# Transmission and Reflection on a potential step

1. Apr 13, 2013

### 71GA

Lets say we have a potential step with regions 1 with zero potential $W_p\!=\!0$ (this is a free particle) and region 2 with potential $W_p$. Wave functions in this case are:

\begin{align}
\psi_1&=Ae^{i\mathcal L x} + B e^{-i\mathcal L x} & \mathcal L &\equiv \sqrt{\frac{2mW}{\hbar^2}}\\
\psi_2&=De^{-i\mathcal K x} & \mathcal K &\equiv \sqrt{\frac{2m(W_p-W)}{\hbar^2}}
\end{align}

Where $A$ is an amplitude of an incomming wave, $B$ is an amplitude of an reflected wave and $D$ is an amplitude of an transmitted wave. I have sucessfuly derived a relations between amplitudes in potential step:

\begin{align}
\dfrac{A}{D} &= \dfrac{i\mathcal L-\mathcal K}{2i\mathcal L} & \dfrac{A}{B}&=-\dfrac{i \mathcal L - \mathcal K}{i \mathcal L + \mathcal K}
\end{align}

I know that if i want to calculate transmittivity coefficient $T$ or reflexifity coefficient $R$ i will have to use these two relations that i know from wave physics.

\begin{align}
T &= \frac{j_{trans.}}{j_{incom.}} & R &= \frac{j_{ref.}}{j_{incom.}}
\end{align}

In above equations $j$ is a probability current which i also know how to derive (nice way is described here):

\begin{align}
j = -\frac{\hbar i }{2m}\left(\frac{d \psi}{dx}\psi^* - \frac{d\psi^*}{dx}\psi\right)
\end{align}

OK so far so good, but in our lectures professor somehow derived below equations which i can't derive. Could anyone please tell me how i can use all my knowledge described above to derive it?

\begin{align}
\boxed{R = \frac{(\mathcal{L - K})^2}{(\mathcal{L + K})^2}} && \boxed{T=\frac{4\mathcal{LK}}{\mathcal{(L+K)^2}}}
\end{align}

Last edited: Apr 13, 2013
2. Apr 13, 2013

### Fightfish

Some slight remarks first:
It is usually convenient to give your answers in the reciprocal way instead (ie B and D in terms of A), because A is the experimental parameter that you control.
I presume you meant $j_{ref.}$ for R?

What you have now:
(1) Formula for calculating probability current
(2) The incoming, reflected and transmitted wavefunctions
(3) The relationship between R, T and the probability currents

Can you use (1) and (2) to compute the probability currents for the incoming, reflected and transmitted wavefunctions?

Can you then substitute the obtained probability currents into (3) to obtain R and T?

3. Apr 13, 2013

### 71GA

I think this is possible i will do it and post results.

But before i continue. Can i rightfully say that function $\psi_I$ consists of two wavefunctions $\psi_{I~in} = Ae^{i\mathcal L x}$ (incomming waves) and $\psi_{I~re} = Be^{-i \mathcal L x}$ (reflected waves)?

4. Apr 13, 2013

### Fightfish

Yes, $\psi_I$ is a superposition of the incoming and reflected waves.

5. Apr 15, 2013

### 71GA

my try

So here i go. 1st i write down an equation for reflection, plug in the probability currents (please confirm if my currents are ok) and try to calculate reflection $R$.
\begin{align}
R &= \dfrac{-\frac{\hbar i}{2m} \left( \dfrac{d\psi_{1re}^*}{dx} \psi_{1re} - \dfrac{d \psi_{1re}}{dx} \psi_{1re}^* \right) }{-\frac{\hbar i}{2m} \left( \dfrac{d\psi_{1in}^*}{dx} \psi_{1in} - \dfrac{d \psi_{1in}}{dx} \psi_{1in}^* \right) } = \dfrac{ \frac{d}{dx}\left(Be^{i\mathcal L x}\right) Be^{-i\mathcal L x} - \frac{d}{dx} \left( Be^{-i\mathcal L x}\right) Be^{i\mathcal L x}}{ \frac{d}{dx}\left(Ae^{-i\mathcal L x}\right) Ae^{i\mathcal L x} - \frac{d}{dx} \left( Ae^{i\mathcal L x}\right) Ae^{-i\mathcal L x} } = \\
&=\dfrac{i\mathcal L B e^{i\mathcal L x} B e^{-i \mathcal L x} - \left(- i\mathcal L B e^{-i \mathcal L x} Be^{i \mathcal L x}\right)}{-i \mathcal L A e^{-i\mathcal L x} Ae^{i \mathcal L x} - i \mathcal L A e^{i \mathcal L x}Ae^{-i \mathcal L x} }=\dfrac{i\mathcal L B^2 + i\mathcal L B^2}{-i \mathcal L A^2 - i \mathcal L A^2}=\dfrac{2 i \mathcal L B^2}{-2i\mathcal L A^2} = -\frac{B^2}{A^2}
\end{align}

Question 1: If i look here reflection is defined as a square of absolute values $T = |B|^2/|A|^2$ while i only got squares and a minus sign $T = - B^2/A^2$. Why do we have to use absolute values?

