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| classes:2009:fall:phys4101.001:lec_notes_0916 [2009/09/20 19:07] – yk | classes:2009:fall:phys4101.001:lec_notes_0916 [2009/09/20 19:08] (current) – yk |
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| * <del>Giving</del> //If you calculate// <math>\sigma^2 \equiv <\hat{H}^2> - <\hat{H}>^2 </math>//you will find it to be zero. If you don't see it immediately, you should think about this until you get zero.// | * <del>Giving</del> //If you calculate// <math>\sigma^2 \equiv <\hat{H}^2> - <\hat{H}>^2 </math>//you will find it to be zero. If you don't see it immediately, you should think about this until you get zero.// |
| * <del>Since <math>\sigma_E^2 = 0</math>,</del> This implies that there is no fluctuation: You get the **same** measurement every time. | * <del>Since <math>\sigma_E^2 = 0</math>,</del> This implies that there is no fluctuation: You get the **same** measurement every time. |
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| == How to find C_i's <del>Linear Combination of Stationary States</del> == | == How to find C_i's <del>Linear Combination of Stationary States</del> == |
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| * //I would say, this is analogous to finding the //x// component of some vector //<math>\vec{r}</math>, which we can obtain by calculating <math>\vec{r}\cdot\hat{x}</math>//. This method may sound overly complex because when //<math>\vec{r}=\[x, y, z \]</math>//, there is no need to come up with a more complex way to figure out what their x, y and z components are. They are obviously x, y and z. However, we sometimes encounter an example of a vector space where no basis vectors are defined (wave functions are one such example) so that we can express the vector as //<math>\vec{r}=\[x, y, z \]</math>//. The method we are discussing right now works even in such a case, or when the basis vector used to define the vector space and the basis along which you want to decompose a vector are different, as long as the dot product is defined, which is always the case (since the dot product is the way we define the lengths of vectors.)// | * //I would say, this is analogous to finding the //x// component of some vector //<math>\vec{r}</math>, which we can obtain by calculating <math>\vec{r}\cdot\hat{x}</math>//. This method may sound overly complex because when //<math>\vec{r}=\[x, y, z \]</math>//, there is no need to come up with a more complex way to figure out what their x, y and z components are. They are obviously x, y and z. However, we sometimes encounter an example of a vector space where no basis vectors are defined (wave functions are one such example) so that we can express the vector as //<math>\vec{r}=\[x, y, z \]</math>//. The method we are discussing right now works even in such a case, or when the basis vector used to define the vector space and the basis along which you want to decompose a vector are different, as long as the dot product is defined, which is always the case (since the dot product is the way we define the lengths of vectors.)// |
| * //Suppose you have succeeded in decomposing a vector to its components, the following should hold:// <math>\vec{r} = x\hat{x}+y\hat{y}+z\hat{z}</math> | * //Suppose you have succeeded in decomposing a vector to its components, the following should hold:// <math>\vec{r} = x\hat{x}+y\hat{y}+z\hat{z}</math> |
| * This equation is a vector equation, which means that there are effectively 3 (or whatever the dimension of the vector space is) equations. To find the values of //x//, //y// and //z//, it may be easier if we convert it into individual equations. Taking a dot product with this vector equation is one way to accomplish this goal. If we take dot product with <math>\hat{x}</math>, for example, we will get <math>(\hat{x}\cdot \vec{r} = \hat{x}\cdot(x\hat{x}+y\hat{y}+z\hat{z})</math>. | * This equation is a vector equation, which means that there are effectively 3 (or whatever the dimension of the vector space is) equations. To find the values of //x//, //y// and //z//, it may be easier if we convert it into individual equations. Taking a dot product with this vector equation is one way to accomplish this goal. If we take dot product with <math>\hat{x}</math>, for example, we will get <math>\hat{x}\cdot \vec{r} = \hat{x}\cdot(x\hat{x}+y\hat{y}+z\hat{z})</math>. |
| * If one realizes that <math>\hat{x}\cdot \hat{x}=1</math>, <math>\hat{x}\cdot \hat{y}=0</math> and <math>\hat{x}\cdot \hat{z}=0,</math>, this is just //x//. //i.e.// <math>x=\hat{x}\cdot \vec{r}</math> | * If one realizes that <math>\hat{x}\cdot \hat{x}=1</math>, <math>\hat{x}\cdot \hat{y}=0</math> and <math>\hat{x}\cdot \hat{z}=0,</math>, this is just //x//. //i.e.// <math>x=\hat{x}\cdot \vec{r}</math> |
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