Why is the integral of the Dirac Delta potential related to its properties?

[alejandrito29]

why in the problem of dirac delta potential, the integral

\int^{\epsilon}_{-\epsilon}\phi''(x)dx is equal to \phi'(\epsilon)-\phi'(-\epsilon)?

but \int^{\epsilon}_{-\epsilon}\phi(x)dx is equal to 0

if, for example\phi(x)=e^x

then \phi(x)''=\phi(x)

but, the firts integral is e^{\epsilon}-e^{-\epsilon}
and the second integral would be to zero

i don't understand

 

[George Jones]

why in the problem of dirac delta potential, the integral

\int^{\epsilon}_{-\epsilon}\phi''(x)dx is equal to \phi'(\epsilon)-\phi'(-\epsilon)?

Do you know why this is required?

but \int^{\epsilon}_{-\epsilon}\phi(x)dx is equal to 0

If \phi is continuous, this is true.

In this situation, you want to work with wavefunctions that satisfy the above two conditions.

if, for example\phi(x)=e^x

then \phi(x)''=\phi(x)

but, the firts integral is e^{\epsilon}-e^{-\epsilon}
and the second integral would be to zero

i don't understand

If \phi \left( x \right)=e^x, then \phi \left( x \right) does not satisfy the above two conditions, and thus \phi \left( x \right)=e^x is not a possible wavefunction.

 

[alejandrito29]

Do you know why this is required?


If \phi is continuous, this is true.

In this situation, you want to work with wavefunctions that satisfy the above two conditions.


If \phi \left( x \right)=e^x, then \phi \left( x \right) does not satisfy the above two conditions, and thus \phi \left( x \right)=e^x is not a possible wavefunction.

then :
¿ only if: \phi(x) is continuous then \int^{\epsilon}_{-\epsilon}\phi(x)dx=0?

if \phi(x) is discontinuous then the integral is not zero?

 

[George Jones]

then :
¿ only if: \phi(x) is continuous then \int^{\epsilon}_{-\epsilon}\phi(x)dx=0?

There is a subtle difference between this and what I wrote. I did not write "only if", I wrote "if".

if \phi(x) is discontinuous then the integral is not zero?

I am not sure what you mean.

Consider a function somewhat related to the problem, \phi \left( x \right) = \left| x \right|. What are: \phi' \left( x \right); \phi'' \left( x \right);

\lim_{\epsilon \rightarrow 0} \int^\epsilon_{-\epsilon} \phi'' \left( x \right) dx?

 

[alejandrito29]

Consider a function somewhat related to the problem, \phi \left( x \right) = \left| x \right|. What are: \phi' \left( x \right); \phi'' \left( x \right);

\lim_{\epsilon \rightarrow 0} \int^{\epsilon}_{-\epsilon} \phi'' \left( x \right) dx?

I think:

\int^{\epsilon}_{-\epsilon} \phi''(x) =\phi'(x)|^{\epsilon}_{-\epsilon}

=(\theta(x)-\theta(-x))|^{\epsilon}_{-\epsilon}

Heaviside function:
\theta(x)=1; x>0 \theta(x)=0; x<0

then
\int^{\epsilon}_{-\epsilon} \phi''(x) =2\theta(\epsilon)-2\theta(-\epsilon)=2-0=2

I'm not sure....

 

[Petar Mali]

\frac{d^2\psi}{dx^2}+\frac{2m}{\hbar^2}[E-V(x)]\psi(x)=0

V(x)=-\alpha\delta(x),\alpha>0

You will have two region; region 1 for x<0 and region 2 for x\geq 0. In the boundary of this two regions wave function must me continuous. Integrate equation

-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}\psi(x)-\alpha\delta(x)\psi(x)=E\psi(x) from -\epsilon to \epsilon you will get

\psi'(0^+)-\psi'(0^-)=-\frac{2m\alpha}{\hbar^2}\psi(0)

 

[alejandrito29]

\frac{d^2\psi}{dx^2}+\frac{2m}{\hbar^2}[E-V(x)]\psi(x)=0

V(x)=-\alpha\delta(x),\alpha>0

You will have two region; region 1 for x<0 and region 2 for x\geq 0. In the boundary of this two regions wave function must me continuous. Integrate equation

-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}\psi(x)-\alpha\delta(x)\psi(x)=E\psi(x) from -\epsilon to \epsilon you will get

\psi'(0^+)-\psi'(0^-)=-\frac{2m\alpha}{\hbar^2}\psi(0)

thanks , but my main question is about the properties of integrals:

¿¿when

\lim_{\epsilon \rightarrow 0} \int^{\epsilon}_{-\epsilon} \phi(x)'' dx,\lim_{\epsilon \rightarrow 0} \int^{\epsilon}_{-\epsilon} \phi(x)' dx,\lim_{\epsilon \rightarrow 0} \int^{\epsilon}_{-\epsilon} \phi(x) dx,

are zero or non zero?

 

[DrDu]

That is a question of definition. The delta function alone is not a well defined operator.
However, the kinetic energy operator with a delta function can be shown to be one.
To define an operator, you have to specify the domain of functions on which it is defined. In one dimension you can either choose psi to be continuous with discontinous first derivative, which then is called a delta function type potential, or to be discontinous, in the latter case, one speaks of a delta' interaction.