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For bounded linear operators, define $C(\beta )=[A,e^{-\beta B}]$. The product rule gives $C^{\prime }+BC=[B,A]e^{-\beta B}$ and $C(0)=0$. Multiplying by the integrating factor $e^{\beta B}$ yieldsThe identity is useful when the two linear operators do not commute: it preserves their order rather than incorrectly differentiating an operator exponential as if all its factors were scalars. For unbounded linear operators, the products and differentiations need a common justified operator domain.

Two linear operators $A$ and $B$ commute when $AB=BA$, with all products taken on the stated operator domains. For finite matrices, or bounded linear operators on a Banach space, commutation impliesThe binomial theorem applies because the factors commute, and absolute convergence of the operator exponential series permits regrouping its terms. For unbounded operators, a formal equality on one common operator domain alone does not justify this exponential identity; the existence and commutation of the corresponding exponential operators require separate hypotheses.

For a bounded skew-adjoint operator on a complex Hilbert space, the operator exponential converges in norm because its term norms are bounded by $\Vert T{\Vert }^n/n!$. Continuity of the adjoint operator gives $(e^T{)}^{\ast }=e^{T^{\ast }}=e^{-T}$. Absolute convergence permits the Cauchy product of $e^T$ and $e^{-T}$: its degree-$n$ coefficient is $T^n{\sum }_{j=0}^n(-1{)}^j/[j!(n-j)!]$, zero for $n>0$ and the identity for $n=0$. Both multiplication orders therefore give the identity, proving unitarity. The boundedness hypothesis makes this elementary power-series proof valid without domain issues for unbounded generators.

For the bounded rank-one operator $Kx=a\phi (x)$, insert the powers of a rank-one operator into the convergent operator exponential power series. Writing $c=\phi (a)$ gives the displayed formula for $c\ne 0$. At $c=0$ the quotient has limit $t$, so $e^{tK}=I+tK$. This includes the zero operator and the nonzero nilpotent operator cases without dividing by zero.

Time dependence in quantum mechanics describes how a prepared quantum state changes and thereby changes the probabilities of later measurements. For an isolated system, the Schrödinger picture assigns this evolution to a state vector $|\psi (t)⟩$, governed by the Time-dependent Schrödinger equationThe Hamiltonian is a self-adjoint operator; it has units of energy, while $\hslash$ has units of action, so $\hat{H}/\hslash$ has units of inverse time. For a particle with potential energy $V(\mathbf{r},t)$, this becomesGiven a suitable initial wavefunction and the physical boundary conditions, this first-order equation determines its subsequent propagation. Its linearity means that the evolution of a quantum superposition is the corresponding quantum superposition of evolved states.

For a real potential energy, this equation and its complex conjugate imply the local continuity equationHere $\rho$ is the probability density and $\mathbf{j}$ the probability current. Vanishing net boundary flux preserves wavefunction normalization, as required by the Born rule.

The time-evolution operator packages the solution for every initial state:Because ${\hat{H}}^{†}=\hat{H}$, differentiation givesTherefore $U^{†}U=I$: this is unitary time evolution, preserving inner products and all state norms. It is reversible, with $U(t,t_0{)}^{-1}=U(t_0,t)=U(t,t_0{)}^{†}$, and obeys the composition law $U(t_2,t_0)=U(t_2,t_1)U(t_1,t_0)$.

For a time-independent Hamiltonian, the solution is the operator exponentialIf the Hamiltonian varies with time, the analogous simple exponential of its integral is valid when $[\hat{H}(t),\hat{H}(t^{\prime })]=0$ for all relevant times. Otherwise time ordering is essential:The meaning is explicit in the first terms of the ordered series,the later-time operator stands on the left. In the position representation, the same propagation is expressed by the quantum-mechanical propagatorIts initial value is the Dirac delta function ${\delta }^3(\mathbf{r}-{\mathbf{r}}^{\prime })$.

