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In Griffiths' Intro to QM [1] he gives the eigenfunctions of the Hermitian operator $\hat{x}=x$ as being

$$g_{\lambda}\left(x\right)~=~B_{\lambda}\delta\left(x-\lambda\right)$$

(cf. last formula on p. 101). He then says that these eigenfunctions are not square integrable because

$$\int_{-\infty}^{\infty}g_{\lambda}\left(x\right)^{*}g_{\lambda}\left(x\right)dx ~=~\left|B_{\lambda}\right|^{2}\int_{-\infty}^{\infty}\delta\left(x-\lambda\right)\delta\left(x-\lambda\right)dx ~=~\left|B_{\lambda}\right|^{2}\delta\left(\lambda-\lambda\right) ~\rightarrow~\infty$$

(cf. second formula on p. 102). My question is, how does he arrive at the final term, more specifically, where does the $\delta\left(\lambda-\lambda\right)$ bit come from?

My total knowledge of the Dirac delta function was gleaned earlier on in Griffiths and extends to just about understanding

$$\tag{2.95}\int_{-\infty}^{\infty}f\left(x\right)\delta\left(x-a\right)dx~=~f\left(a\right)$$

(cf. second formula on p. 53).

References:

  1. D.J. Griffiths, Introduction to Quantum Mechanics, (1995) p. 101-102.
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3 Answers 3

up vote 20 down vote accepted

You need nothing more than your understanding of $$ \int_{-\infty}^\infty f(x)\delta(x-a)dx=f(a) $$ Just treat one of the delta functions as $f(x)\equiv\delta(x-\lambda)$ in your problem. So it would be something like this: $$ \int\delta(x-\lambda)\delta(x-\lambda)dx=\int f(x)\delta(x-\lambda)dx=f(\lambda)=\delta(\lambda-\lambda) $$ So there you go.

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To me, that seems even clearer (in so far as anything involving this bizarre function can be clear). Apologies twistor59, I've now accepted this answer. –  Peter4075 Dec 31 '12 at 12:14
2  
It should be stressed that in the conventional mathematical theory of distributions, the eq. (2.95) is a priori only defined if $f$ is a test-function. In particular, it is not mathematically rigorous to use eq. (2.95) (with $f$ substituted with a distribution) to justify the meaning of the integral of the square of the Dirac delta distribution. –  Qmechanic Sep 28 '13 at 12:31

Well, the Dirac delta function $\delta(x)$ is a distribution, also known as a generalized function.

One can e.g. represent $\delta(x)$ as a limit of a rectangular peak with unit area, width $\epsilon$, and height $1/\epsilon$; i.e.

$$\tag{1} \delta(x) ~=~ \lim_{\epsilon\to 0^+}\delta_{\epsilon}(x), $$ $$\tag{2} \delta_{\epsilon}(x)~:=~\frac{1}{\epsilon} \theta(\frac{\epsilon}{2}-|x|) ~=~\left\{ \begin{array}{ccc} \frac{1}{\epsilon}&\text{for}& |x|<\frac{\epsilon}{2}, \\ \frac{1}{2\epsilon}&\text{for}& |x|=\frac{\epsilon}{2}, \\ 0&\text{for} & |x|>\frac{\epsilon}{2}, \end{array} \right. $$

where $\theta$ denotes the Heaviside step function with $\theta(0)=\frac{1}{2}$.

The product $\delta(x)^2$ of the two Dirac delta distributions does strictly speaking not$^1$ make mathematical sense, but for physical purposes, let us try to evaluate the integral of the square of the regularized delta function

$$\tag{3} \int_{\mathbb{R}}\! dx ~\delta_{\epsilon}(x)^2 ~=~\epsilon\cdot\frac{1}{\epsilon}\cdot\frac{1}{\epsilon} ~=~\frac{1}{\epsilon} ~\to~ \infty \quad \text{for} \quad \epsilon~\to~ 0^+. $$

The limit is infinite as Griffiths claims.

It should be stressed that in the conventional mathematical theory of distributions, the eq. (2.95) is a priori only defined if $f$ is a smooth test-function. In particular, it is not mathematically rigorous to use eq. (2.95) (with $f$ substituted with a distribution) to justify the meaning of the integral of the square of the Dirac delta distribution. Needless to say that if one blindly inserts distributions in formulas for smooth functions, it is easy to arrive at all kinds of contradictions! For instance,

$$ \frac{1}{3}~=~ \left[\frac{\theta(x)^3}{3}\right]^{x=\infty}_{x=-\infty}~=~\int_{\mathbb{R}} \!dx \frac{d}{dx} \frac{\theta(x)^3}{3} $$ $$\tag{4} ~=~\int_{\mathbb{R}} \!dx ~ \theta(x)^2\delta(x) ~\stackrel{(2.95)}=~ \theta(0)^2~=~\frac{1}{4}.\qquad \text{(Wrong!)} $$

--

$^1$ We ignore Colombeau theory. See also this mathoverflow post.

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I guess I should flag my answer as a plausibility argument rather than a proof! –  twistor59 Dec 30 '12 at 16:46
    
This is a little advanced for me as I'm not familiar with Heaviside step functions. Twistor59's answer is more my level though I'm still trying to think it through. –  Peter4075 Dec 30 '12 at 19:46
    
More on square of Dirac distribution: math.stackexchange.com/q/12944/11127 –  Qmechanic Nov 3 at 23:47

Suppose I want to show $$\int \delta(x-a)\delta(x-b) dx = \delta(a-b) $$ To do that , I need to show $$\int g(a)\int \delta(x-a)\delta(x-b) dx da = \int g(a)\delta(a-b) da$$ for any function $g(a)$. $$LHS = \int \int g(a) \delta(x-a)da \ \delta(x-b) dx $$ $$=\int g(x)\delta(x-b)dx $$ $$=g(b) $$ But RHS clearly = $g(b)$ too.

The result follows putting $a=b=\lambda$

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That's plausible enough for me. Out of interest, why isn't this a proof? –  Peter4075 Dec 31 '12 at 8:29
2  
Well, as Qmechanic pointed out, these delta functions are distributions, so you have to be really careful about verifying that the usual manipulations are valid - for example you should really specify the space of test functions, and check convergence etc. Having said that, Dirac, when he introduced them in his "Principles of Quantum Mechanics" was also a little cavalier about manipulating them. –  twistor59 Dec 31 '12 at 9:04

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