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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 $b_k|0\rangle = 0$.

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  • 2
    $\begingroup$ $\uparrow$ Yes. $\endgroup$ – AccidentalFourierTransform Jul 30 at 13:20
  • $\begingroup$ Thank you very much! $\endgroup$ – ewf Jul 30 at 14:08

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