Question: Why is the inner product of momentum eigenstates the Dirac delta distribution?
the momentum eigenstate $\lvert k\rangle$, proportional to $e^{ip\cdot x}$, has definite momentum $\hbar k$ and completely unknown position.
with discrete vectors, the dot product of two eigenstates of a hermitian matrix is zero if they’re distinct and one if they’re the same. i.e. $\langle \psi_{k^\prime}\rvert\psi_k\rangle = \delta_{k^\prime k}$ which is a distribution over the discrete states given by $k$.
however, the momentum eigenstate describes a free particle so $k$ can have any value whatsoever. we can still think of $\lvert\psi_k\rangle \sim e^{ik\cdot x}$ as a vector, but one with one entry for each value in the continuum, i.e. a function. this generalizes the dot product sum $ \langle \psi_{k^\prime}\rvert\psi_k\rangle = \sum_k \psi_{k^\prime}^j\psi_k^j$ to an integral
\[\begin{align} \langle\psi_{k^\prime}\rvert\psi_k\rangle &= \int\text{d}x\, \langle\psi_{k^\prime}\rvert x\rangle\langle x\rvert \psi_k\rangle \\ &= \int\text{d}x\, e^{i\left(k-k^\prime\right)x}. \end{align}\]on its face, it isn’t obvious that this reproduces the orthogonality of distinct states. yet, momentum is a hermitian operator, and the $\lvert\psi_k\rangle$ are its eigenvectors, so oughtn’t they be orthogonal? let’s take a look.
What is the generalized dot product?
first, let’s set set the bounds of the space as $-R$ and $+R$. we will take these to be infinite eventually, but for the purpose of seeing how the inner product gets to its final form, it is very helpful to start with it finite.
\[\begin{align} \langle\psi_{k^\prime}\rvert\psi_k\rangle &= \int\limits_{-R}^{+R}\text{d}x\, e^{i\left(k-k^\prime\right)x} \\ &= \frac{e^{i\left(k-k^\prime\right)R} - e^{-i\left(k-k^\prime\right)R}}{i\left(k-k^\prime\right)} \\ &= \frac{2\sin \left(k-k^\prime\right) R}{\left(k-k^\prime\right)} \\ &= \frac{2\sin \Delta k R}{\Delta k}. \end{align}\]for ease of writing let’s call the gap in wavenumber $\Delta k = k - k^\prime$ as $g$ so that
\[\langle\psi_{k^\prime}\rvert\psi_k\rangle = \frac{2\sin Rg}{g}.\]first of all, this quantity is oscillating. as $R$ gets bigger, it oscillates faster and faster. if we averaged any smooth function against this oscillation, it would sample the function at much greater frequency than any change in the function so that every point on the function averages to zero.
if that were the whole story, then this would be a very boring distribution. but something interesting happens at the origin. as $\sin Rg$ approaches the origin, we can apporoximate it by $Rg$, which makes the quantity equal to $Rg/g = R$ at the origin.
so, as $R$ gets very big, the whole function oscillates fast enough to average any function to zero everywhere except exactly at the origin where the distribution is $R.$ in other words, the distribution is zero everywhere except for the origin where it is infinite.
you can play with the distribution below, seeing what happens as $R$ transitions from small to large.
this means that if we average any function $f(k)$ against this distribution, we will get back $f(0)$.
now, this distribution is not normalized, and may not be unless it has a non-zero, finite area.
