In my previous Phys.SE question, I asked for why $ \newcommand{\k}[2]{\langle #1|#2 \rangle} c_1,c_2,c_3,c_4$ in $$\psi_{sp^3}= c_1\psi_{2s}+ c_2\psi_{2p_{x}} + c_3\psi_{2p_y}+ c_4\psi_{2p_{z}}$$ would add to 1 when their squares are summed in order to ascertain whether hybridisation is a quantum superposition.

Now, I know that hybridised orbital is a quantum superposition which implies

$$c_1= \k{\psi_{2s}}{\psi_{sp^3}}, \\ c_2= \k{\psi_{2p_x}}{\psi_{sp^3}},\\ c_3= \k{\psi_{2p_y}}{\psi_{sp^3}},\\ c_4= \k{\psi_{2p_z}}{\psi_{sp^3}};$$ that is, each coefficient is the amplitude for the electron to go from $|\psi_{sp^3}\rangle$ to the respective atomic orbitals.

But, they are more than just amplitudes as said by Peter Atkins in his explanation:

The coefficients in the hybrid have been chosen to give the correct directional properties of the hybrid. The squares of the coefficients give the proportion of each atomic orbital in the hybrid.

Also, the diagram below also shows how for different values of coefficients, the spatial orientation is changed.

orientation of orbitals

The snapshot above is taken from Principles of Physical Chemistry By Hans Kuhn, Horst-Dieter Försterling, David H. Waldeck

I haven't found any deduction as to how the coefficients determine the spatial orientation. I thought they are just amplitudes to go from the hybridised state to the pure orbital state; how can they direct the spatial orientation? In fact, why would they have to direct the direction of the orbitals. They are probability amplitudes, aren't they? I just want to know how to deduce that these probability amplitudes are actually influencing the direction. How can it be deduced mathematically?

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    $\begingroup$ You have a question, but what's puzzling to me is that the answer is in the text that you pasted in from the book. Other puzzles: why do you use the words "to go from" in describing the role of the coefficients? What do you imagine is going, and from where to where is it going? The basis functions have shapes, and when you add several together you get some other shape. Can you clarify your difficulty? $\endgroup$ – garyp Oct 11 '15 at 15:41
  • $\begingroup$ @garyp: What I meant to say is that $\k{b}{a}$ is the amplitude of an electron in state $a$ to get into state $b$. $\endgroup$ – user36790 Oct 11 '15 at 15:46
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    $\begingroup$ "Get into". This language sounds like a recipe for problems. What exactly does that mean? The coefficients are exactly what they look like: the amplitude ("amount") that a basis function contributes to a wave function. $\endgroup$ – garyp Oct 11 '15 at 17:16
  • $\begingroup$ @garyp: Hmmm.... I was comfortable with this language, If it bothers you, then sorry. Even Feynman in his lecture uses the same phrase. $\endgroup$ – user36790 Oct 11 '15 at 17:23
  • $\begingroup$ Here we have static basis functions, and a time-independent wave function. Nothing is going or getting anywhere. In Feynman's example, atoms are going and getting places. But in any event, the interpretation of the amplitude is the same: the amplitude for finding a system in one state given that it has been prepared in another. Mathematically, in this case, it's the amplitude for the contribution of a basis state to the actual state. Feynman is a little loose with language, but I'll bet if you read his QM book you will find that he will be more clear. $\endgroup$ – garyp Oct 11 '15 at 17:34

You start with:

$$ \psi_{sp^3}= c_1\psi_{2s}+ c_2\psi_{2p_{x}} + c_3\psi_{2p_y}+ c_4\psi_{2p_{z}} \tag{1} $$

Since $\psi_{sp^3}$ is normalised we know that:

$$ \langle \psi_{sp^3} | \psi_{sp^3} \rangle = 1 $$

and if we use equation (1) to substitute for $\psi_{sp^3}$ we get:

$$ \langle c_1\psi_{2s}+ c_2\psi_{2p_{x}} + c_3\psi_{2p_y}+ c_4\psi_{2p_{z}} | c_1\psi_{2s}+ c_2\psi_{2p_{x}} + c_3\psi_{2p_y}+ c_4\psi_{2p_{z}} \rangle = 1 $$

and expanding gives:

$$ \langle c_1\psi_{2s} | c_1\psi_{2s} \rangle + \langle c_1\psi_{2s} | c_2\psi_{2p_{x}} \rangle \, + ... + \, \langle c_4\psi_{2p_{z}} | c_3\psi_{2p_{y}} \rangle + \langle c_4\psi_{2p_{z}} | c_4\psi_{2p_{z}} \rangle = 1 $$

I've skipped the middle twelve terms because, well, life's too short and it should be obvious what they are. We can take the constants $c$ out of the brackets to get:

$$ c_1^2\langle \psi_{2s}|\psi_{2s} \rangle + c_1c_2\langle\psi_{2s}|\psi_{2p_{x}} \rangle \, + ... + \, c_4c_3\langle\psi_{2p_{z}} | \psi_{2p_{y}} \rangle + c_4^2\langle \psi_{2p_{z}} | \psi_{2p_{z}} \rangle = 1 $$

But the $2s$ and $2p$ orbitals are all orthonormal which means that $\langle\psi_i|\psi_j\rangle = \delta^i_j$, so our equation becomes:

$$ c_1^2 + c_2^2 + c_3^2 + c_4^2 = 1 $$

Which I think is what you were asking.