If i use the correct equation (and not the one i got - i still need some explaination on the one i got) $T = |B|^2/|A|^2$ and plug in the expression $B=\frac{i\mathcal L + \mathcal K}{i\mathcal L - \mathcal K} A$ i get the solution which shows that all of the incomming waves gets reflected.
\begin{align}
T= \frac{|i \mathcal L + \mathcal K|^2 }{|i \mathcal L - \mathcal K|^2} \frac{A^2}{|A|^2} = \frac{|i \mathcal L + \mathcal K|^2 }{|i \mathcal L - \mathcal K|^2} = \frac{(i \mathcal L + \mathcal K)(-i \mathcal L + \mathcal K) }{(i \mathcal L - \mathcal K)(-i \mathcal L - \mathcal K)} = \frac{\mathcal L^2-i \mathcal L \mathcal K+i\mathcal L \mathcal K+\mathcal K^2}{\mathcal L^2+i\mathcal L \mathcal K-i\mathcal L \mathcal K + \mathcal K^2} = \frac{\mathcal L^2+ \mathcal K^2}{\mathcal L^2 + \mathcal K^2} = 1
\end{align}

The result allso looks ok for $T$ (i noticed that $\psi_2 = De^{-\mathcal K x}$ and NOT $\psi_2 = De^{- i\mathcal K x}$ as i wrote in my 1st post). If i use these functions result is:.

\begin{align}
T&=\dfrac{-\frac{\hbar i}{2m} \left( \dfrac{d\psi_{1tr}^*}{dx} \psi_{1tr} - \dfrac{d \psi_{1tr}}{dx} \psi_{1tr}^* \right) }{-\frac{\hbar i}{2m} \left( \dfrac{d\psi_{1in}^*}{dx} \psi_{1in} - \dfrac{d \psi_{1in}}{dx} \psi_{1in}^* \right) } = \dfrac{ \frac{d}{dx}\left(De^{-\mathcal K x}\right) De^{-\mathcal K x} - \frac{d}{dx} \left( De^{-\mathcal K x}\right) De^{-\mathcal K x} }{ \frac{d}{dx}\left(Ae^{-i\mathcal L x}\right) Ae^{i\mathcal L x} - \frac{d}{dx} \left( Ae^{i\mathcal L x}\right) Ae^{-i\mathcal L x} } = \\
&=\dfrac{-\mathcal K D e^{-\mathcal K x} D e^{-\mathcal K x} - (-\mathcal K D e^{-\mathcal K x} D e^{-\mathcal K x})}{-i \mathcal L A e^{-i\mathcal L x} Ae^{i \mathcal L x} - i \mathcal L A e^{i \mathcal L x}Ae^{-i \mathcal L x}} =\frac{0}{-2i \mathcal L A^2} = 0
\end{align}

Question 2: Why do i get $T=0$? Does this mean that i cannot penetrate the finite potential if it is infinitely long?

Last edited: Apr 15, 2013
6. Apr 15, 2013

### Fightfish

Right, I went back to your original derivations and found that there were several mistakes that I didn't pick up.

First, check your expressions for D/A and B/A. I got different answers from you when I worked it out explicitly. Next $R = |j_{ref}|/|j_{inc}|$ is a ratio of the magnitudes and hence is always a positive quantity. Making these corrections should enable you to arrive at the desired results.

With regards to the form of $\psi_{2}$, it depends on whether the energy of the wave is greater or less than the step height. If the step height is greater than the energy of the wave, then of course you get a transmission coefficient of 0 as you have worked out.

7. Apr 15, 2013

### 71GA

This would be clear to me if i could figure out why i get negative sign in $T = - B^2/A^2$. Is it maybee that my probability current is somehow messed up?

8. Apr 15, 2013

### Fightfish

I presume you meant R. R is defined as $R =|j_{ref}| / |j_{inc}|$ instead of just $j_{ref}/j_{inc}$. The minus sign appears when you use the latter instead because the probability currents of the reflected and incoming waves are in opposite directions.

