# Double delta potential

Summary:: I have a problem with a particle, which gets scatterd at a double delta-potential

Hello, I am really stuck with the floowing problem:

A particle moves from the left along the x-axis and gets scatterd at a one-dimensional potential V(x)=a[dirac delta of x) +b [dirac delta of x-c], where a,b,c are positive numbers. The question is now: What is the transmission-amplitude t of the particle and the transmission-probability?

You could really help me with this. Thank you in advance!

Staff Emeritus
Since a Dirac delta function is zero except at a point, the solution to the Schrodinger equation for this situation will be just a superposition of plane waves. So ##A e^{i k x} + B e^{-i k x}##. So the challenge is to break the entire x-axis into three regions, the region to the left, the region between the delta-functions, and the region to the right. Each region will have different values of ##A## and ##B##. First find out what those coefficients are in each region.

vanhees71
First of all thank you for you answer! But I want to solve the problem in general, since i can then apply it to future problems, so I got the three regions 1: x<0 A⋅exp(ikx) +B⋅exp(-ikx) ; 2: 0<x<a C⋅exp(ikx) +D⋅exp(-ikx) ; 3: a<x F⋅exp(ikx). But now I don't know, how I get a general solution to the transmission amplitude, any ideas?

Gold Member
2022 Award
You need to employ the conditions to the wave function at the two points, where the ##\delta## distributions "peak".

You mean like, that the wavefunction and its derivative need to be continues at the "peaks"? I tried that and just came to 0=0, which i could not really use...

Gold Member
2022 Award
It's not that simple because you deal with a distribution-valued potential.

Hint: Consider first one ##\delta(x)## potential as the corresponding limit of a box potential,
$$V_{\epsilon}(x)=\Theta(-\epsilon/2<x<\epsilon/2),$$
where
$$\Theta(-\epsilon/2<x<\epsilon/2)=\begin{cases} 1 & \text{for} \quad |x|<\epsilon/2, \\ 0 &\text{for} \quad |x| \geq \epsilon/2. \end{cases}$$
To that end integrate the time-independent Schrödinger equation over the interval ##[-\epsilon/2,\epsilon/2]## and then derive the corresponding boundary condition in the limit ##\epsilon \rightarrow 0^+##. You'll get a "jump condtion" for ##\psi'(x)/\psi(x)## at ##x=0##. Since ##\psi## is continuous at ##x=0## (1st condition) you get a jump for ##\psi'## at ##x=0## (2nd condition), which you need to match the solutions left and right of the singularity of the potential at ##x=0##.

You can now generalize this to the case of two ##\delta##-singularities. Another nice exercise is the "Dirac Comb", i.e., infinitely many periodic ##\delta## distributions.

First of all thank you for your help, I really appreciate it!

If I apply the said and integrate the Schrödinger equation I come to ψ2(0)' - ψ1(0)' = h^2/2ma and the same for x=a, but with Ψ3 and ψ2. Where "h" represents h bar, but I do not know how to write it here.

But from there on, how do I derrive the transmission amplitude? I tried making a system of equations out of the boundary conditions but got like 2 pages of terms as a result, which does not seem right...

You can now generalize this to the case of two ##\delta##-singularities. Another nice exercise is the "Dirac Comb", i.e., infinitely many periodic ##\delta## distributions.

And also thank you for this recommendation! The "Dirac Comb" seems really interesting!

Homework Helper
2022 Award
If I apply the said and integrate the Schrödinger equation I come to ψ2(0)' - ψ1(0)' = h^2/2ma and the same for x=a, but with Ψ3 and ψ2. Where "h" represents h bar, but I do not know how to write it here.
That is not correct. When you integrate over each delta function you get a kink in the derivative proportional to the strength of the ##\delta## potential. Do us all a favor and use capital letters ie $$V(x)=A\delta (x)+B\delta(x-c)$$ Of course you could do a big favor and learn LaTeX. Anyhow try again.

vanhees71
Thank you for your help! I am sorry, but I have never learned Latex and I am afraid I won't do it over night...

I do not quite undestand why I am wrong and how it would be correct... if I integrate HΨ=EΨ, over a Interval from -e to e and then look at the limit of e to 0 I came to Ψ2(0)' -Ψ1(0)'=2mA/h^2 (I see that I made an error there in my last post), where ' means the derivative with respect to x.

I would be verry thankfull if you could tell me, how it is done right!

Homework Helper
2022 Award
Write it out line by line. The potential multiplies ##\Psi##. Get a fresh piece of paper and proceed carefully. When you integrate near the distributions you will pick up the potential strength times the wavefn at x=0 and at x=c.