Lecture 2: Quantum States, Operators, and Approximation

Lecture 1 showed why quantum matter requires structured state spaces, collective variables, and effective descriptions with declared limits. Those claims depend on quantum objects that ordinary calculations often leave implicit: rays rather than coordinate columns, operators rather than their matrices, projectors rather than basis labels, and approximations with declared error scales. This lecture reconstructs the parts of quantum mechanics needed to audit later changes of basis, sector, and scale. A representation enables calculation, whereas a physical claim must survive every representation change that leaves the underlying object unchanged.

Lecture 1 used a local spin model to show that an exponentially large state space can possess simple collective organization. It closed by naming what still had to be distinguished: a state from its coordinates, an operator from its matrix, and a justified reduction from a plausible truncation. Undergraduate quantum mechanics supplies the words for each of these, and a two-level calculation is routine. The words are enough until a calculation changes representation. Take one electron shared between two localized orbitals, as in a double quantum dot, and prepare it on the left. A charge measurement finds it on the left with probability one, and any correct calculation must return one in any basis. Rotate the basis, redo the same arithmetic, and it can return a probability smaller than one. Call a coupling “small,” drop a level, and the retained energies carry an error that nothing in the notation records. Familiar notation conceals which object is held fixed, which array is allowed to change, and which approximation has been compared with an exact answer.

The governing question is:

Which pieces of undergraduate quantum mechanics must be reconstructed and used to analyze quantum solids reliably?

The answer is built on that two-orbital system. Each of those words receives an exact object with a declared space, and every prediction is calculated from those objects and then tested: by changing basis, by imposing a symmetry, and by comparing an approximation with the exact answer.

From memory, write one sentence for each of the following: state vector, eigenvalue, expectation value, commutator, unitary operator, and perturbative denominator. Beside each sentence, record whether you are confident. High-confidence wrong answers are the main target of this reconstruction; low-confidence omissions can usually be repaired from the definitions.

States and Operators Need Mathematical Habitats

Rays, operators, spectral projectors, and invariant subspaces are the objects that survive a change of coordinates; column vectors and matrices are the representatives used to calculate with them. The two-orbital system is the smallest on which each of these objects has an immediate use, and every identity derived on it in finite dimension is exact. The payoff is the ability to say which statements survive a change of representation and when an exact diagonalization must replace an expansion. The lecture’s one reduction, eliminating a remote level, is where that control must be earned: the closing section states the gap ratio and error target that license it and shows where, near degeneracy, it fails.

For one electron in two localized orbitals, labeled left and right, the state space of the system is the two-dimensional complex Hilbert space

$\mathcal H_{\mathrm{orb}} =\operatorname{span}_{\mathbb C}\{|L\rangle,|R\rangle\} \cong\mathbb C^2,$

with inner product ${\langle\phi|\psi\rangle\in\mathbb C}$ . We use the physics convention: the inner product is conjugate-linear in its first argument and linear in its second. The ordered pair ${\mathcal B=(|L\rangle,|R\rangle)}$ is an orthonormal basis.

Let ${\mathcal H}$ be a complex Hilbert space. A pure state is a ray $\Psi=[\psi] =\{c|\psi\rangle:c\in\mathbb C\setminus\{0\}\} \in\mathbb P(\mathcal H)$ represented by any nonzero vector ${|\psi\rangle\in\mathcal H}$ . Calculations normally use a normalized representative, ${\langle\psi|\psi\rangle=1}$ . Multiplying that representative by a global phase ${\mathrm e^{\mathrm i\alpha}}$ does not change the ray [Mac80, p. 612].

A column such as ${\boldsymbol\psi_{\mathcal B}={(\psi_L,\psi_R)}^{\mathsf T}}$ is not the abstract state. It is the coordinate vector of one representative in the basis ${\mathcal B}$ :

$|\psi\rangle=\psi_L|L\rangle+\psi_R|R\rangle.$

The map from an abstract vector to coordinates depends on the basis. The inner product does not.

Operators act on states. The next three definitions fix what an operator is, when its adjoint exists, and which operators represent measurable quantities.

A linear operator on a Hilbert space is a linear map $\hat A:\mathcal D(\hat A)\subseteq\mathcal H\longrightarrow\mathcal H.$ Its domain ${\mathcal D(\hat A)}$ is part of the object. In finite dimension, every everywhere-defined linear operator is bounded and one ordinarily has ${\mathcal D(\hat A)=\mathcal H}$ . For unbounded operators such as position or momentum on ${L^2(\mathbb R^d)}$ , the domain cannot be omitted. Whenever an adjoint is used below, the operator is assumed densely defined [Gie00, arXiv v2 evidence copy, §§2.2.1–2.2.3; pagination and numbering collated by work identity, not by a line-by-line journal comparison].

Measurement outcomes are real numbers, and the operators whose eigenvalues can serve as outcomes are singled out by their relation to the inner product.

For a densely defined operator ${\hat A}$ , the adjoint domain ${\mathcal D(\hat A^\dagger)}$ consists of the vectors ${|\phi\rangle\in\mathcal H}$ for which there exists a vector ${|\chi\rangle\in\mathcal H}$ satisfying $\langle\phi|\hat A\psi\rangle=\langle\chi|\psi\rangle \quad\text{for every }|\psi\rangle\in\mathcal D(\hat A).$ The representing vector is unique; define ${\hat A^\dagger|\phi\rangle:=|\chi\rangle}$ . An operator is self-adjoint when ${\mathcal D(\hat A)=\mathcal D(\hat A^\dagger)}$ and ${\hat A|\psi\rangle=\hat A^\dagger|\psi\rangle}$ throughout that domain. In finite dimension this reduces to the familiar Hermitian-matrix condition in every orthonormal basis.

The definition of an observable adds the physical reading: outcomes are eigenvalues, the alternatives a measurement distinguishes are eigenspaces, and neither is the apparatus.

In the finite-dimensional sharp-observable setting used in the core, a quantum observable is represented by a self-adjoint operator ${\hat A:\mathcal H\to\mathcal H}$ , equivalently by its projection-valued spectral alternatives ${\{(a,\hat P_a)\}}$ in ${\hat A=\sum_a a\hat P_a}$ . This mathematical observable is not the apparatus, a realized pointer or recorded outcome, or a statistic computed from repeated outcomes; connecting these objects requires a measurement model and an explicit calibration/readout map. Generalized observables represented by POVMs lie outside the present core treatment [Gri12, secs. 1.5, 2.2, 3.5–3.9; BL08, pp. 1–4; Met08, chaps. 2.23–C.2.26].

The double-dot Hamiltonian and a charge-difference observable are

$\hat H(\epsilon) =\frac{\epsilon}{2}\hat\sigma_z+t\hat\sigma_x, \qquad \hat Q=\hat\sigma_z,$

both maps from ${\mathcal H_{\mathrm{orb}}}$ to itself. Their units differ: ${\hat H}$ has energy units, whereas ${\hat Q}$ is dimensionless in this normalization.

Choose ${\hat\sigma_z|L\rangle=|L\rangle}$ and ${\hat\sigma_z|R\rangle=-|R\rangle}$ . Relative to their common mean-energy zero, the uncoupled orbital energies are ${+\epsilon/2}$ and ${-\epsilon/2}$ : the real detuning ${\epsilon}$ is the left-orbital energy minus the right-orbital energy. The real parameter ${t=\langle L|\hat H|R\rangle}$ is the interorbital mixing matrix element, the tunneling energy, in this phase convention. At ${t=0}$ the orbitals do not mix; at ${\epsilon=0}$ their uncoupled energies coincide. The observable ${\hat Q}$ gives the normalized occupancy difference, with outcomes ${+1}$ and ${-1}$ , not a charge in coulombs.

Spectral Projectors Turn Alternatives into Probabilities

Projectors and spectral alternatives

The Born rule needs the alternatives that a measurement of ${\hat A}$ can distinguish. For a self-adjoint operator on a finite-dimensional space those alternatives are its eigenspaces, and the following theorem guarantees that they exhaust the space.

Let ${\mathcal H}$ be a finite-dimensional complex Hilbert space and let ${\hat A:\mathcal H\to\mathcal H}$ be self-adjoint. Then every eigenvalue of ${\hat A}$ is real, eigenspaces belonging to distinct eigenvalues are orthogonal, and their direct sum is ${\mathcal H}$ . Equivalently, ${\hat A}$ admits the degeneracy-safe projector decomposition in Equation (2.1). The statement does not by itself cover an unbounded operator or a continuous spectrum [Gri12, secs. 3.5–3.8].

The corresponding operator identities are

(2.1)

$\hat A=\sum_a a\hat P_a, \qquad \hat P_a^\dagger=\hat P_a, \qquad \hat P_a\hat P_b=\delta_{ab}\hat P_a, \qquad \sum_a\hat P_a=\hat I.$

For an eigenvalue ${a}$ of a finite-dimensional self-adjoint operator ${\hat A}$ , the spectral projector ${\hat P_a}$ is the orthogonal projector onto the full eigenspace ${\ker(\hat A-a\hat I)}$ . If ${{\left\{|a,\alpha\rangle\right\}}_{\alpha=1}^{m_a}}$ is any orthonormal basis of that eigenspace, then $\hat P_a=\sum_{\alpha=1}^{m_a} |a,\alpha\rangle\langle a,\alpha|.$ The operator is independent of the chosen basis inside a degenerate eigenspace.

