- #1
bjnartowt
- 284
- 3
Homework Statement
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").
Homework 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]
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...