What is the use of Dirac delta function in quantum mechanics?

[manimaran1605]

If you ask me define Dirac delta function, i can easily define it and prove its properties using laplacian or complex analysis method. But still i don't understand what is the use of DIRAC DELTA FUNCTION in quantum mechanics. As i have done some reading Quantum mechanics from Introduction to Quantum mechanics By griffiths (completed 1st chapter only) but still i haven't see a word like delta function. Please help.

 

[hilbert2]

You probably know the probability density interpretation of the wavefunction. The relative probability of finding the particle in the small interval $[x,x+dx]$ is $|\psi(x)|^2 dx$. Now, if the wavefunction of the particle is the Dirac delta, $\psi(x)=\delta(x-x_{0})$, all of the probability density is concentrated at the point $x=x_{0}$. Therefore, this wavefunction describes a state where the particle is at a certain point with 100% probability (a position eigenstate). That is why we need the Dirac delta function in QM.

 

[nomadreid]

In very broad terms: you use it to fudge to get discrete values out of a continuous distribution. It's a bit better explained here: http://en.wikipedia.org/wiki/Dirac_delta_function#Quantum_mechanics, and gives other examples here: http://en.wikipedia.org/wiki/Dirac_delta_function#Quantum_mechanics

 

[WannabeNewton]

IBut still i don't understand what is the use of DIRAC DELTA FUNCTION in quantum mechanics.

Just think about the uses of $\delta_{ij}$ but for continuous instead of discrete spectra in order to see where and why we use $\delta(x - x')$.

 

[vanhees71]

To avoid trouble later on, it is however very important to realize that the ideas provided in #2 are wrong and one should call it the Dirac \delta distribution rather than calling it a function. A distribution is a kind of "generalized function" and defined as a linear functional on an appropriate subspace of the Hilbert space of square-integrable functions. The functions in the subspace are called test functions. Youl can choose them as all smooth functions of compact support. Then it's defined by the relation
\int_{\mathbb{R}} \mathrm{d} x \delta(x-x_0) f(x)=f(x_0)
for all functions f in the domain of the test functions.

The \delta distribution is thus not a function in the usual sense and particularly it cannot represent a physical state in the sense of a wave function. It's square doesn't make any sense at all!

In quantum theory the \delta distribution is used mostly in the sense WannabeNewton has mentioned in the previous posting. For self-adjoint operators with a con'inuous spectrum they are generalized "eigenfunctions" (where generalized function has to be read as distribution in the above given sense). Then, the generalized eigenfunctions of the position operator indeed satisfy
\langle x|x_0 \rangle=\delta(x-x_0).

 

[hilbert2]

The \delta distribution is thus not a function in the usual sense and particularly it cannot represent a physical state in the sense of a wave function. It's square doesn't make any sense at all!

Well, the wavefunction corresponding to state vector $|\psi>$ is $\psi(x)=<x|\psi>$, and if $|\psi>$ is a position eigenstate $|x'>$, we have $\psi(x)=<x|x'>=\delta(x-x')$. As far as I know, the delta function is just a way to represent a continuous probability distribution where the variable has a definite value. If we have problems with the rigorous definition of a "function", we can just think of the delta function as a very narrow gaussian spike.

 

[WannabeNewton]

Well, the wavefunction corresponding to state vector $|\psi>$ is $\psi(x)=<x|\psi>$, and if $|\psi>$ is a position eigenstate $|x'>$, we have $\psi(x)=<x|x'>=\delta(x-x')$.

That doesn't change what vanhees said. It isn't a physical state-it's not even normalizable under the $L^2$ norm-you'll get a nonsensical square of a distribution in this case and infinity in the case of momentum eigenstates of a free particle. Physical states are by definition elements of the projective space of unit rays of the subspace of $L^2$ of interest. $\delta(x-x')$ and other eigenfunctions that show up in special cases e.g. momentum eigenstates of a free particle are what one calls generalized eigenstates and they are elements of an associated Rigged Hilbert space.

 

[Nugatory]

Y'know, I don't think we're still helping OP here...

 

[bhobba]

But still i don't understand what is the use of DIRAC DELTA FUNCTION in quantum mechanics.

Its used so you can have a function that is the eigenfunction of position. The position operator is x f(x) where f(x) is the wave function ie the expansion of the state in terms of position eigenstates. By definition the eigenfunction would be the function δ(x-x') such that x δ(x-x') = x' δ(x-x'), and ∫x δ(x-x') = x'.

Added later:
The above is bit loose. Here is a better way of looking at it. For discreet eigenvalues if yi are the outcomes ∑ yi <bj|bi> = yj because of the orthogonality of the |bi>, <bj|bi> = δij (δij the Kronecker delta - ∑ δij = 1. δij can be taken to be of the form δ(i-j) where δ(x) = 0 if x ≠ 0. 1 if x = 0). If we go over to a continuum then intuitively yi goes to y(x), δij to a function δ (x-x') such that ∫δ (x-x') =1, δ(x) = 0 if x ≠ 0, ∫y(x) δ(x-x') = y(x'). Actually in doing that because we are integrating over dx, we take δij = (δij/dx)dx and δij/dx → Dirac Delta function as dx→0, which is why its infinite when x = 0. This is the definition of the Dirac delta function - which of course doesn't really exit as a function in the usual sense.

Y'know, I don't think we're still helping OP here...

For starting out in QM simply think of the Dirac Delta function like you do dx when starting out in calculus. You think, at an intuitive level, of dx as a very small increment in x, so small you can ignore it in calculations, but not zero so you don't have the dreaded divide by zero in dy/dx. The rigorous development requires analysis, the concept of limit etc, but you can get a long way in practice with this intuitive idea.

Same with the Dirac Delta function. Simply think of it as a continuously differentiable function, δ(x) that has the property ∫δ(x)f(x) = f(0). No function does that - but a function does exist that's so close to it that for all practical purposes it has that property. Like dx you can get a long way in practice with that view.

Later you can study the correct theory, just like you study analysis a bit later to understand the correct basis of calculus (here is a link to my favourite book on it):
https://www.amazon.com/dp/0521558905/?tag=pfamazon01-20

After that you can look into how its used in QM via Rigged Hilbert Spaces. Start out with Chapter 2 in Ballentine - QM - A Modern Development. Then proceed to Quantum Mechanics For Mathematicians by Hall.

I got caught up in this stuff early on in my QM education and can assure you to start out don't worry about it. Leave sorting out the exact mathematics for later.

Thanks
Bill