What is the origin of the factor 3?

In summary, the three-phonon interactions Hamiltonian is given by a sum over indices and is symmetric. To derive the Heisenberg equation of motion, we take the commutator of ##b_i## with the Hamiltonian. After simplification, we get a result with a factor of 3, which can be explained by a simplification in the previous step.
  • #1
July Zou
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Homework Statement
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Relevant Equations
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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 $$\sum_{ijk}M_{ijk}\delta_{i,-i}=3\sum_{jk}M_{-ijk}?$$
 
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  • #2
July Zou said:
$$
\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]\\

\end{aligned}
$$
You should not use ##i## as a summation index if you are getting the equation of motion for ##b_i## with some fixed ##i##.
 
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Related to What is the origin of the factor 3?

What is the origin of the factor 3 in physics equations?

The factor 3 often appears in physics equations due to the three spatial dimensions in our universe. For example, in statistical mechanics, the factor 3 appears in expressions involving kinetic energy because particles can move in three independent directions: x, y, and z.

Why does the factor 3 appear in the equation for the degrees of freedom in a gas?

The factor 3 in the degrees of freedom for a monatomic gas arises because each atom can move in three perpendicular directions (x, y, and z). This leads to three translational degrees of freedom, each contributing to the energy of the system.

How does the factor 3 relate to the equipartition theorem?

The equipartition theorem states that each degree of freedom contributes (1/2)kT to the total energy, where k is the Boltzmann constant and T is the temperature. For a monatomic gas, there are three translational degrees of freedom, resulting in a total energy contribution of (3/2)kT per particle, hence the factor 3.

What is the significance of the factor 3 in the context of the Stefan-Boltzmann law?

The factor 3 does not directly appear in the Stefan-Boltzmann law, which relates the total energy radiated per unit surface area of a black body to the fourth power of its temperature. However, the law's derivation involves integrating over all possible directions of radiation in three-dimensional space, where the three spatial dimensions play a crucial role.

Can the factor 3 be explained through dimensional analysis?

Yes, dimensional analysis can often reveal why a factor of 3 appears in certain equations. For example, in the context of kinetic theory, the factor 3 arises from considering the three dimensions of space when calculating quantities like pressure or energy density, ensuring that the units are consistent across the equations.

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