# 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!

## 1 Answer

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.