Delta Normalization

Delta normalization states an identity between distribution kernels used inside spectral integrals. It does not assign an infinite norm to a Hilbert-space vector. Its normalization depends on the integration measure: changing momentum to another label changes both the coefficients and the generalized kets by square-root Jacobians.

Required background. Momentum Kets fixes the Fourier kernel $k_p(x)=(2\pi\hbar)^{-1/2}e^{ipx/\hbar}$. Distributions as Generalized Vectors explains why arbitrary generalized vectors have no mutual Hilbert inner product.

Orthogonality is a tested kernel identity

The notation $\langle p|p'\rangle=\delta(p-p')$ abbreviates

$$\frac1{2\pi\hbar}\int_{\mathbb R} e^{i(p'-p)x/\hbar}\,dx=\delta(p-p').$$

The integral is not an ordinary convergent integral. To give it meaning, insert $e^{-\epsilon x^2}$ with $\epsilon>0$. Gaussian integration yields, for $u=p-p'$,

$$K_\epsilon(u) =\frac{1}{2\hbar\sqrt{\pi\epsilon}} \exp\left(-\frac{u^2}{4\epsilon\hbar^2}\right).$$

This is an approximate identity: $K_\epsilon\ge0$, its integral over $u$ equals one, and its mass outside every fixed neighborhood of zero tends to zero. Therefore, for a Schwartz test function $f$,

$$\int K_\epsilon(u)f(u)\,du\longrightarrow f(0).$$

That limiting statement is the delta identity. The height $K_\epsilon(0)$ diverges, but it is regulator-dependent and is not a number called $\delta(0)$. The normalization is fixed by the integral against test functions, not by a diagonal value.

Wavepacket normalization uses the kernel only after integration

For a sufficiently regular amplitude $a(p)$, the generalized expansion defines a Hilbert state. The kernel identity then gives

$$\int\!\int\overline{a(p)}\, \delta(p-p')a(p')\,dp\,dp' =\int|a(p)|^2dp.$$

For arbitrary square-integrable amplitudes, Plancherel supplies the rigorous extension. The double-integral notation should not be used to justify rearrangements for amplitudes outside the required function spaces.

In particular, setting $a(p)=\delta(p-p_0)$ does not produce a unit vector. It attempts to square a distribution in the normalization formula. Generalized basis coefficients and physical square-integrable amplitudes are different objects.

Relabeling requires a square-root Jacobian

Let $q=q(p)$ be a smooth one-to-one change of label on an interval, with nonzero derivative. A basis normalized relative to $dq$ can be chosen as

$$|q\rangle =\sqrt{\left|\frac{dp}{dq}\right|}\,|p(q)\rangle.$$

Indeed, the delta change-of-variable rule gives

$$\delta(p(q)-p(q')) =\frac{\delta(q-q')}{|dp/dq|},$$

with the Jacobian evaluated at the coincident label. The two square-root factors therefore restore $\langle q|q'\rangle=\delta(q-q')$. The amplitude transforms by the same real factor:

$$\begin{gathered} b(q)=\sqrt{|dp/dq|}\,a(p(q)),\\ |b(q)|^2dq=|a(p)|^2dp. \end{gathered}$$

A label-dependent phase of the kets changes the amplitude by the opposite phase and leaves this probability relation unchanged. If the relabeling is not one-to-one, separate branches are required. The two energy branches of a free particle are worked out in Generalized Eigenfunction Expansions; suppressing a branch cannot be repaired by a Jacobian alone.

Finite-box normalization describes a different spectral problem

For periodic boundary conditions in a box of length $L$, normalized modes are $L^{-1/2}e^{ip_nx/\hbar}$, with $p_n=2\pi\hbar n/L$. Their Kronecker normalization is an ordinary Hilbert inner product. Writing $\Delta p=2\pi\hbar/L$, the mode functions satisfy

$$\frac{e^{ip_nx/\hbar}}{\sqrt L} =\sqrt{\Delta p}\,k_{p_n}(x)$$

inside the box. This factor explains how continuum integrals become Riemann sums in an appropriate large-box wavepacket limit. Discrete coefficients then scale like $\sqrt{\Delta p}$ times continuum amplitudes.

It does not equate a periodic-box state with a line state for every purpose. Boundary conditions and the chosen approximation must still be controlled. The mnemonic $\delta(p_n-p_m)\leftrightarrow\delta_{nm}/\Delta p$ belongs to that conversion between sums and integrals; it is not pointwise evaluation of a distribution.

Exercise: logarithmic momentum bins

Restrict to $p>0$ and use the dimensionless label $q=\log(p/p_*)$, where $p_*>0$. Find the normalized ket and amplitude for the $dq$ measure. Why would using $a(p_*e^q)$ without another factor give the wrong probabilities?

Solution

Since $p=p_*e^q$ and $dp/dq=p$, the ket is $|q\rangle=\sqrt p\,|p\rangle$ and the amplitude is $b(q)=\sqrt p\,a(p)$. Thus $|b(q)|^2dq=|a(p)|^2dp$. Omitting the factor would integrate against $dp/p$ rather than $dp$, changing the relative weights of equal logarithmic bins. This is a change in the measured distribution, not merely in notation.

References

• Hall, Brian C., Quantum Theory for Mathematicians, Graduate Texts in Mathematics 267, Springer, 2013, doi:10.1007/978-1-4614-7116-5 — Fourier normalization, Gaussian transforms, and spectral representations.
• Teschl, Gerald, Mathematical Methods in Quantum Mechanics: With Applications to Schrödinger Operators, 2nd ed., Graduate Studies in Mathematics 157, American Mathematical Society, 2014, author’s edition — Fourier kernels and rigorous spectral transforms.