Response to comment:

The coefficients don't rotate anything, they are just numbers. What the diagram shows is that if you take two vectors:


then you can get a vector $\psi$ in any direction you want using:

$$ \psi = c_1\psi_1 + c_2 \psi_2 $$

with the normalisation condition $c_1^2 + c_2^2 = 1$. For example:

$$ \psi = \frac{1}{\sqrt{2}}\psi_1 + \frac{1}{\sqrt{2}}\psi_2 $$

is at 45º. More generally:

$$ \psi = \sin(\theta)\psi_1 + \cos(\theta)\psi_2 $$

gives you a vector at an angle $\theta$ to the horizontal, and it's automatically normalised since $\sin^2 + \cos^2 = 1$.

The diagram is just showing this applies to the $2p$ orbitals. Combining $2p_x$ and $2p_y$ in different proportions gives an orbital at different angles.

  • $\begingroup$ +1;Thanks sir, for paying heed to this question & thanks for your effort; yes I wanted to know the proof of this but actually I've asked another thing also; you can see in the snapshot that for different values of the coefficient, the orientation is changed; now how can I prove that the coefficients indeed rotate the superposed orbitals? $\endgroup$ – user36790 Oct 13 '15 at 16:38
  • $\begingroup$ @user36790: I've extended my answer to respond to your comment. $\endgroup$ – John Rennie Oct 13 '15 at 17:09

I honestly don't see why you're so mystified by this. Keep in mind that, in the end, you're adding up states which represent functions over real space, and these functions each have their own spatial dependence:

s, px, py, pz hydrogenic wavefunctions

(Image source)

The different colours represent different signs: blue is positive, red is negative. If you have a mixed wavefunction like, say, $$ \frac1{\sqrt{2}}\left(\psi_{2s}(\mathbf r)+\psi_{2p_z}(\mathbf r)\right) $$ then at $z>0$, where $\psi_{2p_z}(\mathbf r)>0$, the probability amplitude will increase, while at $z<0$ you have $\psi_{2p_z}(\mathbf r)<0$ so it detracts from $\psi_{2s}(\mathbf r)$ and the probability amplitude will decrease. The overall effect, then, is that the wavefunction's lump is pushed towards positive $z$. (Similarly, the orthogonal wavefunction $\tfrac1{\sqrt{2}}\left(\psi_{2s}(\mathbf r)-\psi_{2p_z}(\mathbf r)\right)$ is pushed towards negative $z$.

For the more general case of $sp^3$ mixing, with wavefunctions of the form $$\psi_{sp^3}(\mathbf r)= c_1\psi_{2s}(\mathbf r)+ c_2\psi_{2p_{x}}(\mathbf r) + c_3\psi_{2p_y}(\mathbf r)+ c_4\psi_{2p_{z}}(\mathbf r),$$ the $c_i$ act as probability amplitudes (in an abstract Hilbert-space view) but they also act as mixing coefficients between the different wavefunctions. Add some $\psi_{2p_{z}}$ and you push the orbital up; subtract some $\psi_{2p_{z}}$ and add some $\psi_{2p_{x}}$ and you push it down and forwards; and so on.

  • $\begingroup$ Why didn't you post it before Rennie? I would definitely accept it as it is very explicit & lucid in its content but I don't like unaccepting an answer & accepting another; I don't know but I feel a bit uneasy to do that; however you do deserve +1:) $\endgroup$ – user36790 Oct 15 '15 at 17:57
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    $\begingroup$ Well, I had time now and I didn't have time before. Having one's answer unaccepted isn't the nicest thing in the world but it's not the end of the universe either, and it's considered OK if it does improve the site for future visitors. Your call, really. (I'll go upvote some other of John's excellent posts if it makes you feel better.) $\endgroup$ – Emilio Pisanty Oct 15 '15 at 18:01
  • $\begingroup$ Just one point; I know these coefficients are the probability amplitudes but how do they act as mixing coefficients? I think there is an emphasis on some, you used in the last sentence, isn't it? Think these amplitudes scale the part to be contributed by an orbital in the superposition. $\endgroup$ – user36790 Oct 15 '15 at 18:05
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    $\begingroup$ (i) The difference between "probability amplitude" and "mixing coefficient", particularly in this setting, is entirely a matter of viewpoint. (ii) Yes, if you increase e.g. $c_3$, then that directly increases the role played by $\psi_{2p_y}$. Don't take the 'add some X' too seriously - it's an idiomatic expression for 'some as-yet-undefined amount of'. See here for more examples. $\endgroup$ – Emilio Pisanty Oct 15 '15 at 18:39

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