In an hydrogen-like atoms the orbitals are solutions to the Schrodinger equation suitable for the problem.

They describe the regions where an electron can be found.

So, why don't they have spherical symmetric?

As in this picture:

enter image description here

Why, in a certain distance from the nucleus, are there points where an electron can be found, and points where it can't?

EDIT: Is there any intuation about the symmetry breaking?

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    $\begingroup$ Why should they have spherical symmetry? The classical gravitational potential is spherical symmetric, too, but orbits are not fully spherical symmetric, are they? $\endgroup$ – ACuriousMind Apr 8 '15 at 19:47
  • $\begingroup$ Why not? What is the difference between two points whitin the same distance from the nucleus? $\endgroup$ – eranreches Apr 8 '15 at 19:49
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    $\begingroup$ In general, adding angular momentum is going to break spherical symmetry. $\endgroup$ – zeldredge Apr 8 '15 at 19:54
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    $\begingroup$ I don't have time, but a complete answer would include the fact that selected sets of the orbitals do sum to a spherically symmetric distribution. And in particular all closed shells are spherically symmetric. This is essentially impossible to see in the visualizations because of the information they throw away. Just another reason to treat visualizations with a certain wary regard. $\endgroup$ – dmckee --- ex-moderator kitten Apr 8 '15 at 20:09
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    $\begingroup$ But the polynomials, which are a complete basis function set, are not all spherically symmetric - they can't be and still be a basis set. Your intuition that all the functions must be spherically symmetric is leading you astray - that just isn't so, and the math is telling you that. $\endgroup$ – Jon Custer Apr 8 '15 at 20:19

My answer has a lot of overlap with the earlier answer by NoEigenvalue, but grew too long for a comment.

The orbitals are segregated by the amount of angular momentum that they carry. No angular momentum, $\ell=0$, are called $s$-orbitals, and are spherically symmetric. The first two examples in your figure are $s$-orbitals with $n=1$ and $n=2$.

Angular momentum explicitly breaks spherical symmetry, so the orbitals with $\ell\neq0$ can't be spherically symmetric. If you define some axis as your $z$ direction, the wavefunction with $m=0$ has value zero in the $x$-$y$ plane, one sign above, the other sign below. The wavefunction with $m=+1$ has its maximum magnitude on a ring in the $x$-$y$ plane, with the phase of the wavefunction increasing as you circle the $z$-axis going one way; the wavefunction with $m=-1$ has its phase increasing as you circle the $z$-axis the other way. This corresponds to $m=\pm1$ wavefunctions representing particles which circle the $z$-axis, and $m=0$ wavenfunctions representing particles with angular momentum vectors lying in the $x$-$y$ plane.

Somewhat confusingly, most of the discussion about orbitals online carries a pretty heavy chemistry bias, where the direction of the angular momentum vector isn't usually so critical as the spatial distribution of the charge. It turns out that if you just add the two orbitals with $m\neq0$ you end up with a probability distribution that looks just like the $m=0$ orbital, but oriented along the $x$-axis instead of the $z$-axis; this is usually called a $p_x$ orbital, contrasted with $p_z$. Likewise if you add the orbitals with a relative phase of $i=\sqrt{-1}$, you end up with $p_y$. That's what's in your figure: $p_x, p_y, p_z$. Each of them has $\ell=1,m=0$, just along a different axis, which makes the angular momentum interpretation more confusing.

You'll be satisfied to know that if I have a particle with $\ell=1$ but absolutely no information about the orientation of that angular momentum, then my particle is in an equal mixture of $p_x, p_y, p_z$ and its probability distribution does become spherically symmetric.

You ask in a comment for a little more detail about signs and phases. We can be explicit: apart from a normalization coefficient, the $\ell=1$ spherical harmonics are \begin{align} Y_1^{-1}(\theta,\phi) &\propto e^{-i\phi} \sin\theta &&= \frac{x-iy}{r} \\ Y_1^{0}(\theta,\phi) &\propto \cos\theta &&= \frac{z}{r} \\ Y_1^{+1}(\theta,\phi) &\propto -e^{i\phi} \sin\theta &&= -\frac{x+iy}{r} \end{align} Here $x,y,z$ are the usual position coordinates, $r$ is the length of a vector $\vec r$ whose components are $(x,y,z)$, the angle $\theta$ is between $\vec r$ and the $z$-axis, and the angle $\phi$ is between the projection of $\vec r$ onto the $x$-$y$ plane and the $x$-axis. The complete wavefunction also includes the radial part $$ \psi_{n\ell m} = R_{n\ell}(r) Y_\ell^m(\theta,\phi) $$ where the radial wavefunctions have the form $$ R_{n\ell}(r) \propto r^{n-1} e^{-r/r_0}. $$ This gives us, for $\ell=1$: \begin{align} \psi_{1,1,-1} - \psi_{1,1,+1} &\propto 2x e^{-r/r_0} \\ -\psi_{1,1,-1} - \psi_{1,1,+1} &\propto 2iy e^{-r/r_0} \\ \psi_{1,1,0} &\propto z e^{-r/r_0} \end{align} This motivates the $p_x,p_y,p_z$ description above.

