How unitary change of basis related to Trace?

[Shing]

Homework Statement

Shanker 1.7.1
3.)Show that the trace of an operator is unaffected by a unitary change of basis (Equivalently, show [itex]TrΩ=TrU^{\dagger}ΩU[/itex]

Homework Equations

I can show that via Shanker's hint, but I however can't see how a unitary change of basis links to [itex]TrΩ=TrU^{\dagger}ΩU[/itex], (and it really giving me a headache!) Would anyone be kind enough to explain to me?

 

[Square]

There is some identity which tells you that [itex]\mathrm{Tr}(AB) = \mathrm{Tr}(BA)[/itex] (more generally one could state that the trace is invariant under cyclic permutations). Use it and your problem should be as good as solved.

 

[Shing]

Forgive my ignorance,
But shouldn't "unaffected by a unitary change of basis" be expressed as[itex]TrUΩ=TrΩ[/itex]

 

[Dick]

Forgive my ignorance,
But shouldn't "unaffected by a unitary change of basis" be expressed as[itex]TrUΩ=TrΩ[/itex]

No. A matrix A is a change of basis of a matrix B if A=T^(-1)BT for a nonsingular matrix T.

 

[Shing]

Thanks for reply,
So here is my understanding: we map A to B, A and B represent the same thing in different bases, and their mathematical relation is [itex]A=T^{-1}AT[/itex]
am I right?
I have a rough picture now, but Sadly, I still can't see the physical picture, neither the mathematics.
, and shanker surely was telling the truth! One should know a bit more of linear algebra before embrace into it!

Thanks guys, anyway :)

 

[Shing]

Now I got it a bit!
For column vector (1,0) -> (0 ,1)
And T is {(0,1),(1,0)} :)
Did I get it right?

 

[Dick]

Now I got it a bit!
For column vector (1,0) -> (0 ,1)
And T is {(0,1),(1,0)} :)
Did I get it right?

If T is unitary then sure that's an option. So T takes (1,0) -> (0,1) and (0,1) -> (1,0). T^(-1) (which happens to be the same as T, but that's usually not the case) does the opposite. So to figure out what A is 'equivalent' (not equal!) to B, you use T to rotate a vector to the basis of B, then let B act on it, then undo the rotation with T^(-1), so A=T^(-1)BT. I know this is vague. But none of this vagueness should stop you from being able to show Tr(B)=Tr(A)=Tr(T^(-1)BT). That change of bases don't change the trace.