefore $T_i=1/(0.15^2{R}_i)$, givingSince the settings are acquired independently rather than simultaneously,Nearly three quarters of the exposure is spent at the latest, lowest-rate setting. Proper time in picoseconds labels the decay distribution; it is not the laboratory acquisition time in days. This estimate assumes counting statistics dominate, with no additional background-subtraction uncertainty or acquisition overhead.

A chi-squared test compares the sum of squared standardized regression residuals with the appropriate chi-squared distribution. With both line parameters fitted, the data give ${\chi }^2=4.6537$ on $6-2=4$ degrees of freedom, with upper-tail P-value about $0.325$: the fitted exponential decay model describes the data adequately.

First consider fixing the alternative lifetime at its central value ${\tau }_a=0.478\mathrm{ps}$. Its slope is $m_a=-1/{\tau }_a=-2.09205{\mathrm{ps}}^{-1}$. One must refit the normalization rather than retain the earlier intercept. Minimization over the intercept gives $c_a=\bar{y}-m_a\bar{x}=7.29773$, andon $6-1=5$ degrees of freedom, with upper-tail P-value about $0.00130$. Thus that fixed central lifetime gives a substantially worse fit than $\hat{\tau }=0.4216\mathrm{ps}$. Equivalently, the increase after profiling over the intercept isThis fixed-value comparison does not yet include the uncertainty of the alternative measurement.

For the compatibility of independent lifetime measurements, include both errors. Under the approximately Gaussian lifetime-error model, equality of the two true lifetimes is tested byThis has one degree of freedom and P-value approximately $0.032$, equivalent to a $2.15$-standard deviation discrepancy. The two measurements are therefore in mild tension: equality is rejected at a $5\%$ significance threshold, but not at $1\%$. There is no unique yes-or-no compatibility verdict without choosing such a threshold. Treating the alternative central value as exact would overstate the disagreement between the experiments.

The lifetime that provides the best fit to these six rates remains $\boxed{\tau =0.4216\mathrm{ps}}$, not $0.478\mathrm{ps}$. The compatibility calculation above assumes the two experiments have independent uncertainties; a shared systematic error would require its covariance.

Assume independent normally distributed angle errors with known standard deviation ${\sigma }_y=0.002$. The relevant chi-squared test uses the sum of squared standardized residuals,For the model through the origin, the predictions and residuals, expressed in milliradians, areConsequently ${\chi }_0^2=30.0280$. There are $4-1=3$ statistical degrees of freedom. The upper-tail P-value is approximately $1.36\times 10^{-6}$; equivalently the value greatly exceeds the $5\%$ critical value $7.815$. The zero-intercept model is strongly rejected under the stated errors.

For the alternative linear regression, minimize ${\sum }_i(y_i-c^{\prime }{x}_i-d{)}^2$. Setting both partial derivatives to zero gives the normal equations, whose solution isFor these data $\bar{x}=2508.83699{\mathrm{m}}^{-2}$, $\bar{y}=0.0475$, and $S_{xx}=4.830232\times 10^6{\mathrm{m}}^{-4}$. This yieldsThe new predictions are $(66.6548,55.4171,41.4315,26.4966)\mathrm{mrad}$, with residuals $(-0.6548,-0.4171,2.5685,-1.4966)\mathrm{mrad}$. ThereforeThe affine model is compatible with the data. Allowing for its additional fitted parameter, the decrease $\Delta {\chi }^2=27.668$ has a chi-squared distribution with one statistical degree of freedom under the nested zero-intercept hypothesis, corresponding to $p\simeq 1.44\times 10^{-7}$. Thus the model with a fitted intercept provides the better description, rather than merely lowering the residual by adding a parameter.

With the measurement variance treated as known, the coefficient uncertainties areHence $d=(13.11\pm 2.49)\mathrm{mrad}$, about $5.26$ standard deviations above zero. A positive constant systematic error in the angular zero, or an additional nearly constant torque, could account for it. Its presence is evidence against the ideal zero-intercept model; four points alone do not identify the physical cause uniquely. A calibration error in $\kappa$ only rescales the inferred gravitational constant and cannot remove this residual pattern.

Using the fitted slope after allowing for this offset would give $G^{\prime }=\kappa {\hat{c}}^{\prime }/(mML)=(7.27\pm 0.51)\times 10^{-11}{\mathrm{m}}^3\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{s}}^{-2}$, with the period and slope errors combined in quadrature. This is conditional on the constant-offset model.