What’s the area?
so, let’s go after
\[\text{area} = 2\int\limits_{-\infty}^{+\infty} \text{d}g\, \frac{\sin R g }{g}.\]first of all, we can change variables to $u = gR$ which shows this integral is, happily, independent of $R$
\[\begin{align} \text{area} &= 2\int\limits_{-\infty}^{+\infty} \text{d}u\, \frac{\sin u }{u} \\ &= 4\int\limits_0^{+\infty} \text{d}u\, \frac{\sin u }{u}. \end{align}\]there is not an easy way to attack this with substitutions, or integration by parts, but the Feynman trick of differentiating under the integral sign does a wonderful job. we need to introduce a factor that we can differentiate to cancel the $g$ in the denominator, leaving us with something nice that we can integrate by parts. we can try $e^{-\beta u}$ so that our integral becomes
\[\int\limits_0^{\infty} \text{d}u\, e^{-\beta u}\frac{\sin u }{u}.\]let’s call the integral, for a given value of $\beta$, $a(\beta)$.
the derivative of $a(\beta)$ with respect to $\beta$ is
\[\frac{\text{d}a(\beta)}{\text{d}\beta} = -\int\limits_0^{\infty} \text{d}u\, e^{-\beta u}\sin u.\]to integrate by parts, we can use $\frac{\text{d}}{\text{d}u}\left[e^{-\beta u} \sin u\right] = -\beta e^{-\beta u}\sin u + e^{-\beta u}\cos u$. rearranging and inserting to our integral, we get
\[\begin{align} \frac{\text{d}a(\beta)}{\text{d}\beta} &= \frac{1}{\beta}\int\limits_0^{\infty} \text{d}u\, \frac{\text{d}}{\text{d}u}\left[e^{-\beta u} \sin u\right] - \frac{1}{\beta}\int\limits_0^{\infty} \text{d}u\, e^{-\beta u} \cos u \\ &= - \frac{1}{\beta}\int\limits_0^{\infty} \text{d}u\, e^{-\beta u} \cos u \end{align}.\]the first piece is zero since the bracketed stuff is zero at both endpoints. it may look like, after another crank of the integration by parts wheel, we will get back the original integral and find ourselves in a futile cycle. but alas, crank two has a nonzero contribution from the boundary term and will give us a nice differential equation for $a(\beta)$.
we look once again for guidance from the product rule $\frac{\text{d}}{\text{d}u}\left[e^{-\beta u} \cos u\right] = -\beta e^{-\beta u}\cos u - e^{-\beta u}\sin u$. making the substitution, we get
\[\begin{align} \frac{\text{d}a(\beta)}{\text{d}\beta} &= \frac{1}{\beta^2}\int\limits_0^{\infty} \text{d}u\, \frac{\text{d}}{\text{d}u}\left[e^{-\beta u} \cos u\right] + \frac{1}{\beta^2}\int\limits_0^{\infty} \text{d}u\, e^{-\beta u} \sin u \\ &= -\frac{1}{\beta^2} - \frac{1}{\beta^2}\frac{\text{d}a(\beta)}{\text{d}\beta} \end{align}.\]solving for $\text{d}a(\beta)/\text{d}\beta$ get us
\[\frac{\text{d}a(\beta)}{\text{d}\beta} = -\frac{1}{1+\beta^2}.\]now, $a(\infty) - a(0) = -\int_0^\infty \frac{\text{d}\beta}{1+\beta^2}$ and we know that $a(\infty) = 0$ due the exponential term, so all we have left is to evaluate the integral on the right. the integral of $1/(1+\beta^2)$ is $\arctan \beta$ so we get $a(0) = \arctan \infty - \arctan 0 = \tfrac12\pi.$
The dust settles
putting everything back together, this means that $\text{area} = 2\pi.$ in turn that means that the normalization constant on the momentum eigenstates must be $1/\sqrt{2\pi},$ therefore
\[\lvert \psi_k\rangle = \frac{1}{\sqrt{2\pi}}e^{ik\cdot x}\]and
\[\langle \psi_{k^\prime}\rvert\psi_k\rangle = \delta(k-k^\prime)\]the Dirac delta function distribution.
this shows that the momentum eigenstates are properly orthonormal, and give rise to an amazing distribution that is zero everywhere, except the origin, where it is $\infty.$ nonetheless, this distribution has a nonzero, and non-infinite area of $1$. a proper distribution.
this has many fascinating properties, but the point of this writeup is just to show that it rises naturally from momentum eigenstates so we will leave it at that.