Equivalent Formulations

Quantum mechanics can be written in several languages that look strikingly different. A pure state may be a ray, a column of amplitudes, or a wavefunction. Time dependence may sit in the state or in the observables. A transition amplitude may be computed from an evolution operator or from a time-sliced sum over histories. An algebraic treatment may begin with observables and expectation-value functionals rather than with a preselected Hilbert-space representation.

These descriptions are equivalent only after a translation dictionary and a domain of validity have been specified. The statement is physical, not merely notational:

Corresponding preparations, transformations, and measurements must give the same outcome statistics.

The different mathematical senses of equivalence matter when deciding which translations preserve a physical prediction. Detailed coordinate translations live in the Representation Translation Table; pictures, propagators, path integrals, and algebraic dynamics live in Quantum Dynamics and Formulations.

Helpful background. Bases and representations explains coordinate choices; minimal postulates identifies the rules whose predictions must agree.

Translations that preserve probabilities

Suppose two formulations describe states by sets $\mathcal S_1$ and $\mathcal S_2$, and measurements by sets $\mathcal M_1$ and $\mathcal M_2$. An equivalence claim requires maps

$$\Phi_{\mathrm S}: \mathcal S_1\longrightarrow\mathcal S_2, \qquad \Phi_{\mathrm M}: \mathcal M_1\longrightarrow\mathcal M_2.$$

For every preparation $s$, measurement $M$, and measurable outcome event $X$, the translated probabilities must agree:

$$p_1(X\mid s,M) = p_2\!\left( X\mid \Phi_{\mathrm S}(s), \Phi_{\mathrm M}(M) \right).$$

If dynamics are included, their translations must also agree. Schematically, if $T_t$ and $\widetilde T_t$ are the evolution rules,

$$\Phi_{\mathrm S}\!\left(T_t(s)\right) = \widetilde T_t\!\left(\Phi_{\mathrm S}(s)\right).$$

For a full operational equivalence, agreement must extend beyond isolated one-time probabilities. It must preserve whatever the stated domain includes:

• normalization and convex mixtures of preparations;
• expectation values and complete outcome distributions;
• transition amplitudes when their phases are operationally relevant;
• sequential-measurement probabilities and conditional updates;
• composition of systems and transformations;
• boundary conditions, operator domains, and approximation choices.

Agreement of one expectation value does not establish full equivalence. For full equivalence, the translations must cover the physically distinct states and procedures on both sides, up to representational redundancy; a one-way embedding alone can compare only a restricted sector.

Several Meanings of Equivalent

The word equivalent carries different mathematical strengths in common quantum-mechanics usage.

Comparison	Translation	What is preserved	Important qualification
abstract vectors and basis components	unitary coordinate map	all inner products and operator matrix elements	basis and measure must be translated together
Schrödinger and Heisenberg pictures	time-dependent unitary conjugation	all corresponding expectation values and histories	states and observables carry complementary time dependence
pure vectors and density operators	$\lvert\psi\rangle\mapsto\lvert\psi\rangle\langle\psi\rvert$	all pure-state statistics	this is an embedding into the rank-one sector, not a representation of every mixed state by a vector on the same space
evolution operators and path integrals	time slicing and a continuum limit	propagators and amplitudes	measure, ordering, boundary data, and regularization are part of the claim
Hilbert-space and algebraic language	observables represented as operators; states as positive functionals	expectation values on the chosen algebra	finite matrix systems are especially direct; infinite systems can have inequivalent Hilbert-space representations

This taxonomy prevents two opposite errors. One should not mistake a basis change for a new theory, but one should also not call a truncation, coarse-graining, or uncontrolled formal manipulation an exact equivalence.

State-Vector Formulation

The state-vector formulation begins with a complex Hilbert space $\mathcal H$. A pure physical state is a ray. A normalized vector $\lvert\psi\rangle$ is a representative, and

$$\lvert\psi\rangle \sim e^{i\alpha}\lvert\psi\rangle$$

represents the same ray for every real $\alpha$.