A time-independent Hamiltonian has energy eigenstates satisfying $\hat{H}|n⟩=E_n|n⟩$. A single such initial state evolves asIt is a stationary state: its global phase changes, but its probability density and the distributions of time-independent observables do not. Its energy is sharp. Any quantum superposition entirely within one degenerate energy eigenspace also acquires one common global phase and is stationary. Stationarity does not require zero probability current; for instance, $u_n(\varphi )=e^{in\varphi }/\sqrt{2\pi }$ on the ring has a constant density and persistent circulation when $n\ne 0$.

An arbitrary normalized state can instead contain different energy eigenvalues:A continuous spectrum requires corresponding integrals. For the discrete case, an energy measurement gives $E_n$ with fixed probability $|c_n{|}^2$ (summed over degeneracies), but relative quantum phases change with time. ConsequentlyThe interference terms can oscillate at angular frequencies $(E_m-E_n)/\hslash$. For two populated energy levels, this supplies a characteristic period $2\pi \hslash /|E_m-E_n|$ when the interference term is nonzero. Thus fixed energy populations do not imply a stationary spatial distribution.

For a normalized state and an observable $\hat{A}(t)$, its expectation value isThis is the statistical mean over repeated preparations and measurements. In an eigenbasis of energy eigenstates for a time-independent Hamiltonian and a time-independent $\hat{A}$,Only matrix elements connecting different energies can generate these oscillations. If $[\hat{A},\hat{H}]=0$, its expectation value is constant for every initial state, provided it has no explicit time dependence.

Differentiating the expectation value and substituting the Schrödinger equation for both the bra and ket proves the general Ehrenfest theorem:Equivalently,These manipulations require the state to lie in the relevant operator domains. Taking $\hat{A}=\hat{H}$ gives $d⟨H⟩/dt=⟨{\partial }_t\hat{H}⟩$, so energy is conserved under a time-independent Hamiltonian.

For $\hat{H}={\hat{p}}^2/(2m)+V(\hat{x})$, the canonical commutation relation givesHence the particularly useful position-and-momentum form of Ehrenfest theorem isIt resembles Newton's second law, but the mean force generally differs from the force at the mean position. If a narrow wave packet samples a region where the force varies little, $⟨V^{\prime }(x)⟩\simeq V^{\prime }(⟨x⟩)$ gives approximate classical motion. For a quadratic potential energy this replacement is exact. In particular the quantum harmonic oscillator obeysThe centroid follows the classical harmonic oscillator, although this does not specify the full evolving wavefunction or its width.

The Heisenberg picture expresses the same physics by assigning evolution to observables instead:The expectation value is unchanged, and differentiation yields $d{\hat{A}}_H/dt=(i/\hslash )[{\hat{H}}_H,{\hat{A}}_H]+U^{†}({\partial }_t{\hat{A}}_S)U$. Thus state propagation, evolving expectation values, and the classical correspondence in Ehrenfest theorem are mutually consistent descriptions of the same unitary time evolution.

For a Hermitian operator $Q$, the operator exponential $C_{\lambda }=e^{i\lambda Q}$ is a unitary operator, with $C_{\lambda }^{†}=e^{-i\lambda Q}$. Differentiation and $[Q,C_{\lambda }]=0$ giveWhen $[Q,A]=cA$, this is the first-order ordinary differential equation $A^{\prime }(\lambda )=icA(\lambda )$ with $A(0)=A$. Multiplication by $e^{-ic\lambda }$ makes its derivative zero, provingEquivalently, the commutator expansion for exponential conjugation has successive nested commutators $c^{r}A$, whose exponential series sums to this expression.

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

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

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

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

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

For a free particle on the line or a circle, $H_0=P^2/(2m)$ commutes with the momentum operator $P$. The commuting operator exponentials in the title therefore factorize. Since $P=-i\hslash {\partial }_x$, the factor $e^{isPt/\hslash }=e^{st{\partial }_x}$ translates a wavefunction to $f(x+st)$, as is verified on each momentum eigenstate. Thus the probability density translates with velocity $-s$ relative to the $H_0$ evolution. A real constant $C$ changes only the global phase. In particular a superposition in one degenerate energy eigenspace translates rigidly under the added term $-sP$.