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classes:2009:fall:phys4101.001:q_a_1002 [2009/10/02 07:35] – x500_razi0001classes:2009:fall:phys4101.001:q_a_1002 [2009/10/04 06:12] (current) – yk
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-====John Galt====+====John Galt 9/28 18:06====
  
 What sections is the quiz covering again?  What sections is the quiz covering again? 
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 I believe it is just up to and including section 2.3. I believe it is just up to and including section 2.3.
 +
 ====poit0009 9/30 16:20==== ====poit0009 9/30 16:20====
  
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 You can find <math>\psi_1</math> first and go from there.  You'll just have to keep track of the <math>\frac{1}{\sqrt{n}}</math> terms. You can find <math>\psi_1</math> first and go from there.  You'll just have to keep track of the <math>\frac{1}{\sqrt{n}}</math> terms.
  
-=== Andromeda ===+=== Andromeda 09/30 19:33 ===
 you can do it both ways. you can operate on <math>\psi_0</math> once and get <math>\psi_1</math> and operate on <math>\psi_1</math> to get <math>\psi_2</math> or you can use (a+)^2 (which expands to give you 4 terms) and operate on <math>\psi_0</math>.  you can do it both ways. you can operate on <math>\psi_0</math> once and get <math>\psi_1</math> and operate on <math>\psi_1</math> to get <math>\psi_2</math> or you can use (a+)^2 (which expands to give you 4 terms) and operate on <math>\psi_0</math>. 
 ====poit0009 10/1 10:27==== ====poit0009 10/1 10:27====
  
 Back for another question.  How do we find the probability of getting a specific energy (last problem of the practice quiz)? Back for another question.  How do we find the probability of getting a specific energy (last problem of the practice quiz)?
-===Andromeda===+===Andromeda 10/1 11:54===
 The probability of finding the particle to have the energy corresponding to a specific state is <math>C_n^2</math>.   The probability of finding the particle to have the energy corresponding to a specific state is <math>C_n^2</math>.  
 ===Hydra 10/1 3:50pm=== ===Hydra 10/1 3:50pm===
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 Thank you.  That was my first inclination, but I thought it was asking for something more complex. Thank you.  That was my first inclination, but I thought it was asking for something more complex.
  
-====Andromeda====+====Andromeda 10/1 12:28====
 Where is the square root of one comes from when calculating <x> in problem 2.13 in the solution?  Where is the square root of one comes from when calculating <x> in problem 2.13 in the solution? 
 ===Hydra 10/1  4pm=== ===Hydra 10/1  4pm===
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 I like your idea, Green Suit, it's easiest for me to think of it that way, too.  I'll get a bit more specific: I like your idea, Green Suit, it's easiest for me to think of it that way, too.  I'll get a bit more specific:
  
-Start with the time-dependent Schrodinger equation: <math>i\hbar\frac{\partial\Psi}{\partial x} = -\frac{\hbar^2}{2m}\frac{\partial^2\Psi}{\partial x^2} + V\Psi</math>.  Assuming that the wavefunction <math>\Psi</math> can be described by a product of two one-variable functions gives us separable solutions:+Start with the time-dependent Schrodinger equation: <math>-i\hbar\frac{\partial\Psi}{\partial x} = -\frac{\hbar^2}{2m}\frac{\partial^2\Psi}{\partial x^2} + V\Psi</math>.  Assuming that the wavefunction <math>\Psi</math> can be described by a product of two one-variable functions gives us separable solutions:
  
 <math>\Psi(x,t) = \psi(x)\varphi(t)</math> <math>\Psi(x,t) = \psi(x)\varphi(t)</math>
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 Other random thoughts: Other random thoughts:
  
-The general solution for any time-independent potential is a linear combination of all of the stationary states (<math>\phi_n(x)</math>'s).  You can also add on the time function and take a linear combination of all <math>\Psi_n(x)</math>'s to get the time-dependent solution.+The general solution for any time-independent potential is a linear combination of all of the stationary states (<math>\phi_n(x)</math>'s - //when you think about a linear comb of solutions, it makes sense only for time-dependent Schrodinger equation, not time-independent version because for the latter, each solution has unique and different //<math>E_n's</math>// on the RHS of the equation, so adding solutions with different //<math>E_n's</math>// won't satisfy the Schrodinger equation.  i.e. linear combination of eigenfunctions are not an eigenfunction (of the Hamiltonian).  Therefore, this should be //<math>\Psi_n(x)</math>'s - //Yuichi//).  <del>You can also add on the time function and take a linear combination of all <math>\Psi_n(x)</math>'s to get the time-dependent solution.</del>
  
 ==== Mercury 10/02/2009 1:11am ==== ==== Mercury 10/02/2009 1:11am ====
 Does anybody know what the operators for x and p are in terms of the raising and lowering operators? I didn't write it down and forgot what the constants in front were. Does anybody know what the operators for x and p are in terms of the raising and lowering operators? I didn't write it down and forgot what the constants in front were.
  
-==== Super Hot Guy 10/02/2009 7:03 in the AM====+=== Super Hot Guy 10/02/2009 7:03 in the AM===
  
 Eq. 2.69 in the book shows them as: Eq. 2.69 in the book shows them as:
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 Sorry, haven't had time to figure out math formatting yet.  By the way how can I get this response to indent? Sorry, haven't had time to figure out math formatting yet.  By the way how can I get this response to indent?
 +
 ==== Super Hot Guy 10/02/2009 6:38am ==== ==== Super Hot Guy 10/02/2009 6:38am ====
  
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 The answer should say Cn=0 for n even except for n=2. The answer should say Cn=0 for n even except for n=2.
  
-Notice how the final eq. for Cn has an n^2-4 term in the denominator, so it doesn't work for n=2, since the denominator would be zero. So get Cn for n=2 you can plug n=2 into the original expression for Cn, which winds up being pretty straightforward since you get a sin squared term right away instead of sina * sinb.+Notice how the final eq. for Cn has an n^2-4 term in the denominator, so it doesn't work for n=2, since the denominator would be zero. To get Cn for n=2 you can plug n=2 into the original expression for Cn, which winds up being pretty straightforward since you get a sin squared term right away instead of sina * sinb.
 In fact, the second level, n=2, winds up having the largest contribution. In fact, the second level, n=2, winds up having the largest contribution.
  
classes/2009/fall/phys4101.001/q_a_1002.1254486927.txt.gz · Last modified: 2009/10/02 07:35 by x500_razi0001

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