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classes:2009:fall:phys4101.001:q_a_1014 [2009/10/14 19:21] – x500_hakim011classes:2009:fall:phys4101.001:q_a_1014 [2009/10/21 11:06] (current) – x500_sohnx020
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    \int f_n*f\, dx    \int f_n*f\, dx
    \end{Bmatrix}</math>    \end{Bmatrix}</math>
-====Andromeda 10/14 7:00Pm==== 
-I was going over mondays lecture and was wondering how did we get D=-A and C=-B?  was it just the fact that we were looking for odd solution which is Antisymmetric? and since we know this we could eliminate some unknowns this way. and another thing is that the imaginary k still bothers me a little and it came up today as well. what does it really (physically) mean to have an imaginary p?  
-====Andromeda 10/14 7:16Pm==== 
-one more thin; we say that because the potential is symmetric we can reduce the problem and look at the continuity of psi and its derivative only at one point (a or -a). would this work if the potential was -V between 0 and 2a rather than -v between a and -a. (and 0 everywhere else) ? 
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 ====John Galt 10/14  6:21 AM==== ====John Galt 10/14  6:21 AM====
 Also in response to Physics4dummies' question, with a question of my own:  Also in response to Physics4dummies' question, with a question of my own: 
 From what I understand, this problem is supposed to crudely represent an electron's state when positioned around two nuclei. Since the energy of the electron in the ground state goes up as the separation between the wells goes up, and since it wants to exist at lower energy, a force exists to push the two nuclei (wells) together. In the first excited state, the energy is lowered as the separation increases, so the force acts to pull the nuclei apart. My question is why does the energy of the electron go down as the wells are pulled apart in the second state, but up when it is in the first state? From what I understand, this problem is supposed to crudely represent an electron's state when positioned around two nuclei. Since the energy of the electron in the ground state goes up as the separation between the wells goes up, and since it wants to exist at lower energy, a force exists to push the two nuclei (wells) together. In the first excited state, the energy is lowered as the separation increases, so the force acts to pull the nuclei apart. My question is why does the energy of the electron go down as the wells are pulled apart in the second state, but up when it is in the first state?
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 +==== Aspirin 10/17/09 2:45 pm ====
 +It is a probably fundamental question.. 
 +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". 
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 +=== 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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classes/2009/fall/phys4101.001/q_a_1014.1255566093.txt.gz · Last modified: 2009/10/14 19:21 by x500_hakim011

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