I am an organic chemistry student learning how atomic orbitals interfere to give rise to molecular obitals. The image below suggests that each of the hydrogens' atomic orbitals interfere both constructively AND destructively. How is this possible? More fundamentally, how does the 1s atomic orbital of a hydrogen atom have both positive and negative phase in the first place? I realize that there is a similar question on this topic but I wasn't able to understand the answer to that post due to my limited background in quantum mechanics and was hence was looking for someone who could explain in simple terms.

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    $\begingroup$ As there are two atomic orbitals (a0) to begin with there are two MO's, we make the hypothesis that they are made up as linear combinations $ao_1\pm ao_2 $ as you have drawn. The total energy is the same as the two ao's, one MO up in energy the other lower, which makes a difference when only two electrons are added but not when four are. Most phys. chem. textbooks explain this in detail but that is the main idea. $\endgroup$
    – porphyrin
    May 10, 2021 at 12:25
  • $\begingroup$ While this does not explain why the orbitals combine that way, it might still be helpful if you take in mind that the molecule is hold by only one. In this sense the molecule is not formed by both constructive and destructive interference, that concerns the orbitals as functions, which include l a phase like in the sketch. $\endgroup$
    – Alchimista
    May 10, 2021 at 13:23

3 Answers 3


What does MO formation entail?

A very common misconception is that the formation of MOs involve addition, or subtraction, of two physical objects. And the confusion arises because, how can the same physical object have two different phases when added or subtracted? However, this is not an accurate picture. MO construction is not a physical process which occurs between two real objects which each possess a definite phase. There are at least two reasons for this:

  1. Orbitals are not objects. The electrons that go into orbitals are objects, but the orbitals themselves are wavefunctions.

  2. Wavefunctions do not have definite phases. When we talk about the phase of an orbital, what it refers to is a constant $k$ which the entire wavefunction is multiplied by. It turns out that a fundamental postulate of quantum mechanics is that that the wavefunction $\psi(x)$ and the wavefunction $k\cdot\psi(x)$ represent exactly the same state, so the phase $k$ is actually completely meaningless. The 1s orbital does not possess a particular phase; it can possess any phase and still be a 1s orbital.

So, what does it really entail? I think a better description is that orbitals are possibilities for where the electrons can go. Let's say you have a hydrogen atom with one electron. This electron can either be in a 1s orbital, or (if you excite the electron) it can be in a 2s orbital, or a 2p orbital, etc. etc. So, the electron has many different options as to where it can go. Of course, in ground-state hydrogen, the eventual choice is the 1s orbital.

Alternatively, let's say you have two hydrogen nuclei, but only one electron. (So there is a net positive charge.) This electron could be in the 1s orbital of hydrogen atom A, or the 1s orbital of hydrogen atom B, or the 2s orbital of hydrogen atom A, etc. etc. These different atomic orbitals represent the possibilities which the electron can take.

In this regard, the orbitals are not objects or boxes to be inhabited by the electron, but they represent a set of choices. If we adopt this mindset, then it already becomes clearer how both bonding and antibonding orbitals form: these are simply the two "possibilities" which can happen. It is not a question of how can they both be true at the same time, because that isn't the point: the point is that MOs themselves are different possibilities for an electron to adopt, and it isn't necessary that both of them be true at the same time. Indeed, they can't be, because an electron can't simultaneously inhabit two MOs.


I hesitated to include this analogy in the original answer, but will include it now, since it may prove helpful to some. There are some problems with it, however, so don't take it as a literal description, but rather a conceptual one.

Consider, then, that you're standing on a hill overlooking a two-lane motorway which runs north to south. However, you don't have a compass on hand, so you don't actually know which way is north or south. (This is analogous to not knowing the actual phase of an orbital $k$: you can't tell whether it's $+1$ or $-1$, for example.)

Suppose you see two cars driving along the motorway, one on each lane. If you focus on one car at a time, what information can you give about each of the two cars? You can't tell which way they're headed, because you don't know north from south. However, you can tell which lane they are driving on: for example, because one lane is closer to you than the other. So you can furnish two pieces of information:

  1. There is a car on the nearer lane. It is going either north or south, but we don't know which.

  2. There is a car on the further lane. It is also going either north or south, but we don't know which.

This situation is exactly analogous to having two hydrogen atoms, with one electron in each 1s orbital. In other words, this corresponds to the "pre-MO" description of the $\ce{H2}$ system:

  • There is a 1s orbital on hydrogen atom A. It has some phase, but we don't know which.

  • There is a 1s orbital in hydrogen atom B. It has some phase, but we don't know which.

Now, let's return to the cars. You might think that there actually is another way we can describe their movement, and that is based on not just zooming onto one of them at a time, but rather by comparing their relative motion. So, again we have two possibilities:

  1. There are two cars on either lane, and they are moving in the same direction. (North/north or south/south, but we don't know.)

