Why we use Dirac delta function? (in 1 Dimension & 3 Dimesions)

[Nasbah BM]

I want to understand why and where exactly we use dirac delta function? what is its exact use?

 

[andrewkirk]

Given a quantum system and a physical quantity that can be measured for the system, such as position or momentum, the Dirac delta is used to define a simple basis that relates to that quantity, for the Hilbert Space of the system. That in turn allows calculations to be performed that makes predictions about the evolution of the system and measurements that might be made.

 

[Nasbah BM]

Can you kindly explain it in a bit more simple terms?

 

[andrewkirk]

Unfortunately, quantum mechanics is not simple, so I can't think of any easier way to express the above. All I can think of to add is that a Hilbert Space is a vector space that has a special additional property of having an inner product. In quantum mechanics, a Hilbert Space is used to represent a physical system so that calculations can be done to make predictions about that system.

Maybe Dirac deltas are used in some other field that is easier to explain. Unfortunately I have only used them in Quantum mechanics, so I can't give you an example of another use. But maybe another poster will know of an easier application of them and can post to tell you about it.

 

[Nasbah BM]

ok and thank you :)

 

[blue_leaf77]

I want to understand why and where exactly we use dirac delta function? what is its exact use?

Actually, a Dirac delta function comes into play whenever an expression of the form
$$
\int_{-\infty}^\infty e^{ikx} dx
$$
appears in the analysis. It can be found in almost every field of physics: quantum mechanics, electromagnetism, optics, and probably classical mechanics.
In some instances, it also appear in the so-called Green's function method for solving linear differential equation.

 

[Nasbah BM]

Okay .well ! at one place it is said that for a point charge its divergence del.v=0 but its surface integral is 4pi n again its volume integral is zero, under such scenario how one can explain that why we have to use delta function ? I mean to say what logic, one is applying to explain the use of delta function here?

 

[Nick Heller]

In terms of quantum mechanics, the Dirac delta function δ(x) represents the wave function with minimum uncertainty in position. Since the Dirac delta function is basically an infinitely small spike at some position in one dimensional space, the probability of finding the particle at that point (|Ψ(x)|²) is arbitrarily close to 100%.
Similarly, you can use delta functions to model large concentrations of potential. The potential of the H₂ quantum system can be modeled quite well by two delta functions a certain distance apart.
​

 

[Khashishi]

The Dirac delta "function" is a way of expressing point-like objects in continuous space. For example, let $\rho(x,y,z)$ be the charge density in some region of space. For a macroscopic object, we can usually average over the fine details, and express $\rho$ as some smooth function. But at a microscopic scale, matter is made up of point-like objects. Let's forget about quantum uncertainty for a moment and imagine there is an electron at $(0,0,0)$. The charge is $-e$, but how do you express the charge density, which is a function over space? The charge density is zero everywhere except at $(0,0,0)$, where it blows up. You can't write this kind of distribution as a normal function.

Now, if the electron has a finite size, you could write down a function with a small region where the charge density is nonzero. For example, if you assume an electron is a uniform ball with radius r, then the charge density would be $\frac{-3e}{(4\pi r^3)}$ inside the ball and $0$ outside.

You get a 3D version of a Dirac delta function if you change the radius of the ball to 0. (Basically you get $\delta(x) \delta(y) \delta(z)$, which is frequently written $\delta^3(\mathbf{r})$.) Note that the shape is unimportant when you change the radius to 0. You can start with a Gaussian distribution and change the width to 0, or a cube and change the length, width, and height to 0, and you still get the same Dirac delta functions, since shape is irrelevant for a point.

 

[Nasbah BM]

Thank you so much . :)

 

[blue_leaf77]

If you are talking about the divergence theorem, the tricky part is an additional care must be taken when applying this theorem to a vector field having singularities, like the electric field due to a point charge. In order to obtain a consistency with this theorem, the singularities must be excluded from the volume over which the theorem will be applied. For an inverse square, radial vector field, an example of appropriate volume is that bounded by two concentric spheres with say, the outer radius $R_2$ and inner radius $R_1 \to 0$ (the inner radius shrinks toward zero but never equals zero). Calculating the volume integral of the divergence in one side of the equation and the surface integral in the other side will give you a consistent equation $0=0$. In this case, the surface integral vanishes (instead of being equal to $4\pi$ like in your calculation) because the flux of an inverse square, radial vector field on a centered spherical surface is constant, independent on the radius of the surface. Therefore, since in this example there are two concentric surfaces bounding the volume over which the divergence is integrated, the flux in the outer surface is exactly canceled out by the flux in the inner surface.
In the above example, you are not using the delta function because the singularity has been excluded, which is the way I believe the right way to deal with the divergence theorem for vector fields with singularities.

 

[Nasbah BM]

Thank you so much :)