Scattering theory (some confusion)

[amjad-sh]

We know that the solution of the schroedinger equation for a free particle in position representation is a plane wave:
\psi(x)=Ae^{ikx}+Be^{-ikx} which means that the particle has a probability to move either to the left or to the right.
"Now let's take the potential step problem.
V(x)=0 for x<0
V(x)=V for x>0
The dynamics of the particle is regulated by the schroedinger equation which is given in those two regions by:
(\frac{d^{2}}{dx^{2}}+k_{1}^{2})\psi_{1}(x)=0 (x<0)
(\frac{d^{2}}{dx^{2}}+k_{2}^{2})\psi_{2}(x)=0 (x>0)
where k_{1}^{2}=2mE/\hbar^{2} and K_{2}^{2}=2m(E-V)/\hbar^{2}
The most general solutions of these two equations are plane waves:
\psi_{1}(x)=Ae^{ik_{1}x}+Be^{-ik_{1}x} (x<0)
\psi_{2}(x)=Ce^{ik_{2}x}+De^{-ik_{2}x} (x>0)
Where Ae^{ik_{1}x}and Ce^{ik_{2}x}represents waves moving in the positive x direction, but Be^{-ik_{1}x}andDe^{-ik_{2}x} corresponds to waves moving in the negative x direction.We are interested in the case where the particles are initially incident on the potential step from the left:they can be reflected and transmitted at x=0.Since no wave is reflected from the region x>0 to the left,the constant D must vanish"(zetteli text).
I have two questions here:
1)Why we are talking about reflection here? I mean even in case where there is no potential
like in the free particle case, its wave function is divided into two components one that moves to the right and the other moves to the left.
2)"constant D must vanish" why there is no reflected wave in the region x>0?
Thanks!

 

[mfb]

If there is no potential step, nothing gets reflected. If there is a potential step, some part (or all, depending on the step) gets reflected.

2)"constant D must vanish" why there is no reflected wave in the region x>0?

A reflection of what? $D\neq 0$ would be a particle coming from the right side moving towards the step. You cannot get this if you require the particle to come from the left side.

 

[drvrm]

like in the free particle case, its wave function is divided into two components one that moves to the right and the other moves to the left.

in the free particle situation you are writing a general solution and it has two parts one is particle going from left to right and another equivalent solution when the particle is moving from right to left-they are independent solutions -not a reflection. a general solution has all the possibilities and the boundary conditions give rise to 'real' solutions.

 

[amjad-sh]

A reflection of what? D≠0D≠0D\neq 0 would be a particle coming from the right side moving towards the step. You cannot get this if you require the particle to come from the left side.

If the particle crosses the region where we have potential V,and the particle is moving to the right now. why it is wrong if the particle goes to the left after some time t while staying in the same region where the potential is V?

in the free particle situation you are writing a general solution and it has two parts one is particle going from left to right and another equivalent solution when the particle is moving from right to left-they are independent solutions -not a reflection.

So the particle may be going from left to right or from right to left.Can't it go from left to right then from right to left?

 

[drvrm]

So the particle may be going from left to right or from right to left.Can't it go from left to right then from right to left?

these scattering processes are guided by the physical situation- suppose the beam of incident particle is from left then -that has to be taken ,if it gets partly reflected by a potential the a reflected wave will be there ,
if it gets partly transmitted the transmitted part will be there -if the potential is such that it has another wall ,so again the reflected and transmitted part will be there and the intensities will be related by currents .

An identical treatment can be done for beam incident from the rt. hand side analogously.

 

[amjad-sh]

these scattering processes are guided by the physical situation- suppose the beam of incident particle is from left then -that has to be taken ,if it gets partly reflected by a potential the a reflected wave will be there ,
if it gets partly transmitted the transmitted part will be there -if the potential is such that it has another wall ,so again the reflected and transmitted part will be there and the intensities will be related by currents .

An identical treatment can be done for beam incident from the rt. hand side analogously.

What goes wrong if that happens in a region that has the same potential V. Can't the particle in this case move from left to right then from right to left?