I noticed that you edited your post, so I shall address some of those queries here too.
First of all, your expression for the coefficients B and D in terms of A are not correct.
Next, you got T = 0, because when you took $\psi_{2} = D e^{-\mathcal{K} x}$, you essentially assumed that $W < W_{p}$.

9. Apr 15, 2013

### 71GA

Yes thats what i thought. Iam sorry for the mistake. Why are those terms not correct? I know that my sign was wrong...

10. Apr 15, 2013

### Fightfish

The general solution for an incoming wave approaching the step potential is:
$$\psi_{I} = A e^{i \mathcal{L} x} + B e^{-i \mathcal{L} x} \quad , \quad x < 0\\ \psi_{II} = D e^{i \mathcal{K} x} \quad , \quad x > 0$$
By continuity, we demand that
$$\psi_{I}(0) = \psi_{II}(0) \quad \Rightarrow \quad A + B = D\\ \frac{d\psi_{I}}{dx}|_{x=0} = \frac{d\psi_{II}}{dx}|_{x=0} \quad \Rightarrow \quad \mathcal{L}A-\mathcal{L}B = \mathcal{K} D$$
Solving simultaneously,
$$\frac{D}{A} = \frac{2 \mathcal{L}}{\mathcal{L}+\mathcal{K}}\\ \frac{B}{A} = \frac{\mathcal{L} - \mathcal{K}}{\mathcal{L}+\mathcal{K}}\\$$

11. Apr 15, 2013

### 71GA

Yes but if i understood right this is only for a particle with energy $W>W_p$, otherwise we would have to define $\psi_{II} = D e^{-\mathcal{K} x}$ like i did and thats why i got solutions for a particle with energy $W<W_p$. Am i right?

12. Apr 17, 2013

### Fightfish

Unfortunately I'm getting some internal server error thingy when I attempt to use the advanced reply options, but anyway let me just give you a quick reply here.

(1) Oh okay, you are right that your solution implicitly assumes that W < Wp and that is why you got the results that you got, namely that T = 0 and R = 1. It is not hard to see why this must be the case because conservation of energy will be violated if the particle could penetrate the potential step (key word being step, as opposed to barrier).

(2) The expressions obtained by your lecturer for T and R, however, are true only for W > Wp.

(3) The solution that I cited actually holds generally for plane waves incident upon the barrier, regardless of whether the energy of the incoming wave is lower or higher than the barrier, provided that the $\mathcal{K}$ is defined to be $\sqrt{2m(W - W_{p})}/\hbar$.
Imagine now that the energy of the wave is less than the barrier potential. Then $\mathcal{K}$ will be imaginary. We can write $\mathcal{K}$ as $i \mathcal{K}'$ where $\mathcal{K}' \equiv \sqrt{2m(W_{p} - W)}/\hbar$ is real.
From there you can obtain the expression $\psi_{II} = e^{-\mathcal{K}'x}$ that you have.

The point of this exercise is that one can obtain a general solution for R and T before deciding whether to pick W < Wp or W > Wp.

13. Apr 17, 2013

### 71GA

Thank you very much for all your help. I think i understand now.

14. Apr 20, 2013

### 71GA

I did calculate the reflectivity coefficient $R$ now but i dont know if my last two steps are legit. Could you please check and tell me if i can swap $|~|$ with $(~)$ and change order of $\mathcal L$ and $\mathcal K$ like i did that.

