Question about the Dirac delta function

[SeM]

Hi, if I have an interval on the x-axis, defined by the parameter L, can this, interval be transformed to a Dirac delta function instead, on the x-axis?

Thanks!

 

[BvU]

Wouldn't it simply become a point ? Like in: the interval $[x,x+\epsilon]$ becomes the point $x$ for $\epsilon \downarrow 0$

 

[SeM]

Wouldn't it simply become a point ? Like in: the interval $[x,x+\epsilon]$ becomes the point $x$ for $\epsilon \downarrow 0$

Good "point"! Thanks, I will give it a thought!
Cheers

 

[FactChecker]

You can do it with integration, if that is ok. An indicator of the interval (a,b) would be f(x) = ∫_-∞^xδ(t-a)dt - ∫_-∞^xδ(t-b)dt.

PS. I'm not sure about the value at the interval endpoints, a and b.

 

[SeM]

Thank you! This is indeed a great suggestion. However, L is included in a function I have, f(x), and currently it is a parameter. It would be useful to use your form somehow, as you write:

\begin{equation}
\int_{-\infty}^x \delta(t-a)dt - \int_{-\infty}^x \delta(t-b)dt
\end{equation}

Is L actually a-b here, and t is the variable of my function, f(x)?

If e^(ix)+L is my function, would the equivalent be :

\begin{equation}
\int_{-\infty}^x \delta(e^{it}-a)dt - \int_{-\infty}^x \delta(e^{it}-b)dt
\end{equation} ?

Thanks!

 

[FactChecker]

Thank you! This is indeed a great suggestion. However, L is included in a function I have, f(x), and currently it is a parameter. It would be useful to use your form somehow, as you write:

\begin{equation}
\int_{-\infty}^x \delta(t-a)dt - \int_{-\infty}^x \delta(t-b)dt
\end{equation}

Is L actually a-b here,

For an interval (a,b) where a<b, that would make L negative. If 'L' stands for length, do you mean L = b-a?
If so, you could change the definition of the indicator function (call it I(x) to distinguish it from your f(x)). Then
\begin{equation}
I(x) = \int_{-\infty}^x \delta(t-a)dt - \int_{-\infty}^x \delta(t-(a+L))dt
\end{equation}

and t is the variable of my function, f(x)?

t is just a dummy variable for the integration, so it does not appear as an input or parameter of the indicator function.

If e^(ix)+L is my function, would the equivalent be :

\begin{equation}
\int_{-\infty}^x \delta(e^{it}-a)dt - \int_{-\infty}^x \delta(e^{it}-b)dt
\end{equation} ?

I'm confused about what you are trying to do here.

 

[SeM]

[QUOTE\I'm confused about what you are trying to do here.[/QUOTE]


Me too. I am unfamiliar with the Dirac delta function. It just says \delta and it gives therefore no idea to me on how it looks like a function.

Nevertheless, you say the dirac f is a dummy function. The thing I am not sure about is that because L is part of the existing function, f(x), say:

\begin{equation}
f(x) = e^{-ipk}/L
\end{equation}

how would the Dirac variant of this look like, where L is represented by the integral?

Thanks

 

[FactChecker]

It just says \delta and it gives therefore no idea to me on how it looks like a function.

'It just says'? Where are you seeing this reference to the delta function? You need to back up and explain more about what your original problem is and why you want to use the delta function.

Nevertheless, you say the dirac f is a dummy function.

Who said that it is a 'dummy' function? I never said that.

The thing I am not sure about is that because L is part of the existing function, f(x), say:

\begin{equation}
f(x) = e^{-ipk}/L
\end{equation}

how would the Dirac variant of this look like, where L is represented by the integral?

Why would you want to replace L with a delta function? Is L a constant?

 

[SeM]

Why would you want to replace L with a delta function? Is L a constant?

L is a parameter, the width of the space the particle is in.