Proof of a property of the trace of a product of two operators, using Dirac bra-ket notation

Question

I want to show that Tr(ΩΛ)=Tr(ΛΩ). My attempt: $$\mathrm{Tr}(\Omega\Lambda)=\sum_{i}(\Omega\Lambda)_{ii}=\sum_{i}(\sum_{k} \Omega_{ik}\Lambda_{ki})=\sum_{i}(\sum_{k}\langle i|\Omega |k\rangle\langle k|\Lambda |i\rangle)=\sum_{i}(\sum_{k}\langle k|\Lambda| i\rangle\langle i|\Omega |k\rangle)= \sum_{k}(\sum_{i}\langle k|\Lambda| i\rangle\langle i|\Omega |k\rangle)=\sum_{k}\langle k|\Lambda\mathbb{I}\Omega |k\rangle=\sum_{k}\langle k|\Lambda\Omega |k\rangle=\sum_{k}(\Lambda\Omega)_{kk}=\mathrm{Tr}(\Lambda\Omega)$$ The commutation of the fourth equality is because they are scalars and scalars can commute. Then I changed the order of summation but I'm not pretty sure that this is correct. Could you give some advice to improve my proof ? Note: This exercise belongs to a chapter titled Active and Passive Transformations, which says something about unitary operators, I don't see how to use them in order to verify the above relation, though.

Answer 1

Let $A=(a_{ij})$ be an $n \times n$ matrix.

Then, the sum of elements of $i$-th row of $A$ is $\sum_j\limits a_{ij}$.

Thus the sum of all elements of $A$ is $\sum\limits_i \sum\limits_j a_{ij}$.

Similarly, the sum of elements of $j$-th columm of $A$ is $\sum\limits_i a_{ij}$.

Thus the sum of all elements of $A$ is $\sum\limits_j \sum\limits_i a_{ij}$.

Hence,

$$\sum\limits_i \sum\limits_j a_{ij} = \sum\limits_j \sum\limits_i a_{ij}.$$

Answer 2

Your proof is correct, although you can directly conclude $\Omega_{ik} \Lambda_{ki} = \Lambda_{ki} \Omega_{ik}$ without any bra-kets! (This then boils down to the proof given by Doyun). You might also want to place some parentheses to indicate how you obtain the identity matrix at step 6.

(And some mathematicians might object to the lack of ranges on your indices.)