A measurement event is represented, in the projective case, by a projector $P(X)$. Its probability is

$$p(X\mid\psi) = \langle\psi\lvert P(X) \rvert\psi\rangle.$$

For a closed system, a unitary evolution operator transports the state:

$$\lvert\psi(t)\rangle = U(t,t_0)\lvert\psi(t_0)\rangle.$$

For states in the appropriate Hamiltonian domain, with the required time regularity,

$$i\hbar \frac{d}{dt} \lvert\psi(t)\rangle = H(t)\lvert\psi(t)\rangle.$$

This abstract form separates the physical state from any coordinate choice. It is especially economical for pure closed systems, symmetry arguments, and finite-dimensional problems. State Vectors explains the ray structure, while Minimal Postulates places pure vectors inside the broader operational postulate package.

A mixed reduced state or an unread measurement output need not be representable by a vector on the same Hilbert space. Some individual measurement branches also remain mixed. Those cases motivate density operators, not a rejection of state vectors.

Wavefunction Formulation

A wavefunction is a coordinate representation of a state. In the position representation,

$$\psi(x) = \langle x\mid\psi\rangle.$$

The map from an abstract state to its position wavefunction preserves the inner product:

$$\langle\phi\mid\psi\rangle = \int \phi(x)^*\psi(x)\,dx.$$

Consequently, normalization becomes

$$\int \lvert\psi(x)\rvert^2\,dx = 1,$$

and the probability of finding the particle in a region $R$ is

$$\Pr(x\in R) = \int_R \lvert\psi(x)\rvert^2\,dx.$$

An abstract operator $A$ becomes an operator $A_x$ on wavefunctions:

$$(A_x\psi)(x) = \langle x\mid A\mid\psi\rangle.$$

For the familiar one-dimensional Hamiltonian,

$$H = \frac{p^2}{2m}+V(x),$$

the position representation is

$$H_x = -\frac{\hbar^2}{2m} \frac{d^2}{dx^2} +V(x),$$

and the abstract Schrödinger equation becomes

$$i\hbar \frac{\partial\psi(x,t)}{\partial t} = H_x\psi(x,t).$$

The notation $\lvert x\rangle$ is generalized-eigenvector notation. In a careful functional-analytic treatment, physical states are square-integrable equivalence classes and position eigenkets belong to a rigged extension rather than to $L^2$ itself. Boundary conditions and the domain of $H_x$ are part of the operator definition. The wavefunction equation is therefore not determined by a differential expression alone.

The complete postulate-level presentation is Wave-Mechanics Postulates.

Discrete coefficients and continuous wavefunctions

Let $\{\lvert n\rangle\}$ be an orthonormal basis, and define

$$c_n = \langle n\mid\psi\rangle, \qquad \phi_n(x) = \langle x\mid n\rangle.$$

Then the same state has the two representations

$$\lvert\psi\rangle = \sum_n c_n\lvert n\rangle, \qquad \psi(x) = \sum_n c_n\phi_n(x).$$

For the following calculation, take the state and operator to have finite expansions in this orthonormal family. The position-space kernel of the operator with matrix elements $A_{mn}$ is

$$\begin{aligned} A(x,x') &= \langle x\mid A\mid x'\rangle \\ &= \sum_{m,n} \phi_m(x) A_{mn} \phi_n(x')^*. \end{aligned}$$

Using orthonormality twice gives

$$\begin{aligned} \langle\psi\mid A\mid\psi\rangle &= \int dx\,dx'\, \psi(x)^* {}\times A(x,x')\psi(x') \\ &= \sum_{m,n} c_m^* A_{mn} c_n \\ &= c^\dagger A c. \end{aligned}$$

The integral and matrix formulas are the same pairing. General operators can have distributional kernels, and infinite sums or exchanges of sums and integrals need separate convergence arguments; the finite-sum calculation avoids these complications. Finite rank alone does not make a matrix expansion finite in an arbitrary basis.

Matrix Mechanics

Choose an orthonormal basis $\{\lvert n\rangle\}$. The state and operator become

$$c_n = \langle n\mid\psi\rangle, \qquad A_{mn} = \langle m\mid A\mid n\rangle.$$

The Schrödinger equation is then a system of coupled equations:

$$i\hbar\dot c_m = \sum_n H_{mn}c_n.$$

This is matrix mechanics in a chosen basis. The basis can be finite, countably infinite, or a finite truncation used for approximation. Only the first two can represent the full system exactly; a truncation requires a separate error analysis.

