# Unitary transformation of bosonic operators

Consider two sets of bosonic operators $$\{b^{\dagger}_k, b_k\}$$ and $$\{B^{\dagger}_i, B_i\}$$ satisfying $$[b_k,b^{\dagger}_{k^{\prime}}]=\delta_{kk^{\prime}}$$ and $$[B_i,B^{\dagger}_j]=\delta_{ij}$$. The ground state (of the kth mode) is defined according to the action of the annihilator $$b_k$$

$$b_k |0\rangle =0$$

on the bosonic Fock space vacuum $$|0\rangle$$. Consider now the unitary transformation

$$b_k=\displaystyle\sum_i t_{ki}B_i, \hspace{0.5cm} B_i=\displaystyle\sum_k t^{\star}_{ik}b_k \, .$$

Is it correct to show the invariance of the bosonic many-particle ground state $$b_k|0\rangle=0$$ under the unitary transformation introduced above, i.e., $$B_i|0\rangle=0$$ according to the following reasoning

$$B_i| 0 \rangle =\displaystyle\sum_k t^{\star}_{ik}b_k|0\rangle = 0 \, ?$$

The last equality should follow from $$b_k|0\rangle = 0$$.

• $\uparrow$ Yes. – AccidentalFourierTransform Jul 30 at 13:20
• Thank you very much! – ewf Jul 30 at 14:08