I am aware that the square of the Wavefunction gives the probability density of finding an electron at a particular point in space. I have also heard that it's a complex number but since it's a function I am unsure as to how that could be the case (perhaps someone could please clear that up for me as a sub-question). Moreover, I have seen it described as the amplitude of something but I literally have no idea what is meant by that. However, the crux of the question is what is the Wavefunction itself (i.e not what the square of it is). I have read in the book "Why Chemical Reactions Happen" that it is essentially synonymous with the word/concept: orbital. I have also seen in various places w(x,y,x) or w(r,theta,thi) [where "w" represents the sign for a wavefunction] so, from what I can gather it is a function in three dimensions that represents the shape of a particular orbital where the function of (x,y,z) or (r,theta,thi) would, of course, be different for the s,p,d and f orbitals. However, in the book "Why Chemical Reactions Happen" it goes onto say that wavefunctions of different atoms interfere with each other to form molecular orbitals and the atomic orbitals somehow have a phase?! For my idea of a wavefunction to be correct, the idea of interference of these wavefunctions must be an idea that was thought of to try and rationalize the process of the formation of molecular bonding and anti-bonding orbitals from atomic ones to make it easier to understand and conceptualize. If I am right about the wavefunction being a purely mathematical representation of the space where an electron is likely to reside then the molecular wavefunctions are just as curious in terms of their shape and suchlike as the shapes of the atomic orbitals they came from (i.e nobody really knows why they are that shape but when atoms or molecules are arranged in a particular way, orbitals "just are" that shape). Also the idea of phase seemed to bring my argument down but then I thought that perhaps phase is simply defined as being opposite relative to another part of an orbital if it is the other side of an angular node (I suspect this is wrong but could someone please shed some light on this?)

Going back to when I said - "the molecular wavefunctions are just as curious in terms of their shape and suchlike as the shapes of the atomic orbitals they came from" - I thought this because I found it bizarre that empty orbitals could "interfere" with an orbital from another atom. For example one p orbital is empty yet all three interact with the four hydrogen s orbitals. Unless, the electrons can jump between the the degenerate p orbitals at such a frequency that all can be considered occupied I find it incomprehensible that just a portion of space can interfere with an orbital. As a result, I came to the conclusion that I quoted at the start of the paragraph. Basically meaning that when atoms come together to form molecules the electrons find the lowest possible energy spaces to reside; or in other words, a molecuar orbital / molecular wavefunction forms which is equally as curious as the atomic wavefunctions with their strange shapes.

Any help with this concept would be greatly appreciated - I am off to Oxford university this October and Molecular Orbital Theory is a small part of the suggested reading list before arrival into the first year. Also, on that note, please refrain from heavily mathematical answers because I simply won't understand as I have only done A-level Maths and A-level Chemistry. Thanks.

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    $\begingroup$ Too many (mostly trivial) questions. Either try to separate the and ask one-by-one, or better go and read some book on introductory quantum mechanics. $\endgroup$
    – Wildcat
    Sep 18, 2014 at 18:26
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    $\begingroup$ Or, maybe, you are trying to learn this just "too soon". I mean, you are just entering the university, where they surely will teach you quantum mechanics. Maybe not the first year, maybe not even the second year, but they will teach you the subject. So, do not jump the gun. $\endgroup$
    – Wildcat
    Sep 18, 2014 at 18:44
  • $\begingroup$ MO theory can be studied with little math, but at least basics of functional analisys and vector operations are very,very useful. $\endgroup$
    – permeakra
    Sep 19, 2014 at 9:07

5 Answers 5


Let me see if I can get at some of your questions. As mentioned above, it's much easier when you ask individual specific questions.

One problem with books on introductory quantum mechanics is that, put simply, the language of quantum mechanics is math. Specifically, most people use the Schrödinger equation which involves second derivatives and differential equations.

When I teach quantum chemistry to undergraduates, there are typically problems due to the mathematical nature.

There are multiple "interpretations" of quantum mechanics. That gets into philosophy. Wikipedia has some nicely-written descriptions.

As you mention, the most widely used interpretation (the Copenhagen interpretation) of the wave function centers on the square of the wave function, or rather $|\psi^*\psi|$ as a probability density. As you mention, the wave function could be imaginary or complex, so this notation indicates a mathematical way of getting a real number for a probability density.

You ask about the shape and why atomic orbitals have particular shapes.

One thing that's amazing about quantum mechanics is that we put in some relatively simple equations to describe the motion of an electron in a hydrogen atom, and the solutions are exactly these atomic orbitals (s, p, d, f, etc.).

So I disagree that "no one knows" why atomic orbitals have particular shapes. The answer is that they are the mathematical solution to the Schrödinger equation. That is, we want to know the energies of the system and there are only certain quantized solutions.

