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Basically I have a set of vectors of unit length, $\{\nu_i\}$, describing the movement of phonons (all orthogonal to each other), $\{\omega_i\}$. Lets say I only have two atoms, $m_1$ and $m_2$. In this case with only two atoms I have $N=3\times2=6$ modes. In the following it really doesn't matter how many modes I have.

The displacement of a single atomic mode would then be $\nu_i\times \sqrt{\frac{\hbar}{m_1\omega}}=r$, from the harmonic oscillators characteristic length, $l_i=\sqrt{\frac{\hbar}{m_i\omega}}$, see Quantum Harmonic oscillator, if it were a single mode.

Thus I have a matrix of orthogonal mode vectors, $V=\{\nu_i\times l_i\}$, which can be used to create a displacement in cartesian coordinates, $R=\{r_i\}$ by knowing the vector of phonon displacements, $U=\{u_i\}$. This means that solving the linear equation: \begin{equation} V\cdot U=R \end{equation} I can find the displaced coordinates by knowing the phonon displacement $U$.

For same mass $m_1=m_2$, I have suspected that I can use the reduced mass for both atoms, thus yielding $l_i=\sqrt{\frac{\hbar}{m_i\omega_i/\sqrt2}}$.
But what about $m_1\neq m_2$? I suspected I could use the reduced mass again: \begin{equation} \mu=\frac{m_1m_2}{m_1+m_2}, \end{equation} however, how to apply it for two different mass objects?

If you have a general application $N$ particles I would be much obliged.

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