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Proving Tr(XY) = Tr(YX) (Sakurai, p. 59, prob. 1.4)

  1. Jul 19, 2010 #1
    1. The problem statement, all variables and given/known data
    You know [itex]{\rm{Tr}}(XY) = \limits^{???} {\rm{Tr}}(YX)[/itex], but prove it, using the rules of bra-ket algebra, sucka. (The late Sakurai does not actually call his reader "sucka").


    2. Relevant equations
    [tex]{\mathop{\rm Tr}\nolimits} (X) \equiv \sum\nolimits_{a'} {\left\langle {a'} \right|X\left| {a'} \right\rangle } [/itex]

    [tex]XY = Z = \left\langle {a''} \right|Z\left| {a'} \right\rangle = \left\langle {a''} \right|XY\left| {a'} \right\rangle = \sum\nolimits_{a'''} {\left\langle {a''} \right|X\left| {a'''} \right\rangle \left\langle {a'''} \right|Y\left| {a'} \right\rangle } [/tex]


    3. The attempt at a solution

    Obviously: XY ≠ YX, but Tr(XY) = Tr(YX). Therefore: there must be something about the trace that allows for this, and not about “X” and “Y”. We are therefore prompted to consider the definition of trace:

    [tex]{\mathop{\rm Tr}\nolimits} (X) \equiv \sum\nolimits_{a'} {\left\langle {a'} \right|X\left| {a'} \right\rangle } [/tex]

    We are also reminded of what matrix multiplication “looks like”:
    [tex]XY = Z = \left\langle {a''} \right|Z\left| {a'} \right\rangle = \left\langle {a''} \right|XY\left| {a'} \right\rangle = \sum\nolimits_{a'''} {\left\langle {a''} \right|X\left| {a'''} \right\rangle \left\langle {a'''} \right|Y\left| {a'} \right\rangle } [/tex]

    So: the trace of this is the sum of the diagonal elements: I now use the First equation in "relevant equations" to say:
    [tex]{\mathop{\rm Tr}\nolimits} (XY) = {\mathop{\rm Tr}\nolimits} \left( {\sum\nolimits_{a'''} {\left\langle {a''} \right|X\left| {a'''} \right\rangle \left\langle {a'''} \right|Y\left| {a'} \right\rangle } } \right) = \sum\nolimits_{b'} {\left( {\sum\nolimits_{a'''} {\left\langle {b'} \right|X\left| {a'''} \right\rangle \left\langle {a'''} \right|Y\left| {b'} \right\rangle } } \right)} [/tex]

    Now: we calculated Tr(XY), but if I calculated Tr(YX) the same way, it seems the traces being equal would be thwarted by XY not being the same as YX.

    Hmmm....
     
  2. jcsd
  3. Jul 19, 2010 #2
    Take your last line and put the <b'|X|a'''> term on the right side, and flip the order of the sums.
     
  4. Jul 19, 2010 #3
    Oh! You can flip the two <||> because <b'|X|a'''> and <a'''|Y|b'> are scalars, is that why?
     
  5. Jul 19, 2010 #4
    This seems a little longer than it should be:

    [tex]\mathrm{Tr}(XY) = \sum_{a'}\langle a'|XY|a'\rangle=\sum_{a',a''}\langle a'|X|a''\rangle\langle a''|Y|a'\rangle = \sum_{a',a''}\langle a''|Y|a'\rangle\langle a'|X|a''\rangle=\sum_{a''}\langle a''|YX|a''\rangle = \mathrm{Tr}(YX)[/tex]

    I think your biggest problem is that you use

    [tex]Z\rightarrow \langle a''|XY|a'\rangle[/tex]

    and somehow you come up with

    [tex]{\sum_{a'''} \left\langle {a''} \right|X\left| {a'''} \right\rangle \left\langle {a'''} \right|Y\left| {a'} \right\rangle } } = \sum_{b'} {\left( {\sum\nolimits_{a'''} {\left\langle {b'} \right|X\left| {a'''} \right\rangle \left\langle {a'''} \right|Y\left| {b'} \right\rangle } } \right)} [/tex]

    Which doesn't seem very good notation.
     
  6. Jul 19, 2010 #5
    Oh, that works very nicely. Thank you. I am trying to prepare for Fall 2010's grad QM course.
     
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