# How is a molecular orbital a 'quantum superposition' of the atomic orbitals?

Let me quote from Feynman's lectures the concept of superposition:

(1)The probability of an event in an ideal experiment is given by the square of the absolute value of a complex number $ϕ$ which is called the probability amplitude: \begin{aligned} P&=\text{probability},\\ \phi&=\text{probability amplitude},\\ P&={|\phi|}^2. \end{aligned} (2)When an event can occur in several alternative ways, the probability amplitude for the event is the sum of the probability amplitudes for each way considered separately. There is interference: $$ϕ=ϕ_1+ϕ_2.$$

The second point is the core of quantum superposition.

Also, in Dirac's language from The principles of Quantum Mechanics,

[...]It requires us to assume that between these states there exists peculiar relationships such that whenever the system is definitely in one state we can consider it as being partly in each of two or more other states.[...]

These all imply that when a system is in a state of superposition, it is 'partly' in both of the states that contribute to the superposition( Here 'partly' means possibility which implies the system has the possibility to exist in both states that contribute in the superposition).

For instance, let's take $\ce{H_2^+}$ to illustrate the concept of superposition. Firstly, the words of Feynman:

A positively ionized hydrogen molecule consists of two protons with one electron worming its way around them. If the two protons are very far apart, what states would we expect for this system? The answer is pretty clear: The electron will stay close to one proton and form a hydrogen atom in its lowest state, and the other proton will remain alone as a positive ion. So, if the two protons are far apart, we can visualize one physical state in which the electron is “attached” to one of the protons. There is, clearly, another state symmetric to that one in which the electron is near the other proton, and the first proton is the one that is an ion. We will take these two as our base states, and we’ll call them $|1⟩$ and $|2⟩.$ They are sketched in Fig.

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Now, the general state of $\ce{H_2^+}$, say, $|\psi_{\ce{H_2^+}}\rangle$ is in quantum superposition of the two base-states $|1\rangle$ & $|2\rangle.$

Now using the point no.(2) of Feynman, we can prove the superposition of the states as :

What should be the amplitude for finding the electron in, say, $x'$ from the center of the molecule $\ce{H_2^+}?$ (We would use each-coordinate as our new base-state, that is, $|x\rangle$ is a base-state; this means we would now treat continuum of base states).

We would've to find $\langle x'|\psi_{\ce{H_2^+}}\rangle.$ This can happen in two ways & thus applying the point no.(2), we get $$$$\langle x'|\psi_{\ce{H_2^+}}\rangle\\=\int \langle x'|x\rangle \langle x|\psi_{\ce{H_2^+}}\rangle dx\\= N\left(\int \langle x'|x\rangle \langle x|1\rangle dx +\int \langle x'|x\rangle \langle x|2\rangle dx\right)$$ ;$$ where $N= \text{normalisation constant}.$ From this, we can easily shew, $$|\psi_{\ce{H_2^+}}\rangle = N(|1\rangle + |2\rangle).$$ Thus, quantum superposition is used when there is more than one way an event can happen; in this case, our event was finding the electron at $x'$ ; we superpose the alternative amplitudes to get the total amplitude of the event of finding the electron at that coordinate.

Let's consider $\ce{H_2}$; we describe first the VB description; again starting with Feynman's words:

As our next two-state system we will look at the neutral hydrogen molecule $\ce{H_2}.$ It is, naturally, more complicated to understand because it has two electrons. Again, we start by thinking of what happens when the two protons are well separated. Only now we have two electrons to add. To keep track of them, we’ll call one of them “electron a” and the other “electron b.” We can again imagine two possible states. One possibility is that “electron a” is around the first proton and “electron b” is around the second, as shown in the top half of Fig. We have simply two hydrogen atoms. We will call this state $|1⟩.$ There is also another possibility: that “electron b” is around the first proton and that “electron a” is around the second. We call this state $|2⟩.$

