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

[tex]{\mathop{\rm Tr}\nolimits} (X) \equiv \sum\nolimits_{a'} {\left\langle {a'} \right|X\left| {a'} \right\rangle } [/itex]<br /> <br /> $$XY = Z = \left\langle {a''} \right|Z\left| {a'} \right\rangle = \left\langle {a''} \right|XY\left| {a'} \right\rangle = \sum\nolimits_{a'''} {\left\langle {a''} \right|X\left| {a'''} \right\rangle \left\langle {a'''} \right|Y\left| {a'} \right\rangle }$$<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 /> $${\mathop{\rm Tr}\nolimits} (X) \equiv \sum\nolimits_{a'} {\left\langle {a'} \right|X\left| {a'} \right\rangle }$$<br /> <br /> We are also reminded of what matrix multiplication “looks like”:<br /> $$XY = Z = \left\langle {a''} \right|Z\left| {a'} \right\rangle = \left\langle {a''} \right|XY\left| {a'} \right\rangle = \sum\nolimits_{a'''} {\left\langle {a''} \right|X\left| {a'''} \right\rangle \left\langle {a'''} \right|Y\left| {a'} \right\rangle }$$<br /> <br /> So: the trace of this is the sum of the diagonal elements: I now use the First equation in "relevant equations" to say:<br /> $${\mathop{\rm Tr}\nolimits} (XY) = {\mathop{\rm Tr}\nolimits} \left( {\sum\nolimits_{a'''} {\left\langle {a''} \right|X\left| {a'''} \right\rangle \left\langle {a'''} \right|Y\left| {a'} \right\rangle } } \right) = \sum\nolimits_{b'} {\left( {\sum\nolimits_{a'''} {\left\langle {b'} \right|X\left| {a'''} \right\rangle \left\langle {a'''} \right|Y\left| {b'} \right\rangle } } \right)}$$<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...[/tex]

 

[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:

$$\mathrm{Tr}(XY) = \sum_{a'}\langle a'|XY|a'\rangle=\sum_{a',a''}\langle a'|X|a''\rangle\langle a''|Y|a'\rangle = \sum_{a',a''}\langle a''|Y|a'\rangle\langle a'|X|a''\rangle=\sum_{a''}\langle a''|YX|a''\rangle = \mathrm{Tr}(YX)$$

I think your biggest problem is that you use

$$Z\rightarrow \langle a''|XY|a'\rangle$$

and somehow you come up with

$${\sum_{a'''} \left\langle {a''} \right|X\left| {a'''} \right\rangle \left\langle {a'''} \right|Y\left| {a'} \right\rangle } } = \sum_{b'} {\left( {\sum\nolimits_{a'''} {\left\langle {b'} \right|X\left| {a'''} \right\rangle \left\langle {a'''} \right|Y\left| {b'} \right\rangle } } \right)}$$

Which doesn't seem very good notation.

 

[bjnartowt]

This seems a little longer than it should be:

$$\mathrm{Tr}(XY) = \sum_{a'}\langle a'|XY|a'\rangle=\sum_{a',a''}\langle a'|X|a''\rangle\langle a''|Y|a'\rangle = \sum_{a',a''}\langle a''|Y|a'\rangle\langle a'|X|a''\rangle=\sum_{a''}\langle a''|YX|a''\rangle = \mathrm{Tr}(YX)$$

[/tex]

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.