The definition of "delta function normalization" says a basis of eigenfunctions of a particle in free space are orthonormal when $$\int_{-\infty}^{\infty}\phi_n^*(\vec{r})\phi_m(\vec{r})\mathrm{d}\vec{r}=\delta_{n,m}$$ where $\delta_{n,m}$ is the Kronecker delta function.
Therefore, consider a particle in one-dimensional free space. The particle's position eigenfunctions are in the form of $\phi_{position}=\delta_n(x-x_n)$, and its momentum eigenfunctions $\phi_{momentum} =e^{ik_nx}$.
I understand that position eigenfunctions are orthonormal, as one can use the sifting property of the delta functions in the following formula, and show that indeed position eigenfunctions are orthonormal in the sense of delta function normalization. In other words$$\int_{-\infty}^{\infty}\delta_{n}^*(x-x_n)\delta_{m}(x-x_m)\mathrm{d}x=\delta_{n}^*(x-x_n)\delta_{m}(x-x_m)=\delta(x_n-x_m)=\delta_{n,m}$$
However, I'm having a hard time applying the same definition to momentum eigenfunctions as
$$\int_{-\infty}^{\infty}(e^{ik_nx})^*\ e^{ik_mx}\mathrm{d}x=\int_{-\infty}^{\infty}e^{-i(k_n-k_m)x}\mathrm{d}x$$
where if $n=m$, then the equation becomes $\int_{-\infty}^{\infty}1\ \mathrm{d}x=\infty$; and if $n\ne m$, then the equation remains oscillatory and does not converge.
In both cases, it is not clear that if the integral satisfies the definition of delta function normalization. For the $n=m$ case, $\int_{-\infty}^{\infty}1\ \mathrm{d}x = \infty$ could be any multiple of the delta function, i.e. $\delta$, $2\delta$, $3\delta$, etc., as the definition of infinity is vague. On the other hand, when $n\ne m$, the integral does not vanish to zero, and thus does not satisfy the definition.
I'm pretty sure that the confusion I have here is related to the definitions of integrating generalized functions, and fourier transform. Because modes in fourier transform are orthogonal to each other too. But as an electrical engineer, I just took fourier transform for granted, and forgot how to prove orthogonality.
What am I missing here? How does one prove $\int_{-\infty}^{\infty}1\ \mathrm{d}x = \infty = \delta$ and $\int_{-\infty}^{\infty}(e^{ik_nx})^*\ e^{ik_mx}\mathrm{d}x=\delta_{n,m}$?
Best Answer
The delta function is not really a function, it is a distribution, In the strict sense both $\delta (x)$ and $e^{ikx}$ are not normalizable when $n=m$
One way to prove your equations is to use fourier transforms
Using Placherels theorem the fourier transform $F([f(x)]k)$ for the function $f(x)$ is given by
$$F[\delta(x-x_0)](k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\delta(x-x_0)e^{-ikx}\mathrm dx= \frac{1}{\sqrt{2\pi}}e^{-ikx_0}$$
consider fourier inverse $$F^{-1}[e^{-ikx_0}]=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{ik(x-x_0)}dk$$
clearly this is $\infty$ but there's trick or a work around for this, we can be sure that for any function $f(x)$ if
$$f(x) \xrightarrow{\text{fourier transform}} F[f(x)](k)$$ then
$$F[f(x)](k) \xrightarrow{\text{ inv fourier transform}} f(x)$$
So we can write $$\int_{-\infty}^{\infty}e^{ik(x-x_0)} \mathrm dk= 2\pi\delta (x-x_0)$$
For momentum eigen functions, replacing using the above equation and replacing $x$, $x_0$, $k$ by $k_m$, $k_n$ , $x$ we have
$$\int_{-\infty}^{\infty}e^{ix(k_m-k_n)} \mathrm dx= \int_{-\infty}^{\infty}e^{ixk_m}(e^{ixk_n})^* \mathrm dx=2\pi\delta (k_m-k_n)$$
So the "normalised" eigen functions of momentum become
$$\delta(k_m-k_n)$$
What is important is that the functions $\delta(x)$ and $e^{ikx}$ are complete and orthogonal and that is all we need if we were to use them as basis vectors for our Hilbert space
Also refer to Griffths- Introduction to Quantum mechanics, Chapter-3- Formalism