To build the spectral projectors of ${\hat H}$ without first solving for eigenvectors, introduce the classical coefficient vector

$\mathbf h=(t,0,\epsilon/2)\in\mathbb R^3, \qquad h:=|\mathbf h| =\sqrt{t^2+\epsilon^2/4}, \qquad \mathbf n:=\frac{\mathbf h}{h},\quad h>0.$

The bold upright ${\mathbf h}$ is a classical vector of coefficients. The hatted vector ${\hat{\boldsymbol\sigma}=(\hat\sigma_x,\hat\sigma_y,\hat\sigma_z)}$ has operator-valued components. With these conventions, ${\hat H=\mathbf h\mathbin{\cdot}\hat{\boldsymbol\sigma}}$ .

For ${\mathbf a,\mathbf b\in\mathbb R^3}$ , the Pauli product rule gives

$(\mathbf a\mathbin{\cdot}\hat{\boldsymbol\sigma}) (\mathbf b\mathbin{\cdot}\hat{\boldsymbol\sigma}) =(\mathbf a\mathbin{\cdot}\mathbf b)\hat I +\mathrm i(\mathbf a\times\mathbf b) \mathbin{\cdot}\hat{\boldsymbol\sigma}.$

Setting ${\mathbf a=\mathbf b=\mathbf n}$ makes the cross product zero and yields

${\left(\mathbf n\mathbin{\cdot} \hat{\boldsymbol\sigma}\right)}^2=\hat I,$

and this single identity is all that the energy projectors need.

Degeneracy and complete energy alternatives

The hypothesis ${h>0}$ matters. At ${h=0}$ , both ${t}$ and ${\epsilon}$ vanish, so ${\hat H=0}$ has the single eigenvalue zero and the sole spectral projector ${\hat I}$ . There is then no canonical rank-one ${\pm}$ split. The following formulas apply only when ${h>0}$ .

Because ${{(\mathbf n\mathbin{\cdot}\hat{\boldsymbol\sigma})}^2=\hat I}$ , define the two operators

(2.2)

$\hat P_\pm =\frac12\left( \hat I\pm\mathbf n \mathbin{\cdot}\hat{\boldsymbol\sigma} \right)$

They satisfy ${\hat P_\pm^2=\hat P_\pm}$ , ${\hat P_\pm^\dagger=\hat P_\pm}$ , ${\hat P_+\hat P_-=0}$ , and ${\hat P_++\hat P_-=\hat I}$ . Moreover,

$\hat H\hat P_\pm =\pm h\hat P_\pm,$

so they are the spectral projectors and

$\hat H=h\hat P_+-h\hat P_-.$

Because the projectors are built from ${\mathbf n}$ alone, the only division is by the norm ${h}$ , which vanishes only at the excluded point ${h=0}$ ; an eigenvector written as a ratio of components fails wherever the component in its denominator vanishes.

Born probabilities, expectations, and repeatability

For a normalized state representative ${|\psi\rangle}$ , the probability of the energy alternative ${E_\pm=\pm h}$ is

(2.3)

$p_\pm(\psi) =\langle\psi|\hat P_\pm|\psi\rangle ={\left\lVert\hat P_\pm|\psi\rangle\right\rVert}^2.$

Equation (2.3) is the Born rule expressed with projectors [Bor26, pp. 865–866]. Positivity follows from the norm, and completeness gives ${p_++p_-=1}$ .

Prepare ${|L\rangle}$ and take ${t=4u}$ , ${\epsilon=6u}$ , where ${u>0}$ is an energy unit. Find the probabilities for an immediate energy measurement and, in a separate identically prepared trial, for a charge measurement, using ${\hat H=(\epsilon/2)\hat\sigma_z+t\hat\sigma_x}$ in the ordered localized basis ${(|L\rangle,|R\rangle)}$ .

Projectors select the alternatives of the measured operator; they need not select the same alternatives for these two questions.

First keep the parameters symbolic and use ${\hat P_\pm=(\hat I\pm\hat H/h)/2}$ with ${h=\sqrt{t^2+\epsilon^2/4}>0}$ ; the readout rule is ${p_\pm(L)=\langle L|\hat P_\pm|L\rangle}$ . For the prepared vector ${|L\rangle}$ , use ${\langle L|\hat\sigma_z|L\rangle=1}$ and ${\langle L|\hat\sigma_x|L\rangle=0}$ :

$p_\pm(L) =\frac12\left(1\pm\frac{\epsilon/2} {\sqrt{t^2+\epsilon^2/4}}\right).$

At large positive detuning, ${p_+(L)\to1}$ ; at zero detuning, ${p_+(L)=p_-(L)=1/2}$ . The result connects a preparation to energy alternatives, not to thermal populations.

For the specified values, ${h=5u}$ and the localized-basis projectors are $\mathsf P_+=\frac1{10}\begin{pmatrix}8&4\\4&2\end{pmatrix},\qquad \mathsf P_-=\frac1{10}\begin{pmatrix}2&-4\\-4&8\end{pmatrix}.$ Their first diagonal entries give ${p(E=5u)=4/5}$ and ${p(E=-5u)=1/5}$ . Charge instead uses ${\mathsf P_L=\operatorname{diag}(1,0)}$ and ${\mathsf P_R=\operatorname{diag}(0,1)}$ , yielding ${p_L=1}$ , ${p_R=0}$ . The two probabilities refer to separate experiments on the same initial state, not to a sequential charge readout after an energy measurement.

Direct multiplication gives ${\mathsf P_+^2=\frac1{100}\begin{pmatrix}80&40\\40&20\end{pmatrix} =\mathsf P_+}$ , while addition gives ${\mathsf P_++\mathsf P_-=\mathsf I}$ . The listed probabilities are nonnegative and sum to one for each observable. The symbolic zero-detuning limit gives equal energy weights, while at ${t=\epsilon=0}$ the only energy projector is ${\mathsf I}$ and the energy is certainly zero. The latter case must not be evaluated by dividing by ${h=0}$ . Energy values carry units of ${u}$ ; projectors and probabilities are dimensionless. A readout prediction therefore requires both a preparation and a specified observable.

For an observable ${\hat A}$ and a normalized vector ${|\psi\rangle\in\mathcal D(\hat A)}$ , the expectation value and variance are ${\langle\hat A\rangle}_\psi :=\langle\psi|\hat A|\psi\rangle, \qquad \operatorname{Var}_\psi(\hat A) :={\left\lVert \left(\hat A-{\langle\hat A\rangle}_\psi\hat I\right)|\psi\rangle \right\rVert}^2.$ The expectation is an average prediction for repeated identical preparations; it need not be one of the possible single-shot outcomes. It has the units of ${\hat A}$ , while the variance has their square. Writing the variance instead as an expectation of ${\hat A^2}$ requires the stronger condition ${|\psi\rangle\in\mathcal D(\hat A^2)}$ .

The prepared state has a sharp charge: for ${|L\rangle}$ and ${\hat Q=\hat\sigma_z}$ , ${\langle\hat Q\rangle}_L=1, \qquad \operatorname{Var}_L(\hat Q)=0.$ For an energy eigenstate at ${\epsilon=0}$ , with nonzero tunneling, ${\langle\hat Q\rangle=0}$ but ${\operatorname{Var}(\hat Q)=1}$ . Zero expectation therefore does not mean that every measurement returns zero; here the outcomes ${\pm1}$ are equally probable.

Return to the energy experiment of Example 2.1: prepare ${|L\rangle}$ independently before each readout and keep ${t>0}$ fixed. Write each energy outcome as ${E_{\mathrm{out}}=hx}$ , where ${h=\sqrt{t^2+\epsilon^2/4}>0}$ and ${x=\pm1}$ . This energy label is not the charge observable ${\hat Q}$ .

For a record of ${N}$ outcomes, the empirical mean is ${\bar x=N^{-1}\sum_i x_i}$ , and the empirical variance is ${v_x=N^{-1}\sum_i{(x_i-\bar x)}^2=1-\bar x^2}$ . The variance here uses denominator ${N}$ . Both statistics are dimensionless; multiplying them by ${h}$ and ${h^2}$ , respectively, gives the empirical energy mean and variance. For the dimensionless statistics, the theoretical counterparts are ${2p_+-1}$ and ${1-{(2p_+-1)}^2}$ .

Set ${\epsilon=0}$ and prepare ${|L\rangle}$ independently before each of ${N=20}$ energy readouts. Predict ${p_+}$ and ${p_-}$ . Must the record contain exactly 10 outcomes of each sign, and must its empirical energy mean vanish? Make the prediction before inspecting the record.

Sample energy readouts of independently prepared ${|L\rangle}$ states; each sign labels ${E_{\mathrm{out}}=\pm h}$ . At zero detuning both signs have probability one half, but twenty trials need not split ten and ten. Step through a record, then change the seed. The running fraction can equal ${p_+}$ , move away from it, and return; convergence over many trials is not monotonic. Raise the detuning ratio and compare the changed probability with the empirical mean ${\bar x}$ and variance ${v_x}$ . The displayed record is simulated, not experimental data.

Now change the experiment before inspecting Figure 2.1: prepare a zero-detuning energy eigenstate with nonzero tunneling independently before each trial and measure the charge observable ${\hat Q}$ , not the energy. For ${N=20}$ , predict the probabilities and decide whether exactly 10 left and 10 right outcomes are required. The energy sampler above does not model this charge sensor; the two experiments happen to have equal probabilities at zero detuning despite their different preparations and observables.