To see basis independence explicitly, let $\{\lvert e_n\rangle\}$ and $\{\lvert f_a\rangle\}$ be orthonormal bases, and define

$$S_{an} = \langle f_a\mid e_n\rangle.$$

With this convention,

$$d = Sc, \qquad A_f = S A_e S^\dagger.$$

The expectation value is unchanged:

$$\begin{aligned} d^\dagger A_f d &= c^\dagger S^\dagger \left(S A_e S^\dagger\right) S c\\ &= c^\dagger A_e c. \end{aligned}$$

The cancellation uses $S^\dagger S=I$. Transforming the state column without transforming the operator matrix is not a new formulation; it is an inconsistent calculation.

For infinite matrices, convergence and operator domains matter. A formally unitary change of basis must map the relevant domains correctly. A finite matrix obtained by projecting onto selected basis states is generally an effective model, not a unitarily equivalent copy of the original unbounded operator. Change of Basis and Operator Representations develop these distinctions.

Density-Operator Formulation

A density operator is a positive trace-one operator:

$$\rho\geq0, \qquad \operatorname{Tr}\rho=1.$$

Every normalized pure vector defines a rank-one density operator,

$$\rho_\psi = \lvert\psi\rangle\langle\psi\rvert.$$

The global phase disappears automatically:

$$\lvert e^{i\alpha}\psi\rangle \langle e^{i\alpha}\psi\rvert = \lvert\psi\rangle\langle\psi\rvert.$$

For a measurement effect $E(X)$, the trace rule reproduces the vector Born rule:

$$\begin{aligned} \operatorname{Tr}(\rho_\psi E(X)) &= \operatorname{Tr}\!\left( \lvert\psi\rangle \langle\psi\rvert E(X) \right)\\ &= \langle\psi\mid E(X)\mid\psi\rangle. \end{aligned}$$

Under closed-system unitary dynamics,

$$\rho(t) = U(t,t_0) \rho(t_0) U(t,t_0)^\dagger.$$

Thus vector and density-operator formulas agree exactly on the pure-state sector.

An embedding, not a bijection onto all states

The map

$$\text{pure ray} \longmapsto \text{rank-one density operator}$$

is one-to-one, but not every density operator has rank one. A mixed state such as

$$\rho = \frac12\lvert0\rangle\langle0\rvert +\frac12\lvert1\rangle\langle1\rvert = \frac{I}{2}$$

cannot be represented by a vector on the same two-dimensional Hilbert space. Indeed,

$$\operatorname{Tr}(\rho^2) = \frac12,$$

whereas every pure-state density operator satisfies $\operatorname{Tr}(\rho_\psi^2)=1$.

A mixed state can be purified as a vector on a larger Hilbert space, but the purification is nonunique and the original state is recovered only after a partial trace. This is an equivalence of subsystem statistics after an extension and reduction, not an identification of $\rho$ with a unique vector on the original space.

The density-operator formulation also accommodates channels and instruments:

$$\rho \longmapsto \mathcal E(\rho),$$

where $\mathcal E$ is completely positive and trace preserving for a quantum channel. The integrated postulate package is Density-Matrix Formulation; the detailed state theory begins with Density Operators.

Heisenberg-Picture Formulation

The Schrödinger and Heisenberg pictures redistribute time dependence between states and observables. Let $U(t,t_0)$ be unitary. For density operators, one may write

$$\begin{aligned} \rho_S(t) &= U(t,t_0)\rho_H U(t,t_0)^\dagger, \\ \rho_H &= \rho_S(t_0). \end{aligned}$$

The corresponding Heisenberg observable is

$$A_H(t) = U(t,t_0)^\dagger A_S(t) U(t,t_0),$$

where $A_S(t)$ may also have explicit time dependence. For bounded observables, cyclicity of the trace gives

$$\begin{aligned} \operatorname{Tr}\!\left( \rho_S(t)A_S(t) \right) &= \operatorname{Tr}\!\left( \rho_H U^\dagger A_S(t)U \right)\\ &= \operatorname{Tr}\!\left( \rho_H A_H(t) \right). \end{aligned}$$

