# Scattering by a Delta Function Well in 1D [closed]

Let us consider a scattering process by a delta function well in 1D: $$V(x) = -\alpha \, \delta(x), \quad \alpha > 0.$$

I solve the Schrödinger equation for the scattering states and get the followings:

$$\psi_1(x) = c_1 e^{ikx} + c_2 e^{-ikx}, \quad x < 0, \quad k^2 = \frac{2mE}{\hbar^2}, \quad E > 0,$$

and,

$$\psi_2(x) = c_3 e^{ikx}, \quad x > 0, \quad k^2 = \frac{2mE}{\hbar^2}, \quad E > 0.$$

Here $$c_1 e^{ikx}$$ is the incoming wave from the left, $$c_2 e^{-ikx}$$ is the reflected wave and, $$c_3 e^{ikx}$$ is the transmitted wave.

Questions

1. What is scattering in 1D quantum mechanics and what is the scattered wave here?
2. How to calculate the scattering phase shift $$\delta_0$$ for this scattering process?
3. Is the scattering phase shift $$\delta_0$$ is given by the phase of $$\frac{c_3}{c_1}$$?
4. The reflected wave should also have a phase shift which is different from the scattering phase shift $$\delta_0$$ . Is it correct?

## closed as off-topic by Chris♦, Aaron Stevens, ZeroTheHero, user191954, Cosmas ZachosNov 18 '18 at 17:18

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$$\let\d=\delta \def\de{\d_{\mathrm e}} \def\do{\d_{\mathrm o}} \def\ps#1{\psi_{\mathrm{#1}}} \def\sgn{{\mathrm{sgn}}}$$ About the meaning of scattering phaseshift. I too never saw this quantity used in 1D scattering. In 3D phaseshifts for central potential are related to angular momentum: there is one phaseshift for each $$l$$ quantum number. This is because angular momentum is conserved and a scattering process can be separately analized for each $$l$$.

Moreover phase factor $$e^{i\d_l}$$ is defined between ingoing and outgoing wave. Actually $$e^{2i\d_l}$$ is an eigenvalue of scattering matrix, which is diagonal in this representation. Ingoing wave is defined as the one solution having asymptotic behaviour $$e^{-ikr}\!/r$$, outgoing $$e^{ikr}\!/r$$.

In 1D we can't use angular momentum, but if potential is an even function of $$x$$ then parity is a good quantum number and makes sense to study even and odd states separately. Using Hanting Zhang's notations an even state has $$A=G$$, $$B=F$$ whereas an odd state has $$A=-G$$, $$B=-F$$. Furthermore an ingoing state has $$B=F=0$$, an outgoing one has $$A=G=0$$.

That is: \eqalign{ \ps{+i} &= A\,e^{-ik|x|} \cr \ps{+o} &= B\,e^{ik|x|}\cr \ps{-i} &= A\,\sgn(x)\,e^{-ik|x|}\cr \ps{-o} &= B\,\sgn(x)\,e^{ik|x|}.\cr}

From Hanting Zhang I take $$F - G - A + B = 2 i z\,(A + B).$$ For even states $$-A + B = i z\,(A + B)$$ $$B = A\,{1 + iz \over 1 - iz}$$ and defining $$\de$$ by $$e^{2i\de} = {B \over A}$$ we find $$e^{2i\de} = {1 + iz \over 1 - iz}$$ $$\de = \arctan z.$$

For odd states $$B = -A$$ $$\do = \pi/2.$$

Caution 1: I don't know what is meant for $$\d_0$$ in the questions. Almost certainly the meaning differs from the one I assumed.

Caution 2: I didn't thoroughly check my equations. Possible wrong signs or other errors.

What is scattering in 1D quantum mechanics and what is the scattered wave here?

Scattering in quantum mechanics usually refers to how a wavefunction behaves when its energy is greater than the potential $$V(x)$$ as $$x \rightarrow \pm\infty$$, normally called a scattering state. Normally we choose $$V(x)$$ to go to zero at infinity so that a state is scattering simply if it has positive energy.

For the delta function well you've given, the scattered wave is just the incoming wave for the left, which scatters into a reflected wave and a transmitted wave.

How to calculate the scattering phase shift $$\delta_0$$ for this scattering process?

Is the scattering phase shift $$\delta_0$$ is given by the phase of $$\frac{c_3}{c_1}$$?

Explicitly, let $$\psi(x) = Ae^{ikx} + Be^{-ikx}, \kern{3mm} x<0$$ $$\psi(x) = Fe^{ikx} + Ge^{-ikx}, \kern{3mm} x>0$$ Continuity of $$\psi(x)$$ requires that, $$A + B = F + G.$$ For $$\psi(x)$$'s derivative, the delta well requires a discontinuity proportional to its strength: $$ik(F-G-A+B) = \left.\Delta\left(\frac{d\psi}{dx}\right)\right|_0 = -\frac{2m\alpha}{\hbar^2}(A+B).$$

Now we kill off incoming waves from the right, and set $$G = 0$$, and rearrange to get:

$$B = \frac{iz}{1-iz}A, \kern{3mm}F = \frac{1}{1-iz}A,$$ where $$z = \frac{m\alpha}{\hbar^2k}$$. The the phase shift is just the phases of the coefficients in front of $$A$$.

The reflected wave should also have a phase shift which is different from the scattering phase shift $$\delta_0$$. Is it correct?

It would seem that yes, the reflected wave does have a different phase.

• The reflected and transmitted both the waves are going through a phase shift. According to your explanation, the phase shift of the reflected wave with respect to the incident wave is the scattering phase shift (which is given by the phase of the factor in front of $A$). That leaves the phase shift of the transmitted wave out of the discussion for the scattering phase shift. – omehoque Nov 18 '18 at 4:22
• Sorry, mind if you clarify what "scattering phase shift" means? We might be talking about different things. – Hanting Zhang Nov 18 '18 at 4:26
• Apparently, I am having trouble understanding the concept of scattering phase shift in 1D. If is there anything called 'scattering phase shift in 1D', what would be the definition then? – omehoque Nov 18 '18 at 4:29
• Oh. I see. I haven't worked a lot with scattering, but I do know that "scattering phase shift" isn't usually used for 1-D scattering. As you mentioned, it becomes an ambiguity since there are 2 different waves. Generally, we use it in 3-D since that ambiguity goes away. (In 3-D the phase shift is also linked to the boundary conditions, and helps to reduce the problem). In 1-D, phase shifts don't really offer anything new. We mostly just want to know the new amplitudes. – Hanting Zhang Nov 18 '18 at 4:38