\begin{align*}
R &= \dfrac{\big|j_{re}\big|}{\big|j_{in}\big|} \!=\! \Bigg|\dfrac{\dfrac{\hbar }{2mi} \left( \dfrac{d\overline{\psi}_{re}}{dx}\, \psi_{re} - \dfrac{d\psi_{re}}{dx}\, \overline{\psi}_{re} \right)}{\dfrac{\hbar}{2mi} \left( \dfrac{d\overline{\psi}_{in}}{dx}\, \psi_{in} - \dfrac{d\psi_{in}}{dx}\, \overline{\psi}_{in} \right) }\Bigg|\!=\! \Bigg| \dfrac{\frac{d}{dx}\big(\overbrace{Be^{i\mathcal L x}}^{\text{konjug.}}\big) Be^{-i\mathcal L x} - \frac{d}{dx} \left( Be^{-i\mathcal L x}\right) \overbrace{Be^{i\mathcal L x}}^{\text{konjug.}}}{ \frac{d}{dx}\big(\underbrace{Ae^{-i\mathcal L x}}_{\text{konjug.}}\big) Ae^{i\mathcal L x} - \frac{d}{dx} \left( Ae^{i\mathcal L x}\right) \underbrace{Ae^{-i\mathcal L x}}_{\text{konjug.}}}\Bigg|= \nonumber\\
&=\Bigg| \dfrac{i\mathcal L B e^{i\mathcal L x} B e^{-i \mathcal L x} - \left(- i\mathcal L B e^{-i \mathcal L x} Be^{i \mathcal L x}\right)}{-i \mathcal L A e^{-i\mathcal L x} Ae^{i \mathcal L x} - i \mathcal L A e^{i \mathcal L x}Ae^{-i \mathcal L x} } \Bigg| = \Bigg| \dfrac{i\mathcal L B^2 + i\mathcal L B^2}{-i \mathcal L A^2 - i \mathcal L A^2} \Bigg|=\Bigg|\dfrac{2 i \mathcal L B^2}{-2i\mathcal L A^2}\Bigg| = \frac{|B|^2}{|A|^2} = \nonumber\\
&=\frac{|\mathcal{K-L}|^2}{|\mathcal{K+L}|^2} = \frac{(\mathcal{L-K})^2}{(\mathcal{L+K})^2} \longrightarrow \boxed{R = \frac{(\mathcal{L-K})^2}{(\mathcal{L+K})^2}}
\end{align*}

Which expression is better the one with $|~|$ or with $(~)$?

EDIT:
I allso did this for transmission coefficient and would need a confirmation on this one as well.

\begin{align}
T &= \dfrac{|j_{tr}|}{|j_{in}|} \!=\! \Bigg|\dfrac{\dfrac{\hbar }{2mi}\! \left( \dfrac{d\overline{\psi}_{tr1}}{dx}\, \psi_{tr1} - \dfrac{d \psi_{tr1}}{dx}\, \overline{\psi}_{tr1} \right)}{\dfrac{\hbar}{2mi} \!\left( \dfrac{d\overline{\psi}_{in}}{dx}\, \psi_{in} - \dfrac{d\psi_{in}}{dx}\, \overline{\psi}_{in} \right) }\Bigg|\!=\! \Bigg|\dfrac{\frac{d}{dx}\big(\overbrace{Ce^{-i\mathcal K x}}^{\text{konjug.}}\big) Ce^{i\mathcal K x} - \frac{d}{dx} \left( Ce^{i\mathcal K x}\right)\! \overbrace{Ce^{-i\mathcal K x}}^{\text{konjug.}}}{ \frac{d}{dx}\big(\underbrace{Ae^{-i\mathcal L x}}_{\text{konjug.}}\big) Ae^{i\mathcal L x} - \frac{d}{dx} \left( Ae^{i\mathcal L x}\right)\! \underbrace{Ae^{-i\mathcal L x}}_{\text{konjug.}}}\Bigg|\! = \nonumber\\
&=\Bigg|\dfrac{-i\mathcal K C e^{-i\mathcal K x} C e^{i \mathcal K x} - i\mathcal K C e^{i \mathcal K x} Ce^{-i \mathcal K x}}{-i \mathcal L A e^{-i\mathcal L x} Ae^{i \mathcal L x} - i \mathcal L A e^{i \mathcal L x}Ae^{-i \mathcal L x} }\Bigg|=\Bigg|\dfrac{-i\mathcal K C^2 - i\mathcal K C^2}{-i \mathcal L A^2 - i \mathcal L A^2}\Bigg|=\Bigg|\dfrac{-2 i \mathcal K C^2}{-2i\mathcal L A^2}\Bigg|= \frac{\mathcal K}{\mathcal L}\frac{|C|^2}{|A|^2} = \nonumber\\
&=\frac{\mathcal K}{\mathcal L}\frac{|2\mathcal{L}|^2}{|\mathcal{K+L}|^2} = \frac{\mathcal K}{\not{\mathcal L}}\frac{4\mathcal L^{\not{2}}}{(\mathcal{L+K})^2} \longrightarrow \boxed{T = \frac{4\mathcal L \mathcal K}{(\mathcal{L+K})^2}}
\end{align}

I hope these derivations will come handy to someone besides me.

Last edited: Apr 20, 2013
15. Apr 22, 2013

### Fightfish

You can only swap || with () if you are sure that the expressions within the || are real.

16. Apr 22, 2013

### 71GA

Thank you for this note!