Hybridised orbitals are linear combinations of atomic orbitals of same or nearly-same energies. Atomic orbitals interfere constructively or destructively to give rise to a new orbital which is what we call hybridised orbital.

This is the definition I'm quite acquainted with. But I couldn't understand one thing. What are $c_1,c_2,c_3,\ldots?$ For instance,

$$\psi_{sp^3}= c_1\psi_{2s}+ c_2\psi_{2p_{x}} + c_3\psi_{2p_y}+ c_4\psi_{2p_{z}}.$$

I've read many books one of which state that these coefficients determine the directional properties of the hybrid while other sources tell these coefficients are normalizing constants that is; $$c_1^2 + c_2^2 + c_3^2 + \cdots = 1.$$

But what is the necessity of the sum of the square of the coefficients to be equal to $1?$

Here is the quote:

\begin{align} ψ_1 &= c_{1,1} φ_1 + c_{1,2} φ_2 + ... + c_{1,n} φ_n\\ ψ_2 &= c_{2,1} φ_1 + c_{2,2} φ_2 + ... + c_{2,n} φ_n\\ \vdots\\ ψ_n &= c_{n,1} φ_1 + c_{n,2} φ_2 + ... + c_{n,n} φ_n \end{align} Here $n$ atomic orbitals (with their wave functions $φ_1, φ_2, ..., φ_n$) are used to construct n hybrid orbitals ($ψ_1, ψ_2, ..., ψ_n$) through a linear combination, where the coefficients $c_{1,1}, c_{1,2}, ..., c_{n,n}$ are normalization constants that must fulfil some requirements:

Hybrid orbitals must be normal: $$ c_{1,n}^2 = c_{1,1}^2 + c_{1,2}^2 + ... + c_{1,n}^2 = 1$$

I then compared the above with these to quantum superposed state $$|\psi\rangle= |1\rangle c_1 + |2\rangle c_2$$ where $|1\rangle,|2\rangle$ are orthogonal states. Here $c_1^2 + c_2^2= 1.$

So, is hybridization a superposition?

Can anyone please explain what these coefficients are actually meant for? Why should their square add to $1?$


3 Answers 3


Yes, hybridization is superposition.

Assuming the basis functions are orthonormal, which we usually take to be the case, $\langle\phi_i\,|\,\phi_j\rangle = \delta_{i,j}$ the condition for normalization of $\psi$ is that the sum of the squares of the coefficients equal unity. More precisely, $$c_1^*c_1+c_2^*c_2+\cdots = 1$$ You should check the math to see that, indeed,$$\langle\,\psi|\,\psi\rangle=1$$

  • $\begingroup$ Do you think MO is also quantum superposition? $\endgroup$
    – user36790
    Feb 27, 2016 at 8:53
  • $\begingroup$ What do you mean by MO? Why are you making me guess? $\endgroup$
    – garyp
    Feb 27, 2016 at 13:36
  • 1
    $\begingroup$ Really sorry. I mean Molecular Orbital by MO. $\endgroup$
    – user36790
    Feb 27, 2016 at 14:47
  • $\begingroup$ That's a good question. I've waited some time hoping someone else would answer. I would say that it hybridization is MO, but we tend to think of it differently. MO theory tries to find wave functions that yield good agreement with experiment, and in doing so the wave functions become quite complex. Hybridization has the same goal, but we normally think of hybridization as involving a small number of (nearly) degenerate atomic orbitals. $\endgroup$
    – garyp
    Mar 4, 2016 at 14:50
  • $\begingroup$ Thanks, sir! I wish I could have a chat with you on this. Tell me if you have time. $\endgroup$
    – user36790
    Mar 4, 2016 at 14:57

Yes, hybridization is just that the hybrid state $\psi$ is a superposition of the different orbital states $\phi_1,\dots,\phi_n$.

Since the orbital states $\phi_i$ are assumed to be normalized ($\lvert\lvert \phi_i \rvert\rvert^2 = 1$) and orthogonal to each other, for the hybrid state to be normalized the squares of the coefficients must sum to $1$.


This is quantum mechanics, my friend.

The statement simply says that one hybridized orbital consists of many "pure" orbitals. In your first equation, one hybrid orbital has four pure orbitals $2s, 2p_x, 2p_y, 2p_z$. The coefficients in front of each term can be thought of as how much of one particular kind of pure orbital can be found in the final hybrid orbital.

But the coefficient itself has no physical meaning. Its square does. The square of a coefficient gives the probability of finding you hybrid state in that one particular pure orbital.

For a concrete example, consider a simpler statement.

$\psi_H = \frac{1}{\sqrt{2}}(\psi_{2s}+\psi_{2p_x})$

If you make a measurement, half of the time (square of $\frac{1}{\sqrt{2}}$) you will get $2s$ and the other half $2p_x$.

Since the sum of all probabilities should be unity, you have the sum of the square of all coefficients to be one.