So, we can write the general state $|\psi_{\ce{H-H}}\rangle$ as the superposition of $|1\rangle$ & $|2\rangle.$ We can again prove the superposition from point no.(2) but this time we've to use $|x_1,x_2\rangle$ instead of $|x\rangle$ that is, $$$$\langle x'_1,x'_2|\psi_{\ce{H-H}}\rangle\\=\int \langle x'_1,x'_2|x_1,x_2\rangle \langle x_1,x_2|\psi_{\ce{H-H}}\rangle dx\\= N'\left(\int \langle x'_1,x'_2|x_1,x_2\rangle \langle x_1,x_2|1\rangle dx +\int \langle x'_1,x'_2|x_1,x_2\rangle \langle x_1,x_2|2\rangle dx\right)$$ ;$$ where $N'=\text{normalisation constant}.$ This evidently proves the superposition $$|\psi_{\ce{H-H}}\rangle= \frac{1}{\sqrt 2}(|1\rangle + |2\rangle)$$ where $N'= \frac{1}{\sqrt 2}.$

Come to the MO description of $\ce{H_2}$(only the bonding orbital is concerned here to shorten the query). This is where my problem begins...

This is quoted from J.D.Lee's Concise Inorganic Chemistry:

Consider two atoms $A$ and $B$ which have atomic orbitals described by the wavefunctions $\psi_{(A)}$ and $\psi_{(B)}.$ If the electron clouds of these two atoms overlap when the atoms approach, then the wavefunction for the molecule(molecular orbital $\psi_{(AB)}$) can be obtained by a linear combination of the atomic orbitals $$\psi_{(AB)}= N(c_1 \psi_{(A)} + c_2\psi_{(B)})$$

[...]Suppose the atoms $A$ and $B$ are hydrogen atoms ; then the wavefunctions $\psi_{(A)}$ and $\psi_{(B)}$ describe the $1s$ atomic orbitals on the two atoms. [...]

Now, wavefunctions are nothing but probability amplitudes; since _molecular orbital is a quantum superposition of the atomic orbitals; it must satisfy the point no.(2) like all other superposition described above.

But I'm not understanding what base states are used in this case; in VB theory, $|1\rangle$ represented the state of having one electron around one $\ce{H}$ atom & the other electron around the other $\ce H.$ So, $\langle x_1,x_2|1\rangle$ meant the amplitude to find electron $a$ at $x_1$ & electron $b$ at $x_2.$

But what should I use in this case? How to represent $\langle x_1,x_2|\psi_{\ce{H-H}}\rangle$ using point no.(2) in MO theory?

I'm not getting how to apply the point no.(2) here; $\psi_{(A)}$ represents the amplitude of only one electron; same with $\psi_{(B)}.$

For applying point no.(2), there must be two ways for the event to find the electrons at $x'_1,x'_2$_ so that point no.(2) yields the result $\psi_{(\ce{H-H})_{(g)}}= N( \psi_{(H)_{1s}} + \psi_{(H)_{1s}}).$ But I'm not getting how $\psi_{(A)}$ and $\psi_{(B)}$ represent the amplitudes of the same final event of finding the electrons at $x'_1,x'_2$; if they were not representing the amplitudes of getting to the same final event, how could they be superposed by point no.(2)? In VB theory, the base states are defined so that the alternative amplitudes define the same final event & so they were superposed; however how should I interpret the base-states in MO theory, so that $\psi_{(A)}$ and $\psi_{(B)}$ represent the amplitudes to the same final event? How should I apply point no.(2) in case of molecular orbital to prove the superposition?

Same story as with the previous question: quantum superposition is always expressed mathematically as a linear combination, but the converse is not necessarily true. Not each and every linear combination expression has something to do with real physical quantum superposition. In many cases it is just a mathematical trick.

In the question about resonance we've already faced a situation when we could only formally think about a linear combination as a superposition of some imaginary states. Same story here: molecular orbitals are just linear combination of atomic orbitals. As for the case of resonance, the states you think are superimposed (atomic orbitals this time) are not even states of electrons in a molecule. The only "real" (though approximate) states of electrons in a molecule are molecular orbitals, not the atomic one, thus, strictly speaking, there is no superposition out there.

