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The even, spherically symmetric hydrogen atom ground state has no permanent electric dipole, so a uniform electric field gives no first-order energy shift. The electric-dipole selection rules for hydrogen allow odd-parity excited states to mix in at second order, yielding a negative shift quadratic in the field.

Here there is a nondegenerate energy eigenvalue, so second-order nondegenerate perturbation theory applies. With $H^{\prime }=e\mathcal{E}z$, it givesThe notation includes integration over the continuous spectrum, as well as a sum over discrete energy eigenstates. All denominators are negative because $1s$ is the ground state. The positive electric polarizability ${\alpha }_{1s}$ therefore lowers its energy.^[ref]

An independent way to see the coefficient avoids truncating that sum. Define $E_{\mathrm{h}}=e^2/(4\pi {ϵ}_0{a}_0)$ and $f=e{a}_0\mathcal{E}/E_{\mathrm{h}}$. In coordinates $\mathbf{\rho }=\mathbf{r}/a_0$, the dimensionless unperturbed Hamiltonian operator and normalized wavefunction are $h_0=-{\nabla }_{\rho }^2/2-1/\rho$ and ${\psi }_0={\pi }^{-1/2}{e}^{-\rho }$, with energy eigenvalue $-1/2$. Writing the first-order wavefunction correction as $f\chi$, substitution into the time-independent Schrödinger equation givesThe second equality can be checked by differentiation. Its parity makes $\chi$ orthogonal to ${\psi }_0$. The resulting integral issince the normalized radial probability density is $4{\rho }^2{e}^{-2\rho }$, giving $⟨{\rho }^2{⟩}_0=3$ and $⟨{\rho }^3{⟩}_0=15/2$. Converting back to SI units gives the stated electric polarizability. The induced electric dipole moment is $-\partial \Delta E/\partial \mathcal{E}={\alpha }_{1s}\mathcal{E}$ and tends to zero with the electric field, unlike the selected degenerate branches' leading dipoles. Both calculations describe weak-field level shifts; a spatially uniform static electric field also allows eventual quantum tunnelling and ionization, which these real energy corrections do not calculate.

The sudden approximation treats a rapid, finite change of a Hamiltonian operator as leaving the quantum state approximately unchanged during the switch. The energy measurement can nevertheless change immediately: the same quantum state has different probability amplitudes in the new eigenbasis. Compute their inner products with the new energy eigenstates, then apply the Born rule. The two-level sudden quench below makes this distinction quantitative without assuming the change is small.

An ideal finite step has exact quantum state continuity at a finite Hamiltonian step; a finite-duration implementation has a small evolution error under the bounded-model conditions below. An impulsive quantum kick is different: its Hamiltonian operator grows as its duration shrinks, retaining a finite integrated action and a finite change of quantum state. A sudden approximation therefore concerns short evolution during switching, not conservation of energy during externally driven change. The subsequent unitary time evolution can be substantial.^[ref]

For a particle with mass $m$ on $(0,L)$, put $K_0(x;\beta )=\sqrt{m/(2\pi {\hslash }^2\beta )}{e}^{-m{x}^2/(2{\hslash }^2\beta )}$. The method of images gives $K_D(q_f,q_i;\beta )={\sum }_r[K_0(q_f-q_i+2rL;\beta )-K_0(q_f+q_i+2rL;\beta )]$. The diagonal integral isThe reflected intervals tile the real line, giving the subtraction $1/2$. The Poisson summation formula converts this to ${\sum }_{n\ge 1}{e}^{-\beta {\hslash }^2{\pi }^2{n}^2/(2m{L}^2)}$, proving equality between the image and energy eigenstate calculations.

For a time-independent Hamiltonian operator $H$, an energy eigenstate satisfiesFor a one-dimensional particle of mass $m$ in a real potential energy $U(x)$,Its full time-dependent wavefunction is $\Psi (x,t)=\psi (x)e^{-iEt/\hslash }$: substitution in the Schrödinger equation checks both the minus sign and the factor $\hslash$. Thus an energy eigenstate has stationary position probability density, even though its quantum phase evolves. The boundary conditions and operator domain are part of this eigenvalue problem, selecting physically allowed solutions; the infinite square well and Radial Schrodinger equation for the hydrogen atom give complementary confined examples.

