Hessian Matrix is a square matrix containing the elements as the second-order partial derivatives of energy-function of a molecule; the derivative is done with respect to geometric coordinates of the molecule.

However, I couldn't get why Hessian matrix is referred to as force-constant matrix; as is done in the book I'm following viz. Computational Chemistry: Introduction to the Theory and Applications of Molecular and Quantum Mechanics by Errol G. Lewars and in many other sources.

If the energy function is a quadratic function which is mostly the case for a molecule following 1D PES; the function which a Hookean spring follows for small displacement, then the second derivative does indeed give the value of force constant, that is $$\begin{align}E- E_0&= k(q-q_0)^2\\ \implies \frac{\mathrm{d}^2 E}{\mathrm{d}q^2}&= 2k\;.\end{align}$$

However, for molecules following multidimensional PES, most of the time, the energy function is not simple quadratic; how can the second order derivative give the force constant? Second order derivatives vary from point to point unlike the simple case above.

So, can anyone tell why Hessian matrix is called force-constant matrix?

  • $\begingroup$ I didn't know what to tag it: since I got the issue related to computational chemistry, I tagged it so. I wanted to tag it as soft-question; but it doesn't exist here:( $\endgroup$
    – user5764
    Commented Jan 21, 2016 at 13:58
  • $\begingroup$ Excuse me if this comment is more of a question than a comment, my background is math rather than chem, and, in math, Hessian matrix has a far more abstract definition than it is for chem, guessing by your question. So for the outsiders, how is your Hessian matrix defined? (Link preferred above explanation). $\endgroup$ Commented Jan 21, 2016 at 18:20
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    $\begingroup$ @GyroGearloose bioinfo.rpi.edu/bystrc/courses/biol4550/lecture20.pdf (please see slide 20-29) $\endgroup$ Commented Jan 21, 2016 at 21:12
  • $\begingroup$ Thanks, so "your" Hessian is about the energy of a molecule by displacement of the atoms, right? $\endgroup$ Commented Jan 22, 2016 at 17:42
  • $\begingroup$ The Hessian is calculated in the equilibrium geometry when used for force constants and such. The fact that the reality is non quadratic, it does not influence the fact that we use quadratic models (or else it wouldn't be constant, would it be) - most approximations still use the harmonic approach for the force fields. $\endgroup$
    – Greg
    Commented Feb 17, 2016 at 8:21

1 Answer 1


You are right that multidimensional PES is not simple quadratic, but close to the PES well minima it can be approximated as harmonic oscillator. Also if you really want to include the real PES you can include anharmonicity in your calculation. In that case you have to include an anharonic constant $x_e\;.$ Your PES can be written as $$E=\left(n+\frac{1}{2}\right)h\nu-x_e\left(n+\frac{1}{2}\right)^2h\nu$$

Usually in computational chemistry the energy can be expanded into a Taylor series $$E=E_0+\frac{\mathrm{d}E}{\mathrm{d}x}(x-x_0)+\frac{1}{2}\frac{\mathrm{d}^2E}{\mathrm{d}x^2}(x-x_0)^2+\textrm{higher order terms}\;.$$ And you can usually neglect those first order term and higher order terms for most of the practical problems. And then your PES will follow the trajectory of a harmonic oscillator.

  • $\begingroup$ Thanks for the answer.... could you please tell how you came up with the formula above the expansion? $\endgroup$
    – user5764
    Commented Jan 21, 2016 at 17:13
  • $\begingroup$ Please see the attached link: chemwiki.ucdavis.edu/Physical_Chemistry/Quantum_Mechanics/… $\endgroup$ Commented Jan 21, 2016 at 17:16
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    $\begingroup$ @mamun The Taylor expansion is missing a 1/2 $\endgroup$
    – Greg
    Commented Feb 17, 2016 at 8:24
  • $\begingroup$ yes, you are right. I am changing it. $\endgroup$ Commented Feb 17, 2016 at 15:07

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