Figure 2.1. An opened RF sample package has five selected bonds labeled W1-W5, matched to an oblique subsurface view and registered cutaway of one semiconductor device. W1 reaches the QPC drain, W2 its source and common, and W3-W5 the left, barrier and right surface gates. Dashed L/R regions and the separate QPC are at the same buried electronic depth, with electrostatic sensing but no electron-transfer path between target and sensor. The cutaway omits overlying material for illustration and exaggerates depth threefold; the surface normal remains upright in both views. The RF chain feeds a within-trial calibration rule and an illustrative twenty-trial record with eleven L and nine R outcomes.

Panel (a) shows a representative centimeter-scale RF package, not an engineering drawing. The die maps to the separately scaled oblique view and registered cutaway in panel (b), which show neighboring regions of one semiconductor device. The labels identify connections across distinct physical carriers, not single wires that run unchanged into the microscopic device. A lead meets a feedthrough pin, a PCB metal route reaches a package bond pad, a fine wire bond joins that pad to a die pad, and patterned on-chip metal reaches a gate or an ohmic contact. The short die fan-in in (a) ends at a schematic scale break; the matching labels in (b) identify its device terminals.

W1 connects the coax center conductor through the series inductor to the QPC drain. W2 connects the QPC source to the PCB common plane through a pad and via; the coax shield and gate-voltage reference share this common. The matching capacitor is a separate shunt to common. W3, W4 and W5 connect to the left plunger, interdot barrier and right plunger metal gates, respectively, not to the dashed charge-site circles. This selected RF circuit has zero applied DC source–drain bias. Loading, confinement and QPC-tuning connections are omitted; the five bonds are not a complete device pinout.

The gold surface gates shape the potential in a buried electron layer. The upper oblique view projects the electronic regions through the surface; its dashed outlines mark possible localization sites in gaps between the depletion gates: ${\mathrm L=(1,0)}$ and ${\mathrm R=(0,1)}$ are the two one-electron charge-basis configurations and possible ideal charge outcomes. The circles do not depict two simultaneously present particles, nor do they exclude a premeasurement superposition of the charge-basis states. The QPC lies beside the double dot in the same electronic layer. Depleted regions separate their conducting regions: the dotted links denote electrostatic sensitivity, not wires or electron transfer into the detector. The A-A′ cut boundary registers the two views. Removing the front surface and its gate metal exposes the common electronic depth; this is an illustrative cutaway, not an etched trench. Its depth is exaggerated by three. For this representative GaAs/AlGaAs structure, the surface-to-electron-layer depth is selected as 100 nm from Reilly’s device description, not asserted for Petersson’s specimen; lateral gate shapes are schematic, not a fabrication mask [RMHG07b, pp. 162101–1 and Fig. 1(a)]. The QPC responds differently to the two charge positions. External gate-voltage controls are distinct from the reflected RF readout path in panel (c); its unarrowed common connection is a shared reference/return, not another adjustable sensor-bias source.

For trial ${i}$ , a selected phase-referenced quadrature ${v_i(\tau)}$ is averaged over one integration window to form ${s_i}$ . This is a within-trial average, not an average over the twenty preparations. Calibration fixes the signal sign and threshold ${s_*}$ : in the convention drawn, ${s_i<s_*}$ is assigned ${\mathrm L}$ , otherwise ${\mathrm R}$ . This rule produces the instrument’s assigned outcome ${\omega_i\in\{\mathrm L,\mathrm R\}}$ . Connecting that label to the Born probabilities assumes ideal projective charge readout; finite bandwidth, charge evolution during integration and assignment errors are omitted here. The cited experiments ground the device and sensing principle, not this idealized zero-detuning twenty-trial protocol.

Panel (d) shows an illustrative sequence constructed for this figure; it is not measured data or a simulation of the analog sensor noise. Its 11 left and 9 right records give the estimators ${p_{\mathrm L,\mathrm{est}}=11/20}$ and ${p_{\mathrm R,\mathrm{est}}=9/20}$ . Unequal finite counts do not contradict the model probabilities ${p_{\mathrm L}=p_{\mathrm R}=1/2}$ . The Born probabilities describe the distribution sampled by repeated identical preparations; they do not prescribe the exact composition of one finite data set [PPLG10b, arXiv v2 evidence copy, pp. 1–2, pagination collated to the published work identity; RMHG07b, arXiv v2 evidence copy, pp. 1–3, pagination collated to the published work identity; Vig+23, arXiv v1 evidence copy, §§II–III, VI.B.1, numbering collated to the published work identity].

The figure therefore separates five objects that are often collapsed in speech: the prepared ray, the charge projector, the analog sensor record, the calibrated outcome, and the relative-frequency estimator. The apparatus is not ${\hat Q}$ , and ${p_{\mathrm L,\mathrm{est}}}$ is not the single-shot result.

Hide Equation (2.2) and reconstruct it from ${{\left(\mathbf n\cdot\hat{\boldsymbol\sigma}\right)}^2=\hat I}$ . Then prove idempotence, completeness, and ${p_++p_-=1}$ . If a sign fails, locate it before proceeding; the gap ${2h}$ and the projector geometry return in the section on operators and their matrices.

Time Evolution from the Spectral Projectors

The same projectors also predict motion. The question changes from what a measurement returns at one instant to how the state moves between measurements. The coordinate comparison in the section on operators and their matrices uses only the static preparation and projector probabilities already established, and a reader may pass directly to it. For a time-independent self-adjoint Hamiltonian, the evolution operator is $\hat U(\tau)=\exp\!\left(-\frac{\mathrm i}{\hbar}\hat H\tau\right).$ It acts on the orbital Hilbert space; time is written ${\tau}$ so that ${t}$ can continue to denote the tunneling energy. Inserting the spectral decomposition gives

(2.4)

$\hat U(\tau) =\mathrm e^{-\mathrm i h\tau/\hbar}\hat P_+ +\mathrm e^{+\mathrm i h\tau/\hbar}\hat P_-.$

Orthogonality and completeness imply ${\hat U^\dagger\hat U=\hat I}$ : the evolution preserves the state norm. Norm preservation is the unitary property; the section on unitary operations and symmetry gives its formal definition.

Prepare ${|L\rangle}$ at ${\tau=0}$ , set ${\epsilon=0}$ , and determine the probability of finding the electron on the right at time ${\tau}$ under ${\hat H=t\hat\sigma_x}$ , with real tunneling energy ${t}$ . For ${t\ne0}$ , find the first complete-transfer time.

Use the spectral phases already established, ${\hat U(\tau)=\mathrm e^{-\mathrm i h\tau/\hbar}\hat P_+ +\mathrm e^{\mathrm i h\tau/\hbar}\hat P_-}$ . For ${t\ne0}$ , ${h=|t|}$ and ${\hat P_\pm=(\hat I\pm(t/|t|)\hat\sigma_x)/2}$ , so adding the two phase factors gives $\hat U(\tau)=\cos(t\tau/\hbar)\hat I -\mathrm i\sin(t\tau/\hbar)\hat\sigma_x.$ Apply this operator to the preparation, using ${\hat\sigma_x|L\rangle=|R\rangle}$ : $\hat U(\tau)|L\rangle =\cos(t\tau/\hbar)|L\rangle-\mathrm i\sin(t\tau/\hbar)|R\rangle.$ The squared right amplitude is therefore $p_{L\to R}(\tau)=\sin^2(t\tau/\hbar).$ The left and right probabilities add to one, independently checking norm preservation in these coordinates. At ${\tau=0}$ the right probability vanishes, and the continuous limit ${t=0}$ gives ${\hat U=\hat I}$ and no transfer without evaluating undefined rank-one projectors. The argument ${t\tau/\hbar}$ is dimensionless. The first positive maximum of this sine squared satisfies ${|t|\tau/\hbar=\pi/2}$ . For nonzero tunneling, the first complete transfer occurs at ${\pi\hbar/(2|t|)}$ ; the probability’s angular frequency ${2|t|/\hbar}$ agrees with the energy gap divided by ${\hbar}$ . The formula predicts the statistics of repeated readouts of a coherently prepared state, not a thermal occupation or an irreversible transport rate.

From coherent transfer to a measured charge record

The worked example predicts a probability, not the output of a detector. Before modeling the readout, ask which experimental control changes that probability and how the detector estimates it. The generic double-dot geometry in Figure 2.1, panels (a) and (b), supplies the spatial relation between two localized states and a nearby charge sensor. The reference device differs from that schematic in geometry and in readout: its detector is a quantum point contact, not the RF sensor chain of panel (c).

Petersson and collaborators prepared one electron in a GaAs/AlGaAs double dot, pulsed the detuning toward resonance, and inferred the left-dot occupation from a quantum-point-contact (QPC) signal averaged over repeated pulse cycles. Figure 2.2 separates this control and measurement sequence.

