My current understanding of orthogonal wavefunctions is: two wavefunctions that are perpendicular to each other and must satisfy the following equation: $$\int\psi_1\, \psi_2\, \mathrm{d}\tau =0$$

From this, it implies that orthogonality is a relationship between 2 wavefunctions and a single wavefunction itself can not be labelled as 'orthogonal'. They must be orthogonal with respect to some other wavefunction. However I have seen some textbooks refer to single wavefunctions as being orthogonal.

Am I missing some crucial thing in my definition of orthogonal wavefunctions? Could anyone provide me with a more complete definition of orthogonal wavefunctions from a chemistry point of view?

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    $\begingroup$ Could you provide a little bit more context i.e. examples of said claims in said textbooks? $\endgroup$ Nov 1, 2015 at 5:38
  • $\begingroup$ @orthocresol Well, it said something along the lines, 'all wavefunctions are orthogonal', and in this question, chemistry.stackexchange.com/questions/7029/…, they talk of single wavefunctions as being orthogonal $\endgroup$
    – user22450
    Nov 1, 2015 at 5:56
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    $\begingroup$ It is not really talking of a single wavefunction as being orthogonal, but rather a set of wavefunctions. By having a set of orthogonal wavefunctions $\psi_1 , \psi_2 \ldots$ it means that any pair of wavefunctions that you pick will be orthogonal to each other. This will always happen if the wavefunctions are eigenfunctions of a hermitian operator (it is one of the standard "elementary" proofs in QM). I think the question that you linked does use the word a little loosely, although I still understood it to refer to a set of wavefunctions. $\endgroup$ Nov 1, 2015 at 6:26
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    $\begingroup$ @orthocresol Right , that makes a lot more sense. So is my understanding of what orthogonal wavefuntions correct? Is there more of a chemistry way at explaining orthogonal wavefunction, rather than using mathematical expressions? $\endgroup$
    – user22450
    Nov 1, 2015 at 6:37
  • $\begingroup$ Somewhat. Have you done MO theory? Say in HF, the H 1s orbital doesn't overlap with the F 2px or 2py orbitals. That's because those two wavefunctions are orthogonal to each other. $\endgroup$ Nov 1, 2015 at 11:10

3 Answers 3


The notion of orthogonality in the context of the question referrers to the very well-known general concept of linear algebra, the branch of mathematics that studies vector spaces. Instead of going deep into the mathematics (that requires at least 50 textbook pages) let's just clear some OP's doubts.

First, a small (but important) correction: two wave functions $\psi_1$ and $\psi_2$ are called orthogonal to each other if $$ \int \overline{\psi}_1 \psi_2 \, \mathrm{d} \tau = 0 \, , $$ where the first function is complex-conjugated as indicated by a bar on top of it. Wave functions are complex-valued functions and complex-conjugation of the first argument is important.1

So, yes, orthogonality is a not a property of a single wave function. It either refers to a pair of them being orthogonal to each other as described above, or, in general, to a set of them, being all mutually orthogonal to each other, i.e. to a set $\{ \psi_i \}_{i=1}^{n}$ such that for any $i \neq j$ $$ \int \overline{\psi}_i \psi_j \, \mathrm{d} \tau = 0 \, . $$ In the last case it is said that the whole set $\{ \psi_i \}_{i=1}^{n}$ is orthogonal.

1) More precisely, one of the two functions has to be complex-conjugated in this expression, where which one is the matter of convention: in physical literature it is often the first function, while in mathematically oriented literature it is usually the second one.

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    $\begingroup$ I wanted to add one other important thing in my answer but you were faster; feel free to include if you think its important ;). It often occurs that textbooks write orthogonal set but mean orthonormal set and normalisation is indeed a property of a wavefunction on it's own. $\endgroup$
    – mcocdawc
    Nov 1, 2015 at 13:00
  • $\begingroup$ @Wildcat Does $\int \overline{\psi}_1 \psi_2 \, \mathrm{d} \tau = 0 \Rightarrow \int \overline{\psi}_2 \psi_1 \, \mathrm{d} \tau = 0 $? $\endgroup$
    – ado sar
    Aug 19, 2020 at 15:13
  • $\begingroup$ @adosar yes you can conjugate the expression. $\endgroup$
    – Silas
    Feb 17, 2023 at 7:56
  • $\begingroup$ @adosar, yes. In fact, $\int \overline{\psi}_i \psi_j \, \mathrm{d} \tau = \int \overline{\psi}_j \psi_i \, \mathrm{d} \tau$. $\endgroup$
    – Wildcat
    Feb 17, 2023 at 10:47

As described in Wildcat's answer, a single wave function cannot be orthogonal, but a set of wave functions can all be mutually orthogonal.

To address the second part of the OP's question, the physical meaning of orthogonality is that a pair of mutually orthogonal wave functions are mutually exclusive; observing one precludes the possibility of observing the other unless the system is changed.


The same wave functions can not be said to be orthogonal. the comparison must be between two different function.

  • $\begingroup$ This is not true. You can measure the overlap of any wavefunction with itself. If it's normalized, it should equal 1. $\endgroup$ Jan 6, 2018 at 20:23

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