Direct differentiation gives ${\psi }^{\prime \prime }/\psi =x^2/a^4-1/a^2$. Inserting this into the Schrödinger equation and separately matching the $x^2$ and constant terms yieldsFor $V_0=m{\omega }_0^2/2$, $a^2=\hslash /(m{\omega }_0)$ and $E=\hslash {\omega }_0/2$. The normalizable Gaussian has no nodes, as expected for the nondegenerate one-dimensional ground state of this confining potential.

The normalization integral is $|C{|}^{2}a\sqrt{\pi }=1$. Choose its irrelevant global phase so thatFor the oscillator ground-state density in a comparison wavelength, choose a free comparison particle with the same total energy. It has $p=\sqrt{2mE}=\hslash /a$, hence comparison wavelength $\lambda =h/p=2\pi a$. ConsequentlyThe bound density is localized on scale $a=\lambda /(2\pi )$, with standard deviation $\lambda /(2\pi \sqrt{2})$, rather than being spatially uniform like the density of a single free plane wave. The comparison does not assign constant free-particle momentum to the bound state; its total energy also includes potential energy.

For the same massless spring of spring constant $k$, Hooke's law gives the vertical simple harmonic motion about the displaced equilibrium position:Gravity shifts the equilibrium extension but does not change this frequency. Use the measured sample means in the ratio estimate. For small measurement errors, logarithmic differentiation givesAssuming the two frequency estimates have independent errors, their variances add. Writing $ϵ=\delta \nu /\overline{\nu }$ for their common fractional standard uncertainty, monomial uncertainty propagation yieldsThis is a first-order uncertainty formula, rather than an exact identity for an arbitrary noisy ratio. It assumes that errors in the two measured frequencies have zero covariance.

There is a statistical distinction in the wording: the standard deviation of individual repetitions is not the standard error of their sample mean. If the supplied $\sigma ({\nu }_i)$ describes individual trials, the frequency means have errors $\sigma ({\nu }_i)/\sqrt{n}$, and the ratio formed from those means instead satisfiesThus the printed formula describes either a single-pair ratio or a convention in which $\delta \nu$ already denotes the uncertainty of each averaged frequency. Under either consistent interpretation, independent repeated measurements reduce statistical error in proportion to $n^{-1/2}$. Halving it requires $4n$ trials in total for each mass, or $3n$ additional trials. Repetition alone does not reduce a fixed systematic error.

Treat the repeated angles as independent measurements of one fixed angle. Their sample mean, sample standard deviation and standard error of the mean areFor $y=A\cos \theta$, uncorrelated standard uncertainties giveInterpreting the stated 1% amplitude accuracy as a standard uncertainty,Use radians in the derivative. This is uncertainty in the result using the mean angle, not the spread of individual angle measurements; the latter convention would use $s_{\theta }$ and give about $2.34\%$. An accuracy bound or systematic angular calibration error would require a different uncertainty model.

The central value is $S=150\cos 45^{\circ }=106.07$. Treating the stated errors as independent standard uncertainties, convert the angular uncertainty to radians and propagate derivatives:The two measured values differ by $10.8/\sqrt{6.0^2+5.6^2}=1.32$ combined standard deviations. Our prediction differs from them by $0.19$ and $1.07$ combined standard deviations respectively. All three values are statistically compatible. If the input errors are strict bounds instead, the first-order bound on the calculated value is about $\pm 9.0$; consistency remains possible, but a Gaussian significance test then has no stated justification.

Two applications of optical interferometry are precision displacement or strain measurement, and Fourier transform spectroscopy, which recovers a spectrum from an interferogram.

For the round-trip phase in a Michelson interferometer, take a lossless 50:50 beam splitter with $t=1/\sqrt{2}$ and $r=i/\sqrt{2}$, and mirror reflection amplitude $-1$. At the depicted detector, each returning path has one beam-splitter transmission and one reflection. Their amplitudes are $-i{A}_0{e}^{2ikD}/2$ and $-i{A}_0{e}^{2ik(D+\delta )}/2$. Consequently,The factor 2 in the phase is the return journey: the labelled $\delta$ is a one-way arm-length difference. The other output port has the complementary intensity. The printed $|A_0{|}^2(1+2\cos k\delta )$ cannot be an intensity: it becomes negative, and does not match the depicted path difference. No choice of normalization fixes that negative minimum. The expression above is the physically consistent result for the drawn output port.

