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Equations with bras and kets

  1. Sep 14, 2014 #1

    dyn

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    When re-arranging equations with bras and kets I was under the impression that the order of the bras and kets had to be maintained unless they formed an inner product ie. just a complex number in which case they could be moved around ? Is this the case ? As I am confused about the following equation I found
    Tr |u><u|v><v| = <v|u><u|v>
    In this example the bra <v| seems to have jumped from the end to the start.
     
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  3. Sep 14, 2014 #2

    Fredrik

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    $$\operatorname{Tr}|u\rangle\langle u|v\rangle\langle v| =\sum_i \langle i|u\rangle\langle u|v\rangle\langle v|i\rangle = \sum_i \langle u|v\rangle\langle v|i\rangle\langle i|u\rangle = \langle u|v\rangle\langle v|\bigg(\sum_i |i\rangle\langle i\bigg)|u\rangle =\langle u|v\rangle\langle v|u\rangle$$
     
  4. Sep 14, 2014 #3

    dyn

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    Thanks for that but i'm confused about the first step. At first I thought I was the imaginary number but it looks like an integer you are summing over but I don't understand how it can be brought into the equation.
     
  5. Sep 14, 2014 #4

    Fredrik

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    Sorry about that. Yes, it's an integer. I should at least have started with this statement: Let ##\{|i\rangle\}_{i=1}^\infty## be any orthonormal basis.

    If the Hilbert space is finite-dimensional, replace ##\infty## with ##\dim\mathcal H## (where ##\mathcal H## denotes the Hilbert space). The first step in that calculation is just the definition of the trace of a linear operator.

    You may want to also take a look at https://www.physicsforums.com/showthread.php?t=694922 [Broken] about the relationship between linear operators and matrices.
     
    Last edited by a moderator: May 6, 2017
  6. Sep 14, 2014 #5

    dyn

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    Thanks. I understand it now. Could you just confirm for me that my original statement is correct , ie the order of bras and kets does matter and they can only be rearranged in order if they form an inner product ?
     
  7. Sep 15, 2014 #6

    Fredrik

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    Yes, it's correct. If you move them around, you may end up with an entirely different object than the one you started with, as in this case, where |u><u|v><v| is a linear operator and <u|v><v|u> is a number.
     
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