And yes, the whole notion of chemistry is that electrons (and hence orbitals) on different atoms interact. Otherwise, there is no bonding.

I'll leave you with this. Without the math, it can be exceptionally hard to understand quantum mechanics. When you do understand the math, I think the results are quite beautiful. The shapes of orbitals aren't arbitrary. This is actually just a reflection of the mathematical nature of quantum mechanics.

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    $\begingroup$ The problem is that many programs teach the required math necessary for quantum over the course of too many semesters. There are way too many credit hours to take in order to gain the appropriate math background by the time quantum is supposed to be taken. This either discourages students and they switch majors or they enter quantum having to learn the math as they go leading to the problematic disconnect. $\endgroup$ Sep 18, 2014 at 22:06
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    $\begingroup$ @LordStryker Indeed! We recently created a "math for chemistry majors" course, since students entered quantum without differential equations. Hopefully this concept will spread. $\endgroup$ Sep 18, 2014 at 22:08
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    $\begingroup$ And at my school the professor says "orbitals aren't real" and QM is bull**** and QM-inclined faculty have been "popping too many math pills." $\endgroup$
    – Dissenter
    Sep 19, 2014 at 4:21
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    $\begingroup$ @Wildcat I believe that a lot of math is required for quantum. From personal experience, there was debate in the department about requiring yet another math class because students going into quantum was having problems understanding the material. A closer look revealed that by requiring another math class would have pushed quantum to a senior semester class just so all the math could be covered before hand. Its atrocious. Maybe a focused line of math classes strictly for chemistry students would fix the problem. I don't know. $\endgroup$ Sep 19, 2014 at 11:49
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    $\begingroup$ @GeoffHutchison, sounds great! Chemistry had changed completely in the last century, it, in a sense, finally emerged as a scientific discipline and not just a random collection of recipes. Some even argue that chemistry is over (the laws that govern chemical species are known), but in my opinion it is only beginning: we finally know how to do chemistry on a solid theoretical ground. The only problem is, we need much more powerful computers. :D $\endgroup$
    – Wildcat
    Sep 19, 2014 at 13:01

I know, I would better write a comment, but suddenly comments are limitted in size, so I will write my thoughts (rather than actual answer) here.

Also, on that note, please refrain from heavily mathematical answers because I simply won't understand as I have only done A-level Maths and A-level Chemistry.

That is the root of your problems. But do not be afraid, you are not alone! As I pointed out in comment to Geoff Hutchison answer, many chemistry students (and let us be honest here, not only students) are strongly reluctant to the following two simple very well established and no longer questioned facts:

  1. In order to understand chemistry you have to understand quantum chemistry. In other words, at the very fundamental level, there is no any other chemistry except for quantum chemistry.

  2. In order to understand quantum chemistry you have to understand quantum mechanics which requires a lot of math.

So, the right way to proceed is to learn some required math first (at least calculus), then physics (at least classical and quantum mechanics in the form of wave mechanics), and only then quantum chemistry. Thus, my advice is: do not put the cart before the horse! If you have some free time, learn some math.


Mathematician's answer

Wavefunction is quantum mechanical conterpart to trajectory in classical mechanics. Unfortunately, since we interact with the world in a way exposing to us mostly classical newtonian physics, there is no way to find pleasant, easily understandable everyday analogy. However, you may consider atomic orbitals as building blocks for 'trajectory' of electrons. They don't have to be filled to be used as building blocks.

To truly understand the role wavefunction plays in quantum mechanics, at least rudimentary understanding of quantum mechanics is required. In includes understanding of what is

  • coordinate space
  • matrix operator
  • eigenvalues and eigenvectors
  • Knowledge of at least one example of basis in some function space, say, fourier series.

To understand MO theory, a very rudimentary understanding of Hartree-Fock formalism may help greatly.

AFAIK, wikipedia provides basic knowledge on the topics I mentioned. It is OK to have very vague understanding of concepts I mentioned until 3rd year in university. There is no need to have deep understanding of quantum mechanics to successfully apply MO theory.

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    $\begingroup$ This is physicist's answer. The mathematician's would be that the wave function is an element of the corresponding $L^2$ space which represent the state space of a quantum system, i.e. it is an equivalence class of functions that are square (quadratically) integrable on a particular domain. $\endgroup$
    – Wildcat
    Aug 10, 2016 at 9:13

Short answer:

Wavefunctions are functions that represent the physical state of certain quantum systems.

