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

[bjnartowt]

Homework Statement

You know {\rm{Tr}}(XY) = \limits^{?} {\rm{Tr}}(YX), but prove it, using the rules of bra-ket algebra, sucka. (The late Sakurai does not actually call his reader "sucka").

Homework Equations

{\mathop{\rm Tr}\nolimits} (X) \equiv \sum\nolimits_{a&#039;} {\left\langle {a&#039;} \right|X\left| {a&#039;} \right\rangle } [/itex]<br /> <br /> XY = Z = \left\langle {a&amp;#039;&amp;#039;} \right|Z\left| {a&amp;#039;} \right\rangle = \left\langle {a&amp;#039;&amp;#039;} \right|XY\left| {a&amp;#039;} \right\rangle = \sum\nolimits_{a&amp;#039;&amp;#039;&amp;#039;} {\left\langle {a&amp;#039;&amp;#039;} \right|X\left| {a&amp;#039;&amp;#039;&amp;#039;} \right\rangle \left\langle {a&amp;#039;&amp;#039;&amp;#039;} \right|Y\left| {a&amp;#039;} \right\rangle }<br /> <br /> <br /> <h2>The Attempt at a Solution</h2><br /> <br /> 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:<br /> <br /> {\mathop{\rm Tr}\nolimits} (X) \equiv \sum\nolimits_{a&amp;#039;} {\left\langle {a&amp;#039;} \right|X\left| {a&amp;#039;} \right\rangle }<br /> <br /> We are also reminded of what matrix multiplication “looks like”:<br /> XY = Z = \left\langle {a&amp;#039;&amp;#039;} \right|Z\left| {a&amp;#039;} \right\rangle = \left\langle {a&amp;#039;&amp;#039;} \right|XY\left| {a&amp;#039;} \right\rangle = \sum\nolimits_{a&amp;#039;&amp;#039;&amp;#039;} {\left\langle {a&amp;#039;&amp;#039;} \right|X\left| {a&amp;#039;&amp;#039;&amp;#039;} \right\rangle \left\langle {a&amp;#039;&amp;#039;&amp;#039;} \right|Y\left| {a&amp;#039;} \right\rangle }<br /> <br /> So: the trace of this is the sum of the diagonal elements: I now use the First equation in &quot;relevant equations&quot; to say:<br /> {\mathop{\rm Tr}\nolimits} (XY) = {\mathop{\rm Tr}\nolimits} \left( {\sum\nolimits_{a&amp;#039;&amp;#039;&amp;#039;} {\left\langle {a&amp;#039;&amp;#039;} \right|X\left| {a&amp;#039;&amp;#039;&amp;#039;} \right\rangle \left\langle {a&amp;#039;&amp;#039;&amp;#039;} \right|Y\left| {a&amp;#039;} \right\rangle } } \right) = \sum\nolimits_{b&amp;#039;} {\left( {\sum\nolimits_{a&amp;#039;&amp;#039;&amp;#039;} {\left\langle {b&amp;#039;} \right|X\left| {a&amp;#039;&amp;#039;&amp;#039;} \right\rangle \left\langle {a&amp;#039;&amp;#039;&amp;#039;} \right|Y\left| {b&amp;#039;} \right\rangle } } \right)}<br /> <br /> 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.<br /> <br /> Hmmm...

 

[chrispb]

Take your last line and put the <b'|X|a'''> term on the right side, and flip the order of the sums.

 

[bjnartowt]

Take your last line and put the <b'|X|a'''> term on the right side, and flip the order of the sums.

Oh! You can flip the two <||> because <b'|X|a'''> and <a'''|Y|b'> are scalars, is that why?

 

[jdwood983]

This seems a little longer than it should be:

\mathrm{Tr}(XY) = \sum_{a&#039;}\langle a&#039;|XY|a&#039;\rangle=\sum_{a&#039;,a&#039;&#039;}\langle a&#039;|X|a&#039;&#039;\rangle\langle a&#039;&#039;|Y|a&#039;\rangle = \sum_{a&#039;,a&#039;&#039;}\langle a&#039;&#039;|Y|a&#039;\rangle\langle a&#039;|X|a&#039;&#039;\rangle=\sum_{a&#039;&#039;}\langle a&#039;&#039;|YX|a&#039;&#039;\rangle = \mathrm{Tr}(YX)

I think your biggest problem is that you use

Z\rightarrow \langle a&#039;&#039;|XY|a&#039;\rangle

and somehow you come up with

{\sum_{a&#039;&#039;&#039;} \left\langle {a&#039;&#039;} \right|X\left| {a&#039;&#039;&#039;} \right\rangle \left\langle {a&#039;&#039;&#039;} \right|Y\left| {a&#039;} \right\rangle } } = \sum_{b&#039;} {\left( {\sum\nolimits_{a&#039;&#039;&#039;} {\left\langle {b&#039;} \right|X\left| {a&#039;&#039;&#039;} \right\rangle \left\langle {a&#039;&#039;&#039;} \right|Y\left| {b&#039;} \right\rangle } } \right)}

Which doesn't seem very good notation.

 

[bjnartowt]

This seems a little longer than it should be:

\mathrm{Tr}(XY) = \sum_{a&#039;}\langle a&#039;|XY|a&#039;\rangle=\sum_{a&#039;,a&#039;&#039;}\langle a&#039;|X|a&#039;&#039;\rangle\langle a&#039;&#039;|Y|a&#039;\rangle = \sum_{a&#039;,a&#039;&#039;}\langle a&#039;&#039;|Y|a&#039;\rangle\langle a&#039;|X|a&#039;&#039;\rangle=\sum_{a&#039;&#039;}\langle a&#039;&#039;|YX|a&#039;&#039;\rangle = \mathrm{Tr}(YX)

[/tex]

Which doesn't seem very good notation.

Oh, that works very nicely. Thank you. I am trying to prepare for Fall 2010's grad QM course.