Sequences, Series, and Convergence

Convergence is the statement that a sequence, series, or approximation scheme approaches a limiting object in a specified sense. In quantum mechanics, the phrase “the expansion converges” is incomplete until the convergence type is stated.

The practical rule is:

Pointwise convergence, uniform convergence, and Hilbert-space norm convergence answer different questions.

This distinction matters whenever wavefunctions are approximated by basis expansions, Fourier transforms, numerical grids, perturbation series, or limiting wave packets.

Sequences

A sequence is an ordered list

$$a_1,a_2,a_3,\ldots.$$

It converges to $a$ if, for every tolerance $\varepsilon>0$, there is an integer $N$ such that

$$n\ge N \quad\Longrightarrow\quad \lvert a_n-a\rvert<\varepsilon.$$

The number $N$ may depend on $\varepsilon$. The definition says that after some point the sequence stays within any chosen tolerance of the limit.

In a normed vector space, replace absolute value by a norm:

$$\lVert v_n-v\rVert\to0.$$

In a Hilbert space, this is the convergence notion supplied by the inner product. It is the default notion behind completeness, basis expansions, and physical state approximation.

Cauchy Sequences and Completeness

A sequence is Cauchy if its terms eventually become close to one another:

$$m,n\ge N \quad\Longrightarrow\quad \lVert v_m-v_n\rVert<\varepsilon.$$

Completeness means every Cauchy sequence converges to an element of the space. This is why Hilbert Spaces are the natural state spaces for quantum mechanics: limits of physically meaningful approximations should remain in the allowed state space.

For example, an infinite basis expansion is usually defined as the limit of its partial sums. Without completeness, the limiting state might fall outside the space.

Series

A series is a sum of infinitely many terms:

$$\sum_{n=1}^{\infty}a_n.$$

It converges if the partial sums

$$S_N = \sum_{n=1}^{N}a_n$$

converge as a sequence. Thus every question about a series is really a question about a sequence of approximations.

A series converges absolutely if

$$\sum_{n=1}^{\infty} \lvert a_n\rvert < \infty.$$

Absolute convergence is stronger than convergence and usually permits safer rearrangements. Conditional convergence can be delicate: changing the order of terms may change the result or destroy convergence.

Function Sequences

For a sequence of functions $f_n(x)$, there are several possible convergence notions.

Pointwise convergence means that for each fixed $x$,

$$f_n(x)\to f(x).$$

The required $N$ may depend on both $\varepsilon$ and $x$.

Uniform convergence means that a single $N$ works for all $x$ in the domain:

$$n\ge N \quad\Longrightarrow\quad \sup_x \lvert f_n(x)-f(x)\rvert < \varepsilon.$$

Uniform convergence is stronger. It is often the condition that allows limits to preserve continuity and supports safer interchange of limiting operations with integrals or derivatives, subject to the relevant theorem. The underlying continuity and integrability language is summarized in Real Analysis Essentials.

Norm Convergence

Norm convergence measures the size of the difference as an element of a normed space. For wavefunctions in $L^2(\mathbb R)$,

$$\lVert\psi_n-\psi\rVert_2^2 = \int_{-\infty}^{\infty} \lvert\psi_n(x)-\psi(x)\rvert^2\,dx \to0.$$

This is the main convergence notion for state vectors. If normalized states converge in Hilbert-space norm, then inner products against fixed normalizable states converge:

$$\langle\phi\vert\psi_n\rangle \to \langle\phi\vert\psi\rangle.$$

Norm convergence does not require pointwise convergence at every point. Conversely, pointwise convergence does not imply norm convergence.

For quantum mechanics this is not a defect. Physical probabilities are usually computed from inner products, projectors, and integrals, not from the value of a wavefunction at one exact point.

Wavefunction Expansions

Let $\{e_n\}$ be an orthonormal basis of a Hilbert space and let

$$c_n = \langle e_n\vert\psi\rangle.$$

The partial sums

$$S_N = \sum_{n=1}^{N}c_n e_n$$

converge to $\psi$ in Hilbert-space norm:

$$\lVert\psi-S_N\rVert^2 = \sum_{n>N} \lvert c_n\rvert^2 \to0.$$

This is the precise meaning of a complete orthonormal basis expansion in the Hilbert-space sense. It does not automatically say that the series converges pointwise or uniformly as a function of position.

For the basis theorem and Parseval relation, see Completeness and Orthonormal Bases.

Why the Type Matters

Different convergence notions preserve different operations.

Pointwise convergence is often too weak to justify:

• moving a limit inside an integral;
• differentiating a series term by term;
• applying an unbounded operator to a limiting state;
• replacing a boundary condition on each term by a boundary condition on the limit.

Norm convergence in $L^2$ is strong enough for Hilbert-space approximation, but it still may not justify pointwise statements or derivatives. A sequence of smooth functions can converge in $L^2$ to a function that is not differentiable in the ordinary sense.

Uniform convergence controls values everywhere, but it is not the natural topology for most quantum state spaces. Many perfectly good $L^2$ wavefunction expansions are not uniformly convergent.

Applying Operators to Limits

Bounded operators behave well with norm limits. If $B$ is bounded and $\psi_n\to\psi$ in norm, then

$$B\psi_n\to B\psi$$

in norm.

Unbounded operators require more care. If $A$ is an unbounded operator, $\psi_n\to\psi$ does not by itself imply that $A\psi_n\to A\psi$, or even that $\psi$ lies in the domain of $A$.

