| Both sides previous revisionPrevious revisionNext revision | Previous revision |
| classes:2009:fall:phys4101.001:lec_notes_0916 [2009/09/20 11:39] – yk | classes:2009:fall:phys4101.001:lec_notes_0916 [2009/09/20 19:08] (current) – yk |
|---|
| * <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. |
| | == How to find C_i's <del>Linear Combination of Stationary States</del> == |
| |
| == Linear Combination of Stationary States == | * In general//, if we accept the idea that a set of stationary state wave functions forms a "complete" set, any wave functions can be expressed as// <math>\psi(x) = \sum_n c_n\psi_n</math> |
| | * //A big questions may be, "How can we find out what //<math>c_i</math>//'s are, knowing what the function// <math>\psi(x)</math> // is?"// |
| | * //To answer that question, Griffiths talks about "Fourier's trick."// |
| | * //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> |
| | * 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> |
| |
| * In general: <math>\psi(x) = \Sigma c_n\psi_n</math> | * //We have not learned this yet, but it turns out that in QM, doing this: //<math>\int\psi_n^* ({\rm wave function})dx</math>// to the wave function// is <del>essentially</del> the dot product, where the subscript //n// implies the particular component (ie, x, y, z, 1, 2, 3, etc) |
| * Thus, we can say: <math>\int\psi_m^*(\psi_n(x))dx = \int\psi_m^*(\Sigma c_n\psi_n)dx</math> | * //For example, the dot product between //<math>\psi_n(x)</math> //and// <math>\psi_m(x)</math>// would be //<math><m|n>=\int\psi^*_n(x)\psi^_m(x)dx=0</math> //unless// <math>m=n</math>. //Note that in QM, a notation, //<math><m|n></math>// is used to represent the dot product between two wave functions. Vectors are represented only by the indeces, m and n, here. We say that the two wave functions are orthogonal. The dot product of two orthogonal vectors is zero, too.// |
| * Since <math>\Sigma c_n</math> is a set of constants, we can pull it out such that: | * //Borrowing the idea used in the vector decomposition above, we will take a dot product of the vector equation above and //<math>\psi_m</math>// and get //<math>\int\psi_m^*(\psi(x))dx = \int\psi_m^*(\sum_n c_n\psi_n)dx</math> |
| * For <math>m \neq n</math>, we have <math>\Sigma c_n\int\psi_m^*\psi_ndx = 0</math> for different indices | * //Since //<math>\sum_n c_n</math>// is a set of constants, we can pull it out of the integral. Since due to orthogonality:// |
| * For <math>m = n</math>, we have <math>\Sigma c_n\int\psi_{m=n}^*\psi_ndx = 1</math> | * For <math>m \neq n</math>, we have <math>\int\psi_m^*\psi_ndx = 0</math> and |
| * This is due to orthogonality | * For <math>m = n</math>, we have <math>\int\psi_{m=n}^*\psi_ndx = 1</math> |
| * The only terms to survive are those where <math> m = n</math>, leaving us <math> \Sigma_n c_n\int\psi_m^*\psi_ndx = c_m</math> | * The only term in the sum which survives is the term where <math> m = n</math>, leaving us <math> \sum_n c_n\int\psi_m^*\psi_ndx = c_m</math> |
| * Thus, our left hand side of the equation becomes: <math>LHS = \int\psi_m^*\psi(x)dx = c_m</math> | * Thus, our left hand side of the equation becomes: <math>LHS = \int\psi_m^*\psi(x)dx = c_m</math> |
| * With: <math>\psi(x) = \Sigma c_n\psi_n</math> | * With: <math>\psi(x) = \sum_n c_n\psi_n</math> |
| * This is similar to finding the x component of <math>\vec{r}</math>, which we can represent by <math>(\vec{r}\cdot\hat{x})</math> | |
| * Here, <math>\int\psi_n^* (~)dx</math> is essentially the dot product, where the subscript n implies the particular component (ie, x, y, z, 1, 2, 3, etc) | |
| |
| **To go back to the lecture note list, click [[lec_notes]]**\\ | **To go back to the lecture note list, click [[lec_notes]]**\\ |