Why is reflection coefficient defined this way

[Danny Boy]

In Griffith's "Introduction to Quantum Mechanics, second edition" he states: For the delta-function potential, when considering the scattered states (with E > 0), we have the general solutions for the time-independent Schrodinger equation: $$\psi(x) = Ae^{ikx} + Be^{-ikx}~~~~\text{for }x<0$$ and $$\psi(x) = Fe^{ikx} + Ge^{-ikx}~~~~\text{for }x>0.$$ In a typical scattering experiment, particles are fired in from one direction-let's say, from the left. In that case the amplitude of the wave coming in from the right will be zero: $$G=0~~~~~(\text{for scattering from the left}).$$

Then A is the amplitude of the incident wave, B is the amplitude of the reflected wave and F is the amplitude of the transmitted wave. Now the probability of finding the particle at a specified location is given by $$|\Psi|^2,$$ so the relative probability that an incident particle will be reflected back is $$R \equiv \frac{|B|^2}{|A|^2}$$ where R is called the reflective coefficient.

Question:

How does the definition of R follow? Where exactly does this probability come from?

Thanks for any assistance.

 

[DrClaude]

How does the definition of R follow? Where exactly does this probability come from?

It follows from:

Then A is the amplitude of the incident wave, B is the amplitude of the reflected wave

 

[Igael]

tunnelling ...
A similar question here : Derivation of reflection/transmission coefficients

 

[Avodyne]

What we would really like to do in this problem is start with a wave packet on the far left, moving to the right, and then time-evolve it with the time-dependent Schrodinger equation. Then the wave packet hits the barrier, and it's a big mess. But then, at late times, we find two isolated wave packets: one on the left, moving to the left, and one on the right, moving to the right. The probability to be in the left wave packet is $|B|^2/|A|^2$, and the probability to be in the right wave packet is $|F|^2/|A|^2$. This can be shown to hold to a very good approximation, using stationary-phase methods; I once worked it through myself.

An easier way to see why it's correct is to use the time-independent solution, and compute the probability current; we get $j=k(|A|^2-|B|^2)$ on the left and $j=k|F|^2$ on the right (with $G=0$). The current $j$ must be continuous across the boundary.

 

[vanhees71]

Perhaps you like some animations :-)):

http://theory.gsi.de/~vanhees/faq/quant/node35.html

and the following pages :-)).

 

[Danny Boy]

What we would really like to do in this problem is start with a wave packet on the far left, moving to the right, and then time-evolve it with the time-dependent Schrodinger equation. Then the wave packet hits the barrier, and it's a big mess. But then, at late times, we find two isolated wave packets: one on the left, moving to the left, and one on the right, moving to the right. The probability to be in the left wave packet is $|B|^2/|A|^2$, and the probability to be in the right wave packet is $|F|^2/|A|^2$. This can be shown to hold to a very good approximation, using stationary-phase methods; I once worked it through myself.

An easier way to see why it's correct is to use the time-independent solution, and compute the probability current; we get $j=k(|A|^2-|B|^2)$ on the left and $j=k|F|^2$ on the right (with $G=0$). The current $j$ must be continuous across the boundary.

What we would really like to do in this problem is start with a wave packet on the far left, moving to the right, and then time-evolve it with the time-dependent Schrodinger equation. Then the wave packet hits the barrier, and it's a big mess. But then, at late times, we find two isolated wave packets: one on the left, moving to the left, and one on the right, moving to the right. The probability to be in the left wave packet is $|B|^2/|A|^2$, and the probability to be in the right wave packet is $|F|^2/|A|^2$. This can be shown to hold to a very good approximation, using stationary-phase methods; I once worked it through myself.

An easier way to see why it's correct is to use the time-independent solution, and compute the probability current; we get $j=k(|A|^2-|B|^2)$ on the left and $j=k|F|^2$ on the right (with $G=0$). The current $j$ must be continuous across the boundary.

Thanks for your response. Using the definition of probability current $j :=\frac{\hbar}{2mi}\bigg(\psi^*\frac{\partial \psi}{\partial x}-\psi\frac{\partial \psi^*}{\partial x} \bigg)$, I get $j = \frac{\hbar k}{m}\bigg( |A|^2-|B|^2\bigg)$ on the left and $j = \frac{\hbar k}{m}|F|^2$ on the right. I have a couple of questions at this point. Firstly, is $j$ required to be continuous simply because the probability going into the boundary $x = 0$ must equal that coming out? Secondly, I assume the continuity then implies that $|A|^2-|B|^2 = |F|^2$. Then clearly $1 = \frac{|B|^2}{|A|^2} + \frac{|F|^2}{|A|^2}$, which is something necessary for probability. But how does this answer my original question as to why the reflection coefficient is described as the probability that an incident particle will be reflected back? Could you give me some detail as to the connection to probability?