For mutually incoherent spectral components with incident power density $S(k)$,The plot labels are ${\lambda }_1=589.59\mathrm{nm}$ and ${\lambda }_2=590.00\mathrm{nm}$; these printed centres are used here. In the narrow-line, equal-power limit, each line has incident intensity $J$ andThere are rapid carrier fringes with approximately $0.2949\mu \mathrm{m}$ mirror-displacement spacing and a slowly varying Interferometric visibility envelope. The first visibility zero is at $0.2121\mathrm{mm}$, and successive visibility revivals are separated byFor the unequal-line visibility in a spectral doublet, with incident line powers $J_1,J_2$, replace the intensity by $[J_1+J_2+J_1\cos (2k_1\delta )+J_2\cos (2k_2\delta )]/2$. The visibility becomesThe visibility minima rise from zero to $|J_1-J_2|/(J_1+J_2)$. The finite widths drawn in the spectrum also make coherence decay with path difference. For example, Gaussian lines of common wave-number standard deviation ${\sigma }_k$ multiply the oscillatory term by $e^{-2{\sigma }_k^2{\delta }^2}$; the actual linewidth is not specified numerically. Thus the narrow-line plot gives the beat positions, while the true long-delay oscillations die away.^[ref]

The common-arm phase cancellation in a balanced interferometer means there is no smallest positive common mirror distance $D$ in a balanced interferometer. With $\delta =0$ the common factor $e^{2ikD}$ cancels from every intensity, and both paths interfere constructively at this detector for every $D>0$. Their infimum is zero. The $0.4242\mathrm{mm}$ beat length describes changes of arm difference, not of the common arm length. It therefore cannot answer the final request as printed.

For the uncertainty product of oscillator number states, the two means vanish, so the standard deviations are the square roots of the second moments:Consequently,Only the ground state $n=0$ saturates the Heisenberg uncertainty principle in this sequence of energy eigenstates; the excited number states have a larger uncertainty product.

The optical telescope supplies collecting area and spatial selection, the spectral selector supplies a passband, and the photodetector converts received radiant flux into a measured signal. For an extended source, a useful schematic photon count iswhere $A\Omega$ is the etendue, $\tau$ the integration time, $\eta$ the net transmission and quantum efficiency, and $L_{\lambda }$ the spectral radiance. An illustrative independent-noise model givesHere $N_b$ and $N_d$ denote background and dark counts, and ${\sigma }_{\mathrm{read}}$ is readout standard deviation. This makes the roles of the three components explicit.

A larger optical telescope collects more photons at fixed angular pixel size and improves the diffraction-limited resolution. Its limits include optical aberrations, alignment, launch size, stray light and pointing. Conservation of etendue constrains simultaneous collecting area and angular acceptance. Increasing spatial resolution by narrowing a pixel's solid angle reduces its photons.

A narrower spectral passband or higher spectral resolving power separates diagnostic features and rejects unwanted background, but usually reduces photons per channel and hence signal-to-noise ratio. Selector transmission, leakage, slit width, detector format and the required time cadence limit the trade. A wider band helps only insofar as it adds useful signal rather than mostly background.

The photodetector's quantum efficiency, dark current, read noise, dynamic range and response time determine conversion losses and added noise. Cooling can suppress thermal background and dark current, but cannot remove photon shot noise. Longer integration improves a stationary signal's signal-to-noise ratio until motion, saturation, drift or required cadence becomes limiting.

For a general assumed speed ratio $r=v_A/v_B$,When $r=2$, each station estimates $t_0$ by $t_A-\Delta t$. Averaging givesThe individual estimates have sample standard deviation $2.74\mathrm{s}$ and range $11$ to $19\mathrm{s}$. If station clock errors are independent with zero mean, the standard error of their mean is $2.74/\sqrt{13}=0.76\mathrm{s}$. Thus one may quote $15.0\pm 0.8\mathrm{s}$ as a conditional one-standard-error mean, with about $\pm 3\mathrm{s}$ scatter for an individual station. The problem does not bound a common clock bias: adding the same offset to every clock shifts this inferred origin time by that offset without changing any time difference. Therefore no unconditional absolute-time accuracy follows from the supplied data.

The speed-ratio check is an origin-time and speed-ratio regression of $t_A$ on $\Delta t$. It giveswhere the uncertainty is a residual-based standard error of slope. The factor-two hypothesis predicts slope $1$, about $1.13$ standard errors from the fitted value: the ratio 2 is satisfactory at the accuracy allowed by these clocks, but is not precisely verified. Formally $r=1+1/\text{slope}\simeq 2.37$, with a local propagated standard error about $0.44$. This inference assumes clock errors have no systematic dependence on station position or time difference. With completely unrestricted station offsets, a regression cannot identify $r$ at all.