Every translated bounded-observable expectation agrees. In finite dimension (or on a suitable common invariant domain), a time-independent Hamiltonian and an observable with no explicit time dependence give

$$\frac{dA_H}{dt} = \frac{i}{\hbar} [H_H,A_H].$$

The fixed Heisenberg state does not mean that nothing happens. It means that time dependence is encoded in the family of observables used to interrogate that state.

The same duality extends beyond unitary channels. A channel $\mathcal E$ acting on states has an adjoint map $\mathcal E^*$ acting on observables, defined by

$$\operatorname{Tr}\!\left( \mathcal E(\rho)A \right) = \operatorname{Tr}\!\left( \rho\,\mathcal E^*(A) \right).$$

This is the open-system analogue of moving evolution across the state-observable pairing. Heisenberg Picture and Picture Transformations develop the full dynamics.

Path-Integral Formulation Preview

The operator formulation assigns a propagator kernel

$$K(x_f,t_f;x_i,t_i) = \langle x_f\mid U(t_f,t_i) \mid x_i\rangle.$$

Its composition law follows directly from inserting the position resolution of the identity at an intermediate time $t$:

$$K(x_f,t_f;x_i,t_i) = \int dx\, K(x_f,t_f;x,t) {}\times K(x,t;x_i,t_i).$$

Repeating this identity over short time steps leads to the formal path-integral notation

$$K(x_f,t_f;x_i,t_i)=\int_{x(t_i)=x_i}^{x(t_f)=x_f}\mathcal D x\,e^{iS[x]/\hbar}.$$

The factor $e^{iS/\hbar}$ is an amplitude phase, not a probability weight. The integration symbol abbreviates a limiting construction; it is not an ordinary uniform measure over real-time paths. For suitable Hamiltonians, operator product limits justify time slicing. Normalization, potential regularity, operator ordering, boundary conditions, and the meaning of convergence belong to that construction.

This makes path integrals useful for semiclassical approximations, but does not guarantee that every formal expression defines the original operator. From propagators to path integrals gives the construction, and Why path integrals? explains its uses.

Algebraic Formulation Preview

The algebraic formulation begins with a unital $C^*$-algebra $\mathcal A$ of bounded observables. A state is a normalized positive linear functional

$$\omega: \mathcal A\longrightarrow\mathbb C$$

satisfying

$$\omega(I)=1, \qquad \omega(A^*A)\geq0.$$

An effect $E$ satisfies $0\leq E\leq I$, and the probability assigned by the state is

$$p(E\mid\omega) = \omega(E).$$

For the finite matrix algebra $\mathcal A=M_d(\mathbb C)$, every state has a unique density-operator representation:

$$\omega_\rho(A) = \operatorname{Tr}(\rho A).$$

Positivity and normalization of $\omega_\rho$ follow from $\rho\geq0$ and $\operatorname{Tr}\rho=1$. Conversely, finite-dimensional trace duality reconstructs a unique positive trace-one $\rho$ from every state functional. In this setting, the algebraic and density-operator languages carry the same statistical content.

Heisenberg dynamics appears algebraically as a family of automorphisms:

$$\alpha_t(A) = U(t)^\dagger A U(t).$$

The dual state evolution is

$$\omega_t(A) = \omega_0\!\left(\alpha_t(A)\right).$$

The algebraic viewpoint becomes more than a stylistic rearrangement for infinitely many degrees of freedom. Different states or physical sectors can generate unitarily inequivalent Hilbert-space representations. The Gelfand–Naimark–Segal construction associates a representation $(\mathcal H_\omega,\pi_\omega,\Omega_\omega)$ with a state such that

$$\omega(A) = \langle \Omega_\omega \mid \pi_\omega(A) \mid \Omega_\omega \rangle.$$

One should therefore not claim that every infinite-system algebraic model is equivalent to one fixed Hilbert-space representation. The canonical entry points are Algebraic Formulation Overview and States as Positive Linear Functionals.