For a complete discrete orthonormal basis of energy eigenstates, linearity gives $\Psi (x,t)={\sum }_n{c}_n{\psi }_n(x)e^{-i{E}_{n}t/\hslash }$, with ${\sum }_n|c_n{|}^2=1$. Cross terms in $|\Psi {|}^2$ oscillate at angular frequencies $(E_n-E_m)/\hslash$. A continuous spectrum requires the corresponding spectral integrals; for a free particle on the whole line, individual plane waves are generalized eigenfunctions, and a normalized Gaussian wave packet is their integral quantum superposition. Adding a spatially constant real $C$ to $U$ shifts every energy eigenvalue by $C$ and multiplies the full state by $e^{-iCt/\hslash }$, a global phase; energy differences, position probabilities and probability current are unchanged.

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.

A normalized quantum state $\chi =\sqrt{q}{\psi }_1+e^{i\eta }\sqrt{1-q}{\psi }_2$ in two orthonormal energy eigenstates returns to its original one-dimensional measurement subspace with probabilityThis follows by applying unitary time evolution to each energy eigenstate, then the Born rule. The relative phase $\eta$ cancels from the return amplitude. Degenerate energies make the probability identically one.

Take an isolated two-level system, or a spin one-half qubit, with Pauli matrices ${\sigma }_x,{\sigma }_z$. Let ${\sigma }_z|0⟩=|0⟩$ and ${\sigma }_z|1⟩=-|1⟩$. The initial and final Hamiltonian operators areThe energy gaps are ${\Delta }_i,{\Delta }_f$. Prepare the initial ground state $|0⟩$, with energy eigenvalue $-{\Delta }_i/2$. A final orthonormal eigenbasis isDirect matrix multiplication gives $H_f|g_f⟩=-{\Delta }_f|g_f⟩/2$ and $H_f|e_f⟩=+{\Delta }_f|e_f⟩/2$. The unchanged initial quantum state is $|0⟩=c|g_f⟩-s|e_f⟩$. Its new probability amplitudes are the inner products $c,-s$; the Born rule gives $p_g=c^2,p_e=s^2$. The angle between the real amplitude vectors is $\theta /2$, whereas the Hamiltonian axis on the Bloch sphere rotates through $\theta$.

At $\theta =\pi /3$, $c=\sqrt{3}/2$ and $s=1/2$, so the final energy measurement gives $-{\Delta }_f/2$ with probability $3/4$ and $+{\Delta }_f/2$ with probability $1/4$. At $\theta =0$ the initial ground state remains the final ground state; at $\theta =\pi$ it is the final excited energy eigenstate. These statements are about the new energy labels, not an instantaneous physical rotation of the quantum state.

If a trial quantum state is dominated by two energy eigenstates of distinct energies $E_a<E_b$, then its Rayleigh quotient is approximately their probability-weighted mean. Rearranging that mean gives the displayed relative admixture. Neglected states must make small contributions to the energy numerator as well as to wavefunction normalization; small probabilities alone need not suffice if their energies are large.

An energy eigenstate has $⟨x⟩=⟨p⟩=0$, $⟨x^2⟩=\hslash (n+1/2)/(m\omega )$ and $⟨p^2⟩=m\hslash \omega (n+1/2)$. Thus $\Delta x\Delta p=\hslash (n+1/2)$; only the ground state in this sequence minimizes the Heisenberg uncertainty principle.

A time-independent Hamiltonian evolves a state by the unitary operator $U(t)=e^{-iHt/\hslash }$.