For a Hamiltonian with eigenvectors $e_n$ and eigenvalues $E_n$, a vector

$$\psi = \sum_n c_n e_n$$

is in the operator domain of $H$ only when the weighted series is square-summable:

$$\sum_n \lvert E_n\rvert^2 \lvert c_n\rvert^2 < \infty.$$

The ordinary normalization condition

$$\sum_n \lvert c_n\rvert^2 = 1$$

is weaker. This is the convergence reason domain questions appear in Domains of Operators.

Worked Example: Pointwise but Not Norm Convergence

Define functions on $[0,1]$ by

$$f_n(x) = \begin{cases} \sqrt n, & 0<x<1/n,\\ 0, & \text{otherwise}. \end{cases}$$

For every fixed $x>0$, eventually $x>1/n$, so $f_n(x)\to0$. Also $f_n(0)=0$ as defined. Thus $f_n$ converges pointwise to $0$.

But the $L^2$ norm stays fixed:

$$\lVert f_n\rVert_2^2 = \int_0^{1/n} n\,dx = 1.$$

Therefore $f_n$ does not converge to $0$ in $L^2$ norm. Pointwise convergence alone is too weak to control probability weight.

Worked Example: Nonuniform Pointwise Limit

On $[0,1]$, let

$$g_n(x)=x^n.$$

For $0\leq x\lt1$, $g_n(x)\to0$, while $g_n(1)=1$ for all $n$. The pointwise limit is

$$g(x) = \begin{cases} 0, & 0\le x<1,\\ 1, & x=1. \end{cases}$$

Each $g_n$ is continuous, but the limit is discontinuous. The convergence is not uniform. This example is a simple warning that pointwise limits can fail to preserve qualitative properties.

Quantum Examples

Bound-state expansions in an infinite square well converge in $L^2$ norm. That is enough to reconstruct the state vector and compute probabilities for ordinary projective measurements. It is not by itself enough to differentiate the series twice and call the result $H\psi$.

Fourier transforms preserve $L^2$ norm under the usual Plancherel theorem. This is why position-space and momentum-space wavefunctions can represent the same Hilbert-space state even when their pointwise behavior looks different.

Wave packets often approximate idealized plane waves or position eigenstates in a limiting sense. The limiting generalized object may not be in the Hilbert space at all. Continuous-spectrum limits are discussed in Continuous Spectra and Rigged Hilbert Spaces, First Look.

Common Mistakes

• Saying “the series converges” without saying pointwise, uniformly, or in norm.
• Treating pointwise convergence as enough to preserve normalization.
• Assuming an $L^2$-convergent wavefunction expansion can be differentiated term by term.
• Forgetting that completeness concerns Cauchy sequences in a specified norm.
• Confusing convergence of coefficients with convergence of the represented vector.
• Treating perturbation series as automatically convergent rather than sometimes asymptotic.
• Assuming a limiting generalized state remains a normalizable Hilbert-space vector.

Cross-Links

• Norms and Metrics
• Real Analysis Essentials
• Asymptotic Analysis
• Hilbert Spaces
• L2 Spaces
• Completeness and Orthonormal Bases
• Domains of Operators
• Continuous Spectra
• Rigged Hilbert Spaces, First Look
• Ordinary Differential Equations
• Sturm–Liouville Theory
• Fourier Transform
• Wave Packets
• Infinite Square Well

References

• W. Rudin, Principles of Mathematical Analysis, 3rd ed., McGraw-Hill, 1976.
• 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.
• G. B. Arfken, H. J. Weber, and F. E. Harris, Mathematical Methods for Physicists, 7th ed., Academic Press, 2013.
• M. L. Boas, Mathematical Methods in the Physical Sciences, 3rd ed., Wiley, 2005.

Exercises

1. Show that $g_n(x)=x^n$ on $[0,1]$ converges pointwise but not uniformly.

Solution

For $0\leq x\lt1$, $x^n\to0$, while $g_n(1)=1$. Thus the pointwise limit is $0$ on $[0,1)$ and $1$ at $x=1$.

If the convergence were uniform, the uniform limit of continuous functions would be continuous. The pointwise limit is discontinuous at $x=1$, so the convergence is not uniform.

2. For the spike functions

$$f_n(x) = \begin{cases} \sqrt n, & 0<x<1/n,\\ 0, & \text{otherwise}, \end{cases}$$

show that pointwise convergence to zero does not imply $L^2$ convergence to zero.

Solution

For every fixed $x>0$, eventually $x>1/n$, so $f_n(x)=0$. Also $f_n(0)=0$. Hence $f_n\to0$ pointwise.

However,

$$\lVert f_n\rVert_2^2 = \int_0^{1/n}n\,dx = 1.$$

Thus the functions do not converge to zero in $L^2$ norm.

3. Let $\{e_n\}$ be an orthonormal basis and let $\psi=\sum_n c_n e_n$ with $\sum_n\lvert c_n\rvert^2\lt\infty$. What is the norm error after keeping only the first $N$ terms?

Solution

The partial sum is $S_N=\sum_{n=1}^N c_n e_n$. Orthogonality gives

$$\lVert\psi-S_N\rVert^2 = \sum_{n>N} \lvert c_n\rvert^2.$$

This tail tends to zero because the coefficient sequence is square-summable.

4. Suppose $H e_n=E_n e_n$ and $\psi=\sum_n c_n e_n$ is normalized. Why does normalization alone not imply $\psi\in D(H)$?

Solution

Normalization says

$$\sum_n \lvert c_n\rvert^2 = 1.$$

For $\psi$ to lie in the domain of $H$, the stronger weighted condition is needed:

$$\sum_n \lvert E_n\rvert^2 \lvert c_n\rvert^2 < \infty.$$

If the coefficients decay too slowly compared with the growth of $\lvert E_n\rvert$, the state is normalizable but $H\psi$ is not a Hilbert-space vector.