By Peter4075


2012-12-30 15:42:41 8 Comments

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.

3 comments

@twistor59 2012-12-30 16:02:34

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 $g(a)$. \begin{align}\textrm{LHS}& = \int \int g(a) \delta(x-a)\;da \ \delta(x-b) \;dx\\ &=\int g(x)\delta(x-b)\;dx \\&=g(b) \end{align} But $\textrm{RHS}$ clearly $=g(b)$ too.

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

@Peter4075 2012-12-31 08:29:49

That's plausible enough for me. Out of interest, why isn't this a proof?

@twistor59 2012-12-31 09:04:00

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.

@Qmechanic 2012-12-30 16:42:38

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.

@twistor59 2012-12-30 16:46:25

I guess I should flag my answer as a plausibility argument rather than a proof!

@Peter4075 2012-12-30 19:46:54

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.

@Qmechanic 2014-11-03 23:47:03

More on square of Dirac distribution: math.stackexchange.com/q/12944/11127

@Lin 2012-12-31 07:34:57

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.

@Qmechanic 2013-09-28 12:31:16

It should be stressed that in the conventional mathematical theory of distributions, the sifting property (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.

@Ján Lalinský 2014-12-28 00:35:53

Qmechanic is right, $\delta(\lambda-\lambda)$ is a misuse of the formalism. The first integral above makes sense for $f$ that are continuous at $a$ or for $\delta(x-b)$, except the case $b=a$. $\delta(0)$ has no meaning (and should be avoided - if your calculation leads to $\delta(0)$, then something went wrong.)

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