Finite vs Infinite-Dimensional Quantum Mechanics

Finite-dimensional quantum mechanics and infinite-dimensional quantum mechanics use the same conceptual grammar:

• states are rays or density operators;
• observables are represented through self-adjoint operators and spectral measures;
• probabilities come from projectors or effects;
• closed dynamics are unitary;
• composite systems use tensor products.

What changes in infinite dimension is not the Born rule. What changes is the analytic care needed to make the same rules meaningful. Infinite sums and integrals require convergence, important operators can be unbounded, domains become part of an operator’s definition, and spectral values need not have normalizable eigenvectors.

The central bridge is:

A qubit column, an infinite sequence of oscillator amplitudes, and a position-space wavefunction are coordinate descriptions of Hilbert-space states. Their computational styles differ, but they are not different quantum theories.

This page explains where the finite-dimensional mental model is exact, where it remains useful, and where it must be supplemented. The detailed functional analysis belongs to the Infinite-Dimensional Hilbert Spaces chapter.

Comparison at a Glance

Feature	Finite dimension	Infinite dimension
Typical spaces	$\mathbb C^d$	$\ell^2$, $L^2$, Sobolev-type domains
State coordinates	finite column	square-summable sequence or square-integrable function
Linear operators	bounded and everywhere defined	may be bounded or unbounded; domains matter
Sharp observables	Hermitian matrices	self-adjoint operators, often unbounded
Spectrum	finitely many eigenvalues	point, continuous, or mixed spectrum
Spectral theorem	finite sum of eigenspace projectors	projector-valued spectral integral
Density operators	positive trace-one matrices	positive trace-class operators
Main technical issue	algebra and basis bookkeeping	convergence, domains, boundary conditions, and limits

Two cautions belong beside the table:

1. Infinite-dimensional does not mean continuous spectrum. The harmonic oscillator is infinite-dimensional with discrete energy levels.
2. Continuous coordinates do not mean a nonseparable Hilbert space. Standard spaces such as $L^2(\mathbb R)$ have countable orthonormal Hilbert bases even though position is labeled by every real number.

What Dimension Counts

The dimension of a Hilbert space is the number of vectors in an orthonormal Hilbert basis, not the number of possible values of one observable.

A finite-dimensional Hilbert space has an orthonormal basis

$$\left\{ \lvert e_1\rangle, \ldots, \lvert e_d\rangle \right\}.$$

Every vector is a finite linear combination:

$$\lvert\psi\rangle = \sum_{n=1}^{d} c_n\lvert e_n\rangle.$$

An infinite-dimensional separable Hilbert space has a countable orthonormal basis

$$\left\{ \lvert e_1\rangle, \lvert e_2\rangle, \ldots \right\},$$

and every vector is a norm-convergent infinite expansion:

$$\lvert\psi\rangle = \sum_{n=1}^{\infty} c_n\lvert e_n\rangle, \qquad \sum_{n=1}^{\infty} \lvert c_n\rvert^2 < \infty.$$

The partial sums converge in Hilbert-space norm:

$$\lim_{N\to\infty} \left\lVert \lvert\psi\rangle - \sum_{n=1}^{N} c_n\lvert e_n\rangle \right\rVert = 0.$$

Most ordinary quantum-mechanical Hilbert spaces are separable. Abstractly, every infinite-dimensional separable complex Hilbert space is unitarily isomorphic to $\ell^2(\mathbb N)$. That statement does not make all physical systems identical: their distinguished operators, domains, tensor factorizations, Hamiltonians, and symmetries can differ.

Nonseparable Hilbert spaces exist, but they are not required for the standard particle and wave-mechanics examples on this page.

Finite-Dimensional Systems

Finite-dimensional models include:

• qubits and qudits;
• spin-$j$ degrees of freedom, with dimension $2j+1$;
• genuinely finite-level systems;
• finite spin networks and finite-site lattices with finite local dimension;
• effective atomic, molecular, or circuit truncations.

For $\mathcal H=\mathbb C^d$, a normalized state satisfies

$$\sum_{n=1}^{d} \lvert c_n\rvert^2 = 1.$$

Every linear operator is represented by a finite matrix and is bounded. Every Hermitian matrix is self-adjoint on the whole space. A Hermitian operator has a finite spectral decomposition

$$A = \sum_a aP_a,$$

where the sum runs over distinct eigenvalues and the $P_a$ are orthogonal eigenspace projectors.

