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

1. Jul 19, 2010

### bjnartowt

1. The problem statement, all variables and given/known data
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").

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

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:

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

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

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

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. Jul 19, 2010

### chrispb

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

3. Jul 19, 2010

### bjnartowt

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

4. Jul 19, 2010

### 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.

5. Jul 19, 2010

### bjnartowt

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