Diagonalization of Hessian H= PkP$^{-1}$: what do the elements of P matrix physically mean?

This is quoted from *Computational Chemistry* by *Errol J. Lewars*' chapter 2's *Stationary Points & Normal-Mode
Vibrations: ZPE*:



> 
> \begin{align}\mathbf H &=\begin{bmatrix}
>  \dfrac{\partial ^2 E}{\partial q\_1 \partial q\_1} & \dfrac{\partial ^2 E}{\partial q\_1 \partial q\_2} & \dfrac{\partial ^2 E}{\partial q\_1 \partial q\_3} & \dots & \dfrac{\partial ^2 E}{\partial q\_1 \partial q\_9} \\
>  \dfrac{\partial ^2 E}{\partial q\_2 \partial q\_1} & \dfrac{\partial ^2 E}{\partial q\_2 \partial q\_2} & \dfrac{\partial ^2 E}{\partial q\_2 \partial q\_3} & \dots & \dfrac{\partial ^2 E}{\partial q\_2 \partial q\_9}\\
>  \vdots & \vdots & \vdots & \ddots & \vdots \\
>  \dfrac{\partial ^2 E}{\partial q\_9 \partial q\_1} & \dfrac{\partial ^2 E}{\partial q\_9 \partial q\_2} & \dfrac{\partial ^2 E}{\partial q\_9 \partial q\_3} & \dots & \dfrac{\partial ^2 E}{\partial q\_9 \partial q\_9}
> \end{bmatrix}\\ \\ &=\underset{\mathbf P}{\begin{bmatrix}
>  q\_{11} & q\_{12} & q\_{13} & \dots & q\_{19} \\
>  q\_{21} & q\_{22} & q\_{23} & \dots & q\_{29} \\
>  \vdots & \vdots & \vdots & \ddots & \vdots \\
>  q\_{91} & q\_{92} & q\_{93} & \dots & q\_{99}
> \end{bmatrix}} \underset{\mathbf k}{\begin{bmatrix}
>  k\_{1} & 0 & 0 & \dots & 0 \\
>  0 & k\_{2} & 0 & \dots & 0 \\
>  \vdots & \vdots & \vdots & \ddots & \vdots \\
>  0 & 0 & 0 & \dots & k\_{9}
> \end{bmatrix}}\mathbf P^{-1}\end{align}
> 
> 
> The 9 × 9 Hessian for a tri-atomic molecule
>  (three Cartesian coordinates for each atom) is decomposed by diagonalization into a P
>  matrix whose columns are “***direction vectors***” for the vibrations whose force constants
>  are given by the $\bf k$ matrix. Actually, columns 1, 2 and 3 of $\bf P$ and the corresponding $k\_1,k\_2,k\_3$ of $\bf k$ refer to *translational* motion of the molecule (motion of the whole
>  molecule from one place to another in space); these three “force constants” are nearly
>  zero. Columns 4, 5 and 6 of $\bf P$ and the corresponding $k\_4,k\_5,k\_6$ of $\bf k$ refer to 
>  *rotational* motion about the three principal axes of rotation, and are also nearly zero.
>  Columns 7, 8 and 9 of $\bf P$ and the corresponding $k\_7,k\_8,k\_9$ of $\bf k$ are the direction
>  vectors and force constants, respectively, for the normal-mode vibrations: and
>  refer to vibrational modes 1, 2 and 3, while the 7th, 8th, and 9th columns of $\bf P$ are
>  composed of the x, y and z components of vectors for motion of the three atoms in mode
>  1 (column 7), mode 2 (column 8), and mode 3 (column 9).
> 
> 
> 


I couldn't conceive what actually $\bf P$ implies; it is, *of course*, the eigenvector matrix of the Hessian $\bf H $ but what does it represent?


Lewars wrote that columns of $\bf P$ are actually ***direction vectors***; now what did he mean by ***direction vector***? 


I thought $q$ represents the geometric coordinate; I got $k$ is the force-constant; but what are those series of $q$ s in $\mathbf P\;?$ 


Are the $q$ s in $\bf P$ coordinates of the atoms in molecules? If so, aren't they always changing since they are vibrating, rotating...? 


Can anyone please explain this to me what those $q$ s actually represent? What did Lewars want to mean by ***direction vectors***?





