I'm a bit confused by a past paper question I came across for one of my courses, shown below.

The wavefunctions corresponding to the lowest, $\psi_{1}$, and second-lowest, $\psi_{2}$, energy molecular orbitals in 1,3-butadiene are: $$\psi_{1}=0.372\phi_{1} + 0.602\phi_{2}+0.602\phi_{3} +0.372\phi_{4}$$ $$\psi_{2}=0.602\phi_{1} + 0.372\phi_{2}-0.372\phi_{3} -0.602\phi_{4}$$ where $\phi_{i}$ is the $p_{z}$ orbital on atom $i$. Calculate the norm of $\psi_{1}$, and the overlap between $\psi_{1}$ and $\psi_{2}$. Is the result what you would expect? Explain.

Neither calculating norm nor the overlap between 2 orbitals in the way the question asks came up. I assume by norm it means the normalisationg factor, but that makes no sense to me since it's already normalised. I assume the overlap should be 0 since they should be orthonormal, but don't know how to show that. I'd really appreciate some help! Sadly the lecturer for the course has contracted Covid-19 and so of course is not available to answer questions.

  • $\begingroup$ A screenshot or picture of an exercise is not searchable. Please consider rewriting it, so that it can be of help for future visitors. $\endgroup$ Apr 28, 2020 at 12:07
  • $\begingroup$ Please add a citation to where this exercise comes from. Also note that the use $\psi_x$ in the text and $\Psi_x$ in the equation. They should be the same, and should probably be all lower-case. $\endgroup$ Apr 28, 2020 at 12:17
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    $\begingroup$ The "norm" refers to en.wikipedia.org/wiki/Norm_(mathematics)#Euclidean_norm. $\endgroup$ Apr 28, 2020 at 12:25
  • $\begingroup$ As I stated in my question, it comes from a past paper at my university. I take your point about the notation; that's simply how the question is in the exam paper. $\endgroup$
    – atbm
    Apr 28, 2020 at 12:33
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    $\begingroup$ if the $\phi$ are orthonormal then $\langle \phi_i\phi_j\rangle = \delta_{i,j} $ meaning that if $i=j$ the result is 1 else it is zero. So multiply out term by term to get overlap $\langle \psi_1\psi_2\rangle$ $\endgroup$
    – porphyrin
    Apr 28, 2020 at 13:38

1 Answer 1


I'm unsure if you're meant to solve this via the Hückel molecular orbital theory (see Chemistry LibreTexts), however, if you need to calculate the norm of a wave function $\Psi$, this is $|\Psi |$. Given what we know in your post about both $\Psi_{1, 2}$ then \begin{align} |\Psi_1|^2 &= \langle\Psi_1 | \Psi_1\rangle \\ &= 2\cdot(0.602^2)+2 \cdot (0.372^2) \\ &= 1.001576 \\ \to |\Psi_1| &= 1.000788 > 1\\ \end{align} I write that this is greater than one, as, on closer inspection, is down to rounding from the coefficients of each $p_z$ orbital.

The repeat process can be found for the overlap part, which I am interpreting as $\langle\Psi_1 | \Psi_2\rangle$.

  • $\begingroup$ Yes, we did cover Huckel in this course. I suspect that the fact that it's greater than 1 is a rounding error rather than intentional. Thank you - that's really useful $\endgroup$
    – atbm
    Apr 28, 2020 at 12:31
  • $\begingroup$ Great I'm glad to have helped, in terms of thr rounding, you may have a point actually, looking at various other Wave functions for the Energy bounds of 1,3-Butadiene I see close enough numbers that do indeed sum-square to unity. $\endgroup$
    – user82205
    Apr 28, 2020 at 12:54
  • $\begingroup$ @Martin mod - thanks for the improvment in my Bar-Ket notation. $\endgroup$
    – user82205
    Apr 28, 2020 at 13:00
  • $\begingroup$ Happy to help. MathJax (i.e. Latex syntax) is not the most intuitive. There are some guides on our meta, if you want to learn more. $\endgroup$ Apr 28, 2020 at 17:58

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