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classes:2009:fall:phys4101.001:q_a_1014

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classes:2009:fall:phys4101.001:q_a_1014 [2009/10/17 15:06] – x500_choxx169classes:2009:fall:phys4101.001:q_a_1014 [2009/10/21 11:06] (current) – x500_sohnx020
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 When we got the general solution(<math>\Psi_I_I(x)=B*exp(ikx) + C*exp(-ikx))</math> in the region II for the finite square well which is from the lectures on last Mon & Wed,  B = -C since <math/> Psi_II(0) = 0</math>.   I don't understand why <math>\Psi_I_I(x=0)</math> becomes "zero".  When we got the general solution(<math>\Psi_I_I(x)=B*exp(ikx) + C*exp(-ikx))</math> in the region II for the finite square well which is from the lectures on last Mon & Wed,  B = -C since <math/> Psi_II(0) = 0</math>.   I don't understand why <math>\Psi_I_I(x=0)</math> becomes "zero". 
  
 +=== Blackbox 10/21/09 10:45 am ===
 +There are two different approaches about this problem. The first one is Odd case and the other is Even case. In the lecture, Professor considered of Odd case, which is asymmetric. The Shrodinger equation of the second region can be expressed like B`sin(kx). Therefore we will get B = -C since <math/> Psi_I_I(0) = 0</math>.   
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 **Return to Q&A main page: [[Q_A]]**\\ **Return to Q&A main page: [[Q_A]]**\\
 **Q&A for the previous lecture: [[Q_A_1012]]**\\ **Q&A for the previous lecture: [[Q_A_1012]]**\\
 **Q&A for the next lecture: [[Q_A_1016]]** **Q&A for the next lecture: [[Q_A_1016]]**
classes/2009/fall/phys4101.001/q_a_1014.1255810010.txt.gz · Last modified: 2009/10/17 15:06 by x500_choxx169

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