> 
> Are the $q$s in $\mathbf{P}$ coordinates of the atoms in molecules? 
> 
> 
> 


Close, but not quite. The $q\_{i,j}$ represent the *displacements from the equilibrium geometry* of the appropriate atom/coordinate $i$, for each mode *j*. Thus, for example, $q\_{5,9}$ corresponds to the displacement from equilibrium of atom $2$ along axis $2$ (typically the $y$-axis) when the molecule is undergoing vibration in mode $9$.



> 
> If so, aren't they always changing since they are vibrating, rotating...?
> 
> 
> 


There are no changes due to translation or rotation. It is customary to "project out" the translational and rotational motion from the Hessian and the geometry. Qualitatively, this means that we simply ignore the translation and rotation and consider the molecule as "fixed" in its equilibrium configuration (allowing for vibrational 'wiggling'), with that configuration always oriented in the same way relative to the coordinate system.


As for the vibrational motion, though: yes, the $q\_{i,j}$ values do technically also 'wiggle' as the molecule vibrates. However, the magnitudes of the perturbations $\delta q\_{i,j}$ are in many cases very small, and can be neglected. In those cases where they cannot, the harmonic oscillator approximation to the vibrational motion breaks down and is no longer a useful model for vibration along that mode.


See my answers [here](https://chemistry.stackexchange.com/questions/41551/computing-accurate-vibrational-and-rotational-contributions-to-the-free-energy-o/41558#41558) and [here](https://chemistry.stackexchange.com/questions/29688/rotational-degrees-of-freedom-3n-5-and-3n-6/33106#33106) for some links to literature on situations where this occurs (emphasis on the 'internal rotation' topics, though some of the 'anharmonic' and 'Coriolis' references might be somewhat relevant). [Here](https://chemistry.stackexchange.com/questions/464/how-come-trigonal-s-and-p-compounds-do-not-undergo-inversion-at-room-temperature) is a Chem.SE question about [nitrogen inversion](http://en.wikipedia.org/wiki/Nitrogen_inversion) that might be of interest. More generally, a Google search for "[low-frequency vibrational modes](https://www.google.com/webhp?sourceid=chrome-instant&ion=1&espv=2&ie=UTF-8#q=low%20frequency%20vibrational%20modes&start=0)" should turn up lots of interesting reading.


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First one should understand what normal coordinates and (mass-weighted) cartesian coordinates are: They are both a means to specify where in space each single atom of a molecule is located. In cartesian coordinates you do this simply by specifying the x,y and z coordinates of each single atom. Normal coordinates are a little trickier to understand, since they include always several atoms.


A simple example might illustrate this: Suppose you have two masses which are connected by a spring and can move only in one direction (the direction of the spring). Also assume that their mass is $m=1$ so we don't have to distinguish between the usual and mass-weighted cartesian coordinates. Then there are two ways to specify the position of the two masses in space: You can specify the x coordinates of the two masses (cartesian coordinates) or you can specify the center of mass and the distance between the two masses (normal modes).
To be more concrete, the dynamical matrix (Hessian of the potential in mass-weighted coordinates) of the above problem is
\begin{equation}
\mathbf H = \begin{pmatrix}
k & -k \\
-k & k
\end{pmatrix}
\end{equation}
where k is the force constant of the spring.
Diagonalization then yields the force constants 0 and 2k and the matrix
\begin{equation}
\mathbf P = \frac{1}{2} \begin{pmatrix}
1 & -1 \\
1 & 1
\end{pmatrix} \end{equation}
The matrix $\mathbf P$ now tells us how to compute the values of the normal coordinates $Q\_i$ from the values of the cartesian coordinates:
\begin{equation}
Q\_i = \sum\_j q\_i P\_{ij}
\end{equation}
From this we see that the first normal coordinate is proportional to the center of mass and the second one proportional to the distance of the two masses. Note that my example used usual cartesian coordinates as $q\_i$ while in real problems you usually use *displacement* coordinates. So you can see that the matrix $\mathbf P$ just let's you express the values for one set of coordinates in terms of the values for the other.


You can of course also do it the other way round and ask: what is the position/displacement of some atom, when I know that the normal coordinates have the values $Q\_i$. Then you can calculate your cartesian coordinates via
\begin{equation}
q\_i = \sum\_j (P^{-1})\_{ji} Q\_j = \sum\_j P\_{ij} Q\_j
\end{equation}


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