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classes:2009:fall:phys4101.001:lec_notes_1016 [2009/10/16 17:03] – ykclasses:2009:fall:phys4101.001:lec_notes_1016 [2009/10/17 23:50] (current) – x500_nikif002
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   * How the main points fit with the big picture of QM.  Or what is not clear about how today's points fit in in a big picture.   * How the main points fit with the big picture of QM.  Or what is not clear about how today's points fit in in a big picture.
   * wonderful tricks which were used in the lecture.   * wonderful tricks which were used in the lecture.
 +
 ====Andromeda 10/16 3:00==== ====Andromeda 10/16 3:00====
 +===nikif002 10/17 23:11===
 +My notes match Andromeda's closely, so I added my thoughts and questions in //italic//
 ===Main points for concluding chapter 2=== ===Main points for concluding chapter 2===
   *We mostly were learning how to solve schrodinger's time independent equation given a potential.    *We mostly were learning how to solve schrodinger's time independent equation given a potential. 
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 <math>\ f(x) = \sum_n c_n\psi_n</math> can be written as <math>\ f(x) <=>\normalsize \left(\large\begin{array}{GC+23} \\c_{1}\\c_{2}\\c_{n}\end{array}\right)\ {\Large</math> <math>\ f(x) = \sum_n c_n\psi_n</math> can be written as <math>\ f(x) <=>\normalsize \left(\large\begin{array}{GC+23} \\c_{1}\\c_{2}\\c_{n}\end{array}\right)\ {\Large</math>
 where <math>\ c_n = \int f(x)*\psi_n(x) \,\mathrm dx</math>  where <math>\ c_n = \int f(x)*\psi_n(x) \,\mathrm dx</math> 
 +  *//the reason we can do this is because all stationary states are orthonormal. The stationary states in function space are like the basis vectors in //<math>R^n</math>
   *the above vector can have finite or infinite number of components depending on how many stationary states there are.   *the above vector can have finite or infinite number of components depending on how many stationary states there are.
   *in the language of vectors, operators are transformations and are featured by Matrices. (I think one point of the lecture was the following: by operation on a function you get another function analogous to for example when you apply addition rules on a vector you get another vector which belongs to the original vector space)   *in the language of vectors, operators are transformations and are featured by Matrices. (I think one point of the lecture was the following: by operation on a function you get another function analogous to for example when you apply addition rules on a vector you get another vector which belongs to the original vector space)
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   *in general when we have an arbitrarily wave-function f(x), we get:   *in general when we have an arbitrarily wave-function f(x), we get:
 <math> \hat{H} f(x)=\hat{H} \sum_n c_n\psi_n(x)=\sum_n c_n E_n\psi_n<=>\normalsize \left(\large\begin{array}{GC+23} \\c_{1}E_{1}\\c_{2}E_{2}\\c_{n}E_{n}\end{array}\right)\ {\Large</math>  where <math> \sum_n c_n E_n</math>  is a constant. <math> \hat{H} f(x)=\hat{H} \sum_n c_n\psi_n(x)=\sum_n c_n E_n\psi_n<=>\normalsize \left(\large\begin{array}{GC+23} \\c_{1}E_{1}\\c_{2}E_{2}\\c_{n}E_{n}\end{array}\right)\ {\Large</math>  where <math> \sum_n c_n E_n</math>  is a constant.
 +  *//A little more explanation about this one, since steps were skipped in class. The key point in this expanded explanation is that we can move the Hamiltonian operator inside the sum because it is a linear operator. A linear operation on a linear combination of operands is equal to the same linear combination of the operation on each operand//
 +<math>\hat{H} f(x)=\hat{H}\sum_n c_n\psi_n(x)=\sum_n c_n\hat{H}\psi_n(x)=\sum_n c_n E_n\psi_n</math>
 +
 ===Main points from 3.2=== ===Main points from 3.2===
   *observable quantities like position, momentum, energy and etc. are represented by operators.   *observable quantities like position, momentum, energy and etc. are represented by operators.
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   *average value of a Hermitian operator is always real.once you assume this you can show that the operator is Hermitian meaning that you can show the above equality holds.    *average value of a Hermitian operator is always real.once you assume this you can show that the operator is Hermitian meaning that you can show the above equality holds. 
   *physicist's notation for a dot product is <math>\ <\psi|Q\psi> </math>  where <math>\ \psi </math> is a vector and <math>\ Q\psi </math> is another vector.   *physicist's notation for a dot product is <math>\ <\psi|Q\psi> </math>  where <math>\ \psi </math> is a vector and <math>\ Q\psi </math> is another vector.
 +  *<math>\int \psi(x)*Q\psi(x)\,\mathrm dx =\ <\psi|Q\psi> </math> //In general, the definition of the inner product in Hilbert function space is //<math> <f(x)|g(x)>=\int_{-\infty}^{\infty}f^*(x)g(x)</math>
   *since in quantum mechanics we usually are dealing with complex quantities, we might change the value of a dot product if we change the order of the dot product. if the quantity is complex, we get its complex conjugate back,   *since in quantum mechanics we usually are dealing with complex quantities, we might change the value of a dot product if we change the order of the dot product. if the quantity is complex, we get its complex conjugate back,
-<math>\ <\psi|Q\psi>  = <Q\psi|\psi>* ,</math> and if it is real we are not changing anything by changing the order. +<math>\ <\psi|Q\psi>  = <Q\psi|\psi>* ,</math> and if it is real we are not changing anything by changing the order. //This is an easy way to see that a real expectation value is necessary and sufficient to show that the operator is Hermitian.// 
-  *determinate state is a state with no uncertainty and each measurement of some quantity Q gives bak the same result q. +  *determinate state is a state with no uncertainty and each measurement of some quantity Q gives back the same result q. //In other words, the standard deviation of the observable is 0.// 
- +  *//This is something that confused me. Which operators can we define an eigenvalue for? All linear operators? All Hermitian operators? Certainly we can define some non-linear operators that don't have eigenvalues (such as the squaring operator). Looking at the book, it explains it. Determinate states for an operator Q are states that are eigenfunctions of Q. So //<math>\hat{Q}f(x)=q f(x)</math>//, where q is a constant. So we can define eigenvalues for an operator Q if the function it is operating on is a determinate state of Q. As far as I understand, this means that, for example, all solutions to the Schrodinger equation are determinate states of the Hamiltonian operator.// 
- +  *//Another new piece of notation: //<math>|q_n></math> //indicates the wavefunction corresponding to eigenvalue //<math>q_n</math>//. Such that //<math>\hat{Q} |q_n>=q_n|q_n></math>
- +
-====Main points ==== +
- +
  
  
classes/2009/fall/phys4101.001/lec_notes_1016.1255730607.txt.gz · Last modified: 2009/10/16 17:03 by yk

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