Worked Example: One Qubit in Four Presentations

Let

$$H = \frac{\hbar\omega}{2}Z, \qquad \lvert\psi(0)\rangle = \lvert+x\rangle,$$

and ask for the probability of obtaining $+1$ in an $X$ measurement at time $t$. Write $\theta=\omega t$.

State-vector calculation

The evolution operator is

$$U(t) = e^{-i\theta Z/2}.$$

Since

$$\lvert+x\rangle = \frac{ \lvert0\rangle+\lvert1\rangle }{ \sqrt2 },$$

the evolved state is

$$\lvert\psi(t)\rangle = \frac{1}{\sqrt2} \left( e^{-i\theta/2}\lvert0\rangle + e^{i\theta/2}\lvert1\rangle \right).$$

Therefore

$$\begin{aligned} p_{+x}(t) &= \left| \langle+x\mid\psi(t)\rangle \right|^2\\ &= \cos^2\!\left(\frac{\theta}{2}\right)\\ &= \frac{1+\cos\theta}{2}. \end{aligned}$$

Matrix calculation

In the $Z$ basis,

$$c_0 = \frac{1}{\sqrt2} \begin{pmatrix} 1\\ 1 \end{pmatrix}, \qquad U = \begin{pmatrix} e^{-i\theta/2}&0\\ 0&e^{i\theta/2} \end{pmatrix},$$

and

$$P_{+x} = \frac12 \begin{pmatrix} 1&1\\ 1&1 \end{pmatrix}.$$

The same probability is

$$p_{+x}(t) = c_0^\dagger U^\dagger P_{+x} U c_0 = \frac{1+\cos\theta}{2}.$$

Nothing physical was added by the matrices. A basis made every abstract operator and vector explicit.

Density-operator calculation

Initially,

$$\rho_0 = P_{+x} = \frac12(I+X).$$

Schrödinger evolution rotates the Bloch vector:

$$\rho_S(t) = \frac12 \left( I+X\cos\theta+Y\sin\theta \right).$$

Using the trace rule and $\operatorname{Tr}(\sigma_i\sigma_j)=2\delta_{ij}$,

$$\begin{aligned} p_{+x}(t) &= \operatorname{Tr}\!\left( \rho_S(t)P_{+x} \right)\\ &= \frac{1+\cos\theta}{2}. \end{aligned}$$

Heisenberg-picture calculation

Keep $\rho_H=\rho_0$ fixed and evolve the measurement projector:

$$\begin{aligned} P_{+x,H}(t) &= U^\dagger P_{+x}U \\ &= \frac12 \left( I+X\cos\theta-Y\sin\theta \right). \end{aligned}$$

Then

$$\begin{aligned} p_{+x}(t) &= \operatorname{Tr}\!\left( \rho_H P_{+x,H}(t) \right)\\ &= \frac{1+\cos\theta}{2}. \end{aligned}$$

The opposite signs of the $Y$ terms encode opposite rotations of states and observables. The measured probability is invariant because the entire state-observable pairing was translated.

When Equivalence Needs Care

Domains and boundary conditions

An unbounded operator is not specified by a differential expression alone. For example, the formal operator

$$-i\hbar\frac{d}{dx}$$

can have different domains on an interval, and boundary conditions affect self-adjointness and the spectrum. Two coordinate formulas with different domains need not represent the same observable.

Approximation is not equivalence

Projection onto a finite subspace,

$$H \longmapsto P_N H P_N,$$

generally discards states and alters high-energy behavior. It may be an excellent controlled approximation, but it is not invertible and therefore is not an exact change of representation. The same warning applies to coarse-graining, tracing out an environment, replacing a channel by a unitary model, and taking an asymptotic limit.

Measurement translations matter

Transforming only a state is insufficient. If

$$\rho' = V\rho V^\dagger,$$

then the corresponding effect must be transformed consistently:

$$E' = V E V^\dagger.$$

Only then does

$$\operatorname{Tr}(\rho'E') = \operatorname{Tr}(\rho E).$$

A passive basis change and an active physical unitary can use similar formulas. Their interpretation differs, so the surrounding convention must be stated.