  2. There are two cars on either lane, and they are moving in opposite directions. (North/south or south/north, but we don't know.)

This is precisely analogous to the MO description of $\ce{H2}$:

  • There is a bonding orbital formed from two 1s orbitals on either atom, and they have the same phase, although we don't know exactly what that phase is.

  • There is an antibonding orbital formed from two 1s orbitals on either atom, and they have opposite phases, although we don't know exactly what phases those are.

Note that we didn't somehow physically stick the two cars together, or force them to travel in any particular direction. What we have added together are not cars, but rather possibilities: we combined the two "base" descriptions, of one car moving along each lane, into two "possibilities" where they are either going in the same or opposite direction. At any given point in time, whenever you see two cars going along the motorway, either of these possibilities can be realised.

The same is true of MOs. The MOs don't represent physical addition of two orbitals, but rather different possibilities of how these orbitals can combine: they can either combine constructively, or destructively. For any given electron, either of these possibilities can be realised, too: and unlike the motorway, it's not mutually exclusive, because one (or two) electron(s) can be in the bonding orbital and another can be in the antibonding orbital.

Linear algebra

This is a more formal description of how MOs are constructed, and much closer to the truth. There are very few, and possibly no, quick and dirty ways to explain MO theory, in my opinion. However, if you have studied linear algebra before,* then the transition from AOs and MOs is the same as a change of basis. A basis refers to a minimal set of vectors, from which any arbitrary vector may be constructed through linear combination. If you have a vector $(a, b)$, you can express this as: $$\begin{pmatrix}a\\b\end{pmatrix} = a\begin{pmatrix}1\\0\end{pmatrix} + b\begin{pmatrix}0\\1\end{pmatrix}.$$

It doesn't matter what $a$ and $b$ are, you can always express $(a, b)$ as a linear combination of the two basis vectors. You can think of the basis vectors $(1, 0)$ and $(0,1)$ as being the 1s orbitals on both individual atoms.

The MOs correspond to using the basis $(0.5, 0.5)$ and $(0.5, -0.5)$.† Notice that this is still a valid basis, because we can still express any arbitrary vector $(a, b)$ as a linear combination of these:

$$\begin{pmatrix}a\\b\end{pmatrix} = (a+b)\begin{pmatrix}0.5\\0.5\end{pmatrix} + (a-b)\begin{pmatrix}0.5\\-0.5\end{pmatrix}.$$

If we had rejected the "antibonding" combination $(0.5, -0.5)$, and only taken the "bonding" combination $(0.5, 0.5)$, then we would not be able to form a valid basis. For example, we could not express the vector $(1, 2)$ as just a linear combination of $(0.5, 0.5)$.

From this point of view, AOs and MOs are simply different bases.‡ The existence of both the bonding and antibonding MO is not only logical, but also mandatory in order to form a complete basis.

The familiar AOs that you have studied (1s, 2s, 2p, ...) are simply a series of basis states for the hydrogen atom. In the $\ce{H2}$ molecule, there are two sets of basis states, one for each atom. The transition from "2 sets of AOs" to "1 set of MOs" is just a change of basis: we go from two sets of atomic bases to one single molecular basis.


The proper answer is, of course, to pick up a QM textbook and study it. By QM I do indeed mean quantum mechanics, not just MO theory. It is a long journey, and you may not necessarily be fully prepared for it right now, but it will reward you with a better understanding of how this works. MO theory fundamentally relies on quantum mechanics, and to learn MO theory without understanding quantum mechanics is what leads to confusion and questions like yours.


* The analogy I use above is slightly simplified, and might look slightly contrived, but the link between QM and linear algebra is actually very deep. QM is somewhat analogous to linear algebra but on an infinite-dimensional vector space (technically, a Hilbert space).

† Technically this should be $1/\sqrt{2}$, not $0.5$. In QM we are nearly always concerned with orthonormal bases, i.e. ones where each basis vector has a length of $1$ and forms a scalar product of $0$ with every other basis vector. (Mathematically, we have a basis $\{|i\rangle\}$ for which $\langle i | j \rangle = \delta_{ij}$.)

‡ "Bases" being the plural of "basis".

