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$\textbf{Context :}$ If the context helps, then the things motivated me to ask this question are action of squezing operator or displacement operator on density matrix in quantum optics and Bogolioubov transformation acting on thermal state in theory of superconductivity, Lang-Firsov transformation (Polaron transformation for spin-boson problem) in physics of dissipative quantum systems and many such canonical transformations in quantum many body physics.

Suppose we have an operator algebra ($\mathcal{O}$) consisting of operators $\{\hat{a}_{i}^{},\hat{a}_{i}^{\dagger}\}$ and all possible linear combinations (coefficients from the complex field) of arbitrary products of $\{\hat{a}_{i}^{},\hat{a}_{i}^{\dagger}\}$ acting on a Hilbert space ($\mathcal{H}$) defined over complex field.

Let $\mathbb{U}[\{\hat{a}_{i}^{},\hat{a}_{i}^{\dagger}\}]$ be an unitary operator (a function of $\{\hat{a}_{i}^{},\hat{a}_{i}^{\dagger}\}$) acting on $\mathcal{H}$ and further suppose $\mathbb{Z}[\{\hat{a}_{i}^{},\hat{a}_{i}^{\dagger}\}]$ be an arbitrary operator (a function of $\{\hat{a}_{i}^{},\hat{a}_{i}^{\dagger}\}$) also acting on $\mathcal{H}$.

Then when (conditions neccessary for $\mathbb{Z}$ and $\mathbb{U}$) is the following identity in $\mathcal{O}$ is true :

$$\mathbb{U}_{}^{}\mathbb{Z}[\{\hat{a}_{i}^{},\hat{a}_{i}^{\dagger}\}]\mathbb{U}_{}^{\dagger}\stackrel{\textbf{?}}{=}\mathbb{Z}[\{\mathbb{U}_{}^{}\hat{a}_{i}^{}\mathbb{U}_{}^{\dagger},\mathbb{U}_{}^{}\hat{a}_{i}^{\dagger}\mathbb{U}_{}^{\dagger}\}].$$

$\textbf{Note :}$ If $\mathbb{Z}[\{\hat{a}_{i}^{},\hat{a}_{i}^{\dagger}\}]$ is taylor expandable, we can clearly see the above identity.

Qmechanic
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Sunyam
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  • Could you clarify your notation? For instance, is $\mathbb{U}[{\hat a_i,\hat a_i^\dagger}]$ a function of $\hat a_i$ and $\hat a_i^\dagger$? Likewise, presumably your $\mathbb{Z}$ is not completely arbitrary but some function of the arguments? Do you allow for polynomials such as $\hat a_i \hat a_j^\dagger$ or are you limited to “diagonal” operators with $i=j$? – ZeroTheHero Nov 15 '17 at 02:35
  • @ZeroTheHero Yes $\mathbb{U}({\hat{a}{i}^{},\hat{a}{i}^{\dagger}})$ and $\mathbb{Z}({\hat{a}{i}^{},\hat{a}{i}^{\dagger}})$ are functions of ${\hat{a}{i}^{},\hat{a}{i}^{\dagger}}$. I am looking for a general case of $\mathbb{Z}({\hat{a}{i}^{},\hat{a}{i}^{\dagger}})$. In case $\mathbb{Z}({\hat{a}{i}^{},\hat{a}{i}^{\dagger}})$ is taylor expandable, I can proove the above identity. – Sunyam Nov 15 '17 at 04:48
  • @Sunyam Then what does $\mathbb{U}\hat{a_i}\mathbb{U}^\dagger$ mean? – probably_someone Nov 15 '17 at 05:00
  • @probably_someone It means this : $\mathbb{U}{}^{}\hat{a}{i}^{}\mathbb{U}=\mathbb{U}{}^{}[{\hat{a}{i}^{},\hat{a}{i}^{\dagger}}]\hat{a}{i}^{}\mathbb{U}{}^{\dagger}[{\hat{a}{i}^{},\hat{a}_{i}^{\dagger}}]$. – Sunyam Nov 15 '17 at 06:51
  • Would [math.se] be a better home for this question? – Qmechanic Nov 15 '17 at 08:10
  • @Qmechanic I thought about it. I will post it there too. I will leave the question here in the hope to see more canonical transformations relevant in physics and the examples where it doesn't hold. – Sunyam Nov 15 '17 at 10:16
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  • How is $\mathbb{U}$ supposed to be a function of the $a,a^\dagger$ - if it is unitary, there are only very specific "functions" of them that are unitary, and it is not clear what you gain by expressing these unitaries as functions of the $a_i$ instead of just leaving them as is. 2. How are you defining $\mathbb{Z}$ if it is not Taylor expandable in the first place? Since the $a,a^\dagger$ are not self-adjoint, functional calculus via eigenvalues won't work. 3. Please do not crosspost questions (i.e. if you wish to ask this in math.SE, say so and we will move this question there)
    • – ACuriousMind Nov 15 '17 at 10:17
  • @ACuriousMind 1. Example $\mathbb{U}=e^{\sum_{i,j}^{}H_{ij}^{}\hat{a}{i}^{\dagger}a{j}^{}}{}$ with operators ${a{i}^{},a_{i}^{\dagger}}$ satisfying bosonic commutation relationship and matrix $(H_{ij}^{})$ being anti-hermitian. 2. Is it not allowed to define non-analytic function of operators (what restricts in considering only analytic functions of operators)? 3. I had cross posted the question at Mathematics Stackexchange (before seeing your reply) already, will delete it there. Wait for responses it here for few days and then you can migrate the question. 4. Thanks for inputs. – Sunyam Nov 15 '17 at 10:39
  • Ah, but is it relevant for this question that the unitary is such a function? Are you really only interested in the case where the unitary is such a specific function and not in the behaviour under general unitary transformations? 2. Well, sure it is allowed, but you have to specify how you do it - in the absence of a Taylor series, one needs to use things like Borel functional calculus to properly define the application of a function to a (in that case self-adjoint) operator.
    • – ACuriousMind Nov 15 '17 at 10:50
  • @ACuriousMind Thank you for reply. 1. I was looking for general class of unitary operators, the above one is a simple example. 2. I Will have to read more of operator theory and your suggestion of Borel functional calculus. – Sunyam Nov 15 '17 at 10:58
  • Related: https://physics.stackexchange.com/q/194436/2451 and links therein. – Qmechanic Nov 15 '17 at 11:49