I am given to understand that the Dirac delta function is strictly not a function in the conventional sense and it is a "functional or a distribution". The part which I can not understand why the ...
Yesterday a friend asked me what a Dirac delta function really is. I tried to explain it but eventually confused myself. It seems that a Dirac delta is defined as a function that satisfies these
On Wikipedia, the definition of the dirac delta function is given as: Suppose I have a function where at two points, the function goes to infinity. Given that the distance between the two points is...
So, I was reading about the Dirac delta function and how its differentiation works. So, pretty much all texts and online sources I saw, define it using the integral: $$ \int_ {-\infty}^ {+\infty}x\df...
First a small remark : the dirac delta is not strictly speaking a function, it's called a distribution. It's often defined as being the distribution such that $\int f (x) \delta (x) dx = f (0)$. Using that definition, your equality follows from a change the variable in the integral (from x to x−t). This definition also gives you the properties you state. On the other hand, if you start from ...
In quantum mechanics is often useful to use the following statement: $$\int_ {-\infty}^\infty dx, e^ {ikx} = 2\pi \delta (k)$$ where $\delta (k)$ is intended to represent Dirac's Delta Function. I would like to understand this statement, or at least know a justification of it, rather than blindly apply this result.
The Dirac delta is not a function in the traditional sense as no function defined on the real numbers has these properties. The Dirac delta function can be rigorously defined either as a distribution or as a measure.