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?

  • 1
    $\begingroup$ Do you want $U$ to be unitary (as in the first line), or hermitian? The two conditions are not the same. $\endgroup$ Mar 9, 2020 at 19:10

1 Answer 1


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.

  • $\begingroup$ 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 $\bar U_{lj}=U_{jl}^\dagger$ $\endgroup$
    – RicardoP
    Mar 11, 2020 at 15:02
  • $\begingroup$ 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 $$ \bar a^\dagger_j = (U^T)_{jl} a^\dagger_l = U_{lj} a_l$$ $\endgroup$ Mar 11, 2020 at 20:33

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