Petersson and collaborators start with the electron on the right, whereas Example 2.2 starts on the left. At resonance the ideal transition probability is symmetric under that exchange; the calibrated average still need not equal the ideal probability at the end of a pulse. [PPLG10b, pp. 246804-2–246804-3, pulse preparation and time-averaged QPC readout]

Figure 2.2. From detuning control to averaged charge readout. The pulse is idealized; the boxes describe the measurement principle, not a reconstruction of the electronics. [PPLG10b, pp. 246804-2–246804-3]

Now predict what increasing the tunnel splitting should do to the period; the contrast is discussed after the figure. Figure 2.3 shows an original stochastic simulation of this type of experiment, not recovered experimental points. The two input splittings are taken from the paper; pulse imperfections, detuning fluctuations, relaxation and repeated-cycle readout are modeled with reported scales and explicitly assumed nuisance parameters. Circles and plus signs distinguish settings, not uncertainty intervals. A noisy calibrated estimate may be negative even though the underlying occupation probability cannot be. See the reference experiment, Petersson et al. (2010) for the measured traces.

Figure 2.3. Original charge-qubit simulation under assumed experimental conditions. Finite pulses, control noise, relaxation, cycle averaging and detector noise modify the ideal transfer probability. Device and charge-noise scales follow the reference experiment; pulse-transfer and detector settings are assumptions, not a calibrated reconstruction. No published trace is copied or fitted. [PPLG10b, fig. 3(c), p. 246804-3, and Fig. 1(d), p. 246804-2]

Move the tunneling energy and the elapsed time and read ${p_{\mathrm L\to\mathrm R}(\tau)=\sin^2(t\tau/\hbar)}$ against the clock. The default setting reproduces the smaller reported splitting; raise the tunneling to the larger one and compare the periods with the period comparison below. The first complete transfer occurs at ${\pi\hbar/(2|t|)}$ , so a larger splitting transfers the electron sooner, not further.

For the resonant Hamiltonian, the splitting is ${\Delta E=2|t|}$ , so the ideal probability period is ${T=2\pi\hbar/\Delta E}$ . Compare this prediction with the authors’ reported time-domain frequencies. This separate table quotes experimental results from the paper; it is not inferred from the simulation:

Spectroscopic ${\Delta E/(2\pi\hbar)}$	Predicted ${T}$	Reported oscillation frequency	Corresponding period
${3.3\,\mathrm{GHz}}$	${303\,\mathrm{ps}}$	${3.4\,\mathrm{GHz}}$	${294\,\mathrm{ps}}$
${4.5\,\mathrm{GHz}}$	${222\,\mathrm{ps}}$	${4.4\,\mathrm{GHz}}$	${227\,\mathrm{ps}}$

The reported faster oscillation at larger splitting agrees with the two-state prediction at this level of comparison. The comparison is not an uncertainty-based agreement test: the table uses rounded reported values and does not estimate new fit uncertainties. Finite pulse rise time, preparation, relaxation, and cycle averaging affect the contrast and early-time response. In particular, the authors associate the missing first short-pulse oscillation with pulse attenuation. A unit-contrast sine squared is therefore not a justified fit to that detector record. The simulated traces illustrate these measurement effects under declared assumptions; agreement within a simulation is not independent experimental validation. [PPLG10b, fig. 3(c) and discussion, p. 246804-3]

We now return to the static preparation. The time trace changed the physical state; changing coordinates for that same state is a different operation.

Operators and Their Matrices Are Different Kinds of Object

Every calculation so far has been written in the localized basis. To find out which results depend on that choice, separate the operator from the array of numbers that represents it in a basis.

Let ${\hat A:\mathcal H\to\mathcal H}$ be a linear operator on an ${n}$ -dimensional Hilbert space and ${\mathcal B=(|e_1\rangle,\ldots,|e_n\rangle)}$ an ordered orthonormal basis. The matrix representation of ${\hat A}$ in ${\mathcal B}$ is $\mathsf A_{\mathcal B} :={\left[\hat A\right]}_{\mathcal B}\in\mathbb C^{n\times n}, \qquad {\left(\mathsf A_{\mathcal B}\right)}_{ij} =\langle e_i|\hat A|e_j\rangle.$ ${\hat A}$ is the operator; ${\mathsf A_{\mathcal B}}$ is a basis-dependent coordinate array.

Let the columns of a unitary matrix ${\mathsf U}$ be the vectors of a new ordered basis ${\mathcal B'}$ , expressed in ${\mathcal B}$ . Then the coordinates of a state and the matrix of an operator transform as

(2.5)

$\boldsymbol\psi_{\mathcal B'} =\mathsf U^\dagger\boldsymbol\psi_{\mathcal B}, \qquad \mathsf A_{\mathcal B'} =\mathsf U^\dagger\mathsf A_{\mathcal B}\mathsf U.$

Let ${\mathcal B}$ and ${\mathcal B'}$ be ordered orthonormal bases of the same finite-dimensional Hilbert space, with the columns of the unitary matrix ${\mathsf U}$ equal to the ${\mathcal B}$ -coordinates of the ${\mathcal B'}$ basis vectors. Then states and operator matrices transform as in Equation (2.5), and every consistently formed scalar matrix element is unchanged. The proposition concerns passive coordinates; it does not assert that ${\mathsf U}$ is a physical symmetry [Dir27, p. 628, Eq. (4)].

To derive the second relation, write ${\left(\mathsf A_{\mathcal B'}\right)}_{ij} =\langle e_i'|\hat A|e_j'\rangle =\sum_{m,n}U_{mi}^* \langle e_m|\hat A|e_n\rangle U_{nj}.$ The right-hand side is exactly the ${ij}$ entry of ${\mathsf U^\dagger\mathsf A_{\mathcal B}\mathsf U}$ .

The expectation is invariant:

$\begin{aligned} \boldsymbol\psi_{\mathcal B'}^\dagger \mathsf A_{\mathcal B'} \boldsymbol\psi_{\mathcal B'} &= \boldsymbol\psi_{\mathcal B}^\dagger \mathsf U\mathsf U^\dagger \mathsf A_{\mathcal B} \mathsf U\mathsf U^\dagger \boldsymbol\psi_{\mathcal B}\\ &= \boldsymbol\psi_{\mathcal B}^\dagger \mathsf A_{\mathcal B} \boldsymbol\psi_{\mathcal B}. \end{aligned}$

The same scalar therefore appears in every orthonormal basis; it is the number an experiment can test.

Before inspecting Figure 2.4, predict which of the following can change under a basis rotation: matrix entries, coordinate columns, eigenvalue gap, and a consistently calculated probability.

Figure 2.4. Three aligned panels distinguish localized-basis coordinates, spectral projectors and energies, and the coefficient vector of the same two-level operator. The displayed coefficient plane describes this real Hamiltonian, not a state trajectory.

Panel (a) displays coordinates; panel (b) displays spectral alternatives; and panel (c) encodes the operator by the coefficient vector ${\mathbf h}$ . The last panel is an operator-coefficient plane: here ${h_x=t}$ , ${h_z=\epsilon/2}$ , and ${h_y=0}$ . Its normalized direction determines the eigenprojectors for ${h>0}$ ; it is neither a general state nor a classical spin trajectory. Of the four quantities in the prediction, the matrix entries and coordinate columns of panel (a) change under a basis rotation, and with them the components of ${\mathbf h}$ in panel (c); the gap ${2h}$ and the consistently calculated probability do not.

The charge readout gives a concrete test of this distinction. Keep the physical preparation ${|L\rangle}$ and the left-charge projector ${\hat P_L=|L\rangle\langle L|}$ fixed. In the localized basis their arrays are ${\boldsymbol\psi_{\mathcal B}={(1,0)}^{\mathsf T}}$ and ${\mathsf P_{L,\mathcal B}=\operatorname{diag}(1,0)}$ , so ${p_L=1}$ . Now use the Hadamard matrix $\mathsf U=\frac1{\sqrt2} \begin{pmatrix}1&1\\1&-1\end{pmatrix}.$ Its columns are the new basis vectors: ${|e'_1\rangle=(|L\rangle+|R\rangle)/\sqrt2}$ and ${|e'_2\rangle=(|L\rangle-|R\rangle)/\sqrt2}$ . This matrix is real, unitary, and self-adjoint; the general change-of-coordinates rule still uses ${\mathsf U^\dagger}$ , not an assumed equality of a unitary and its adjoint.

Before the next calculation, find the coordinates of ${|L\rangle}$ and the matrix of ${\hat P_L=|L\rangle\langle L|}$ in ${\mathcal B'=((|L\rangle+|R\rangle)/\sqrt2,(|L\rangle-|R\rangle)/\sqrt2)}$ . Predict the charge probability from the physical preparation, then use that prediction to check the matrix calculation.

Prepare ${|L\rangle}$ . Compute its immediate left-charge probability in the localized basis ${\mathcal B=(|L\rangle,|R\rangle)}$ and the basis ${\mathcal B'=( (|L\rangle+|R\rangle)/\sqrt2, (|L\rangle-|R\rangle)/\sqrt2 )}$ . Transform the Hamiltonian ${\hat H=(\epsilon/2)\hat\sigma_z+t\hat\sigma_x}$ , the state, and the readout ${\hat P_L=|L\rangle\langle L|}$ consistently, and verify that the scalar probability is unchanged.

