Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I have a quick question that just came up in my research and I could not find an answer anywhere so I thought I'd try here.

So one of the definitions of the Dirac Delta is the limit of the Lorentzian function with $\epsilon$ going to zero. See here http://hitoshi.berkeley.edu/221a/delta.pdf for the expression on the first page.

My question is, can I define the Dirac Delta just as well with this

$$\delta(t) ~=~ \lim_{\epsilon\rightarrow 0} \frac{1}{\pi}\frac{\epsilon^2}{\epsilon^2+t^2},$$

where I have included an extra $\epsilon$ in the numerator. My hunch is that this is no problem since the limiting behavior looks the same to me.

share|improve this question
1  
Any construction where you have some $f(x,\epsilon)$ where $\int_{-\infty}^{+\infty}f\,dx =1$ for all $\epsilon$ and where $\lim_{\epsilon\rightarrow 0}f$ is zero for all $x\neq 0$ will give you the Dirac delta function. –  Jerry Schirmer May 8 '11 at 5:50

2 Answers 2

up vote 6 down vote accepted

It looks like a delta-function. However, because $\epsilon / (\epsilon^2+t^2)$ - you should omit one $\epsilon$ in the numerator, to get the right integral equal to one, by the way - decreases too slowly as $|t|\to\infty$, as $1/t^2$, it will only work as the Dirac delta distribution for test functions that decrease at infinity or at least increase slower than as $O(t)$. If the test function is $t^2$, for example, the integral $$ \int_{-\infty}^\infty dt\,t^2\,\delta(t) $$ should yield 0 because $t^2=0$ for $t=0$. However, with your definition of the delta function, you will get a divergent answer because the infinite-range integral ultimately beats any $\epsilon$. For this reason, one usually wants approximations of delta functions that decrease faster at $|t|\to\infty$ than the Lorentzian.

Obviously, if you include one extra $\epsilon$, you get $\epsilon\cdot \delta(t) = 0$ regardless of details about the $|t|\to\infty$ behavior.

share|improve this answer
    
thanks for the response. If I understand you correctly, you are also saying that the Lorentzian is a poor choice to define the Delta function by? Anyways, my test function is unity so I suppose I am fine considering 'my' definition to be ok for the Delta function. –  BeauGeste May 8 '11 at 5:20
3  
I have no idea what you are talking about. $t^2$ is not a test function by any means. A test function is either a smooth function with compact support (if you consider all distributions) or a Schwartz class function (if you consider tempered distributions). –  Willie Wong May 8 '11 at 19:55
3  
Maybe in maths, Willie. But physicists integrate $\delta$ functions with any other factors and they almost never encounter functions whose support is strictly compact. This is a physics forum so the relevance of your comment is strictly equal to zero. –  Luboš Motl May 9 '11 at 4:52

The answer is no, the generalized function (=distribution)

$$ \lim_{\epsilon\rightarrow 0} \frac{\epsilon^2}{\epsilon^2+t^2}=0 \qquad\mathrm{a.e.} $$ is almost everywhere (a.e.) equal to the ordinary zero function $0:\mathbb{R}\to\mathbb{R}$ that sends $t\mapsto 0$.

Proof. Consider a test function $f\in C^{\infty}_c(\mathbb{R})$, i.e., an infinitely often differentiable function $f$ with compact support. Then

$$\int_{\mathbb{R}} dt \ f(t)\frac{\epsilon^2}{\epsilon^2+t^2} ~\stackrel{t=\epsilon x}{=}~\epsilon \cdot \int_{\mathbb{R}} dx \ f(\epsilon x)\cdot\frac{1}{1+x^2} $$ $$ \longrightarrow 0\cdot f(0)\cdot\int_{\mathbb{R}} dx \ \frac{1}{1+x^2} = 0 \cdot f(0)\cdot\pi =0 \qquad \mathrm{for} \qquad \epsilon \to 0,$$

because of, e.g., Lebesgue's dominated convergence theorem. $\Box$

The distribution only becomes $\pi\delta(t)$, if we remove one factor of $\epsilon$. Here $\delta(t)$ is the Dirac delta distribution (often called the Dirac delta function).

Instead of using distribution theory, we may simply interpret the formula

$$ \lim_{\epsilon\rightarrow 0} \frac{\epsilon^2}{\epsilon^2+t^2}~=~\delta_{t,0}~=~\left\{\begin{array}{rcl} 1 &\mathrm{for}& t=0 \\ 0 &\mathrm{for}& t\in\mathbb{R}\backslash\{0\} \end{array}\right. $$

as a $t$-pointwise limit. Here $\delta_{t,0}$ is the Kronecker delta function, which should not be confused with the Dirac delta distribution. The former is an ordinary function, while the latter is not. The last formula has the added benefit, that it is true both in a $t$-pointwise sense and in a distribution sense, since the Kronecker delta function is zero almost everywhere (with respect to the Lebesgue measure).

share|improve this answer
2  
The OP should also compare this again Jack's comment (as they are isomorphic in spirit). Using the change of variables formula you can show that as $\epsilon\to 0$ the "total integral" of $\epsilon^2/(\epsilon^2 + t^2)$ decreases to 0. It comes from exactly the "pulling" out of an $\epsilon$ as in the first equality in the Qmechanic's displayed formula. –  Willie Wong May 8 '11 at 19:58
    
thanks for the answer. So since the expression in question is NOT a form for the Delta function, it will be zero even if t=0. Is that right? –  BeauGeste May 8 '11 at 23:28
1  
@BeauGeste: $\int_{-t_{0}}^{t_{0}}\,dt \frac{e^{2}}{e^{2}+t^{2}}$ comes out to $2e\tan^{-1}(\frac{t_{0}}{e})$. If you take the limit of $t_{0}$ to infinity first, this works out to just be $2\,e\pi$, which goes to zero as $e\rightarrow 0$. If you take the limit of $e\rightarrow 0$ first, you find that the limit is zero for all $t_{0} \neq 0$. –  Jerry Schirmer May 9 '11 at 4:37

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.