For a density operator,

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

all traces are finite sums. The maximally mixed state exists:

$$\rho_* = \frac{I_d}{d}.$$

These properties make finite-dimensional quantum mechanics especially clean:

• no linear-operator domain needs to be chosen;
• every spectrum is finite and consists of eigenvalues;
• matrix exponentials define unitary dynamics directly;
• all vector-space norms define the same notion of convergence;
• bounded sets have compactness properties unavailable in infinite dimension.

This mathematical convenience does not guarantee physical exactness. A two-level atom may use a $2\times2$ Hamiltonian while the actual atom has infinitely many bound and continuum states. The matrix calculation can be exact inside the model and still be an approximation to the apparatus.

Finite-Dimensional Postulates owns the complete finite postulate package.

Infinite Sequences Before Wavefunctions

Infinite dimension does not force one to begin with continuous coordinates. The sequence space

$$\ell^2(\mathbb N) = \left\{ (c_0,c_1,\ldots): \sum_{n=0}^{\infty} \lvert c_n\rvert^2 < \infty \right\}$$

is an infinite-dimensional Hilbert space with inner product

$$\langle d,c\rangle = \sum_{n=0}^{\infty} d_n^*c_n.$$

The harmonic oscillator energy basis is the canonical example:

$$\lvert\psi\rangle = \sum_{n=0}^{\infty} c_n\lvert n\rangle.$$

Its Schrödinger equation can be written as an infinite matrix equation,

$$i\hbar\dot c_m = \sum_{n=0}^{\infty} H_{mn}c_n,$$

provided the state lies in the appropriate domain and the series is meaningful.

This is still matrix mechanics, but the matrix is infinite and may represent an unbounded operator. Finite-matrix intuition remains useful, while convergence and domain questions no longer disappear automatically.

Wavefunctions and Function Spaces

For a particle on the line, normalizable position-space wavefunctions belong to

$$L^2(\mathbb R).$$

The norm and inner product are

$$\lVert\psi\rVert^2 = \int_{-\infty}^{\infty} \lvert\psi(x)\rvert^2\,dx,$$

and

$$\langle\phi\mid\psi\rangle = \int_{-\infty}^{\infty} \phi(x)^*\psi(x)\,dx.$$

Two functions that differ only on a set of measure zero represent the same $L^2$ vector. A point value $\psi(x_0)$ therefore is not an invariant property of an arbitrary $L^2$ equivalence class without choosing a suitable representative or adding regularity.

Many familiar wavefunctions are smoother than a generic $L^2$ vector because they lie in the domains of differential operators. That extra smoothness comes from the state and Hamiltonian being considered, not from the definition of $L^2$ alone.

Countable bases inside a continuous representation

Although the coordinate $x$ ranges over an uncountable set, $L^2(\mathbb R)$ is separable. For any complete countable orthonormal basis $\{\phi_n(x)\}$,

$$\psi(x) = \sum_{n=0}^{\infty} c_n\phi_n(x)$$

with convergence in $L^2$ norm.

The coefficient sequence and wavefunction are related by a unitary representation map:

$$c_n = \int_{-\infty}^{\infty} \phi_n(x)^*\psi(x)\,dx.$$

Thus a wavefunction calculation can be converted into an infinite matrix calculation, and an infinite coefficient calculation can be converted into a function representation. Bases and Representations owns the general translation.

Continuous Labels Are Not Ordinary Hilbert Bases

Physicists write

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

and the formal relations

$$\langle x\mid x'\rangle = \delta(x-x'), \qquad I = \int_{\mathbb R} dx\, \lvert x\rangle\langle x\rvert.$$

The exact position kets $\lvert x\rangle$ are not normalizable vectors in $L^2(\mathbb R)$. They are generalized spectral vectors used under integrals and pairings. In particular,

$$\langle x\mid x\rangle = \delta(0)$$

is not a finite norm.

The same warning applies to plane-wave momentum kets. A plane wave has constant magnitude:

$$\lvert e^{ipx/\hbar}\rvert^2 = 1,$$

so it is not square integrable on the full line. Normalizable wave packets are superpositions of these generalized momentum modes.

The rigorous language can be supplied by spectral measures or a rigged Hilbert space. The working rule is modest: continuous-spectrum kets are useful distributional coordinates, not physical finite-norm states. See Generalized Eigenvectors for the canonical treatment.