Extended answer:

  1. Ordinary wavefunctions are complex functions defined in a Hilbert Space (HS), corresponding to pure quantum states. There is not a one-to-one correspondence between pure states and wavefunctions. For instance if $\Psi$ is a valid wavefunction in the energy basis (with Hamiltonian operator $\hat{H}$), then the wavefunction $\Psi´ = e^{i\phi}\Psi$ represents the same state because $$E = \int d^3x \> (\Psi´)^{*} \hat{H} \Psi´ = \int d^3x \> \Psi^{*} \hat{H} \Psi$$

  2. Wavefunctions are used in the wavefunction formulation of quantum mechanics, which is only one of many available formulations. There are other formulations of quantum mechanics, where wavefunctions are used for different purposes and no longer represent a quantum state, and there are formulations where wavefunctions are not used at all. You can find a basic review of nine formulations of quantum mechanics in this article and a new formulation of quantum mechanics without wavefunctions in this article.

  3. In the position representation, wavefunctions are expressed like $\Psi = \Psi(x,t)$, and the square of the wavefunction is often physically interpreted as the probability density of finding the particle at a given position. This is correct for wavefunctions that satisfy the Schrödinger equation, but it is not for Klein-Gordon and Dirac equations. I.e. the square of a Dirac wavefunction does not give the probability density of finding the particle at a given position.
  4. Wavefunctions do not describe mixed quantum states. Mixed quantum states are outside the scope of the wavefunction formulation of quantum mechanics. For instance, an atom in a molecule does not have wavefunction associated to its quantum state, because atoms in molecules are open systems and quantum correlations with neighbor atoms prohibit the description of the properties of the atom with wavefunctions. A heat bath at non-zero temperature does not have wavefunction describing its quantum state either. Mixed quantum states are properly described by other formulations of quantum mechanics.

    Ordinary wavefunctions cannot be used to rigorously describe instable quantum states. Ordinary wavefunctions --sometimes named "Dirac Kets" in the abstract representation of wavefunctions-- can be applied to stable particles, but not to decaying particles or resonances.

    There are attempts to extend the ordinary wavefunction formulation to instable systems. Those generalized states are sometimes named "Gamow kets" and are defined in a Rigged Hilbert Space (RHS). You can find this formalism in this book.

  5. From the above, it looks like if wavefunctions are a kind of analog of the phase space densities $\rho$ used in classical mechanics. This is more than a mere formal analogy and we can develop a phase-space formulation of quantum mechanics where wavefunctions are replaced by quantum phase-space $\rho$ and operators are replaced by phase-space functions, for instance the energy of a quantum system in the phase-space formulation is computed as

    $$E = \int d^3x \int d^3p \> H \rho$$

    which looks as a classical average, but is a proper quantum average.

    Alternatively one can formulate classical mechanics in terms of wavefunctions, which satisfy the classical equation

    $$ i\hbar \frac{\partial \psi_\mathrm{class}}{\partial t} = \left[ -\hbar^2\frac{\nabla^2}{2m} + V + \hbar^2\frac{\nabla^2 \sqrt{\psi_\mathrm{class}^{*}\psi_\mathrm{class}}}{2m \sqrt{\psi_\mathrm{class}^{*}\psi_\mathrm{class}}} \right] \psi_\mathrm{class}$$

    sometimes named the classical Schrödinger equation in the literature. Note that the last term between brackets is absent in quantum mechanics. This classical equation is nonlinear and this nonlinearity kills the superposition principle and other aspects associated to quantum mechanics.


I think that @permeakra's last comments are very important, which is that you don't have to have a deep knowledge of quantum mechanics to be able to apply it and also that a lot of the maths detail can be left until your last undergraduate year.

One other point not mentioned above is that any area has its own often obscure language and notation. Quantum is one of the worst in this respect, e.g. basis sets, orbitals, wavefunctions, operators, hamiltonians, eigenvalues, eigenfunctions, bra's, ket's, stationary states and so on. Often textbooks (and lecturers) assume that one is already familiar with these concepts and if you are not (the general case) this jargon just gets in the way when you start to learn.

As a chemist you have to understand quantum mechanics primarily to help you understand spectroscopy, (vibrational, rotation, & most importantly nmr) as well as bonding.

But to the chemist, QM is a tool we need to use, and a fairly simple one. (If you think QM is hard, consider making your own nmr machine or even a laser from scratch). Although you will need to know how to solve the Schrodinger equation, all the cases you meet as an undergraduate have been solved long ago, with well know solutions; textbook are full of these. It is, however, essential to understand the physical assumptions and limitations that lead the solution, but not necessarily all the nitty-gritty.

What is also fundamentally important is to understand how to formulate a problem, and to identify unphysical solutions you may come across. That way you begin to think like a scientist by clearly focusing on the problem and do not get distracted/overwhelmed by any maths. Believe that solving equations is, at the end of the day, a simple skill!


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