quantum mechanics - Why aren't orbitals symmetric?

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:

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?

Answer

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.

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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.