Operators and Domains

Every linear operator on a finite-dimensional Hilbert space is bounded and defined everywhere. In infinite dimension, important observables are often unbounded.

For example, the momentum differential expression is

$$P = -i\hbar\frac{d}{dx}.$$

Not every $L^2$ function has a square-integrable derivative, so this formula cannot act on all of $L^2$. An operator includes a domain:

$$P: \mathcal D(P) \subset L^2 \longrightarrow L^2.$$

The interval and boundary conditions matter as well. On $[a,b]$, integration by parts gives

$$\begin{aligned} \langle\phi\mid P\psi\rangle - \langle P\phi\mid\psi\rangle = -i\hbar \left[ \phi(x)^*\psi(x) \right]_{a}^{b}. \end{aligned}$$

The boundary term must vanish on the chosen domain for $P$ to be symmetric. Self-adjointness requires the operator and its adjoint to have the same domain, a stronger condition.

This distinction is not a correction to ordinary matrix mechanics. It is what the familiar condition $A^\dagger=A$ means when the adjoint has a nontrivial domain. Hermitian vs Self-Adjoint Operators develops the issue without requiring a full extension theory.

Why self-adjointness matters

A self-adjoint Hamiltonian has a real spectral measure and generates unitary time evolution:

$$U(t) = e^{-iHt/\hbar}.$$

A merely formal differential expression does not guarantee those properties. Boundary conditions can change the spectrum and dynamics even when the interior formula is unchanged.

For most introductory problems, the intended domain is standard and one writes “Hermitian operator” without interruption. The mature habit is to remember where the hidden assumption lives and to inspect it when boundaries, singular potentials, scattering, or rigorous claims make it relevant.

Spectrum Is Not Dimension

Finite dimension implies a finite discrete spectrum for every observable. The converse implications fail.

System or observable	Hilbert-space dimension	Typical spectrum
qubit spin component	$2$	two discrete values
spin-$j$ component	$2j+1$	finitely many discrete values
harmonic oscillator energy	infinite	countably discrete
particle in a finite box energy	infinite	countably discrete
position on the line	infinite	continuous
free-particle momentum	infinite	continuous
finite well or Coulomb energy	infinite	bound levels plus continuum

The harmonic oscillator shows why “discrete” does not mean “finite”:

$$E_n = \hbar\omega \left( n+\frac12 \right), \qquad n=0,1,2,\ldots.$$

There are infinitely many normalizable energy eigenvectors.

A free particle shows the opposite issue. Momentum values fill a continuum, but no exact momentum ket is normalizable on the line. Probabilities are assigned to intervals through spectral projectors or, when a density exists, through integrals.

One Spectral Theorem, Two Forms

For a finite-dimensional Hermitian operator,

$$A = \sum_a aP_a.$$

For a general self-adjoint operator, the spectral theorem uses a projection-valued measure $E_A$:

$$A = \int_{\mathbb R} \lambda\,dE_A(\lambda).$$

The probability that a measurement lies in a measurable set $\Delta\subseteq\mathbb R$ is

$$\Pr(A\in\Delta) = \langle\psi\mid E_A(\Delta) \mid\psi\rangle.$$

This formula covers finite, countable, continuous, and mixed spectra. For an isolated eigenvalue $a$,

$$E_A(\{a\}) = P_a.$$

For a continuous interval, $E_A(\Delta)$ remains a genuine Hilbert-space projector even though the symbolic kets $\lvert\lambda\rangle$ are generalized.

The detailed physics distinction is Discrete and Continuous Spectra. The operator-theory version is Spectral Theorem, Practical Version.

Density Operators in Infinite Dimension

In finite dimension, every positive trace-one matrix is a valid density operator. In infinite dimension, a density operator must be positive and trace class:

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

If $\rho$ has spectral decomposition

$$\rho = \sum_{n=1}^{\infty} p_n \lvert n\rangle\langle n\rvert,$$

then

$$p_n\geq0, \qquad \sum_{n=1}^{\infty} p_n = 1.$$

Not every bounded positive operator can be normalized into a state. The identity on an infinite-dimensional separable space has

$$\operatorname{Tr}I = \infty,$$

so there is no density operator

$$\rho_* = \frac{I}{\operatorname{Tr}I}$$

representing a uniform maximally mixed state over all basis vectors.