 

[vanhees71]

You have to read $j$ as a directed quantity. It's a probability current, i.e., a vector. Of course, here it has only one component, because you deal with particle motion in only one spatial direction. The first part $\propto |A|^2>0$. Since you consider the scattering of a particle coming in from the left, a positive current means it's in the direction of the incoming particle. The 2nd part $\propto -|B|^2<0$ is probability running to the left, i.e., it's the part that is reflected from the potential. The condition $G=0$ in your above ansatz means that there is no incoming particle from the right. Thus you have the following situation, described by my animations in the link above: You start with an incoming wave packet, describing an asymptotic free particle that comes with some momentum from the far left running towards the potential to the right. Then the particle interacts with the potential and thus with some probability gets reflected (which is described by the wave running to the left in the region $x<0$) or goes through to the region $x>0$ (described by the wave running to the right into the region $x>0$). The probability to get reflected is thus $|B|^2$ and for going through $|F|^2$. The total probability of course must be conserved, i.e., $|A|^2 \stackrel{!}{=}1=|B|^2+|F|^2$. This determines the correct normalization conditions for the wave function. It's also important to note that a single particle never gets split somehow in a part being reflected and another part going through the potential but you have always only one particle. The meaning of the wave function is that $|\psi|^2$ is the probability density of its position. The Schrödinger equation implies the continuity equation
$$\partial_t |\psi|^2+\partial_x j=0,$$
which implies that the total probability is always 1, i.e.,
$$\int_{\mathbb{R}} \mathrm{d} x |\psi(t,x)|^2=1=\text{const}.$$

 

[Avodyne]

But how does this answer my original question as to why the reflection coefficient is described as the probability that an incident particle will be reflected back? Could you give me some detail as to the connection to probability?

$R$ is defined to be the probability that an incident particle is reflected.

The problem is, given this definition of $R$, how do we compute $R$? And the answer is, take the wave function to be $Ae^{ikx}+Be^{-ikx}$ on the left and $Fe^{iix}$ on the right, and impose appropriate matching conditions; this gives $B$ and $F$ in terms of $A$. Then the reflection probability $R$ is given by $R=|B|^2/|A|^2$.

 

[Danny Boy]

$R$ is defined to be the probability that an incident particle is reflected.

The problem is, given this definition of $R$, how do we compute $R$? And the answer is, take the wave function to be $Ae^{ikx}+Be^{-ikx}$ on the left and $Fe^{iix}$ on the right, and impose appropriate matching conditions; this gives $B$ and $F$ in terms of $A$. Then the reflection probability $R$ is given by $R=|B|^2/|A|^2$.

Yes I think I didn't state my question well. I understand that #R# is defined in that way so it isn't derived from anything. What I don't understand is why $|B|^2$ is the probability that the particle will get reflected and $|F|^2$ is the probability that it goes through?

 

[vanhees71]

What's unclear about my answer in #7?

 

[Danny Boy]

What's unclear about my answer in #7?

The part "The probability to get reflected is thus $|B|^2$ and $|F|^2$". This might be a trivial question about waves but I have only recently started learning about QM.

 

[vanhees71]

Think in terms of wave packets, which is anyway a more adequate picture for the scattering process (see also the animations, I posted before). The wave packet splits in a reflected and a transmitted part at the potential well. You see this when looking at the current, which in the region $x<0$ has a positive and a negative part. The positive part means the part of the wave traveling to the right, and the negative is the one traveling to the left. The coefficient in the corresponding energy eigenmodes squared give the strengths of these parts. In the region $x>0$ there's only a positive component by construction (according to the initial condition of a asymptotically free wave in the initial state coming from the left). So that's the part that gets transmitted.

 

[Jilang]

Danny Boy, are you asking why the probability is the amplitude squared?

 

[Danny Boy]

Danny Boy, are you asking why the probability is the amplitude squared?

Yes that's what I'm asking.

 

[Jilang]

There was a thread here about that.
https://www.physicsforums.com/threads/why-is-probability-amplitude-squared.724721/

 

[Danny Boy]

There was a thread here about that.
https://www.physicsforums.com/threads/why-is-probability-amplitude-squared.724721/

Thanks, yes this is what I was asking, will have a look at these discussions and associated links.

 

[Avodyne]

That discussion (like most on the topic) quickly gets rather advanced. For a beginner, I think the simplest explanation is that "probability = amplitude squared" is one of the postulates of quantum mechanics, one that has been extensively confirmed by experiment. Postulates are not derived from anything else, they are the starting point.

There is an ongoing discussion in physics and philosophy about whether this postulate can be derived from some other more basic postulate. No consensus has been achieved as yet.

 

[vanhees71]

I'd say it's rather a discussion in philosophy than physics. In physics we just take the Born rule, i.e., that for the position-space wave function this wave function's modulus squared is the probability-distribution function for finding the particle at the position given in its argument. Analogously, for the momentum-space wave function, which is the Fourier transform of the position-space wave function this rule holds (and for any other eigenbasis of a complete set of compatible observables): The modulus of the wave function squared is the probability (discrete eigenvalues) or probability density (continuous eigenvalues) to find that value when measuring the observable. It's such a profound finding that Born got his Nobel prize for that interpretation of the wave function!

All empirical findings are in very good agreement with this postulate, and it's most probably not derivable somehow from the other postulates of quantum theory (defining mostly the dynamics of states and observables).

 

[Danny Boy]

Thanks for all the assistance. One more question, for the transmission coefficient $T$, is $$T \equiv \frac{|F|^2}{|A|^2}$$ a special case for when the incident wave and transmitted waves are traveling at the same speed. Is the more general form of the transmitted coefficient: $$T \equiv \frac{|F|^2v_t}{|A|^2v_i},$$ where $v_t$ and $v_i$ are the speed of the respective incident and transmitted waves? Also is another definition of the transmitted coefficient in terms of the probability current $$T \equiv\frac{J_t}{J_i}?$$

 

[Jilang]

I have seen the second equation where v is the group velocity.

 

[Avodyne]

Yes and yes.