For a closed quantum system, unitary time evolution changes a quantum state through a unitary operator, preserving norms and inner products. For a time-independent self-adjoint Hamiltonian operator $H$, choose the initial time as zero:Here $\hslash =h/(2\pi )$ is the reduced Planck constant. The same $U(t)$ acts on every initial quantum state; it depends on the Hamiltonian operator and elapsed time, rather than the chosen preparation.^[ref]

The self-adjointness of $H$ makes the exponential a unitary operator. In finite dimension, conjugating its matrix exponential givesThe two exponents commute because both are multiples of $H$. Hence, for any two initial vectors,This short proof explains why normalization survives, why orthogonality of initial vectors survives, and why the transition probability between two states evolved by the same $U$ is unchanged. It does not require that a fixed measurement in quantum mechanics have time-independent probabilities: its measurement basis need not move with the quantum states.

For an unbounded self-adjoint operator, define the exponential through the spectral theorem, rather than assuming that its power series acts on every vector. It is a unitary operator on the entire Hilbert space; the strong Schrödinger equation holds for initial vectors in the operator domain $D(H)$. For a density operator, unitary time evolution reads $\rho (t)=U(t)\rho (0)U(t{)}^{†}$. It preserves eigenvalues and purity of a density operator, but can change its entries in a fixed basis.

The useful distinction is between conserving total probability and conserving each measurement's probabilities. A fixed energy spectrum supplies different ticking rates for different probability amplitudes. Their relative quantum phases can move the Bloch vector and create quantum interference even while every energy eigenstate population stays fixed. Qubit precession under a constant Hamiltonian makes this visible; Ramsey interferometry converts the accumulated relative quantum phase into a population difference that a simple readout can detect.

Long imaginary-time evolution multiplies an energy eigenstate by $e^{-T{E}_n}$. After normalization it projects onto the lowest energy component with nonzero initial overlap, or onto the lowest-energy subspace if the vacuum is degenerate. This provides vacuum boundary conditions for a path integral and is related to the Feynman i-epsilon prescription.

The variational method estimates the lowest energy of a Quantum Hamiltonian by choosing a trial wavefunction, evaluating its energy, and optimizing the trial parameters. It uses the quantum variational principle, the Rayleigh-Ritz variational principle expressed for quantum states. Let $H$ be a self-adjoint operator bounded below, and define $E_0=\inf \sigma (H)$, the bottom of its spectrum. Every normalized admissible trial quantum state satisfiesWhen $E_0$ is an energy eigenvalue, it is the ground-state energy; equality holds precisely for a quantum state in the ground-state subspace. Otherwise the bottom of the spectrum need not be attained by any normalizable wavefunction.

Admissible means finite form energy. Choose a real $b\le E_0$ and use the shifted quadratic form of a positive quantum Hamiltonian:The form domain $Q(H)$ is independent of this choice of $b$ and can be larger than the operator domain $D(H)$. For $\psi \in D(H)$, $E_{\psi }=⟨\psi ,H\psi ⟩$ is the usual expectation value; for nonzero unnormalized trials, divide $q_H[\psi ]$ by $\Vert \psi {\Vert }^2$ to get the form version of the Rayleigh quotient. See the quantum variational principle for the general domain justification.

Thus each trial energy is an upper bound on $E_0$, and minimizing within a trial family improves that upper bound. It is not a lower bound on $E_0$. The Gaussian variational energy for the harmonic oscillator reaches $E_0=1/2$ because its family contains an exact ground state. The exact variational solution in a two-level system makes the equality condition visible as vanishing excited-state probability amplitude.

The method is useful when solving the Schrödinger equation exactly is difficult but integrals of a trial wavefunction are tractable. Enlarging a trial family cannot increase its optimal energy, although an optimum need not be attained. The beauty is that even an imperfect trial gives a rigorous bound, provided its form energy is computed exactly: numerical integration errors require separate control. The Gaussian variational estimate for a solvable sextic potential provides an existing example beyond a quantum harmonic oscillator.