Thermal states have the form

$$\rho_\beta = \frac{ e^{-\beta H} }{ Z(\beta) }, \qquad Z(\beta) = \operatorname{Tr} e^{-\beta H},$$

only when the partition function is finite in the model and parameter regime under consideration.

Infinite-dimensional entropies and expectation values can also diverge. A normalized state need not lie in the domain of every unbounded observable, and $\operatorname{Tr}(\rho A)$ need not exist merely because $\operatorname{Tr}\rho=1$.

Convergence Is Part of the Statement

Finite algebra permits many manipulations without comment. Infinite expressions require a convergence mode.

For a basis expansion,

$$\sum_{n=1}^{N} c_n\lvert e_n\rangle \longrightarrow \lvert\psi\rangle$$

usually means norm convergence. In a function representation, norm convergence does not necessarily imply pointwise convergence at every coordinate.

For operators, several notions occur:

• operator-norm convergence controls every unit vector uniformly;
• strong convergence controls each fixed vector;
• weak convergence controls matrix elements between fixed vectors;
• resolvent convergence is often useful for unbounded operators and spectra.

This overview does not require mastering those definitions. It does require not replacing the words “take the limit” with an unspecified formal step. Interchanging sums, integrals, derivatives, traces, and limits needs conditions.

Finite Truncations of Infinite Systems

Let $\{\lvert n\rangle\}_{n=0}^{\infty}$ be an orthonormal basis and define

$$P_N = \sum_{n=0}^{N-1} \lvert n\rangle\langle n\rvert.$$

For a normalized state

$$\lvert\psi\rangle = \sum_{n=0}^{\infty} c_n\lvert n\rangle,$$

the discarded probability is

$$\epsilon_N = \langle\psi\mid (I-P_N) \mid\psi\rangle = \sum_{n=N}^{\infty} \lvert c_n\rvert^2.$$

If $\epsilon_N<1$, the normalized truncated state is

$$\lvert\psi_N\rangle = \frac{ P_N\lvert\psi\rangle }{ \sqrt{1-\epsilon_N} }.$$

Its fidelity with the exact state is

$$\left| \langle\psi\mid\psi_N\rangle \right|^2 = 1-\epsilon_N.$$

For a bounded observable $A$,

$$\left| \langle A\rangle_\psi - \langle A\rangle_{\psi_N} \right| \leq 2\lVert A\rVert \sqrt{\epsilon_N}.$$

The bound explains why small discarded norm controls bounded measurements. It does not supply a uniform bound for an unbounded observable. A tiny amplitude at very high energy can make a non-negligible contribution to an energy moment.

Example: oscillator coherent-state cutoff

A coherent state has number-basis coefficients

$$c_n = e^{-\lvert\alpha\rvert^2/2} \frac{ \alpha^n }{ \sqrt{n!} }.$$

The discarded norm beyond the first $N$ levels is the Poisson tail

$$\epsilon_N = 1 - e^{-\lvert\alpha\rvert^2} \sum_{n=0}^{N-1} \frac{ \lvert\alpha\rvert^{2n} }{ n! }.$$

This gives a direct state-fidelity criterion for selecting an oscillator cutoff. A simulation should also increase $N$ until the specific observables and evolution times of interest stabilize.

Truncation can change operator relations

Projecting an operator gives

$$A_N = P_N A P_N.$$

The projected operators need not obey every exact algebraic relation. For example, no finite matrices can satisfy

$$[X,P] = i\hbar I$$

exactly, because the trace of a finite commutator is zero while $\operatorname{Tr}(i\hbar I)\neq0$. Finite oscillator or grid calculations therefore reproduce canonical commutation relations only approximately on a controlled low-energy sector.

Worked Comparison: Four Systems

Qubit

The Hilbert space is $\mathbb C^2$. Every observable is a $2\times2$ Hermitian matrix, every spectrum has at most two distinct values, and every density operator is a finite positive trace-one matrix.

Spin without position

A fixed spin-$j$ degree of freedom has dimension $2j+1$. This remains finite even when the spin belongs physically to a particle whose motional Hilbert space has been omitted.

If position is included, the full state space becomes

$$\mathcal H = L^2(\mathbb R^3) \otimes \mathbb C^{2j+1},$$

which is infinite-dimensional.

Harmonic oscillator

The energy spectrum is discrete but infinite. Energy-basis vectors form a countable Hilbert basis, position wavefunctions provide a continuous representation, and the position and momentum operators are unbounded.

