Continuous Spectra

A continuous spectrum is a spectral set labeled by a continuous variable rather than by a countable list of normalizable eigenvectors. Position and momentum on the real line are the standard examples. Free-particle and scattering Hamiltonians also have continuous energy spectra.

The working rule is:

Continuous-spectrum eigenstates are generalized eigenvectors. They are useful distributions, not normalizable Hilbert-space vectors.

The precise object behind the notation is the spectral measure from the Spectral Theorem, Practical Version.

Spectrum versus Eigenvectors

For an operator $A$, a spectral value $\lambda$ is not necessarily an eigenvalue with a normalizable eigenvector. In finite dimensions, these ideas coincide. In infinite dimensions, they separate.

A point-spectrum value has a nonzero vector $\psi\in\mathcal H$ such that

$$A\psi=\lambda\psi.$$

A continuous spectral value is detected by the spectral theorem and by approximate eigenstates, but it need not have any normalizable eigenvector. Informally, the operator has states that can be made sharply concentrated near $\lambda$, while no finite-norm state is exactly concentrated at $\lambda$.

This is why the phrase “possible measurement value” is broader than “eigenvalue with a Hilbert-space eigenvector.”

Generalized Eigenvectors

Physicists write generalized eigenvectors as

$$A\lvert\lambda\rangle = \lambda\lvert\lambda\rangle.$$

For continuous spectra, $\lvert\lambda\rangle$ is not usually in $\mathcal H$. It is a distributional object used to represent the spectral transform.

The practical framework that makes this statement more precise is Rigged Hilbert Spaces, First Look, where generalized eigenvectors live in a distribution space attached to the Hilbert space.

For a focused guide to the notation $\lvert x\rangle$, $\lvert p\rangle$, and $\lvert E,\alpha\rangle$, see Generalized Eigenvectors.

The formal completeness relation is

$$I = \int \lvert\lambda\rangle\langle\lambda\rvert\,d\lambda.$$

This should be read through its action on normalizable states. A state $\psi$ is represented by coefficients

$$\psi(\lambda) = \langle\lambda\vert\psi\rangle,$$

and normalizability becomes

$$\int \lvert\psi(\lambda)\rvert^2\,d\lambda = 1$$

when the spectral representation has a simple density.

Delta Normalization

Continuous generalized eigenvectors are usually delta-normalized:

$$\langle\lambda\vert\lambda'\rangle = \delta(\lambda-\lambda').$$

The delta function is not an ordinary number. It is a distribution characterized by

$$\int \delta(\lambda-\lambda')f(\lambda')\,d\lambda' = f(\lambda).$$

Thus $\langle\lambda\vert\lambda\rangle$ is not a finite norm. Delta normalization is a bookkeeping device for spectral integrals, not a statement that $\lvert\lambda\rangle$ is a physical finite-norm state.

For the distribution itself, see Delta Function.

Position Eigenstates

On $L^2(\mathbb R)$, the position operator $X$ acts by multiplication:

$$(X\psi)(x)=x\psi(x).$$

Physicists write

$$X\lvert x\rangle=x\lvert x\rangle, \qquad \langle x\vert x'\rangle=\delta(x-x').$$

The position-space wavefunction is

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

The actual probability of finding position in a set $\Delta$ is not the probability of an exact point. It is

$$\Pr(X\in\Delta) = \int_{\Delta} \lvert\psi(x)\rvert^2\,dx.$$

In spectral-measure language, $E_X(\Delta)$ multiplies the wavefunction by the indicator function of $\Delta$.

Momentum Eigenstates

The momentum operator on the real line is represented formally by

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

Its generalized eigenfunctions are plane waves:

$$\langle x\vert p\rangle = \frac{1}{\sqrt{2\pi\hbar}} e^{ipx/\hbar}.$$

They satisfy

$$P\lvert p\rangle=p\lvert p\rangle$$

in the distributional sense, and

$$\langle p\vert p'\rangle = \delta(p-p').$$

A normalizable state has a momentum-space wavefunction

$$\phi(p)=\langle p\vert\psi\rangle$$

with

$$\int_{-\infty}^{\infty} \lvert\phi(p)\rvert^2\,dp = 1.$$

The connection between $\psi(x)$ and $\phi(p)$ is the Fourier transform, with convention-dependent factors of $2\pi$ and $\hbar$. The representation page is Position and Momentum Representations, and the transform conventions are summarized in Fourier Transform.

Exact Values and Intervals

For an absolutely continuous spectrum, a single exact value usually has probability zero:

$$\Pr(P=p_0)=0.$$

This does not mean momentum measurements are impossible. It means probabilities are assigned to intervals:

$$\Pr(p_1\le P\le p_2) = \int_{p_1}^{p_2} \lvert\phi(p)\rvert^2\,dp.$$

The probability density $\lvert\phi(p)\rvert^2$ is not itself a probability. It must be integrated over a set.

