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 ?

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    $\begingroup$ That's right, 1s orbital has no nodes and no negative part. Higher s-orbitals do have both. $\endgroup$ – Ivan Neretin Aug 7 '16 at 17:25

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 (one of many sources):

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

Since $Z$ and $e$ are both positive numbers, $\psi_{\mathrm{1s}}$ is positive for all values of $(r,\theta,\phi)$, the wavefunction being independent of the two angles.


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).

  • $\begingroup$ Well, what you would write in an answer really depends on how deep down you want to go :) I just went for a superficial one. +1 anyway. $\endgroup$ – orthocresol Aug 10 '16 at 8:01
  • $\begingroup$ But then, when combining 1s and 1s orbital by LCAO, we have both sigma and sigma* molecular orbitals(as in hydrogen molecule). $\endgroup$ – Shoubhik Raj Maiti Oct 20 '16 at 8:53
  • $\begingroup$ @ShoubhikRajMaiti yes, so what? Excuse me, I do not get what you are trying to point out $\endgroup$ – user1420303 Oct 20 '16 at 11:15
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    $\begingroup$ @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 $\endgroup$ – user1420303 Oct 20 '16 at 17:44
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    $\begingroup$ @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? $\endgroup$ – user1420303 Oct 20 '16 at 17:55

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