As far as signs and phases: you can see that $\cos\theta$ is positive if $\vec r$ is near the $z$-axis, negative if $\vec r$ is near the $-z$-axis, and zero on the $x$-$y$ plane; that gives the "opposite-colored blobs" interpretation that you're asking about.

For the $m\neq0$ harmonics, you can see that $\sin\theta$ is positive for all $\theta$ (notice that from the positive $z$-axis to the negative $z$-axis is only a half-turn, not a full turn). The difference between the two states is the sign of the exponent in $e^{\pm i\phi}$. However the $m\neq0$ harmonics are purely imaginary on the $y$-axis, which makes them a bit harder to draw.

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The orbitals are only spherically symmetric for $S$-states, for which the angular momentum number $l$ is zero. So in you picture the first two orbitals are the $s=1$ and the $s=2$ state, while the other three pictures correspond to the $n=2,l=1$ with the three different possibilities of $m_l$.

There are several ways to see that the wavefunctions with $l=0$ are sherically symmetric, one of which is that you can write a wavefunction of the hydrogen atom as

$$\psi_{nlm}=R_{nlm}(r)Y_{lm}(\theta,\phi), $$

where $Y$ are the spherical harmonics which determine the angular dependence of the wavefunction. Then, if you look up the spherical harmonic with $l=0$, you see that it has no angular dependence. The wavefunction is in this case spherically symmetric (and 3D, since it's the "spherical coordinates" - $r$).

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Just wanted to add that it's not totally true that the drawn orbitals are "the regions where an electron can be found". But my answer grew and grew...

Let's take a neutral boron atom where it has filled 1s and 2s shells and one electron in the 2p shell. Suppose it is floating in space, far away from messy interactions with other things. So, you wonder, which 2p orbital is the electron in?

If you physically prepare an atom in a given state, then you have physically introduced a very real asymmetry. For example let's say you prepare a $2p_z$ atom, and you somehow you find a way to image where electrons are. After repeating this on many many identically prepared atoms, and building statistics to build your experimental "probability cloud", you would find lobes on top and bottom (added to the $1s$ and $2s$ spherical clouds).

But of course it does not have to be prepared in an x, y, or z state. While you can in principle prepare a boron atom with a $2p_x$ state, or $2p_y$ state, you can also prepare it in a $\frac{1}{\sqrt{2}}(|2p_x\rangle + |2p_y\rangle)$ superposition state or any other superposition. Those superpositions are also valid orbitals, no lesser than the orbitals you have drawn. So you can also prepare boron atoms so that the hypothetical image of electron clouds, the lobes are oriented 45 degrees from z-axis. Or you could prepare the atoms to have not lobes but instead a fat equator: $\frac{1}{\sqrt{2}}(|2p_x\rangle + i \, |2p_y\rangle)$.

For a randomly chosen boron atom in a gas, we would of course like a mathematical description that restores the spherical symmetry. Here is where the quantum mechanical concept of mixed states, and density matrices, comes in. This complete description of the boron atom, including uncertainty, tells you that for any measurement you make, the randomly chosen atom has no preferred direction. Unfortunately this complete quantum mechanics is mathematically more confusing for the initiate, and it seems we have not invented the proper English to describe mixed states; that is why we say strange things like "the unpaired boron electron is randomly in one of three 2p orbitals" which is quite correct in some sense but may be misleading. It is also equally valid in any other basis of 2p orbitals to say it is randomly in one of the three orbitals. That human choice of a mathematical basis is where the spherical asymmetry comes in, but the basis will never appear in any observable.

Technical note: If you keep digging, you find out that due to spin-orbit coupling, the six spin-orbital states of boron have broken degeneracy. Two form the ground state and the other four have slightly elevated energy (See NIST database). But does this mean the random boron atom is in a mixed state of the two ground state orbitals? Nope! The energy splitting is only 1.9 meV; so, for most experimental temperatures all six spin-orbital states would have equal probability.

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Comments to the question (v2):

The fact that the TISE is invariant under a symmetry group $G$ (in this case the Lie group $G=SO(3)$ of 3D rotations) does not imply that the orbital/wave-function solutions $\psi$ must be $G$-invariant as well.

(Think e.g. on spontaneous symmetry breaking where the governing laws of a physical system are invariant under a symmetry group $G$, but the system chooses a ground state which is not invariant under $G$. See also e.g. this Phys.SE post.)

More generally, a rotated wave-function solution is again a solution to TISE.

In other words, the wave-function solutions $\psi$ do transform in Lie group representations of $G$, although (as said above) not necessarily in the trivial/single representation. See also e.g. this Phys.SE post.

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