Wave Packets

A physical state with a narrow momentum spread is a wave packet:

$$\lvert\psi\rangle = \int \phi(p)\lvert p\rangle\,dp,$$

where $\phi$ is square-integrable. In position representation,

$$\psi(x) = \frac{1}{\sqrt{2\pi\hbar}} \int \phi(p)e^{ipx/\hbar}\,dp.$$

The exact plane wave is the idealized limiting case of a packet whose momentum distribution becomes infinitely sharp. It is useful for calculations, but the normalizable states are wave packets.

Scattering States

Scattering theory uses continuous-spectrum energy and momentum labels. For a free particle,

$$E=\frac{p^2}{2m}$$

is continuous for $E\ge0$. Plane waves diagonalize the free Hamiltonian but are not square-integrable on the line.

In a potential-scattering problem, one often writes generalized energy eigenstates

$$H\lvert E,\alpha\rangle = E\lvert E,\alpha\rangle,$$

where $\alpha$ labels degeneracy such as direction, angular momentum channel, spin, or incoming/outgoing boundary behavior. The normalization may be chosen as

$$\langle E,\alpha\vert E',\beta\rangle = \delta(E-E')\delta_{\alpha\beta}.$$

Physical scattering states are wave packets built from these generalized states. Cross sections and amplitudes are extracted from the spectral and asymptotic structure, not from finite-norm plane waves by themselves.

Spectral Representation View

The rigorous spectral theorem says that a self-adjoint operator can be represented as multiplication by the spectral variable on an appropriate spectral representation. In the simplest continuous case, $A$ becomes

$$(A\psi)(\lambda)=\lambda\psi(\lambda)$$

on an $L^2$ space of spectral wavefunctions.

This is the precise analogue of diagonalizing a finite Hermitian matrix. A diagonal matrix multiplies each component by an eigenvalue; a continuous spectral representation multiplies each spectral wavefunction by $\lambda$.

Common Mistakes

• Treating $\lvert x\rangle$, $\lvert p\rangle$, or $\lvert E\rangle$ as normalizable states.
• Interpreting $\delta(0)$ as a physical norm.
• Assigning nonzero probability to an exact point in an absolutely continuous spectrum.
• Confusing a probability density with a probability.
• Forgetting degeneracy labels in continuous spectra.
• Replacing a normalizable wave packet by a plane wave without tracking the idealization.
• Assuming every self-adjoint operator has only point and absolutely continuous spectrum; singular continuous spectra exist mathematically, even if they are less common in basic quantum models.

Cross-Links

• Spectral Theorem, Practical Version
• Position and Momentum Representations
• Generalized Eigenvectors
• Rigged Hilbert Spaces, First Look
• Delta Function
• Fourier Transform
• Discrete and Continuous Spectra
• Born Rule for Continuous Spectra
• Free Particle
• Scattering Amplitude

References

• P. A. M. Dirac, The Principles of Quantum Mechanics, 4th ed., Oxford University Press, 1958.
• J. von Neumann, Mathematical Foundations of Quantum Mechanics, Princeton University Press, 1955.
• 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.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.

Exercises

1. Show that a plane wave is not square-integrable on the full line.

Solution

For

$$u_p(x) = \frac{1}{\sqrt{2\pi\hbar}} e^{ipx/\hbar},$$

the modulus is constant:

$$\lvert u_p(x)\rvert^2 = \frac{1}{2\pi\hbar}.$$

Therefore

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

The plane wave is a generalized eigenfunction, not a vector in $L^2(\mathbb R)$.

2. What does $\langle p\vert p'\rangle=\delta(p-p')$ mean operationally?

Solution

It means the generalized momentum labels are normalized so that spectral integrals reproduce wavefunctions and inner products. For a normalizable momentum wavefunction $\phi(p)$,

$$\int \delta(p-p')\phi(p')\,dp' = \phi(p).$$

It does not mean that $\lvert p\rangle$ has finite norm.

3. If $\phi(p)$ is normalized in momentum space, write the probability that momentum lies in $[p_1,p_2]$.

Solution

The probability is

$$\Pr(p_1\le P\le p_2) = \int_{p_1}^{p_2} \lvert\phi(p)\rvert^2\,dp.$$

The integrand is a probability density.

4. Why do scattering calculations use plane waves if plane waves are not normalizable?

Solution

Plane waves are generalized eigenstates that diagonalize the free Hamiltonian and make asymptotic momentum labels explicit. Physical scattering states are wave packets assembled from these generalized states. Plane-wave amplitudes are idealized building blocks whose measurable predictions are obtained after normalization conventions, flux factors, and wave-packet or cross-section interpretations are supplied.