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Let $\chi =\chi (r)$, $r=|x|$, and let $f(r)|0⟩$ be a normalized eigenstate of $H_0$ with $E>0$, where $|0⟩$ is the Clifford vacuum. Acting with the supercharge differentiates only its wavefunction:The normalized partner is $E^{-1/2}(f^{\prime }-{\chi }^{\prime }f){\psi }_r^{†}|0⟩$. Indeed $[H,Q]=0$, and $\Vert Qf|0⟩{\Vert }^2=E\Vert f|0⟩{\Vert }^2$; acting with $Q$ again returns $\sqrt{E}$ times the original state. The expression at $r=0$ is interpreted by the regular extension of the original state. Although its coefficient depends only on $r$, its degree-one components contain the radial unit vector.

Separation of variables in a central potential writes a wavefunction as $R_{nl}(r)Y_l^m(\theta ,\varphi )$. With normalized spherical harmonics the radial factor satisfies ${\int }_0^{\infty }{r}^2|R_{nl}{|}^{2}dr=1$. The reduced radial function is $u=rR$, with normalization $\int |u{|}^{2}dr=1$; the two functions have different behavior at the origin. For a regular Coulomb eigenstate, $R_{nl}\propto r^l$ near the origin.

An energy and orbital angular momentum eigenstate for a real two-dimensional central potential can be chosen with integer $k$ and real radial function $f$. Its radial equation isQuantum degeneracy permits other stationary states, including $f(r)\cos (k\varphi )/\sqrt{\pi }$, which are not single angular exponentials. Separation of variables supplies a basis rather than describing every superposition in a degenerate eigenspace.

To identify the physical channel $\rho \mapsto V\rho V^{†}$ of a single-qubit unitary gate, prepare each of the three positive Pauli matrix eigenstates and estimate each output Pauli expectation. Their nine means areThey determine the Bloch vector rotation $\mathbf{r}\mapsto R\mathbf{r}$, hence the channel on all states. If two gates have the same channel, their relative unitary commutes with all three Pauli matrices: commuting with $Z$ forces it diagonal and commuting with $X$ forces equal diagonal entries. Thus they differ by a scalar global phase, proving uniqueness up to that phase. This procedure needs no controlled gate or eigenstate of the unknown gate. Estimating its bounded measurement means to error $ϵ$ with fixed confidence uses order ${ϵ}^{-2}$ repetitions. It determines the relative eigenphases but cannot determine an overall eigenvalue phase from uncontrolled gate access alone.

A spin eigenstate along a specified axis is a simultaneous eigenstate of the squared spin operator $S^2$ and its component $S_{\hat{n}}$. Its eigenvalues are ${\hslash }^{2}s(s+1)$ and $\hslash m$, where $m=-s,-s+1,\dots ,s$. The axis must be specified because components along different axes generally do not commute. For spin one-half, the two field-axis eigenstates have projections $\pm \hslash /2$ and form the branches of a Stern-Gerlach experiment.

For spin one-half, choose the normalized spin eigenstates $|+z⟩,|-z⟩$ with $S_z|\pm z⟩=\pm (\hslash /2)|\pm z⟩$. The spin one-half along an axis pure state in the $xz$ plane isIts norm is one because the two eigenstates are orthogonal and ${\cos }^2(\theta /2)+{\sin }^2(\theta /2)=1$. Applying the Born rule to a $z$ projective measurement givesConsequently the expectation value, second moment, and variance areAt $\theta =\pi /3$, the outcome probabilities are $3/4$ and $1/4$, the mean is $\hslash /4$, and the variance is $3{\hslash }^2/16$. Every individual outcome remains $\pm \hslash /2$. At $\theta =\pi /2$, the mean is zero and the variance is maximal, ${\hslash }^2/4$. At either pole, one outcome has probability one and the variance vanishes.

The diagonal mixed state ${\rho }_{\mathrm{diag}}=p_{+}|+z⟩⟨+z|+p_{-}|-z⟩⟨-z|$ has the same $S_z$ probabilities, expectation value, and variance as $|{\psi }_{\theta }⟩$. It is nevertheless a different quantum state for $0<\theta <\pi$: the displayed pure state has $⟨S_x⟩=\hslash \sin \theta /2$, while ${\rho }_{\mathrm{diag}}$ has $⟨S_x⟩=0$. A single component's mean does not reveal all relative quantum phases or establish purity of a density operator.

