Is showing $\text{Tr}\Omega U = \text{Tr} \Omega$ equivalent to showing $\text{Tr}\Omega = \text{Tr}(U^{\dagger} \Omega U)$?

Question

On pg. 30 of Principles of Quantum Mechanics (by Shankar), we are asked:

Show that the trace of an operator is unaffected by a unitary change of basis $|i \rangle \rightarrow U |i\rangle$. [Equivalently, show $\text{Tr}\Omega = \text{Tr}(U^{\dagger} \Omega U)$].

I have already convinced myself that

$$ \text{Tr} \Omega = \text{Tr}(U^{\dagger} \Omega U) $$

by using the fact that

$$ \text{Tr} ( \Omega \Lambda ) = \text{Tr} ( \Lambda \Omega ) $$

implies that

$$ \text{Tr} \Omega = \text{Tr}(\Omega (U U^{\dagger})) = \text{Tr}((\Omega U) U^{\dagger}) = \text{Tr}(U^{\dagger} (\Omega U)) = \text{Tr}(U^{\dagger} \Omega U) $$

as desired.

Question: But why is showing this equivalent to showing that the trace of an operator is unaffected by a unitary change of basis $|i \rangle \rightarrow U | i \rangle$? It seems to me this means that

$$ \text{Tr} \Omega = \text{Tr}(\Omega U) $$

which is a distinct condition from showing that

$$ \text{Tr} \Omega = \text{Tr}(U^{\dagger} \Omega U) $$

Answer 1

It seems as though you have a fundamental misunderstanding of what a "change of basis" means in this context. In short: as copper.hat says, you need to change the basis in the domain and in the range.

The operator $\Omega$ of interest is represented by a matrix whose entries are $\omega_{ij}$. This means that $$ \Omega |j \rangle = \sum_{i=1}^n \omega_{ij} |i \rangle. $$ In other words, the $j$th column of the matrix encodes the destination of the $j$th basis vector. Correspondingly, if $M$ is the matrix of $\Omega$ relative to the new basis $U |1 \rangle,\dots U|n\rangle$ and has entries $m_{ij}$, then we should have $$ \Omega (U |j \rangle) = \sum_{i=1}^n m_{ij} (U |i \rangle). $$ We can relate this expression to the original as follows: if we multiply both sides by $U^\dagger$, then we have $$ U^\dagger \Omega (U |j \rangle) = \sum_{i=1}^n m_{ij} U^\dagger U|i \rangle \implies\\ (U^\dagger \Omega U) |j \rangle = \sum_{i=1}^n m_{ij} |i \rangle. $$ In other words, saying that $M$ is the matrix of the operator $\Omega$ relative to the new basis $\{U |i \rangle\}$ is the same as saying that $M$ is the is the matrix of the operator $U^\dagger \Omega U$ relative to the original basis $\{|i \rangle\}$.

Answer 2

In an orthonormal basis, an operator $\Omega=\sum_{ij}\Omega_{ij}|i\rangle\langle j|$ has trace $\sum_i\Omega_{ii}=\sum_i\langle i|\Omega|i\rangle$, which becomes transformed to $\sum_i\langle i|U^\dagger\Omega U|i\rangle$.