Let $K=(1/v_B-1/v_A)=\sqrt{M}$. The difference curve determines $K$, $x_0$, and $z_0$, regardless of the particular value assigned to $r$. Changing the ratio to $2.25$ while refitting the individual speeds givesHenceThe conditional standard error of this new mean is $0.72\mathrm{s}$. Its predicted slope $0.8$ is also compatible with the regression. Holding one speed fixed while changing $r$ would alter the depth, but that adds a constraint not present in this question: here both speeds are inferred from the same difference data and chosen ratio.

Assume $a,b>0$, the standard deviations. Substitute the two normal densities into (ii):The last step uses ${\int }_0^{\infty }u{e}^{-A{u}^2/2}du=1/A$. With $c=b/a$ this is $\boxed{f_V(v)=c/[\pi (c^2+v^2)]}$, the centered Cauchy density of scale $c$.

The desired event is $|V|<1$, outside the null event $X=0$. Integrating the density gives

A linear combination of independent normal random variables is normal, with variances adding when the coefficients are $1,-1$. Applying the joint central limit approximation from (i) givesEquivalently the difference minus $p-q$, divided by the displayed standard deviation, converges in distribution to a standard normal under the same regular large-sample regime.

Generate independent standard normal variables $Z_1,Z_2$ and put $\boxed{Y=Z_1/Z_2}$. The event $Z_2=0$ has probability zero. The ratio calculation in M4 with equal standard deviations gives density $1/[\pi (1+y^2)]$, exactly the required $t$ distribution with one degree of freedom. The equivalent form $Z_1/|Z_2|$ has the same law by symmetry.

For a nonzero quantum state $\psi$ and a self-adjoint operator $A$, the expectation value and standard deviation of its measurement areThe squared deviation is also called the variance of the observable. All displayed products are understood on quantum states for which the relevant moments exist. Self-adjointness gives $\overline{⟨\psi ,A\psi ⟩}=⟨A\psi ,\psi ⟩=⟨\psi ,A\psi ⟩$, so $a$ is real. Expanding the squared norm gives the variance expression. A norm is nonnegative and is zero exactly for the zero vector, hence ${\Delta }_{\psi }(A)=0$ precisely when $A\psi =a\psi$. Thus zero dispersion is equivalent to being an eigenstate.

For bounded $H$, the operator exponential is the norm-convergent power seriesFor a general unbounded self-adjoint Hamiltonian operator, the spectral theorem defines the same notation byThis definition does not require the power series to converge on every quantum state. complex conjugation of the spectral multiplier gives $U_t^{\ast }=U_{-t}$, and multiplication of the phases gives $U_{-t}{U}_t=I$. Thus $U_t$ is unitary, preserves the quantum state norm, and has $U_0=I$.

For initial data in the domain of $H$, differentiating the spectral expression yields ${\partial }_t{U}_t{\psi }_0=-(i/\hslash )H{U}_t{\psi }_0$. Consequentlysolves the Schrödinger equation with the prescribed initial quantum state. Uniqueness follows by differentiating $U_{-t}{\psi }_t$: the two terms cancel, leaving the constant ${\psi }_0$.

In the two-dimensional energy basis, expand the initial quantum state as ${\psi }_0=c_1{\Psi }_1+c_2{\Psi }_2$. Each energy eigenvector receives its own phase under the operator exponential, givingThis is the general solution because every initial vector has a unique basis expansion.

In that orthonormal basis the other observable has the real symmetric matrixIts conjugate transpose equals itself, so it is self-adjoint. Also $A^2=I$ and $\mathrm{tr}A=0$, giving eigenvalues $+1$ and $-1$. Direct substitution gives the normalized orthogonal eigenvectorsAn irrelevant global phase may multiply either vector. With the chosen lower eigenstate initially,The Born rule says that the probability for the higher eigenstate is $|⟨v_{+},{\psi }_t⟩{|}^2$. Its amplitude is $\sin \alpha \cos \alpha (e^{-it{E}_2/\hslash }-e^{-it{E}_1/\hslash })$. Using $|e^{-iu}-e^{-iv}{|}^2=4{\sin }^2((u-v)/2)$ givesThis is the transition probability for a rotated two-level observable. Its time dependence comes from the relative phase between two energy eigenstates. If their energies coincide, or if the measured observable has the energy eigenvectors as its eigenvectors, this transition probability is zero at all times.

Let $a={\mathbb{E}}_{\psi }(A)$, which is real. The dispersion, or quantum uncertainty, is the nonnegative standard deviationIndeed, self-adjointness and normalization giveA squared norm is nonnegative, so the square root is defined. It vanishes precisely when $(A-aI)\psi =0$, namely when $\psi$ is an eigenvector of $A$. Conversely, an eigenvector with eigenvalue $\lambda$ has $a=\lambda$ and hence zero dispersion. Thus zero dispersion is equivalent to being an eigenstate.