For a qubit, use the dimensionless Pauli matrices as observables $A={\sigma }_x$, $B={\sigma }_z$. Their Pauli matrix multiplication law givesThese observables have eigenvalues $\pm 1$; the physical spin components are $S_j=(\hslash /2){\sigma }_j$, where $\hslash$ is the reduced Planck constant, so their standard deviations acquire a factor $\hslash /2$.

In the computational basis, prepare $|+y⟩=(|0⟩+i|1⟩)/\sqrt{2}$, an eigenstate of ${\sigma }_y$ with eigenvalue $+1$. Direct multiplication givesBoth Pauli measurements return $+1$ and $-1$ with equal probabilities, and the Robertson uncertainty principle is saturated: $1\cdot 1=1$.

Instead prepare $|+z⟩=|0⟩$. Now the expectation values are $⟨{\sigma }_x⟩=⟨{\sigma }_y⟩=0$, $⟨{\sigma }_z⟩=1$, so $\Delta {\sigma }_x=1$, $\Delta {\sigma }_z=0$. The Pauli measurement of ${\sigma }_z$ is certain, and both sides of the bound vanish. The commutator is still nonzero as an operator: noncommuting observables do not guarantee a positive lower bound independent of the quantum state. This example is useful for distinguishing the algebra of the observables from the expectation values in a particular preparation.

An eigenstate with phase $\varphi$ becomes an eigenstate with phase $n\varphi$ under the tensor product of $n$ copies of the unitary operator. A promise $0\le \varphi <1/n$ prevents modulo-one aliasing. For $n=2^s$ and dyadic $\varphi =x/2^m$, exact quantum phase estimation then needs $m-s$ phase bits and $2^{m-s}-1$ queries to a supplied joint controlled oracle. This is a saving in joint-oracle calls, not in the total number of individual oracle invocations needed to build that joint operation.

No fixed unitary operator maps both eigenstate rays of each of the three Pauli matrices to their orthogonal partners. Such a map would give $U^{†}XU=-X$, $U^{†}YU=-Y$ and $U^{†}ZU=-Z$. But unitary conjugation preserves multiplication. The relation $XY=iZ$ would imply both $U^{†}(XY)U=iZ$ and $U^{†}(iZ)U=-iZ$, a contradiction.

Two axes alone do not suffice: $U=Y$ flips the X- and Z-basis labels while preserving the Y-basis labels. Thus six specified input rays already obstruct a unitary implementation, without requiring tests on every point of the Bloch sphere.

In an orthonormal basis of localized atomic orbitals, a ring of three equivalent sites with real hopping $t$ has Hamiltonian $H=(E_0-t)I+tJ$, where $J$ is the three-by-three all-ones matrix. The normalized symmetric state $(|1⟩+|2⟩+|3⟩)/\sqrt{3}$ has energy $E_0+2t$. Its two-dimensional orthogonal complement has energy $E_0-t$. Thus the symmetric state is the unique spatial ground state for $t<0$, with an energy difference $3|t|$ to the excited eigenstates. Equivalently, the three discrete Bloch states have energies $E_0+2t\cos (2\pi l/3)$, $l=0,\pm 1$. Nonorthogonal atomic orbitals instead require a generalized eigenvalue problem.

A transverse confinement eigenstate that defines a threshold for propagation along a narrow conductor. Above that threshold, its longitudinal dispersion supplies an incident channel from each reservoir. spin can label separate channels; an unresolved spin-degenerate mode contributes twice the single-spin spin-resolved conductance quantum. This physical use of mode is unrelated to the most probable value of a probability distribution.

For two orthonormal energy eigenstates ${\psi }_1,{\psi }_2$, let an observable have distinct nondegenerate eigenstates ${\varphi }_{\pm }=({\psi }_1\pm {\psi }_2)/\sqrt{2}$. Starting in ${\varphi }_{-}$, unitary time evolution gives $(e^{-i{E}_{1}t/\hslash }{\psi }_1-e^{-i{E}_{2}t/\hslash }{\psi }_2)/\sqrt{2}$. Its probability amplitude along ${\varphi }_{+}$ is $(e^{-i{E}_{1}t/\hslash }-e^{-i{E}_{2}t/\hslash })/2$. The Born rule therefore gives the stated transition probability. The energy difference determines the oscillation frequency, while a common energy shift has no effect.

At large but finite magnetic field, opposite-spin admixtures in approximate product eigenstates let nominally spin-forbidden electric-dipole channels borrow intensity. Their amplitudes scale as spin-orbit interaction energy divided by Zeeman energy; their intensities therefore vanish quadratically in that ratio.