Infinite systems can change the problem

For finitely many canonical position–momentum pairs, regular irreducible representations of the Weyl relations, with the same nonzero value of $\hbar$, are unitarily equivalent by the Stone–von Neumann theorem. A formal commutator equation on some domain is not enough to invoke this result. For infinitely many degrees of freedom, inequivalent representations can describe different phases, vacua, temperatures, or superselection sectors. The finite-dimensional intuition that every representation is connected by one unitary matrix no longer applies universally.

Formulation does not choose interpretation

Wavefunctions, density operators, path integrals, and observable algebras can be used within more than one interpretation of quantum mechanics. Rewriting the predictive formalism does not by itself settle ontology, the status of probability, or why one outcome is experienced. What the Postulates Do Not Say keeps those questions separate.

Exercises

1. Let $S$ be unitary, $d=Sc$, and $A_f=SA_eS^\dagger$. Prove that the complete probability distribution for a projective measurement is basis independent, not only its expectation value.

Solution

Let $P_a^{(e)}$ be the projector for outcome $a$ in the original basis. In the new basis,

$$P_a^{(f)} = S P_a^{(e)}S^\dagger.$$

The translated probability is

$$\begin{aligned} p_f(a) &= d^\dagger P_a^{(f)}d\\ &= c^\dagger S^\dagger \left( S P_a^{(e)}S^\dagger \right) S c\\ &= c^\dagger P_a^{(e)}c\\ &= p_e(a). \end{aligned}$$

Because this holds for every outcome projector, the entire distribution is preserved.

2. For finite sums in an orthonormal family, start from $\psi(x)=\sum_n c_n\phi_n(x)$ and $A(x,x')=\sum_{m,n}\phi_m(x)A_{mn}\phi_n(x')^*$. Derive $\langle A\rangle=c^\dagger A c$.

Solution

Insert both expansions:

$$\begin{aligned} \langle A\rangle &= \int dx\,dx'\, \psi(x)^* {}\times A(x,x')\psi(x') \\ &= \sum_{r,m,n,s} c_r^* A_{mn} c_s I_{rm}J_{ns}, \\ I_{rm} &= \int dx\, \phi_r(x)^*\phi_m(x), \\ J_{ns} &= \int dx'\, \phi_n(x')^*\phi_s(x'). \end{aligned}$$

Orthonormality gives $I_{rm}=\delta_{rm}$ and $J_{ns}=\delta_{ns}$, so

$$\langle A\rangle = \sum_{m,n} c_m^*A_{mn}c_n = c^\dagger A c.$$

3. Show that two normalized vectors define the same rank-one density operator if and only if they differ by a global phase.

Solution

If $\lvert\phi\rangle=e^{i\alpha}\lvert\psi\rangle$, their projectors are equal because the two phases cancel.

Conversely, suppose

$$\lvert\phi\rangle\langle\phi\rvert = \lvert\psi\rangle\langle\psi\rvert.$$

Apply both sides to $\lvert\psi\rangle$:

$$\lvert\phi\rangle \langle\phi\mid\psi\rangle = \lvert\psi\rangle.$$

Thus $\lvert\phi\rangle$ is proportional to $\lvert\psi\rangle$. Since both vectors are normalized, the proportionality constant has modulus one and is $e^{i\alpha}$ for some real $\alpha$.

4. Reproduce the qubit result $p_{+x}(t)=(1+\cos\omega t)/2$ by evolving $X$ with the Heisenberg equation rather than by conjugating it with $U(t)$.