The decisive choice is to transform every coordinate array with its own input and output rule while keeping the physical objects fixed. The new basis vectors form the columns of ${\mathsf U=\frac1{\sqrt2}\begin{pmatrix}1&1\\1&-1\end{pmatrix}}$ . Thus ${\boldsymbol\psi_{\mathcal B'}=\mathsf U^\dagger\boldsymbol\psi_{\mathcal B}}$ and ${\mathsf A_{\mathcal B'}=\mathsf U^\dagger\mathsf A_{\mathcal B}\mathsf U}$ for any operator matrix ${\mathsf A}$ . In the localized basis, ${\boldsymbol\psi_{\mathcal B}={(1,0)}^{\mathsf T}}$ and ${\mathsf P_{L,\mathcal B}=\operatorname{diag}(1,0)}$ , so ${\boldsymbol\psi_{\mathcal B}^{\dagger}\mathsf P_{L,\mathcal B} \boldsymbol\psi_{\mathcal B}=1}$ . The state transforms with one factor of ${\mathsf U^\dagger}$ : $\boldsymbol\psi_{\mathcal B'} =\frac1{\sqrt2} \begin{pmatrix}1&1\\1&-1\end{pmatrix} \begin{pmatrix}1\\0\end{pmatrix} =\frac1{\sqrt2}\begin{pmatrix}1\\1\end{pmatrix}.$ The operator has an input and an output coordinate, so both sides transform: $\begin{aligned} \mathsf P_{L,\mathcal B'} &=\mathsf U^\dagger\mathsf P_{L,\mathcal B}\mathsf U\\ &=\frac12 \begin{pmatrix}1&1\\1&-1\end{pmatrix} \begin{pmatrix}1&0\\0&0\end{pmatrix} \begin{pmatrix}1&1\\1&-1\end{pmatrix}\\ &=\frac12\begin{pmatrix}1&1\\1&1\end{pmatrix}. \end{aligned}$ Consequently $\mathsf P_{L,\mathcal B'}\boldsymbol\psi_{\mathcal B'} =\boldsymbol\psi_{\mathcal B'},\qquad p_L=\frac14 \begin{pmatrix}1&1\end{pmatrix} \begin{pmatrix}1&1\\1&1\end{pmatrix} \begin{pmatrix}1\\1\end{pmatrix}=1.$ The vector remains normalized, and the transformed projector is Hermitian and squares to itself. The right-charge projector ${\mathsf I-\mathsf P_{L,\mathcal B'}}$ annihilates this vector, giving ${p_R=0}$ and ${p_L+p_R=1}$ . These checks test different parts of the calculation: state normalization, the measurement alternative, and the scalar probability.

The same coordinate change sends $\mathsf H_{\mathcal B} =\frac{\epsilon}{2}\mathsf\sigma_z+t\mathsf\sigma_x \quad\longmapsto\quad \mathsf H_{\mathcal B'} =t\mathsf\sigma_z+\frac{\epsilon}{2}\mathsf\sigma_x,$ because ${\mathsf U^\dagger\mathsf\sigma_z\mathsf U=\mathsf\sigma_x}$ and ${\mathsf U^\dagger\mathsf\sigma_x\mathsf U=\mathsf\sigma_z}$ . Both Hamiltonian matrices have zero trace. Their determinants are $\det\mathsf H_{\mathcal B} =-\frac{\epsilon^2}{4}-t^2 =\det\mathsf H_{\mathcal B'},$ because the transformed matrix has diagonal entries ${t,-t}$ and off-diagonal entries ${\epsilon/2}$ . The trace and determinant are independent checks on the Hamiltonian transformation. The charge result holds for any ${\epsilon,t}$ in this model because it concerns a fixed preparation and readout. This invariance is what permits a symmetry-adapted basis in the next section: the physical prediction is the same in every basis.

A classmate keeps the localized readout matrix ${\mathsf P_{L,\mathcal B}=\operatorname{diag}(1,0)}$ , transforms the state to ${\boldsymbol\psi_{\mathcal B'}={(1,1)}^{\mathsf T}/\sqrt2}$ , but evaluates $p_{\mathrm{mixed}} =\boldsymbol\psi_{\mathcal B'}^\dagger \mathsf P_{L,\mathcal B}\boldsymbol\psi_{\mathcal B'}.$ Compute this number before continuing. Which line first combined incompatible coordinates, and which matrix must be repaired?

Answer. The mixed expression gives $\tfrac12$ . The first error pairs the new state coordinates with the old readout matrix; replacing that matrix by $\mathsf P_{L,\mathcal B'}$ recovers $p_L=1$ . One agreeing number certifies nothing on its own: the Hamiltonian never enters this expression, so an untransformed Hamiltonian would leave it unchanged.

The mixed expression is $p_{\mathrm{mixed}} =\frac12\begin{pmatrix}1&1\end{pmatrix} \begin{pmatrix}1&0\\0&0\end{pmatrix} \begin{pmatrix}1\\1\end{pmatrix} =\frac12.$ The first error is pairing the new state coordinates with the old readout matrix. Replacing that matrix by ${\mathsf P_{L,\mathcal B'}}$ recovers ${p_L=1}$ . By contrast, changing only the Hamiltonian while leaving both state and readout arrays unchanged would still give this static probability as one: the Hamiltonian does not occur in that expression. Agreement of one number therefore cannot certify that every object has been transformed consistently.

For a second basis change, take $|\chi\rangle=\frac{\sqrt3|L\rangle+\mathrm i|R\rangle}{2}, \qquad \mathsf V=\frac1{\sqrt2} \begin{pmatrix}1&\mathrm i\\ \mathrm i&1\end{pmatrix},$ whose columns again define the new basis. Keeping the same physical left-charge readout ${\hat P_L=|L\rangle\langle L|}$ , calculate the new state coordinates, the readout’s projector matrix, and the probability in both bases. This time ${\mathsf V^\dagger\ne\mathsf V}$ : identify what fails if the state is transformed with ${\mathsf V}$ instead of ${\mathsf V^\dagger}$ . Record the intermediate arrays as well as the final probability so that the conjugation step and the first mixed-coordinate line can be checked.

A passive coordinate change has now altered three arrays while preserving the physical preparation, measurement, and Hamiltonian. An active unitary operation instead acts on a state, and a symmetry preserves the Hamiltonian under that action. The next question is when such a symmetry can organize the state space into invariant sectors.

Unitary Operations and Symmetry-Preserved Sectors

A unitary operator is a linear map ${\hat U:\mathcal H\to\mathcal H}$ satisfying $\hat U^\dagger\hat U =\hat U\hat U^\dagger =\hat I.$ It preserves inner products and norms. A unitary can represent an active physical operation or symmetry; a unitary matrix can also relate passive coordinate choices. The interpretation must be stated [Dir27, p. 628, Eq. (4)].

A unitary symmetry of a Hamiltonian ${\hat H}$ is a unitary operator ${\hat S}$ that preserves the domain of ${\hat H}$ and satisfies $\hat S\hat H\hat S^\dagger=\hat H.$ In the finite-dimensional setting used here this is equivalent to ${[\hat S,\hat H]=0}$ . Not every unitary change of coordinates is a physical symmetry [Gie00, arXiv v2 evidence copy, §§2.2.1–2.2.3, numbering collated to the published work identity; Bar64, sec. 6; SMCS08a, arXiv v2 evidence copy, §II, numbering collated to the published work identity].

A finite-dimensional operator ${\hat S}$ is normal when ${\hat S^\dagger\hat S=\hat S\hat S^\dagger}$ . Every unitary operator is normal, because both products equal ${\hat I}$ . Normality guarantees a complete orthogonal decomposition into eigenspaces, called symmetry sectors; commutation with ${\hat H}$ then makes each sector invariant.

Let ${\hat S}$ and ${\hat H}$ act on a finite-dimensional complex Hilbert space, with ${\hat H}$ self-adjoint and ${\hat S}$ normal, meaning ${\hat S^\dagger\hat S=\hat S\hat S^\dagger}$ . Every unitary ${\hat S}$ is normal because both products equal ${\hat I}$ . If ${[\hat S,\hat H]=0}$ , every eigenspace of ${\hat S}$ is invariant under ${\hat H}$ ; consequently one may choose a basis that resolves the ${\hat S}$ sectors and reduces the Hamiltonian to blocks. This statement supplies invariant subspaces, not a mechanical interpretation of the symmetry eigenvalues.

Why does a commutator help? If ${\hat S|\phi_s\rangle=s|\phi_s\rangle}$ , then $\hat S(\hat H|\phi_s\rangle) =\hat H\hat S|\phi_s\rangle =s(\hat H|\phi_s\rangle).$ Thus ${\hat H|\phi_s\rangle}$ remains in the ${s}$ -eigenspace of ${\hat S}$ . This invariance argument itself uses only commutation and the eigenvector equation. Normality supplies the rest: the symmetry eigenspaces are orthogonal and span the whole space. Commutation and normality together therefore give a complete block basis.

