Schrodinger question

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    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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    (Original post by Airess3)
    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.
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    (Original post by Eimmanuel)
    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.*Name:  new post.jpg
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    (Original post by Airess3)
    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.*Name:  new post.jpg
Views: 24
Size:  501.9 KB
    The wavefunction is different in different regions, so the differentiation should be done on the wavefunction in region A when applying the time-independent Schrodinger equation (TISE).

    You are differentiating a weird total wavefunction and applying the TISE to obtain the energy which is incorrect.. I cannot really read how you differentiate the wavefunction, I suspect your differentiation is incorrect, too.

    Don't mean to be offensive or rude in asking you these question - how good is your calculus?

    Note : You will be doing alot of differentiation and integration in Quantum mechanics if it is dealing with wavefunction (hope you are aware of it).
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    I did not complete the whole solution.
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    For d), it's worth 7 marks but I can't think of any other steps but I got to the right final answer.*Name:  work.jpg
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    (Original post by Airess3)
    For d), it's worth 7 marks but I can't think of any other steps but I got to the right final answer.*Name:  work.jpg
Views: 22
Size:  511.3 KB*oAttachment 586646586648
    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.
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    (Original post by Eimmanuel)
    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?

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    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?
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    (Original post by Airess3)
    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 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_{incide  nt}}| ----- equation (1)

    (Original post by Airess3)
    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?

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    Why did you equate jright to jleft?

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

    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?
    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 jincident.
    2) Find the jtransmitted.
    3) Use equation and simplified the expression
 
 
 
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