  • $\begingroup$ Good explanation. I had a rough time learning about SALC's and AO -> MO transitions, the identity tables etc - but I had never seen the maths before. After learning what linear combination actually IS, and what it basically DOES - it was much easier to understand what is going on. A learning that happened to me much later in life, but was very appreciated in the raw "A-HA!" value. $\endgroup$
    – Stian
    May 11, 2021 at 12:00
  • $\begingroup$ I appreciate the effort. But some sentences simplify to much and could be improved. "Indeed, they can't be, because an electron can't simultaneously inhabit two MOs." You literally show that electrons can be in superpositions of orbitals in your linear algebra example. In the H2 example It also sounds as if the electron could only be at A or B, neglecting the fact that the electron will be in a superposition of orbitals at A and B. Breaking the math down to concepts is hard, I don't think I could have done it better, but perhaps this point could be improved a bit. $\endgroup$
    – Hans Wurst
    Nov 27, 2021 at 10:49
  • $\begingroup$ @HansWurst Please feel free to edit if you like! I did struggle with writing this answer and I honestly don't think it's perfect. But I disagree with your first point: the electron may be in a superposition $2^{-1/2}(|\psi_1\rangle + \mathrm{e}^{\mathrm{i}\theta}|\psi_2\rangle)$, but that's not the same thing as simultaneously inhabitng the two states $|\psi_1\rangle$ and $|\psi_2\rangle$. $\endgroup$ Nov 27, 2021 at 12:02
  • $\begingroup$ @orthocresol I guess it is a problem of semantics but arguing semantics is pointless. Colloquially I would say that a system in a superposition state can be viewed as being partly in each of two or more other states. But in the end all semantics are error-prone in comparison to the correct mathematical expression. I think "simultaneously inhabiting two states" could be used to describe superpositions, the other case where it could be used would be the density operator $\hat\rho=c_1|\psi_1\rangle\langle\psi_1|+c_2|\psi_2\rangle\langle \psi_2|$. Would you use it there ? $\endgroup$
    – Hans Wurst
    Nov 27, 2021 at 12:36
  • $\begingroup$ Orthocresol & @Hans Wurst the superposition should be a state, not really two taken simultaneously. But I admit that all this is rather obscure to me. H2 is a stuff, it is hold by the bond we known. I can't understand this continuous emerging of superposition concepts. Please help. $\endgroup$
    – Alchimista
    Nov 27, 2021 at 13:08

Unfortunately, in quantum mechanics there is rarely an explanation "in simple terms." Quantum mechanics is a mathematical construct that so far seems to predict the results of all experiments that have been done to test it, but explaining those maths in terms of anything we have learned from the larger-than-quantum scale "classical" systems has so far eluded us. In fact, attempting to understand QM by applying intuition from classical physics generally causes more confusion, as seems to be the case here.

It seems that you are considering the two H-atom 1s orbitals to be static standing wave systems that can only combine constructively or destructively, which would be the case if these were classical standing waves. But they aren't. They're quantum mechanical systems that live by different rules. When the two atoms come together, the whole system is changed, and the wave function describing the new system is not just an overlay of the previously separated "waves". Conceptually, it may be helpful to think of the new system as formed by constructive and destructive interference, but only in the vaguest conceptual sense. Pushing that metaphor too far gets you into the confusion you are in now.

As unsatisfying as it will be, it might be best at this point to accept that the conceptual description is qualitatively consistent with the mathematics and so is a guide to intuitively constructing molecular orbitals from atomic orbitals, but it isn't an explanation of why?. If you continue farther in your study of quantum chemistry, you'll learn how to construct these molecular orbitals more quantitatively, and you'll get better answers of how the math works to give certain answers, but you'll never get a good answer of why.


I will try to answer without using any mathematical formula, as your question is just made of words, without any mathematical symbols.

You should know that the wave function of an electron in the $\ce{H}$ atom is normalized. It means that the integral of the wave function over the whole space is equal to $1$. And it must be equal to $1$.

If you add two of these wave functions from two $\ce{H}$ atoms, the integral of this sum will be $2$. This is forbidden. It must be $1$. So you cannot simply add two normalized wave functions, like the $1s$ in the $\ce{H}$ atom. You have to take one-half of the wave function of the first atom and to add it to one-half of the wave function of the second atom. This would produce a new molecular wave function whose integral is equal to $1$. It is nice. This new molecular wave function describes a sigma bond, designed as $\ce{H-H}$ in the Lewis model.

Now you have to use and combine the other half of these wave functions, without adding them. The only thing to do is to subtract them. But such a subtraction produces a point between the two nuclei where the obtained wave function is equal to zero. As a consequence, this molecular orbital is a antibonding orbital. And the integral of this antibonding orbital is also equal to $1$.

Hopefully the pure theorists will pardon me for having simplified the description of the problem. I should have spoken of the factor √2/2 instead of $1/2$. I should have said that it is the square of the wavefunction that should have been integrated. I have tried to be understandable to a beginning student

  • $\begingroup$ Maurice: It is the square of the wavefunction that should integrate to 1. You are mixing a couple of things. I am not a QM expert, but please reconsider double checking the statements. $\endgroup$
    – AChem
    Jul 23, 2021 at 17:17
  • $\begingroup$ @M. Farooq. I know that it is the square of the wave function that should be integrated. I have been teaching it in my classes. But you know. The author of the post is a beginner. I haver tried to be as elementary as possible to give him an answer that he could understand. $\endgroup$
    – Maurice
    Jul 23, 2021 at 20:31

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