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classes:2009:fall:phys4101.001:q_a_1014 [2009/10/14 12:17] – czhangclasses:2009:fall:phys4101.001:q_a_1014 [2009/10/21 11:06] (current) – x500_sohnx020
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 It's as simple as k is chosen to be positive and real by convention.  For the E<0 case, the k you have is real and positive.  For E>0, the second k you have listed is real and positive.  If you are solving the Schroedinger Equation for an area with a nonzero potential, remember to include the V in your determination of whether k is positive or negative. It's as simple as k is chosen to be positive and real by convention.  For the E<0 case, the k you have is real and positive.  For E>0, the second k you have listed is real and positive.  If you are solving the Schroedinger Equation for an area with a nonzero potential, remember to include the V in your determination of whether k is positive or negative.
- 
 ====Liam Devlin 10/13 8:30p==== ====Liam Devlin 10/13 8:30p====
 I have a notation question.  The book gives a fourier trick used to find coefficients for functions that are orthonormal in Hilbert space.  I'm a bit confused by the notation used, it gives  Cn=<fn|f>.  What would this look like in matrix form? I have a notation question.  The book gives a fourier trick used to find coefficients for functions that are orthonormal in Hilbert space.  I'm a bit confused by the notation used, it gives  Cn=<fn|f>.  What would this look like in matrix form?
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 think about Cn is a matrix of constant and fn is a collection of all solutions. think about Cn is a matrix of constant and fn is a collection of all solutions.
 <math>C =  <math>C = 
-\begin{pmatrix}+\begin{Bmatrix}
 c_1\\ c_1\\
 c_2\\ c_2\\
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 .  \\ .  \\
 c_n c_n
-  \end{pmatrix}  </math> +  \end{Bmatrix}  </math>  <math> fn =  
- +\begin{Bmatrix}
-<math> fn =  +
-\begin{pmatrix}+
 f_1\\ f_1\\
 f_2\\ f_2\\
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 .  \\ .  \\
 f_n f_n
-  \end{pmatrix}  </math>+  \end{Bmatrix}  </math>
  
  
-then do corresponding <math> Cn=<f_n|f> =\int f_nf\, dx =   \begin{Bmatrix} +then corresponding each Cn with the element in the matrix, then use orthonormal rule to evaluate each integral   
-    \int f_1*f\, dx          \\ + 
-    \int f_2*f\, dx          \\+ 
 + <math> \begin{Bmatrix} 
 +c_1\\ 
 +c_2\\ 
 +.  \\ 
 +.  \\ 
 +c_n 
 +  \end{Bmatrix} =<f_n|f> =\int f_nf\, dx =   \begin{Bmatrix} 
 +   \int f_1*f\, dx          \\ 
 +   \int f_2*f\, dx          \\
     \int f_n*f\, dx=\delta_n_m\\     \int f_n*f\, dx=\delta_n_m\\
     .\\     .\\
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    \int f_n*f\, dx    \int f_n*f\, dx
    \end{Bmatrix}</math>    \end{Bmatrix}</math>
- 
  
 ====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?
 +
 +
 +
 +==== 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". 
 +
 +=== 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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 **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_1014.1255540661.txt.gz · Last modified: 2009/10/14 12:17 (external edit)

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