Construct invariant blocks for the following three-state model. Let ${E_0,E_1,t,g}$ be real energies, and use ${\mathcal B_3=(|L\rangle,|R\rangle,|C\rangle)}$ and

$\mathsf S_{\mathcal B_3} {=} \begin{pmatrix} 0&1&0\\ 1&0&0\\ 0&0&1 \end{pmatrix}, \qquad \mathsf H_{\mathcal B_3} {=} \begin{pmatrix} E_0&t&g\\ t&E_0&g\\ g&g&E_1 \end{pmatrix}.$

The matrix ${\mathsf S_{\mathcal B_3}}$ exchanges left and right and leaves the central orbital fixed. Direct multiplication verifies ${\mathsf S\mathsf H=\mathsf H\mathsf S}$ , so the operator ${\hat S}$ that this matrix represents commutes with ${\hat H}$ . Because ${\hat S^2=\hat I}$ , its spectral projectors are

$\hat P_\pm=\frac12(\hat I\pm\hat S).$

The odd state $|-\rangle=\frac{|L\rangle-|R\rangle}{\sqrt2}$ spans the ${-1}$ sector and has energy ${E_0-t}$ . The ${+1}$ sector is spanned by $|+\rangle=\frac{|L\rangle+|R\rangle}{\sqrt2}, \qquad |C\rangle.$ The off-diagonal entry follows from adding the two equal paths: ${\langle+|\hat H|C\rangle=(g+g)/\sqrt2=\sqrt2g}$ ; for the odd state they cancel, ${(g-g)/\sqrt2=0}$ . The even sector therefore carries the reduced matrix $\mathsf H_+ {=} \begin{pmatrix} E_0+t&\sqrt2\,g\\ \sqrt2\,g&E_1 \end{pmatrix}.$

In the original ordered coordinates, the unitary with columns ${(|-\rangle,|+\rangle,|C\rangle)}$ is $\mathsf U=\begin{pmatrix} 1/\sqrt2&1/\sqrt2&0\\ -1/\sqrt2&1/\sqrt2&0\\ 0&0&1 \end{pmatrix}.$ It gives ${\mathsf U^\dagger\mathsf H\mathsf U=(E_0-t)\oplus\mathsf H_+}$ . As checks, sector dimensions add to three, and the reduced trace is ${(E_0-t)+(E_0+t)+E_1=2E_0+E_1}$ , the original trace. Multiplying ${\mathsf U[(E_0-t)\oplus\mathsf H_+]\mathsf U^\dagger}$ reconstructs the equal ${LC}$ and ${RC}$ entries ${g}$ and the ${LR}$ entry ${t}$ . At ${g=0}$ , the central orbital decouples and the three energies are ${E_0-t,E_0+t,E_1}$ , as the original matrix also shows. The reduction is exact for the declared symmetry, independently of whether ${g}$ is small. Approximation is a subsequent choice within the retained block.

Three steps carry the reduction: the commutator proves that the subspaces are invariant; the projectors construct them; the reduced matrices then follow. A block-shaped matrix in one basis is the consequence of this argument, not the argument itself. Lecture 3 applies the same logic to lattice translations: translation eigenvalues label invariant Hamiltonian sectors. A symmetry label is not automatically the value of a separately defined mechanical observable.

Controlled Approximation Means Ratio, Output, and Failure Test

An approximation is worth using only when its error can be compared with a target. The exact two-level spectrum makes such a comparison possible here, so we first fix what a controlled approximation must declare.

A controlled approximation specifies:

the exact object or equation being approximated;
a dimensionless expansion parameter;
the target output and parameter regime;
an error order or quantitative diagnostic; and
a condition signaling breakdown.

Calling a coupling “small” without naming the scale to which it is compared is not a controlled approximation.

Lecture 1’s diagonal Ising Hamiltonian left the retained energies unchanged when higher-wall states were excluded. The two-level system gives a contrast. At positive detuning, the lower bare state ${|R\rangle}$ has energy ${-\epsilon/2}$ and couples to ${|L\rangle}$ through ${t}$ . We can calculate the lower energy while including the shift caused by that coupling.

The exact lower eigenvalue at positive detuning is

$E_-(\epsilon) =-\sqrt{\frac{\epsilon^2}{4}+t^2} =-\frac{\epsilon}{2} \sqrt{1+\frac{4t^2}{\epsilon^2}}.$

The binomial expansion $\sqrt{1+x}=1+\frac{x}{2}-\frac{x^2}{8}+O(x^3) \quad\text{as }x\to0$ with ${x=4t^2/\epsilon^2}$ gives

(2.6)

$E_-(\epsilon) =-\frac{\epsilon}{2} -\frac{t^2}{\epsilon} +\frac{t^4}{\epsilon^3} +O\!\left(\frac{t^6}{\epsilon^5}\right), \qquad \left|\frac{t}{\epsilon}\right|\ll1.$

The first correction has energy units. The expansion parameter ${|t/\epsilon|}$ is dimensionless. The approximation is not uniform near ${\epsilon=0}$ , where its denominator vanishes even though the exact eigenvalues remain finite.

Take ${t=0.10\,\mathrm{meV}}$ , ${\epsilon=1.00\,\mathrm{meV}}$ , and approximate the lower energy by ${E_{\mathrm{app}}=-\epsilon/2-t^2/\epsilon}$ . Use the exact eigenvalue ${E_-=-\sqrt{\epsilon^2/4+t^2}}$ as the reference. Does this approximation meet relative-error targets of ${0.1\%}$ and ${0.01\%}$ ? The target is the energy itself, with relative error ${r=|E_{\mathrm{app}}-E_-|/|E_-|}$ , not a probability or the small energy shift.

Compare the approximate energy with the exact two-level eigenvalue, and then test each tolerance. Here ${|t/\epsilon|=0.1}$ , and direct evaluation gives $E_-=-\sqrt{0.26}\,\mathrm{meV}\simeq-0.509902\,\mathrm{meV},\qquad E_{\mathrm{app}}=-0.510000\,\mathrm{meV},$ $|E_{\mathrm{app}}-E_-|\simeq9.80\times10^{-5}\,\mathrm{meV},\qquad r\simeq1.92\times10^{-4}=0.0192\%.$ Thus the first tolerance is met and the second is not; for the second, retain the exact two-level answer or verify a higher-order truncation. The omitted term ${t^4/\epsilon^3=10^{-4}\,\mathrm{meV}}$ independently checks the magnitude and positive sign of the correction to the approximate energy. In the limit ${t\to0}$ , the error vanishes. At ${\epsilon=0}$ , however, this expansion is undefined while the exact lower energy is ${-|t|}$ . The successful detuned comparison cannot be exported to zero detuning. The next figure repeats this target-specific decision over a range of detunings.

Before reading Figure 2.5, predict how the relative error of ${-\epsilon/2-t^2/\epsilon}$ , with ${\epsilon>0}$ , scales for large ${|\epsilon/t|}$ , and decide whether “ ${t=0.10\,\mathrm{meV}}$ is small” is enough to select a valid interval.

Figure 2.5. The exact lower energy and its leading large-detuning approximation converge as the detuning-to-coupling ratio grows; a logarithmic error panel compares the relative error with the selected tolerance (two percent in the static figure).

Panel (a) compares an exact scalar function with an asymptotic scalar function. Panel (b) compares the relative error with the displayed tolerance. The static figure uses two percent; the HTML controls let you change it. The marked region in panel (b) is not “where perturbation theory is false”; it is where this truncation does not meet this tolerance for this output.

This question refers to the accompanying figure.

For positive detuning ${\epsilon>0}$ , move the ratio ${x=|\epsilon/t|}$ and the tolerance, and read the relative error of ${-\epsilon/2-t^2/\epsilon}$ against the exact branch. The error falls as ${x^{-4}}$ , so the scaled error ${x^4r}$ settles at large ratio; a stated tolerance then names the smallest ratio that meets it, which a tunneling energy by itself never does.

For ${\hat H=\hat H_0+\hat V}$ , a nondegenerate expansion about an eigenvalue of ${\hat H_0}$ is controlled for a declared output only when the relevant coupling-to-gap ratios are small enough that the estimated omitted terms meet the stated tolerance. A vanishing denominator, strong state mixing, or a failed exact/error comparison closes that window. The value of a dimensional coupling by itself is not a control parameter [Kat52, pp. 323–337].

Near ${\epsilon=0}$ , the two localized states are degenerate in the unperturbed model, so the coupling must be diagonalized inside their joint subspace; keeping more powers of ${t/\epsilon}$ cannot repair a vanishing denominator. The exact ${2\times2}$ calculation gives the finite separation ${2|t|}$ .

Let ${\hat P}$ project onto states whose mutual splittings are comparable to their couplings, and suppose the complementary ${\hat Q}$ states remain separated by declared nonzero gaps. Diagonalize ${\hat P\hat H\hat P}$ exactly before expanding the remote-state effect. The procedure fails if the retained space omits a near-resonant state or if the ${\hat P}$ -to- ${\hat Q}$ separation is too small for the target accuracy [Low51, pp. 1396–1399; BHJ26, pp. 565–580; Sch26, pp. 441–445, 447–464].

This method choice is the heart of perturbation theory near degeneracy:

if the reference level is isolated and coupling-to-gap ratios are small, expand;
if retained splittings are comparable to retained couplings, diagonalize the retained block exactly; and
if states are remote, treat them perturbatively only after their separation and coupling have been declared.