Finite potential well

The Hilbert space is infinite-dimensional. The Hamiltonian can have a finite set of discrete bound-state energies together with a continuous scattering spectrum. A basis consisting only of bound states is incomplete for general states and dynamics.

These examples separate three independent questions:

1. What is the Hilbert-space dimension?
2. What is the spectrum of the chosen observable?
3. Which representation is being used for calculation?

Choosing the Right Mental Model

Use finite-dimensional intuition confidently when:

• the physical degree of freedom is genuinely finite;
• a finite invariant subspace has been proved;
• a controlled truncation has converged for the target observables;
• the page is teaching the probability and operator grammar before analytic caveats matter.

Switch on infinite-dimensional caution when:

• wavefunctions and differential operators appear;
• an infinite basis or continuum threshold matters;
• exact position, momentum, or scattering kets are used;
• an operator is unbounded;
• boundary conditions can change the spectrum;
• traces, entropies, or energy moments may diverge;
• a finite cutoff is being removed.

A practical audit asks:

1. What is the Hilbert space?
2. Is its dimension finite or infinite?
3. Is the chosen coordinate label discrete or continuous?
4. Is the observable bounded?
5. What is its domain and boundary condition?
6. Is the spectrum point-like, continuous, or mixed?
7. What kind of convergence justifies an expansion or cutoff?
8. Which error measure controls the prediction being reported?

Common Mistakes

• Equating a discrete spectrum with finite dimension.
• Equating a continuous coordinate label with a nonseparable Hilbert space.
• Assuming every infinite-dimensional calculation must use wavefunctions.
• Treating position or momentum kets as normalizable physical states.
• Assuming a differential expression defines an operator without a domain.
• Treating “Hermitian” and self-adjoint as automatically identical for unbounded operators.
• Writing $I/\operatorname{Tr}I$ as a maximally mixed state in infinite dimension.
• Assuming norm convergence of states controls every unbounded expectation value.
• Treating a finite truncation as an exact representation of canonical commutation relations.
• Checking only eigenvalue convergence when time evolution or other observables are the real target.

Connections

• Finite-Dimensional Hilbert Spaces owns the matrix-space mathematics.
• L2 Spaces owns square-integrable wavefunction spaces.
• Wave-Mechanics Postulates gives the continuous-representation postulate package.
• Discrete and Continuous Spectra owns spectral classification and physical examples.
• Hermitian vs Self-Adjoint Operators owns operator domains and self-adjoint boundary conditions.
• Spectral Theorem, Practical Version gives the projection-valued-measure formulation.

References

• M. Reed and B. Simon, Methods of Modern Mathematical Physics, Volume I: Functional Analysis, Academic Press, 1980.
• B. C. Hall, Quantum Theory for Mathematicians, Springer, 2013.
• J. Weidmann, Linear Operators in Hilbert Spaces, Springer, 1980.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• D. J. Griffiths and D. F. Schroeter, Introduction to Quantum Mechanics, 3rd ed., Cambridge University Press, 2018.
• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, Cambridge University Press, 2010.

Exercises

1. Classify the Hilbert-space dimension and the stated observable’s spectrum for: a qubit measured in $Z$; a spin-1 degree of freedom; harmonic-oscillator energy; free-particle momentum on the line; and the energy of a finite square well with both bound and scattering states.

Solution

The qubit Hilbert space has dimension $2$, and $Z$ has two discrete eigenvalues. A spin-1 degree of freedom has dimension $3$, and any spin component has three discrete eigenvalues.

The harmonic oscillator is infinite-dimensional, while its energy spectrum is countably discrete. A free particle on the line is infinite-dimensional, and its momentum spectrum is continuous. A finite square well is infinite-dimensional and has a mixed energy spectrum: discrete bound-state energies and a continuous scattering sector.

2. Show that no sequence with constant nonzero magnitude belongs to $\ell^2(\mathbb N)$. Explain the relation to a “uniform superposition over all basis states.”

Solution

If $\lvert c_n\rvert=c>0$ for every $n$, then

$$\sum_{n=0}^{\infty} \lvert c_n\rvert^2 = \sum_{n=0}^{\infty} c^2 = \infty.$$

The sequence is not square summable and therefore is not a Hilbert-space state. There is no normalized vector with equal nonzero amplitude on every member of a countably infinite orthonormal basis. Finite uniform superpositions may approach useful generalized or distributional objects, but they do not converge to such a normalized vector.

