Schrodinger question

I don't know if I'm doing this right and I don't know the next steps. The end answer should be E = hbar^2 * k^2 / (2m) on the lhs and E = hbar^2 * q^2 / (2m) + V on the rhs. *

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You are given the wavefunction in (b) . Differentiate it twice wrt x and replace the second derivative in the equation. Remember that V is different in different regions.

I did that but somehow I ended up with a strange answer for region A. Didn't finish the one for region B because it will also turn out strange. I thought of substituting the original eqs for the psi symbols but it still wouldn't be the right answers.*

For d), it's worth 7 marks but I can't think of any other steps but I got to the right final answer.**o

You are on the right track of using the current density formula. Why are you not simplifying it. The later part seems that you are just copying your question answer, instead of arriving the correct answer. I have pm you a link on a set of lecture videos which have all the queries that you have. I would really suggest that you talk your lecturer and seek feedback from them to know what skills are you lacking.

I asked the lecturer and he said to put psi-exp(ikx) on the incident current (left hand side) but I don't know why.

I've multiplied out the brackets and subtracted on the left hand side. Do I need to multiply all the right hand side and left hand side by h-bar/2im? or is it ok to leave it like this?

He says he wants me to show a calculation with complex e-functions to get T(c.c)*exp(-iqx)*iq*T*exp(ipx)=|T|^2*iq . How does that come about?

I am not sure which psi are you referring. You really need to know what does each term in the wavefunction mean, if not I really doubt you can even pass your exam - "plug and play" method would not work in quantum mechanics.

The wavefunction in region A has two terms - what do these two terms represent?
The wavefunction in region B has only one term - why?

The particle current (or sometime people would call it current density) j helps to find the transmission coefficient through the following "definition":

$T =| \frac{j_{transmitted}}{j_{incident}}|$ ----- equation (1)



Why did you equate j_right to j_left?

You should apply the formula of j to incident wavefunction and transmitted wavefunction independently.



Not sure what you mean? It does not make sense. How did you get rid of the exp() terms.


To do it problem, it involves three steps:

1) Find the j_incident.
2) Find the j_transmitted.
3) Use equation and simplified the expression

I've watched the set of videos that you sent me about the solutions. I've redone 7d), the final answer is wrong but I'm not sure why.
Region A is jincident and Region B is jtransmitted.

Your working is a bit confusing. In the same equation, it seems that you have two different T that described two different quantities. You are going to confuse whoever is marking your script. It will hinder you from finding the mistake especially if you are new to it.

Why do you rederive the T that you have shown (I think) in part b? T= 2k/(q+k)
Look back to your part (b) working to see how T is derived.

But anyway it seems that you have learned to differentiate stuff properly. Your algebra is still a bit weak or I should say you are not careful enough.

Hope you will keep trying to improve. Quantum mechanics is a difficult field to learn. You will learn more if you make mistakes. However, you should not view mistakes in learning new stuff as you are not good enough or stupid or some negative remarks on yourself.

I'm trying to find out how the transmission coefficient is (q/k)(4k^2/(q+k)^2). As asked in the question. But it didn't come out with the right answer.

The T in your modulus function (|T|) is already found in (b), so don't have to rederive it. It is again called know your stuff.

At the bottom of one of your images, you found that transmission coefficient (I would label it as T_c) to be

T_c = (q/k)*|T|^2 / |I|^2

As mentioned before, T is found to be 2k/(q+k) in (b). So just use it and substitute in T_c. Again this algebra manipulation, so be careful in your working.

ok, I've substituted them.