Generalized Eigenvectors

A generalized eigenvector is the disciplined version of a formal eigenket such as $\lvert x\rangle$, $\lvert p\rangle$, or $\lvert E,\alpha\rangle$ when the corresponding spectral value belongs to a continuous spectrum. It behaves like an eigenvector under pairings and integrals, but it is not usually a normalizable vector in the Hilbert space.

The working rule is:

Generalized eigenvectors are spectral-coordinate distributions. Use them under pairings, integrals, and resolution-of-identity formulas, not as finite-norm physical states.

The broader spectral setting is Continuous Spectra. The rigged-Hilbert-space setting that houses generalized kets as distributions is Rigged Hilbert Spaces, First Look.

Ordinary versus Generalized Eigenvectors

For an operator $A$ on a Hilbert space $\mathcal H$, an ordinary eigenvector is a nonzero vector $\psi\in\mathcal H$ satisfying

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

It has finite norm and can be normalized when $\psi\ne0$.

A generalized eigenvector is different. It is not required to lie in $\mathcal H$. Instead, it acts on a class of well-behaved test vectors and satisfies the eigenvalue relation in a weak or distributional sense.

In physicists’ notation one writes

$$A\lvert\lambda\rangle = \lambda\lvert\lambda\rangle,$$

but the safe reading is:

$$\langle\lambda\vert A\varphi\rangle = \lambda\langle\lambda\vert\varphi\rangle$$

for suitable test vectors $\varphi$.

This is why generalized eigenvectors are useful without being normalizable states.

Test-Space Reading

Choose a dense test-vector space $\Phi\subset\mathcal H$ on which the relevant operators and integrations are controlled. A generalized eigenbra with eigenvalue $\lambda$ is a distributional functional $F_\lambda$ on $\Phi$ such that

$$F_\lambda(A\varphi) = \lambda F_\lambda(\varphi), \qquad \varphi\in\Phi.$$

Dirac notation writes this as

$$F_\lambda(\varphi) = \langle\lambda\vert\varphi\rangle.$$

The choice of $\Phi$ matters. For position and momentum on the real line, the Schwartz space $\mathcal S(\mathbb R)$ is a standard choice because point evaluation, differentiation, Fourier transformation, and plane-wave pairings behave well on it.

For the general distribution language, see Distributions.

Position Kets

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

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

For a fixed point $x_0$, define the evaluation functional

$$\langle x_0\vert\varphi\rangle = \varphi(x_0).$$

Then

$$\langle x_0\vert X\varphi\rangle = (X\varphi)(x_0) = x_0\varphi(x_0) = x_0\langle x_0\vert\varphi\rangle.$$

Thus $\langle x_0\vert$ is a generalized eigenbra of position. The ket $\lvert x_0\rangle$ is the corresponding formal spectral label.

It is not an $L^2$ wavefunction. A delta distribution centered at $x_0$ is not square-integrable, and exact point localization is not a normalizable state.

Momentum Kets

The momentum operator on the line is formally

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

For a fixed real $p$, define

$$\langle p\vert\varphi\rangle = \frac{1}{\sqrt{2\pi\hbar}} \int_{-\infty}^{\infty} e^{-ipx/\hbar}\varphi(x)\,dx.$$

For Schwartz test functions, integration by parts gives

$$\langle p\vert P\varphi\rangle = p\langle p\vert\varphi\rangle.$$

So $\langle p\vert$ is a generalized momentum eigenbra. In position representation, the corresponding generalized eigenfunction is the plane wave

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

This plane wave has constant magnitude, so

$$\int_{-\infty}^{\infty} \left\lvert \frac{1}{\sqrt{2\pi\hbar}} e^{ipx/\hbar} \right\rvert^2 dx = \infty.$$

It is therefore not a vector in $L^2(\mathbb R)$. It is a distributional building block for normalizable wave packets.

Delta Normalization

Generalized eigenvectors for continuous labels are usually delta-normalized:

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

These formulas do not say that the kets have finite norm. They say that the corresponding integral kernels act as identities under integration.

For example, the formal completeness relation

$$I = \int_{-\infty}^{\infty} \lvert x\rangle\langle x\rvert\,dx$$

means that, for suitable states,

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

and

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

Similarly,

$$I = \int_{-\infty}^{\infty} \lvert p\rangle\langle p\rvert\,dp$$

encodes the momentum representation and the Fourier transform relation between $\psi(x)$ and $\phi(p)=\langle p\vert\psi\rangle$.

The delta distribution itself is reviewed in Delta Function.

Spectral Transform View

The most reliable interpretation comes from the spectral theorem. A self-adjoint operator $A$ can often be represented as multiplication by a spectral variable:

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

The generalized ket $\lvert\lambda\rangle$ is notation for the coordinate functional

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

The identity resolution

$$I=\int\lvert\lambda\rangle\langle\lambda\rvert\,d\mu(\lambda)$$

is shorthand for the spectral representation’s inner-product formula

$$\langle\chi\vert\psi\rangle = \int \chi(\lambda)^*\psi(\lambda)\,d\mu(\lambda),$$

with the correct spectral measure and degeneracy labels included.

