# Anti-commutator for annihilation and creation operators: ordering of indices

I'm trying to prove that $$\{\tilde a_i,\tilde a_j^{\dagger} \}=\delta_{ij}$$, by defining $$\tilde a_i=\sum_j \bar U_{ji}a_j$$. U is an unitary matrix and $$a_i$$ refers to an element of the operator $$a$$. Also, the anti-commutator of these operators are defined to be $$\{a_i,a_j^{\dagger}\}=\delta_{ij}$$. So, plugging it it in:

$$\{\tilde a_i,\tilde a_j^\dagger\}=\{\sum_k \bar U_{ki}a_k,\sum_l( \bar U_{lj}a_l)^\dagger \}=\sum_k\sum_l \{U_{ik}^\dagger a_k, (U_{jl}^{\dagger}a_l)^{\dagger} \}$$ We consider the matrix U to be hermitian. Thus: $$\sum_k\sum_l U_{ik}^\dagger U_{jl}\{a_k, a_l^{\dagger} \}=\sum_k\sum_l U_{ik}^\dagger U_{jl}\delta_{kl}=\sum_k U_{ik}^\dagger U_{jl}$$ My problem is the ordering of the indices, which I would like it to be:

$$\sum_kU_{ik}^\dagger U_{kj}=(U^{-1}U)_{ij}=\delta_{ij}$$

Is there some step I've gone wrong?

• Do you want $U$ to be unitary (as in the first line), or hermitian? The two conditions are not the same. Mar 9, 2020 at 19:10

You have $$\sum_l \bar U_{lj} a_l = \sum_l U^\dagger_{jl} a_ l \neq \sum_l \big(U^\dagger_{jl} a_l\big)^\dagger$$
and if you had started from $$\{\bar a_i, \bar a_j^\dagger\}$$, you'd have $$\{\bar a_i, \bar a_j^\dagger\} = \big\{\sum_k \bar U_{ki} a_k, (\sum_l \bar U_{lj} a_l)^\dagger\big\} = \sum_k \sum_l \big\{\bar U_{ki} a_k, (\bar U_{lj} a_l)^\dagger\big\} = \sum_k \sum_l \big\{\bar U_{ki} a_k, U_{lj} a_l^\dagger\big\}$$ In the last equality above we need to remember that as far as hermitian conjugation is concerned $$\bar U_{kl}$$ are just numbers (elements of a matrix), and for $$\lambda$$ being a number and $$A$$ being an operator we have $$(\lambda A)^\dagger = \bar\lambda A^\dagger$$
You could also do with less index notation, if you treat $$a_i$$ as elements of vector $$a$$, and $$\bar a_i$$ as an effect of multiplication of matrix $$U^\dagger$$ with vector $$a$$: $$\bar a_i = \sum_j \bar U_{ji} a_j = \sum_j (U^\dagger)_{ij} a_j \quad \Leftrightarrow \quad \bar a = U^\dagger a$$ Then you also need to rembember that hermitian conjugation switches the order of multiplication $$\bar a^\dagger= (U^\dagger a)^\dagger = a^\dagger U$$ that is $$\bar a^\dagger_j = \sum_l a^\dagger_l U_{lj}$$ Alternatively, you can treat $$a$$ as covector and write $$\bar a = a\bar U$$. The final result is the same.
• Thank you for the answer. Regarding your first statement, I did a mistake when writing my answer in the post. I've now corrected it. But although I agree with your final statement, I thought when doing thelj transpose of $U_{ij}$ one would have to switch the indices, despite being just an element of that matrix, just like when you write $\bar U_{lj}=U_{jl}^\dagger$ Mar 11, 2020 at 15:02
• You could, but then you'd also need to treat matrix $\bar U$ as multiplying vector $a$: $$\bar a_i = \sum_j a_j \bar U_{ji} \quad \Leftrightarrow \quad \bar a = a\bar U$$ and switch the order of multiplication upon hermitian conjugation $$\bar a^\dagger= (a \bar U )^\dagger = U^T a^\dagger$$ that is $$\bar a^\dagger_j = (U^T)_{jl} a^\dagger_l = U_{lj} a_l$$ Mar 11, 2020 at 20:33