Solution

With $H=\hbar\omega Z/2$ and $[Z,X]=2iY$, $[Z,Y]=-2iX$,

$$\frac{dX_H}{dt} = \frac{i}{\hbar}[H,X_H].$$

Write

$$X_H(t) = a(t)X+b(t)Y.$$

The commutators give

$$\dot a = \omega b, \qquad \dot b = -\omega a,$$

with $a(0)=1$ and $b(0)=0$. Hence

$$X_H(t) = X\cos(\omega t) -Y\sin(\omega t).$$

Since

$$P_{+x,H}(t) = \frac12\left(I+X_H(t)\right),$$

and the initial $\lvert+x\rangle$ state has $\langle X\rangle=1$, $\langle Y\rangle=0$,

$$p_{+x}(t) = \frac12 \left( 1+\cos(\omega t) \right).$$

5. Let $\mathcal E$ be a quantum channel and define its adjoint by $\operatorname{Tr}(\mathcal E(\rho)A) =\operatorname{Tr}(\rho\,\mathcal E^*(A))$. Show that trace preservation of $\mathcal E$ implies $\mathcal E^*(I)=I$.

Solution

Set $A=I$. Trace preservation gives

$$\operatorname{Tr}\!\left( \mathcal E(\rho)I \right) = \operatorname{Tr}\rho.$$

By the defining adjoint relation,

$$\operatorname{Tr}\!\left( \rho\,\mathcal E^*(I) \right) = \operatorname{Tr}(\rho I)$$

for every density operator $\rho$. Therefore

$$\operatorname{Tr}\!\left[ \rho \left( \mathcal E^*(I)-I \right) \right] = 0$$

for all states. States separate Hermitian operators, so $\mathcal E^*(I)-I=0$. Thus the observable-picture map is unital.

6. Derive the propagator composition law from $U(t_f,t_i)=U(t_f,t)U(t,t_i)$.

Solution

Take the position matrix element and insert the resolution of the identity:

$$\begin{aligned} K(x_f,t_f;x_i,t_i) &= \int dx\, \langle x_f\mid U(t_f,t)\mid x\rangle {}\times \langle x\mid U(t,t_i)\mid x_i\rangle \\ &\quad= \int dx\, K(x_f,t_f;x,t) {}\times K(x,t;x_i,t_i). \end{aligned}$$

Repeated use of this identity is the operator origin of time slicing.

7. For a finite-dimensional density operator $\rho$, verify that $\omega_\rho(A)=\operatorname{Tr}(\rho A)$ is normalized and positive.

Solution

Normalization is immediate:

$$\omega_\rho(I) = \operatorname{Tr}(\rho I) = \operatorname{Tr}\rho = 1.$$

For positivity, diagonalize

$$\rho = \sum_n p_n \lvert n\rangle\langle n\rvert, \qquad p_n\geq0.$$

Then

$$\begin{aligned} \omega_\rho(A^\dagger A) &= \sum_n p_n \langle n\mid A^\dagger A \mid n\rangle\\ &= \sum_n p_n \lVert A\lvert n\rangle\rVert^2\\ &\geq0. \end{aligned}$$

Thus every density operator defines a normalized positive functional.

8. A harmonic-oscillator Hamiltonian is replaced by its projection onto the first ten energy eigenstates. Is the ten-dimensional model equivalent to the full oscillator? State what is preserved and what is lost.

Solution

The projection is not an exact equivalence because it is not invertible on the full Hilbert space. It removes every state component above the tenth retained level and cannot reproduce arbitrary high-energy observables or transitions.

Inside the retained invariant subspace, the projected Hamiltonian reproduces the exact energies and unitary phases of those ten eigenstates. Calculations whose initial states, observables, and dynamics remain inside that subspace can therefore agree exactly. Once a perturbation couples retained states to discarded levels, or an observable probes the omitted sector, the model is an approximation whose error must be estimated.

References

• O. Bratteli and D. W. Robinson, Operator Algebras and Quantum Statistical Mechanics 1, 2nd ed., Springer, 1987.
• P. A. M. Dirac, The Principles of Quantum Mechanics, 4th ed., Oxford University Press, 1958.
• R. P. Feynman and A. R. Hibbs, Quantum Mechanics and Path Integrals, McGraw-Hill, 1965.
• R. Haag, Local Quantum Physics: Fields, Particles, Algebras, 2nd ed., Springer, 1996.
• B. C. Hall, Quantum Theory for Mathematicians, Springer, 2013.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• L. S. Schulman, Techniques and Applications of Path Integration, Dover, 2005.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• J. von Neumann, Mathematical Foundations of Quantum Mechanics, Princeton University Press, 1955.