I know that the signs '+' and '-' are written in case of lobes of p and d atomic orbitals, which denote the sign of the wave function of that orbital. But s-orbital is also a wave. So, it should also have negative regions. However, most books show the s-orbital with a '+' sign only. What is the significance of this? Again, the 2s, 3s, 4s, ... etc orbitals have some nodes. So does these nodes indicate the seperation of '+' phase and '-' phase? The 1s orbital does not have any node.So where is the '-' part of the 1s orbital ?

  • 5
    $\begingroup$ That's right, 1s orbital has no nodes and no negative part. Higher s-orbitals do have both. $\endgroup$ Aug 7, 2016 at 17:25
  • $\begingroup$ I take a much simpler approach. Since s orbitals can interact in or out of phase (in bonding or anti-bonding interactions), then they could be represented by opposite signs (phases). $\endgroup$
    – Guest
    Nov 1, 2019 at 23:14

3 Answers 3


The sign of the wave function does not have any physical meaning since for every solution $\Psi$ of the Schrödinger equation $-\Psi$ is also a solution. In fact there are infinitely many equivalent solutions that only differ in their overall phase.

So the (meaningless) answer to your question is yes.

It should be noted that $\Psi \Psi^*$, that can be physically interpreted as a probability distribution for finding a particle at a certain point, is always a non-negative real number.


As noted in the comments, the 1s orbital does not have any nodes. In general an orbital with quantum numbers $n$ and $l$ will have $n-1$ nodes, $l$ of which are angular nodes, and $n-l-1$ of which are radial nodes.

The wavefunction for the 1s orbital (in a hydrogenic atom with nuclear charge $Z$) is commonly written as:

$$\psi_{\mathrm{1s}} = \sqrt{\frac{1}{4\pi}}\cdot 2Z^{3/2}e^{-Zr}$$

Since $Z$ and $e$ are both positive numbers, $\psi_{\mathrm{1s}}$ is positive for all values of $(r,\theta,\phi)$; indeed, the wavefunction has no dependence on the two angles. Thus, the most conventional representation of the 1s orbital depicts it as being purely positive.

However, note that if $\psi_{\mathrm{1s}}$ satisfies the Schrödinger equation for the hydrogen atom, so must $-\psi_{\mathrm{1s}}$ (and also any other phase-shifted form $\mathrm{e}^{\mathrm i\alpha}\psi_{\mathrm{1s}}$ for a generic angle $\alpha$; for $\alpha = \pi$, this prefactor evaluates to $-1$). It turns out that the absolute phase of a wavefunction, on its own, has no physical meaning. We cannot measure the wavefunction directly, only indirectly by way of measuring observables, and the values of all observables are independent of the phase of the wavefunction:

$$\langle O \rangle = \langle -\psi | \hat{O} | {-\psi} \rangle = \langle \psi | \hat{O} | \psi \rangle$$

Thus, the choice to display the 1s orbital as being purely positive is not mandated by a physical requirement, but is rather a conventional choice. The difference between the 1s orbital and a 2p orbital, for example, is that the 2p orbital contains regions that have opposite signs to one another: thus, no matter how you choose the phase of a 2p orbital, there will always be a positive and a negative region.

Note that the absolute phase of an orbital does not have the same physical meaning as the relative phases of two orbitals when constructing linear combinations (e.g. in molecular orbitals). For example, the bonding MO of dihydrogen is made up of an in-phase combination of hydrogen 1s orbitals. Here it is physically important, and mandatory, that the two contributing 1s orbitals have the same phase. However, exactly what phase this is does not matter: they can both be positive, or they can both be negative.

  • $\begingroup$ So you want to say that we tend to write 1s orbital with positive sign conventionally however it can be negative as well as that also satisfies the wave equation? $\endgroup$
    – Lalit
    Sep 29, 2021 at 13:48
  • $\begingroup$ @Lllt Yes, and it goes beyond orbitals, it is true of any quantum state. $\psi$ is exactly the same state as $-\psi$. $\endgroup$ Sep 29, 2021 at 16:13

Although the answers are correct, I think that they do not help you because your question shows a misunderstanding. Those are not the signs of the orbitals. The signs often used represent the coefficients of a linear combination of orbitals (look for LCAO method). They have not sense in isolated atoms.

1s orbitals are not special in this sense. Although the orbitals commonly used in practice are real valued functions, they do not need to be at all. Even more, excluding $n$s orbitals they are (before any transformation) complex valued functions, so the meaning of the signs would be very diffuse. Also, orbitals are defined up to a phase factor. If you multiply an orbital by a complex number (eg. -1+0i), it is as valid as before (it is not exactly true (only valid for the wave function), they loose some important property, but the orbitals you use are already transformed).

  • $\begingroup$ But then, when combining 1s and 1s orbital by LCAO, we have both sigma and sigma* molecular orbitals(as in hydrogen molecule). $\endgroup$
    – S R Maiti
    Oct 20, 2016 at 8:53
  • $\begingroup$ @ShoubhikRajMaiti yes, so what? Excuse me, I do not get what you are trying to point out $\endgroup$ Oct 20, 2016 at 11:15
  • $\begingroup$ @user1420303 A sigma orbital forms when orbitals of same sign overlap and sigma* when orbitals of opposite sign overlap. If s-orbital has only positive sign then there would be no antibonding orbital. Shouldn't it? $\endgroup$
    – S R Maiti
    Oct 20, 2016 at 16:52
  • 2
    $\begingroup$ @ShoubhikRajMaiti , In my answer I tried to clarify that it is not true that A sigma orbital forms when orbitals of same sign overlap and sigma when orbitals of opposite sign overlap*, because there it is meaning less to speak about "sign of an orbital". We can say that: $\sigma = c_A \cdot f_{1s_A}+ c_B \cdot f_{1s_B}$ , where $f_{1s_A}$ and $f_{1s_B}$ differ from each other only by the coordinate origin (nuclei A or B). In such case '+' and '-' are related to $C$s coefficients and are not related to the orbitals $f$s. The equality in $f$s is arbitary. You can choose other options, but $\endgroup$ Oct 20, 2016 at 17:44
  • 1
    $\begingroup$ @ShoubhikRajMaiti ... in such case you have to modify $C$s coefficients according. Please note, if you take the more primitive and general form for $f$s, you'll get complex valued functions. In such case you can't speak about "positive" or "-" at all. Also, note that if $f$ is a valid wave function, $z\cdot f$ (with $z$ being a complex number) is as valid as $f$. So, signs in $f$ is meaning less. Even more: Consider the a real valued wavefunction of a particle in a box: hyperphysics.phy-astr.gsu.edu/hbase/quantum/pbox.html Is it negative or positive? $\endgroup$ Oct 20, 2016 at 17:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.