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 $\int_{-\infty}^{\infty}(e^{ik_nx})^*\ e^{ik_mx}\mathrm{d}x=\delta_{n,m}$?

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    $\begingroup$ The equations need to be understood in the sense of distributions. $\endgroup$ Apr 28 '16 at 3:17
  • $\begingroup$ Yes, you are not really understanding what a dirac delta is. It isn't an infinity at zero, it is a distribution defined by its integral properties. So one the first integral expression of the final line is incorrect. $\endgroup$ Apr 28 '16 at 5:11
  • $\begingroup$ @Benjamin Wait. Can you elaborate on why one cannot prove the first integral of the last line? That's exactly why I'm confused. I know that Dirac delta is a generalized function and it only works in the way that its integral is one. But it does not work the other way around and a mere result of infinity is not a Dirac delta. That's why I mentioned it could be anything. Then the question still remains as how to do Dirac function normalization for momentum eigenfunctions. $\endgroup$ Apr 28 '16 at 5:17
  • $\begingroup$ Ah, I see the main confusion going on here... the 1-d free particle wave-function isn't normalizable since it doesn't go to zero at infinity. Here might be a nice source to look at: physicspages.com/2012/09/11/… $\endgroup$ Apr 28 '16 at 5:27
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    $\begingroup$ Related: physics.stackexchange.com/q/47934/2451 , physics.stackexchange.com/q/89958/2451 and links therein. $\endgroup$
    – Qmechanic
    Apr 28 '16 at 7:01

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


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

  • $\begingroup$ I see your point now. You used fourier pairs of the delta function and complex exponentials to prove that the basis of momentum eigenfunctions are orthogonal. Then, you claimed that since they are orthogonal and complete, one can expand any functions in terms of the basis, a claim that I've also read in other textbooks. So far so good. I totally agree with your argument. But still the notion that the basis is not normalizable doesn't seem very physical to me. Do you have any thoughts on this maybe? $\endgroup$ Apr 28 '16 at 7:57
  • $\begingroup$ p.s.: You might want to change your wording. If the orthogonal set of vectors are not normalizable, then they are not orthonrmal and such set of vectors don't meet the requirement to form a basis for a Hilbert space. Just some mathematical rigor here. (Though I'm confused now by how to refer to the set of complex exponentials, since they are not a basis. :-/) $\endgroup$ Apr 28 '16 at 8:01
  • $\begingroup$ Sorry for that, I updated $\endgroup$
    – Courage
    Apr 28 '16 at 8:18
  • $\begingroup$ @user3336365 Hey, you don't really have to accept my answer if you are not satisfied (I know that you are not), that is all most UG level textbooks have to offer, and I have only studied that :) for a deeper answer you could probably wait or see the links provided by Qmechanic :) $\endgroup$
    – Courage
    Apr 30 '16 at 4:15

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