 

[drvrm]

What goes wrong if that happens in a region that has the same potential V. Can't the particle in this case move from left to right then from right to left?

movement of particle is affected by the potential i.e. how the probability density changes -if you can get probability finite for a potential you can have them- the probability current is related to velocity of the particle -and the continuity condition holds true- the total probability has to be one for normalized wave functions. so taking the boundary conditions if there is one potential wall ,and after that its constant -
one can not get a physical solution for a particle getting reflected in the region on the right hand side of the wall and moving from right to left in a constant potential region.there must be some reason for particle to turn back- just like humans.

 

[amjad-sh]

one can not get a physical solution for a particle getting reflected in the region on the right hand side of the wall and moving from right to left in a constant potential region.there must be some reason for particle to turn back- just like humans.

So I can understand from this, that it is impossible to measure a "one" quantum particle at a position x1 at time t1 and then at postion x2 at time t2 (x2>x1) and then at position x3 at time t3 and position x4 at time t4 (x4<x3) (in a region of constant potential V).
Suppose we have a wave function \psi(x,t)=a\psi_{1}(x,t)+b\psi_{2}(x,t)
\psi_{1}(x,t) indicates that the particle is moving to the right and \psi_{2}(x,t) indicates that the particle is moving to the left.
Now the probability to detect the particle moving to the right is a number say a1 and the probability to detect the particle moving to the left is a number say b1. Where a1+b1=1, the wave function \psi(x,t) is normalizable.(all of this is happening in a region of constant potential)
To check if this is true, let's prepare a big chunk of particles( Nparticles), a1*N of the particles will move to the right and b1*N of the particles will move to the left.
Can't this be happen?
If yes, why we choose D=0 above?

 

[mfb]

So I can understand from this, that it is impossible to measure a "one" quantum particle at a position x1 at time t1 and then at postion x2 at time t2 (x2>x1) and then at position x3 at time t3 and position x4 at time t4 (x4<x3) (in a region of constant potential V).

Your measurement changes the particle state. This does not work.

A particle cannot just "turn around" without a reason (like a change in potential).

To check if this is true, let's prepare a big chunk of particles( Nparticles), a1*N of the particles will move to the right and b1*N of the particles will move to the left.

You can prepare such a state (then you would have $D\neq 0$), but the problem in post 1 is discussing a different state.

 

[amjad-sh]

You can prepare such a state (then you would have D≠0D≠0D\neq 0), but the problem in post 1 is discussing a different state.

OK.
I think the idea is more clarified now.

I understood it like this: First the particle is moving to the right (we require it to move to the right), and so it can be represented by Ae^{ik_{1}x}. We can't add Be^{-ik_{1}x} now, because we require the particle to move to the right. After the particle hit the boarder of potential V, the potential changed here. The particle may complete its path to the right with wave function Ce^{ik_{2}x} or may recoil back in the first region where V=0. So in this case, the wave function in the first region may be written now as :\psi(x)=Ae^{ik_{1}x}+Be^{-ik_{1}x} and in region 2 as \psi(x)=Ce^{ik_{2}x}, as the particle can't change its path while potential is the same, so it will remain moving to the right.

But here a question arises: Why the quantum particle can't change its path if there is no reason, like change in potential or other things?
I know that this holds in classical mechanics, is there a proof or a postulate behind this?

 

[drvrm]

Why the quantum particle can't change its path if there is no reason,

now you are asking a question pertaining to metaphysics(philosophy)...

I know that this holds in classical mechanics,

the quantum behavior of particles can always connect to the classical world as all the observers/detectors and labs/beams are still classical.

 

[amjad-sh]

now you are asking a question pertaining to metaphysics(philosophy)...

But aren't there any postulate concerning this(that the the quantum particle can't change its path)?or even experiments that revealed this fact?

 

[drvrm]

But aren't there any postulate concerning this(that the the quantum particle can't change its path)?or even experiments that revealed this fact?

actually you are asking {Why the quantum particle can't change its path if there is no reason,}
so 'no reason' has to be found- and knowledge about a particle taking a particular position is due to its 'probability' in quantum mechanics there is no 'path' concept =the position probability is defined and momentum probability can be estimated. but one can not build a quantum path as in classical mechanics.

 

[mfb]

It is possible to construct theories that would allow such a direction change, but they would massively violate the results of hundreds of thousands of experiments, unless you require really obscure setups for that to occur. While you can never rule out anything completely (we are quite confident all apples fall down on Earth, but we did not test literally every single apple on Earth), some things would be too ridiculous to spend much time on them.

 

[amjad-sh]

Thanks to all for contribution!