The precise probability object is the projection-valued measure $E_A(\Delta)$, not the point ket itself:

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

Energy Labels and Degeneracy

In scattering and free-particle problems, generalized eigenvectors often carry both a spectral value and a degeneracy label:

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

The label $\alpha$ may encode direction, angular momentum channel, spin, incoming or outgoing boundary behavior, or another degeneracy. A common normalization is

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

Changing the spectral label changes the normalization. For a one-dimensional free particle,

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

has two momentum branches for $E>0$:

$$p_\sigma(E)=\sigma\sqrt{2mE}, \qquad \sigma=\pm.$$

If momentum kets satisfy $\langle p\vert p'\rangle=\delta(p-p')$, then energy-normalized branch kets can be chosen schematically as

$$\lvert E,\sigma\rangle = \left( \frac{m}{\lvert p_\sigma(E)\rvert} \right)^{1/2} \lvert p_\sigma(E)\rangle.$$

The factor is a Jacobian. It is the same warning that appears in delta-function changes of variables: generalized eigenvectors are normalization-convention sensitive.

How to Use Them Safely

A calculation with generalized eigenvectors is safest when every formal object can be translated into one of the following:

• a spectral projection $E_A(\Delta)$ acting on a normalizable state;
• a wavefunction or spectral coefficient such as $\psi(x)$ or $\phi(p)$;
• a distributional pairing with a test vector;
• an integral kernel that reconstructs an inner product or a normalizable wave packet;
• a limiting calculation whose final probabilities are assigned to intervals or wave packets.

For example,

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

is safe when it means that a square-integrable momentum wavefunction $\phi(p)$ reconstructs a normalizable state. The individual $\lvert p\rangle$ is not physical by itself; the packet is.

What They Are Not

Generalized eigenvectors are not:

• hidden normalizable states with infinite norm;
• ordinary basis vectors indexed by a continuum;
• permission to write $\delta(0)$ as a physical number;
• a substitute for checking self-adjointness or operator domains;
• unique objects independent of normalization and measure conventions.

The notation is powerful precisely because it compresses a spectral transform into familiar linear-algebra language. The price is that the distributional meaning must be remembered.

Common Mistakes

• Treating $\lvert x\rangle$ or $\lvert p\rangle$ as a physical pure state.
• Reading $\langle x\vert x\rangle$ as a literal finite or infinite norm.
• Dropping degeneracy labels in continuous spectra.
• Changing from momentum normalization to energy normalization without a Jacobian.
• Assigning a nonzero probability to one exact value in an absolutely continuous spectrum.
• Using formal eigenvectors while ignoring the spectral measure that gives probabilities.

Cross-Links

• Continuous Spectra
• Rigged Hilbert Spaces, First Look
• Spectral Theorem, Practical Version
• Position and Momentum Representations
• Delta Function
• Distributions
• Fourier Transform
• Discrete and Continuous Spectra
• Wavefunctions as Representations
• Scattering Amplitude

References

• P. A. M. Dirac, The Principles of Quantum Mechanics, 4th ed., Oxford University Press, 1958.
• I. M. Gel’fand and N. Ya. Vilenkin, Generalized Functions, Volume 4: Applications of Harmonic Analysis, Academic Press, 1964.
• K. Maurin, Generalized Eigenfunction Expansions and Unitary Representations of Topological Groups, Polish Scientific Publishers, 1968.
• 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 the position evaluation functional is a generalized eigenbra of $X$.

Solution

For a test function $\varphi$,

$$\langle x_0\vert X\varphi\rangle = (X\varphi)(x_0) = x_0\varphi(x_0) = x_0\langle x_0\vert\varphi\rangle.$$

This is the eigenvalue equation in distributional form.

2. Verify the generalized momentum eigenvalue equation.

Solution

Using

$$\langle p\vert\varphi\rangle = \frac{1}{\sqrt{2\pi\hbar}} \int e^{-ipx/\hbar}\varphi(x)\,dx,$$

and $P=-i\hbar\,d/dx$,

$$\langle p\vert P\varphi\rangle = \frac{1}{\sqrt{2\pi\hbar}} \int e^{-ipx/\hbar} \left( -i\hbar\frac{d\varphi}{dx} \right)dx.$$

Integration by parts has no boundary term for Schwartz functions. Since

$$\frac{d}{dx}e^{-ipx/\hbar} = -\frac{ip}{\hbar}e^{-ipx/\hbar},$$

the result is

$$\langle p\vert P\varphi\rangle = p\langle p\vert\varphi\rangle.$$

3. Why does $\langle p\vert p'\rangle=\delta(p-p')$ not mean that $\lvert p\rangle$ is normalized to one?

Solution

The delta distribution is not a number. The formula means that momentum labels are normalized so that integrals over momentum reconstruct wavefunctions and inner products. Setting $p=p'$ would produce the meaningless symbol $\delta(0)$, which signals that $\lvert p\rangle$ is not a Hilbert-space vector with finite norm.

4. Suppose $\phi(p)$ is square-integrable. Explain why

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

can represent a physical state even though no single $\lvert p\rangle$ is normalizable.

Solution

The integral is a spectral expansion. The coefficient function $\phi(p)$ is the normalizable momentum-space wavefunction, and the formal kets $\lvert p\rangle$ are distributional basis labels. When $\phi\in L^2(\mathbb R)$ and is normalized, the reconstructed state has finite Hilbert-space norm even though each exact-momentum ket is only a generalized eigenvector.

5. Let $E=p^2/(2m)$ on the positive-momentum branch. What factor converts a momentum delta normalization to an energy delta normalization?

Solution

For $p>0$,

$$\frac{dE}{dp} = \frac{p}{m}, \qquad \frac{dp}{dE} = \frac{m}{p}.$$

If $\langle p\vert p'\rangle=\delta(p-p')$, then choosing

$$\lvert E\rangle = \left( \frac{dp}{dE} \right)^{1/2} \lvert p(E)\rangle = \left( \frac{m}{p(E)} \right)^{1/2} \lvert p(E)\rangle$$

gives $\langle E\vert E'\rangle=\delta(E-E')$ on that branch. With both momentum branches present, one also needs a branch label.