classes:2009:fall:phys4101.001:q_a_0909
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| classes:2009:fall:phys4101.001:q_a_0909 [2009/09/13 22:47] – created yk | classes:2009:fall:phys4101.001:q_a_0909 [2009/09/23 14:01] (current) – x500_baka0010 |
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| === Anaximenes 18:34 - 9/9/09=== | === Anaximenes 18:34 - 9/9/09=== |
| About the relationship between diagonalization and spanning: Griffiths says on page 452 in Appendix A just below equation A.80 that "...a matrix is diagonalizable if and only if its eigenvectors span the space." This is a feature of linear algebra in general, not just as applied to quantum mechanics. I or someone else could probably dig up a formal proof if you want, but think of it this way: each vector in a set that can't be expressed as a linear combination of the other vectors in the set represents a dimension. Your set of vectors spans the space if and only if the set of vectors describes the same number of dimensions as the space. The vectors in a diagnalized matrix clearly can't be expressed as sums of the other elements, and the number of elements is equal to the number of dimensions in the space. | About the relationship between diagonalization and spanning: Griffiths says on page 452 in Appendix A just below equation A.80 that "...a matrix is diagonalizable if and only if its eigenvectors span the space." This is a feature of linear algebra in general, not just as applied to quantum mechanics. I or someone else could probably dig up a formal proof if you want, but think of it this way: each vector in a set that can't be expressed as a linear combination of the other vectors in the set represents a dimension. Your set of vectors spans the space if and only if the set of vectors describes the same number of dimensions as the space. The vectors in a diagnalized matrix clearly can't be expressed as sums of the other elements, and the number of elements is equal to the number of dimensions in the space. |
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| | === Jake22 20:26 - 9/14/09=== |
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| | If an nxn matrix A has less than n distinct eigenvalues, then A may or may not be diagonalizable. Whether or not it is diagonalizable is a question of whether there exists an eigenbasis for A. To find this we have to find the eigenvalues of A and corresponding eigenspaces. A is diagonalizable iff the dimensions of the eigenspaces add up to n. If we then form an eigenbasis S by concatenating the bases of these eigenspaces, the matrix D=(S^-1)AS is diagonal, and its ith entry is the eigenvalue associated with the ith eigenbasis. |
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| ==== Schrödinger's Dog - 08:04 9/8/09==== | ==== Schrödinger's Dog - 08:04 9/8/09==== |
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| As you can see, the uncertainty in p is proportional to the uncertainty in λ. | As you can see, the uncertainty in p is proportional to the uncertainty in λ. |
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classes/2009/fall/phys4101.001/q_a_0909.1252900058.txt.gz · Last modified: 2009/09/13 22:47 (external edit)