- #1

July Zou

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- Homework Statement
- below

- Relevant Equations
- below

We have the three-phonon interactions Hamiltonian $$H_{\mathrm{ph}-\mathrm{ph}}=\sum_{i j k} M_{i j k}\left(b_{-i}^{\dagger}+b_i\right)\left(b_{-j}^{\dagger}+b_j\right)\left(b_{-k}^{\dagger}+b_k\right).$$

We will not need the explicit expression for the $$M_{ijk}$$

here, but only note that it is symmetric in all its indices and fulfill $$M_{i j k}=M_{-i-j-k}^*$$ and further assume $$M_{-iij}=0$$

To get the Heisenberg equation of motion

$$ i\hbar \frac{d}{d t} b_i =\left[b_i, H_{ph-ph} \right]$$, we have derived

\begin{equation}

\begin{aligned}

\left[b_i, H_{ph-ph} \right]&=\Big[ b_i, \sum_{i j k} M_{i j k}\left(b_{-i}^{\dagger}+b_i\right) \left(b_{-j}^{\dagger}+b_j\right)\left(b_{-k}^{\dagger}+b_k\right) \Big]\\

&= \sum_{ijk} M_{ijk} \Big(\left[b_i, b_{-i}^\dagger + b_i\right]\left(b_{-j}^\dagger + b_j\right)\left(b_{-k}^\dagger + b_k\right) \\

&\quad + \left(b_{-i}^\dagger + b_i\right)\left[b_i, b_{-j}^\dagger + b_j\right]\left(b_{-k}^\dagger + b_k\right) \\

&\quad + \left(b_{-i}^\dagger + b_i\right)\left(b_{-j}^\dagger + b_j\right)\left[b_i, b_{-k}^\dagger + b_k\right] \Big)\\

&= \sum_{ijk} M_{ijk} \Big(\delta_{i,-i} \left(b_{-j}^\dagger + b_j\right)\left(b_{-k}^\dagger + b_k\right)\\

&\quad +\delta_{i,-j} \left(b_{-i}^\dagger + b_i\right)\left(b_{-k}^\dagger + b_k\right)+ \delta_{i,-k}\left(b_{-i}^\dagger + b_i\right)\left(b_{-j}^\dagger + b_j\right) \Big)\\

&=\sum_{jk} \Big( M_{-ijk} \left(b_{-j}^\dagger + b_j\right)\left(b_{-k}^\dagger + b_k\right)\\

&\quad +\underbrace{M_{-jjk}}_{0} \left(b_{-i}^\dagger + b_i\right)\left(b_{-k}^\dagger + b_k\right)+ \underbrace{M_{-kjk}}_{0}\left(b_{-i}^\dagger + b_i\right)\left(b_{-j}^\dagger + b_j\right) \Big)\\

&=\sum_{jk} M_{-ijk} \left(b_{-j}^\dagger + b_j\right)\left(b_{-k}^\dagger + b_k\right).

\end{aligned}

\end{equation}

However, the result is $$3\sum_{jk} M_{-ijk} \left(b_{-j}^\dagger + b_j\right)\left(b_{-k}^\dagger + b_k\right)$$, where is the factor 3 from? Should it be $$\sum_{ijk}M_{ijk}\delta_{i,-i}=3\sum_{jk}M_{-ijk}?$$

We will not need the explicit expression for the $$M_{ijk}$$

here, but only note that it is symmetric in all its indices and fulfill $$M_{i j k}=M_{-i-j-k}^*$$ and further assume $$M_{-iij}=0$$

To get the Heisenberg equation of motion

$$ i\hbar \frac{d}{d t} b_i =\left[b_i, H_{ph-ph} \right]$$, we have derived

\begin{equation}

\begin{aligned}

\left[b_i, H_{ph-ph} \right]&=\Big[ b_i, \sum_{i j k} M_{i j k}\left(b_{-i}^{\dagger}+b_i\right) \left(b_{-j}^{\dagger}+b_j\right)\left(b_{-k}^{\dagger}+b_k\right) \Big]\\

&= \sum_{ijk} M_{ijk} \Big(\left[b_i, b_{-i}^\dagger + b_i\right]\left(b_{-j}^\dagger + b_j\right)\left(b_{-k}^\dagger + b_k\right) \\

&\quad + \left(b_{-i}^\dagger + b_i\right)\left[b_i, b_{-j}^\dagger + b_j\right]\left(b_{-k}^\dagger + b_k\right) \\

&\quad + \left(b_{-i}^\dagger + b_i\right)\left(b_{-j}^\dagger + b_j\right)\left[b_i, b_{-k}^\dagger + b_k\right] \Big)\\

&= \sum_{ijk} M_{ijk} \Big(\delta_{i,-i} \left(b_{-j}^\dagger + b_j\right)\left(b_{-k}^\dagger + b_k\right)\\

&\quad +\delta_{i,-j} \left(b_{-i}^\dagger + b_i\right)\left(b_{-k}^\dagger + b_k\right)+ \delta_{i,-k}\left(b_{-i}^\dagger + b_i\right)\left(b_{-j}^\dagger + b_j\right) \Big)\\

&=\sum_{jk} \Big( M_{-ijk} \left(b_{-j}^\dagger + b_j\right)\left(b_{-k}^\dagger + b_k\right)\\

&\quad +\underbrace{M_{-jjk}}_{0} \left(b_{-i}^\dagger + b_i\right)\left(b_{-k}^\dagger + b_k\right)+ \underbrace{M_{-kjk}}_{0}\left(b_{-i}^\dagger + b_i\right)\left(b_{-j}^\dagger + b_j\right) \Big)\\

&=\sum_{jk} M_{-ijk} \left(b_{-j}^\dagger + b_j\right)\left(b_{-k}^\dagger + b_k\right).

\end{aligned}

\end{equation}

However, the result is $$3\sum_{jk} M_{-ijk} \left(b_{-j}^\dagger + b_j\right)\left(b_{-k}^\dagger + b_k\right)$$, where is the factor 3 from? Should it be $$\sum_{ijk}M_{ijk}\delta_{i,-i}=3\sum_{jk}M_{-ijk}?$$

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