To test the third rule on the double dot, add a remote third level: on ${\mathcal H_3=\operatorname{span}\{|L\rangle,|R\rangle,|X\rangle\}\cong \mathbb C^3}$ , in the ordered orthonormal basis ${\mathcal B_3=(|L\rangle,|R\rangle,|X\rangle)}$ , define

${\bigl[\hat H_3\bigr]}_{\mathcal B_3} {=} \begin{pmatrix} \epsilon/2&t&0\\ t&-\epsilon/2&g\\ 0&g&\Delta_{\mathrm{ex}} \end{pmatrix}, \qquad \Delta_{\mathrm{ex}}>0.$

Here ${|X\rangle}$ is a remote state and ${g}$ is its coupling to ${|R\rangle}$ . Let ${\hat P}$ project onto ${\operatorname{span}\{|L\rangle,|R\rangle\}}$ and let ${\hat Q=\hat I-\hat P}$ project onto ${\operatorname{span}\{|X\rangle\}}$ . In the corresponding ordered bases ${\mathcal B_P=(|L\rangle,|R\rangle)}$ and ${\mathcal B_Q=(|X\rangle)}$ , the matrix blocks are $\mathsf H_{PP}:={\bigl[\hat P\hat H_3\hat P\bigr]}_{\mathcal B_P}, \quad \mathsf H_{QQ}:={\bigl[\hat Q\hat H_3\hat Q\bigr]}_{\mathcal B_Q}, \quad \mathsf V_{PQ}:={\bigl[\hat P\hat H_3\hat Q\bigr]}_{\mathcal B_P\leftarrow\mathcal B_Q}, \quad \mathsf V_{QP}:={\bigl[\hat Q\hat H_3\hat P\bigr]}_{\mathcal B_Q\leftarrow\mathcal B_P},$ where the subscript ${\mathcal B_P\leftarrow\mathcal B_Q}$ names the input basis on the right and the output basis on the left. Write the eigenvector in retained and eliminated coordinates. The two block rows of the eigenvalue equation are $\mathsf H_{PP}\boldsymbol\psi_P+\mathsf V_{PQ}\boldsymbol\psi_Q =E\boldsymbol\psi_P,\qquad \mathsf V_{QP}\boldsymbol\psi_P+\mathsf H_{QQ}\boldsymbol\psi_Q =E\boldsymbol\psi_Q.$ If ${E}$ is not an eigenvalue of ${\mathsf H_{QQ}}$ , written ${E\notin\sigma(\mathsf H_{QQ})}$ , the second row can be inverted: $\boldsymbol\psi_Q= {\left(E\mathsf I_Q-\mathsf H_{QQ}\right)}^{-1} \mathsf V_{QP}\boldsymbol\psi_P.$ Substituting this expression into the first row gives the exact retained equation. The plus sign comes from substitution; the negative low-energy shift below comes from the denominator, not a changed block convention:

$\left[ \mathsf H_{PP} +\mathsf V_{PQ} {\left(E\mathsf I_Q-\mathsf H_{QQ}\right)}^{-1} \mathsf V_{QP} \right]\boldsymbol\psi_P =E\boldsymbol\psi_P.$

If ${|E|\ll\Delta_{\mathrm{ex}}}$ , then

${\left(E-\Delta_{\mathrm{ex}}\right)}^{-1} =-\frac1{\Delta_{\mathrm{ex}}} \left(1+\frac{E}{\Delta_{\mathrm{ex}}} +O\!\left(\frac{E^2}{\Delta_{\mathrm{ex}}^2}\right)\right).$

Keeping the leading term produces the low-energy correction

$\delta\mathsf H_{PP} \simeq-\frac1{\Delta_{\mathrm{ex}}} \mathsf V_{PQ}\mathsf V_{QP}.$

For a remote state coupled by ${g}$ only to ${|R\rangle}$ ,

$\delta\mathsf H_{PP} \simeq -\frac{g^2}{\Delta_{\mathrm{ex}}} \begin{pmatrix}0&0\\0&1\end{pmatrix}.$

For a declared target energy resolution ${\delta E_{\mathrm{res}}=5\,\mu\mathrm{eV}}$ , with ${g=0.20\,\mathrm{meV}}$ and ${\Delta_{\mathrm{ex}}=4.0\,\mathrm{meV}}$ , the leading correction, measured by the operator norm of ${\delta\mathsf H_{PP}}$ (here its single nonzero entry ${g^2/\Delta_{\mathrm{ex}}}$ ), is ${0.010\,\mathrm{meV}=10\,\mu\mathrm{eV}}$ . Because ${10\,\mu\mathrm{eV}}$ exceeds the stated resolution, this estimate alone cannot certify that the remote level is negligible for every retained state at that tolerance. The same estimate leaves open whether any particular level shifts by a resolvable amount: for a normalized nondegenerate retained eigenstate ${|n\rangle}$ , the leading eigenvalue shift is ${-g^2|\langle R|n\rangle|^2/\Delta_{\mathrm{ex}}}$ , so a state with little ${R}$ weight may shift much less. Specify the target eigenvalue or operator accuracy, then check the neglected terms (or diagonalize the three-level matrix). Advanced Appendix Section 2.B develops the energy-dependent correction beyond this leading estimate.

Two checks matter. At ${g=0}$ , the correction vanishes. Doubling ${\Delta_{\mathrm{ex}}}$ halves the leading shift. A calculation giving a correction linear in ${g}$ for this setup has skipped the two couplings needed to leave and return to the retained space.

The Working Quantum Interface for the Course

We can now state the quantum-mechanical chain precisely:

$\begin{aligned} \text{preparation} &\longrightarrow \text{state ray} \longrightarrow \text{observable and spectral projectors}\\ &\longrightarrow \text{probabilities and expectations} \longrightarrow \text{representation-invariant scalars}. \end{aligned}$

A unitary symmetry adds $[\hat S,\hat H]=0 \longrightarrow \text{symmetry projectors} \longrightarrow \text{invariant subspaces} \longrightarrow \text{smaller eigenproblems}.$

A controlled approximation adds $\text{coupling-to-gap ratio} \longrightarrow \text{error criterion} \longrightarrow \text{validity window or method change}.$

Lecture 3 will use the second chain with lattice translations: commuting unitary translations decompose a periodic Hamiltonian into sectors labeled by translation eigenvalues. Lecture 4 will reuse the exact-versus-approximate two-level calculation to decide whether a crossing is protected or hybridized. Later lectures will add tensor products, variable particle number, mixed preparations, and response only when their physical questions require them.

Quantum mechanics becomes reliable for this course through compatible mathematical types and operations, and the parts that had to be reconstructed are exactly the links of the three chains above. This lecture worked mainly in finite dimension with pure states. Position and momentum domains, continuous spectra, identical particles, density operators, and statistical weights require additional structure, which later lectures introduce when their questions demand it.

Problems

1. Projectors, probabilities, and evolution

Choose the phase convention ${t\ge 0}$ , take ${\epsilon\in\mathbb R}$ , and let $\hat H=\frac{\epsilon}{2}\hat\sigma_z+t\hat\sigma_x, \qquad \hat Q=\hat\sigma_z, \qquad |\psi(0)\rangle=|L\rangle.$

Declare the domains and codomains of ${\hat H}$ and ${\hat Q}$ , and the Hilbert space and normalization of the state representative. State the units of each operator.
For ${h>0}$ , starting from the Pauli product rule, derive ${\hat P_\pm}$ . Verify self-adjointness, idempotence, orthogonality, completeness, and ${\hat H\hat P_\pm=E_\pm\hat P_\pm}$ . State the spectral projector and explain why the rank-one ${\pm}$ split is unavailable when ${h=0}$ .
Calculate ${p_\pm(L)}$ , their sum, and their limits for ${\epsilon/t\to+\infty}$ , ${\epsilon=0}$ , and ${t\to0}$ at fixed positive ${\epsilon}$ .
For ${\epsilon=0}$ , calculate ${{\langle\hat Q\rangle}_\tau}$ and ${\operatorname{Var}_\tau(\hat Q)}$ after exact evolution for time ${\tau}$ . Check ${\tau=0}$ , ${t=0}$ , probability bounds, and the gap-frequency relation.
Explain why none of these probabilities is a thermal occupation.

2. Building symmetry sectors rather than spotting blocks

Let ${E_0,E_1,t,g,\delta\in\mathbb R}$ , and use $\mathsf S= \begin{pmatrix}0&1&0\\1&0&0\\0&0&1\end{pmatrix}, \qquad \mathsf H= \begin{pmatrix}E_0&t&g\\t&E_0&g\\g&g&E_1\end{pmatrix}.$

Verify ${\mathsf S^\dagger\mathsf S=\mathsf I}$ , ${\mathsf S^2=\mathsf I}$ , and ${[\mathsf S,\mathsf H]=0}$ .
Construct ${\mathsf P_\pm=(\mathsf I\pm\mathsf S)/2}$ and verify all projector identities.
Find orthonormal bases of the two sectors and derive the one-dimensional odd block and two-dimensional even block.
Show that the characteristic polynomial of the full matrix equals the product of the reduced characteristic polynomials.
Change the Hermitian pair ${H_{LC}=H_{CL}}$ from ${g}$ to ${g+\delta}$ while leaving ${H_{RC}=H_{CR}=g}$ . Calculate the commutator and identify which sector argument fails. Do not describe the result merely as “the block disappears.”
State why the eigenvalue of ${\hat S}$ is a symmetry label but is not, from the supplied information alone, a separately defined position or momentum observable.

