The Potential Step - Quantum Mechanics (Additional steps in between needed)?

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    So, I only got 5/7 for this question and I'm not sure where and what the additional steps in between this solution is needed.Name:  1.jpg
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    (Original post by Airess3)
    So, I only got 5/7 for this question and I'm not sure where and what the additional steps in between this solution is needed.Name:  1.jpg
Views: 22
Size:  263.2 KB*Attachment 585736585740*Attachment 585736585740585700
    I guess that it is because you did not show the working of how you combine the two continuity equations meaning solving the two simultaneous equations to obtain T and R respectively.
    It may also be good that make all your workings explicit in future unless you are told by your lecturer otherwise.
    Hope it helps.
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    (Original post by Eimmanuel)
    I guess that it is because you did not show the working of how you combine the two continuity equations meaning solving the two simultaneous equations to obtain T and R respectively.
    It may also be good that make all your workings explicit in future unless you are told by your lecturer otherwise.
    Hope it helps.
    Thanks. Could also explain the (matching psi) and *(matching derivative of psi) part of my answer? I copied that from the board and when I looked back on it, I didn't quite understand how it related to the question.
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    (Original post by Airess3)
    Thanks. Could also explain the (matching psi) and *(matching derivative of psi) part of my answer? I copied that from the board and when I looked back on it, I didn't quite understand how it related to the question.
    I don't quite understand your question.

    Hope the following would help:

    Matching the wavefunction at x = 0:

    wavefunction in A = wavefunction in B

     \psi_A(0) = \psi_B (0)

     \exp( ik(0)) + R \exp(-ik(0)) = T\exp( iq(0))

    Matching the derivative of wavefunction at x = 0:

    derivative of wavefunction in A = derivative of wavefunction in B

     \[{{\left. \frac{d}{dx}{{\psi }_{A}} \right|}_{x=0}}\] =\[{{\left. \frac{d}{dx}{{\psi }_{B}} \right|}_{x=0}}\]

     ik \left( \exp( ik(0)) - R \exp(-ik(0))) \right ) =  iqT\exp( iq(0))

    Imposing these conditions allow us to have a smooth wavefunction.
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    (Original post by Eimmanuel)
    I don't quite understand your question.

    Hope the following would help:

    Matching the wavefunction at x = 0:

    wavefunction in A = wavefunction in B

     \psi_A(0) = \psi_B (0)

     \exp( ik(0)) + R \exp(-ik(0)) = T\exp( iq(0))

    Matching the derivative of wavefunction at x = 0:

    derivative of wavefunction in A = derivative of wavefunction in B

     \[{{\left. \frac{d}{dx}{{\psi }_{A}} \right|}_{x=0}}\] =\[{{\left. \frac{d}{dx}{{\psi }_{B}} \right|}_{x=0}}\]

     ik \left( \exp( ik(0)) - R \exp(-ik(0))) \right ) =  iqT\exp( iq(0))

    Imposing these conditions allow us to have a smooth wavefunction.
    Thanks.
 
 
 
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