A short proof explains the direction. With a complete orthonormal eigenbasis of energy eigenstates, write $\psi ={\sum }_n{c}_n{\varphi }_n$, where $H{\varphi }_n=E_n{\varphi }_n$ and ${\sum }_n|c_n{|}^2=1$. ThenThis is a weighted mean of nonnegative energy differences; equality requires every occupied energy eigenstate to have energy $E_0$. ^[ref]. For continuous spectrum, the spectral theorem replaces the sum by an integral against the scalar spectral measure of $\psi$, with the same nonnegative integrand. This proves the bound without assuming an orthonormal eigenbasis exists.

The historical setting predates quantum mechanics: Rayleigh's The Theory of Sound, volume I, appeared in 1877, and Ritz's paper on solving variational problems of mathematical physics appeared in volume 135 of Crelle's journal, catalogued by its publisher as 1909, pp. 1–61. These are original works associated with the Rayleigh-Ritz method; the dates here identify the publications rather than assert priority. ^[ref]; ^[ref].

Wavefunction normalization makes the total Born rule probability equal to one. In an orthonormal basis of energy eigenstates, $\psi ={\sum }_n{c}_n{\varphi }_n$ is normalized precisely when ${\sum }_n|c_n{|}^2=1$.

For a proposed infinite square well mode $A\sin (n\pi x/L)$, the integral of its squared modulus is $|A{|}^{2}L/2$. Thus wavefunction normalization requires $|A|=\sqrt{2/L}$; choosing $A$ real and positive fixes a convenient global phase. For the equal quantum superposition $({\varphi }_1+{\varphi }_2)/\sqrt{2}$, orthogonality makes the integrated cross term zero, leaving $1/2+1/2=1$. The cross term can nevertheless be nonzero locally: normalization constrains total probability, not the shape of its probability density.

An equal incoherent mixture within a fixed total-$J$ multiplet has density operator $\rho =P_J/(2J+1)$, where $P_J$ is its orthogonal projection. If the applied-field Hamiltonian operator is proportional to $L_z+2S_z$ and no spin-orbit interaction is retained, its energy eigenstates are the uncoupled states $|m_L,m_S⟩$. The Born rule gives their populations by the formula above, or equivalently by squared Clebsch-Gordan decomposition coefficients. Equal populations of coupled states need not give equal populations of magnetic energy eigenvalues.

For $L=1,S=1/2,J=3/2$, the nonextremal coupled states have weights $2/3$ and $1/3$ in their two uncoupled components. The five dimensionless magnetic energies $-2,-1,0,1,2$ then have populations $1/4,1/6,1/6,1/6,1/4$; the zero energy contains two independent states, each with population $1/12$. In a different, weak-field approximation with a dominant spin-orbit interaction, projecting into the isolated multiplet gives the Landé g-factor spectrum instead. The full operator and its projection must not be interchanged.

For $M$ complete energy eigenstates of a one-particle Quantum Hamiltonian for identical noninteracting fermions, order their energies ${ϵ}_1\le \cdots \le {ϵ}_M$. At fixed particle number $0\le N\le M$, the Quantum Hamiltonian and the particle-number constraint areThe ground state occupies the $N$ lowest complete one-particle states. A short minimization argument suffices: moving an occupant from a higher energy to an empty lower energy decreases the total energy. This filling describes the ground state at temperature $T=0$; exclusion by itself also applies to excited quantum states and at positive temperature. The underlying thermal description is the Fermi-Dirac distribution.^[ref]

If the last occupied energy belongs to a group of $g$ one-particle states at the same energy and $q$ of them must be occupied, with all lower states filled, the ground state subspace has dimension $\left(\genfrac{}{}{0ex}{}{g}{q}\right)$. Exclusion does not select a particular basis or a particular quantum superposition in that subspace. The factor of two for an electron spatial orbital assumes both spin projections are included; a model restricted to one spin projection has only one state per spatial orbital. A magnetic field can split the degenerate energy levels, and interactions can change the lowest-energy electron configuration. Thus this finite noninteracting filling rule is not a complete prescription for interacting chemical atoms.