3. A plane wave on the line is $\psi_p(x)=e^{ipx/\hbar}$. Why is it not in $L^2(\mathbb R)$, and how can it still be useful?

Solution

Its squared magnitude is one, so

$$\int_{-\infty}^{\infty} \lvert\psi_p(x)\rvert^2\,dx = \infty.$$

It is not a normalizable state. It is useful as a generalized momentum eigenfunction under Fourier integrals and distributional pairings. Normalizable wave packets are formed by superposing plane waves with a square-integrable momentum amplitude.

4. For $P=-i\hbar\,d/dx$ on $[0,L]$, use integration by parts to find the boundary form. Show that periodic boundary conditions make it vanish for two vectors satisfying the same condition.

Solution

Define the boundary form by

$$B(\phi,\psi) = \langle\phi\mid P\psi\rangle - \langle P\phi\mid\psi\rangle.$$

Integration by parts gives

$$\begin{aligned} B(\phi,\psi) &= -i\hbar \left[ \phi(x)^*\psi(x) \right]_{0}^{L}\\ &= -i\hbar\, \phi(L)^*\psi(L)\\ &\quad+ i\hbar\, \phi(0)^*\psi(0). \end{aligned}$$

If

$$\psi(L)=\psi(0), \qquad \phi(L)=\phi(0),$$

the two endpoint products are equal, so the boundary form vanishes. This establishes symmetry on that domain. Full self-adjointness also requires checking the adjoint domain.

5. Let $A$ have spectral measure $E_A$. Show that $\mu_\psi(\Delta)=\langle\psi\mid E_A(\Delta)\mid\psi\rangle$ is normalized for a normalized state.

Solution

A projection-valued measure satisfies

$$E_A(\mathbb R) = I.$$

Therefore

$$\mu_\psi(\mathbb R) = \langle\psi\mid I\mid\psi\rangle = 1.$$

It is nonnegative because every spectral projector is positive, and it is countably additive on disjoint measurable sets by the defining properties of the spectral measure. Thus it is a probability measure on the spectrum.

6. Why does an infinite-dimensional Hilbert space have no maximally mixed density operator proportional to the identity?

Solution

For an orthonormal basis $\{\lvert n\rangle\}$,

$$\operatorname{Tr}I = \sum_{n=1}^{\infty} \langle n\mid I\mid n\rangle = \sum_{n=1}^{\infty}1 = \infty.$$

No finite constant can normalize $I$ to trace one while assigning equal positive weight to every basis vector. Infinite-dimensional thermal or mixed states must have summable eigenvalues rather than a uniform nonzero spectrum.

7. For the truncation $P_N$ and discarded weight $\epsilon_N$, derive $\lvert\langle\psi\mid\psi_N\rangle\rvert^2=1-\epsilon_N$.

Solution

By definition,

$$\lvert\psi_N\rangle = \frac{ P_N\lvert\psi\rangle }{ \sqrt{ \langle\psi\mid P_N\mid\psi\rangle } }.$$

Since

$$\langle\psi\mid P_N\mid\psi\rangle = 1-\epsilon_N,$$

the overlap is

$$\langle\psi\mid\psi_N\rangle = \sqrt{1-\epsilon_N}.$$

Taking the squared magnitude gives

$$\left| \langle\psi\mid\psi_N\rangle \right|^2 = 1-\epsilon_N.$$

8. Let $\widehat N\lvert n\rangle=n\lvert n\rangle$ and define

$$\lvert\psi_k\rangle = \sqrt{1-\frac1k}\, \lvert0\rangle + \frac1{\sqrt{k}}\, \lvert k\rangle.$$

Show that $\lvert\psi_k\rangle$ converges in norm to $\lvert0\rangle$, but its number expectation does not converge to that of $\lvert0\rangle$. What does this teach about unbounded observables?

Solution

The overlap is

$$\langle0\mid\psi_k\rangle = \sqrt{1-\frac1k},$$

so the fidelity approaches one and the norm distance approaches zero. However,

$$\begin{aligned} \langle\psi_k\mid \widehat N \mid\psi_k\rangle &= \frac1k \langle k\mid \widehat N \mid k\rangle\\ &= \frac1k\,k\\ &= 1. \end{aligned}$$

The limiting state $\lvert0\rangle$ has number expectation zero. Norm convergence of states therefore does not guarantee convergence of expectations for an unbounded observable. Additional control of high-energy or high-number tails is required.