Besides, you don't understand what molecular orbital is. It is not a many-electron wave function that describes all the electrons in a molecular system, rather, it is (approximate) one-electron wave function that describes an individual electron in a molecule. To construct (approximate) many-electron wave function you need to make (antisymmetric) product of $n$ molecular orbitals, where $n$ is the number of electrons. And since each and every molecular orbital describes just one electron, all your analysis of molecular orbitals for two-electron case of a hydrogen molecule is plain wrong.

P.S. And to be honest, I don't understand what you did for one-electron $\ce{H2+}$ case as well. To me it looks like you have too many misconceptions in your head currently. I suspect, you're jumping the gun by trying to learn the applications of quantum mechanics to chemistry having no really solid background in theory itself.

So, let's look on what you're saying on $\ce{H2+}$ case. Following Feynman, we write the general state of the system $\lvert \psi \rangle$ as a superposition of the two basis states $\lvert \psi_1 \rangle$ and $\lvert \psi_2 \rangle$, $$\lvert \psi \rangle = c_1 \lvert \psi_1 \rangle + c_2 \lvert \psi_2 \rangle \, .$$ Fine. And then you decided to "reverify the superposition of the states as". What do you want to "reverify" out there? Superposition is asserted, you don't need to "reverify" it whatever it means.

Yes, you can switch from the state vector formalism to the wave function one by taking the inner product of both sides with position eigenvectors $\lvert x \rangle$, $$\langle x \vert \psi \rangle = c_1 \langle x \vert \psi_1 \rangle + c_2 \langle x \vert \psi_2 \rangle \, , \\ \psi(x) = c_1 \psi_1(x) + c_2 \psi_2(x) \, ,$$ where the wave function is defined as usual, i.e. $\psi(x) = \langle x \vert \psi \rangle$. That's fine, though trivial: you indeed get the (weighted) sum of the corresponding wave functions which are the probability amplitudes. But that is because it just an alternative form of a superposition expression: it tells the same thing as before, just in different words. We didn't "reverify" anything, merely restated the superposition one more time.

Anyway, that would be the thing I at least understand. But look what you did! A complete mess! First, you forget the coefficients in the linear combination. Secondly, you used the resolution of identity for absolutely no reason. Thirdly, normalisation constant just appeared out of the blue. Finally, what you think you get at the end is in fact the starting point of a journey: it is an assertion Feynman did.

So, for me all you algebra appears as random manipulations of symbols you've seen somewhere but don't really understand.

• +1; I am not bothered at you critisizing me; you've every right but what I'm bothered is why you didn't understand the $\ce{H_2^+}$ case; I've just used the point no.(2) & the continuum of base states the same way Feynman explains in chapter 16. It deduces the correct result. So, what is the wrong there?
– user5764
Oct 15 '15 at 11:39
• @user36790, no, I do not criticise you. In fact, I like your questions. You ask a questions on things which many do not bother about but just take for granted. The problem is that your questions are not well suited for Q&A sites, such as Chem.SE: they are too long, not sharp focused, and usually consists of few questions rather then one. Oct 15 '15 at 11:51
• Now, let me explain what in particular I do not understand about $\ce{H2+}$ case right in my answer. Oct 15 '15 at 11:55
• You are right, sir that  you don't understand what molecular orbital is. It is not a many-electron wave function that describes all the electrons in a molecular system, rather, it is (approximate) one-electron wave function that describes an individual electron in a molecule.: in Atkins' Physical Chemistry book, he wrote that MO orbitals are one-electron wavefunction . But I didn't pay heed as every book says that each molecular orbital is occupied by two electrons; so if it were a one-electron wavefunction, how could it be occupied by two electrons?
– user5764
Oct 15 '15 at 11:57
• @user36790, it is a long story. In short there are two kinds of orbitals out there: spin orbitals and spatial orbitals. For now, just think of each spatial orbital as consisting of two spin orbitals. So, when it is said that a particular spatial orbital is doubly-occupied, what is actually meant is that both its consisting spin orbitals are singly-occupied. Oct 15 '15 at 12:31