# Symmetry transformation of the second quantization operator

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!

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.