3. Selecting a method near degeneracy

For ${t>0}$ and ${\epsilon>0}$ , compare $E_-^{\mathrm{exact}} =-\sqrt{\epsilon^2/4+t^2}, \qquad E_-^{(2)} =-\epsilon/2-t^2/\epsilon.$

Derive the expansion through ${t^4/\epsilon^3}$ , including its remainder order. Identify the dimensionless parameter.
For ${\epsilon/t=1,2,4,8}$ , calculate the exact value, approximation, absolute error, and relative error. Determine which points satisfy a two-percent tolerance.
Explain why adding more terms in ${t/\epsilon}$ is not a controlled remedy at ${\epsilon=0}$ . State and execute the appropriate retained-subspace method.
Add a remote state at ${\Delta_{\mathrm{ex}}}$ coupled only to ${|R\rangle}$ by ${g}$ . Derive the leading retained-space correction using the block equation. Check ${g\to0}$ , the sign for positive ${\Delta_{\mathrm{ex}}}$ , dimensions, and the effect of doubling ${\Delta_{\mathrm{ex}}}$ .
With ${g=0.20\,\mathrm{meV}}$ , ${\Delta_{\mathrm{ex}}=4.0\,\mathrm{meV}}$ , and ${5\,\mu\mathrm{eV}}$ resolution, decide whether to keep the correction, shrink the validity window, or expand the exact retained space. State the target output that makes your decision meaningful.

(advanced): Functions, ${L^2}$ Vectors, and Generalized Eigenstates

The main text of this lecture avoids a distinction that matters in continuum quantum mechanics. A wave function may be written as a function ${\psi:\mathbb R^d\to\mathbb C}$ , but a vector in ${L^2(\mathbb R^d)}$ is an equivalence class of functions that agree almost everywhere. A point value ${\psi(\mathbf r)}$ therefore belongs to a chosen representative, while inner products and operator actions are defined on the equivalence class when their domains permit.

The position operator has the formal action $(\hat{\mathbf r}\psi)(\mathbf r)=\mathbf r\,\psi(\mathbf r),$ but its domain is restricted to square-integrable functions for which ${\mathbf r\,\psi(\mathbf r)}$ is also square integrable. The momentum operator has the formal action $(\hat{\mathbf p}\psi)(\mathbf r) =-\mathrm i\hbar\nabla\psi(\mathbf r),$ with a domain and boundary conditions chosen so that the intended self-adjointness statement is true. The differential expression alone is not the complete operator.

One convenient realization is a rigged Hilbert space, a triple $\Phi=\mathcal S(\mathbb R^d) \subset L^2(\mathbb R^d) \subset \Phi^\times=\mathcal S'(\mathbb R^d),$ where ${\mathcal S(\mathbb R^d)}$ is the Schwartz space of smooth, rapidly decreasing test functions and ${\Phi^\times}$ is its space of continuous antilinear functionals, the tempered distributions in the present inner-product convention. A plane wave lies outside ${L^2(\mathbb R^d)}$ . Its generalized ket instead acts on test functions in ${\Phi}$ . With wave-vector normalization $\langle\mathbf r|\mathbf k\rangle ={(2\pi)}^{-d/2}\mathrm e^{\mathrm i\mathbf k\cdot\mathbf r},$ one has, distributionally, $\langle\mathbf k|\mathbf k'\rangle =\delta^{(d)}(\mathbf k-\mathbf k'), \qquad \int \mathrm d^d k\,|\mathbf k\rangle\langle\mathbf k|=\hat I.$ In a periodic box of volume ${V=L^d}$ , the ordinary normalized vectors are ${V^{-1/2}\mathrm e^{\mathrm i\mathbf k\cdot\mathbf r}}$ , with ${\mathbf k=2\pi\mathbf n/L}$ for an integer vector ${\mathbf n\in\mathbb Z^d}$ , and with Kronecker rather than Dirac normalization. The choice ${\Phi=\mathcal S}$ is an explicit example, not a universal domain prescription; boundaries and operator domains can require another test space [Gie00, arXiv v2 evidence copy, §§2.2.1–2.2.3, numbering collated to the published work identity; Sch45, pp. 57–62]. Lecture 3 chooses the finite-periodic route first and states the bulk limit separately.

This appendix distinguishes a function, an ${L^2}$ vector, and a distribution. Other mathematical objects enter only when later geometry requires them: sections of bundles and multivalued functions. None of the finite-dimensional projector, basis, symmetry, or approximation calculations in the main text depends on this appendix.

(advanced): The Schur Complement Beyond Leading Order

For the block Hamiltonian $\mathsf H= \begin{pmatrix} \mathsf H_{PP}&\mathsf V_{PQ}\\ \mathsf V_{QP}&\mathsf H_{QQ} \end{pmatrix},$ let ${d_P:=\dim\operatorname{ran}\hat P}$ and ${d_Q:=\dim\operatorname{ran}\hat Q}$ . The block ${\mathsf V_{PQ}\in\mathbb C^{d_P\times d_Q}}$ maps eliminated coordinates to retained coordinates, whereas ${\mathsf V_{QP}\in\mathbb C^{d_Q\times d_P}}$ maps in the reverse direction. The exact retained equation contains the energy-dependent self-energy $\Sigma_P(E) =\mathsf V_{PQ} {\left(E\mathsf I_Q-\mathsf H_{QQ}\right)}^{-1} \mathsf V_{QP}.$

This elimination is the exact Schur-complement step. Löwdin partitioning applies elimination between retained and excluded spaces to bound-state perturbation theory and commonly expands the resulting energy-dependent effective operator; it is distinct from Löwdin symmetric orthogonalization of a nonorthogonal basis [Low50, pp. 365–372]. Feshbach projection develops a ${P/Q}$ resolvent formalism for reaction and scattering problems, including the boundary-condition structure of the eliminated sector. A Schrieffer–Wolff construction instead chooses an anti-self-adjoint generator, conventionally also written ${\hat S}$ and unrelated to the symmetry operator of the main text, and block-diagonalizes ${\mathrm e^{\hat S}\hat H\mathrm e^{-\hat S}}$ order by order, normally producing an energy-independent effective Hamiltonian while requiring observables to be transformed as well. These methods share block-algebra ancestry, but their names are not interchangeable [BDL11a, arXiv v1 evidence copy, §§2.3, 3.2–3.4, numbering collated to the published work identity; Fes64, pp. 1076–1078; Low51, pp. 1396–1401].

If the spectrum of ${\mathsf H_{QQ}}$ stays at least a distance ${\Delta>0}$ from the energy window of interest, define the aggregate off-diagonal coupling and its induced operator norm by $\mathsf V_{\mathrm{off}} := \begin{pmatrix} 0&\mathsf V_{PQ}\\ \mathsf V_{QP}&0 \end{pmatrix}, \qquad \|\mathsf V_{\mathrm{off}}\| :=\sup_{\|\boldsymbol\psi\|=1} \|\mathsf V_{\mathrm{off}}\boldsymbol\psi\|.$ The matrix ${\mathsf V_{\mathrm{off}}}$ belongs to ${\mathbb C^{(d_P+d_Q)\times(d_P+d_Q)}}$ . The small ratio ${\|\mathsf V_{\mathrm{off}}\|/\Delta\ll1}$ controls remote-state admixture; the exact self-energy has norm at most ${\|\mathsf V_{PQ}\|\|\mathsf V_{QP}\|/\Delta}$ . An expansion in energy requires a separate condition. Choose a real reference energy ${E_*}$ outside the spectrum of the self-adjoint ${\mathsf H_{QQ}}$ , and define $\mathsf R_*={(E_*\mathsf I_Q-\mathsf H_{QQ})}^{-1},\qquad d_*=\operatorname{dist}(E_*,\sigma(\mathsf H_{QQ})),\qquad q=\frac{|E-E_*|}{d_*}.$ Factoring the inverse and using the geometric series gives, for ${q<1}$ , ${(E\mathsf I_Q-\mathsf H_{QQ})}^{-1} =\sum_{j=0}^{m}{[-(E-E_*)]}^j\mathsf R_*^{\,j+1} +\mathsf{Rem}_m,\qquad \|\mathsf{Rem}_m\|\le\frac{q^{m+1}}{d_*(1-q)}.$ For one expansion covering a whole energy window, use the supremum of ${q}$ over that window and choose the order ${m}$ against the target tolerance. Multiplying this remainder bound by ${\|\mathsf V_{PQ}\|\|\mathsf V_{QP}\|}$ bounds the self-energy truncation error. Small coupling alone does not guarantee ${q<1}$ ; the exact Schur expression and the spectral-distance norm bound do not need this expansion. Each self-energy term carries two couplings and remote denominators. The exact expression also shows the failure condition: when ${E}$ approaches an excluded eigenvalue, the inverse becomes large and the state should no longer be treated as remote.

This block elimination is the generic algebra behind low-energy Hamiltonians. Its use is conditional on a declared spectral separation, operator norm or matrix-element scale, target output, and accuracy. The appendix is optional: Lecture 4 will retrieve only the two-level consequence needed for microscopic hybridization.

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Lecture 2: Quantum States, Operators, and Approximation

2.1 States and Operators Need Mathematical Habitats

2.2 Spectral Projectors Turn Alternatives into Probabilities

2.2.1 Projectors and spectral alternatives

2.2.2 Degeneracy and complete energy alternatives

2.2.3 Born probabilities, expectations, and repeatability

2.3 Time Evolution from the Spectral Projectors

2.3.1 From coherent transfer to a measured charge record

2.4 Operators and Their Matrices Are Different Kinds of Object

2.5 Unitary Operations and Symmetry-Preserved Sectors

2.6 Controlled Approximation Means Ratio, Output, and Failure Test

2.7 The Working Quantum Interface for the Course

2.8 Problems