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As we know, under the symmetry operation $U$, the operator $\hat A$ and the state $|\alpha \rangle $act as $$\hat A \longrightarrow U\hat A U^{\dagger}$$ $$|\alpha \rangle \longrightarrow U|\alpha\rangle$$ However, for the second quantization operator, i.e. $c^{\dagger}$ or $b^{\dagger}$, sometimes they transform like a state. For example, suppose we rotate the local spin basis(cf. Fradkin Chapter 2.2): $$c'_\sigma(r)=U_{\sigma\sigma'}c_{\sigma'}(r)$$ where $U$ is a $2 \times 2 $ SU($2$) matrix.

Or, under U($1$) symmetry transform: $$c'_\sigma(r)=e^{i\theta}c_{\sigma}(r)$$ Both of above case implies that the second quantization operator act like a state under symmetry transform.

However, under the transform between the Schrödinger picture and Heisenberg picture: $$c_{(H)}=e^{i\hat Ht} c_{(s)} e^{-i\hat Ht}$$ in this case, the second quantization operator act like a operator under transform. And I know that the above boson/fermion operator sometimes are just the combination of the "normal" operator, e.g. for harmonic oscillator $\hat a^{\dagger} = \sqrt{\frac{m\omega}{2\hbar}} (\hat x+\frac{i}{m\omega}\hat p)$. So I think the second quantization operator should follow the transform style of "normal" operator naturally.

In summary, I wonder that what kind of style of symmetry transformation on the second quantization operator, i.e. operator or state? And are there some mistakes about my argument above? Thanks!

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Creation and Annihilation Operators $a,a^\dagger$ also transform by the rule

$a \mapsto UaU^\dagger, a^\dagger \mapsto Ua^\dagger U^\dagger$!

But: Often creation/Annihilation Operators are applied not solely; mostly These are combined with states. Therefore, if These Operators are combined with states, they seem to behave like $a \mapsto Ua$. Example: Let $|\alpha>$ be a state. After symmetry Transformation it must clearly hold:

$|\alpha> \mapsto U|\alpha>$

Now, the state can be expressed in Terms of 2nd quantization Operators; namely a product of creators, say $a^\dagger$ acting on vacuum state $|0>$. Let $|\alpha> = \prod_j a_j^\dagger|0>$. We can recast above symmetry Transformation for states simply by defining that the vacuum state does not Change after symmetry Transformation, i.e. $U|0>=|0>$ (this is the reason that sometimes second quantization Operators seem to transform like states). When we use the rules for Operators, we will have

$|\alpha> = U a_1^\dagger U^\dagger U a_2^\dagger U^\dagger U a_3^\dagger U^\dagger \dots a_n^\dagger U|0>$

and finally, $UU^\dagger=1$ leads to the desired state Transformation rule.

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