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Solution of Schrödinger's equation for an Hydrogen atom is well known: $$\Psi_{n,l,m} (r, \theta, \phi) = N e^{\frac{-r}{n r_1}} R_n^l (r) P_l^m(cos \theta) e^{im\phi} \,.$$

If we interested with state with nonzeros angular momentum (say, $l=1$), the wave functions are: \begin{align} \Psi_{2,1,0} &= \frac{1}{4\sqrt{2\pi}a_0^{3/2}}\frac{r}{a_0}e^{-\frac{r}{2a_0}}\cos\theta\\ \Psi_{2,1,\pm1} &= \frac{1}{8\sqrt{\pi}a_0^{3/2}}\frac{r}{a_0}e^{-\frac{r}{2a_0}}\sin\theta e^{\pm i\phi} \end{align}

This wave functions descrie the state with certain projection of angular momentum to some Cartesian axes, i.e. it is an eigenfunctions functions of an operator $\hat L_z$. So, if we switch on the magnetic (or electric) field, the atom will be in some state of three possible, and the shape of its cloud will have some form that will not be spherically symmetric (for example, look here here).


But if there is no such direction, then what will be the shape of the hydrogen atom?


In chemistry, not the wave functions themselves are used, but their linear combinations, which are real:

\begin{align} p_z = \Psi_{2,1,0} &= \frac{1}{4\sqrt{2\pi}a_0^{3/2}}\frac{r}{a_0}e^{-\frac{r}{2a_0}}\cos\theta\\ p_x = \frac{1}{\sqrt{2}} (\Psi_{2,1, + 1} + \Psi_{2,1, - 1} ) &= \frac{1}{8\sqrt{2\pi}a_0^{3/2}}\frac{r}{a_0}e^{-\frac{r}{2a_0}}\sin\theta \cos\phi\\ p_y = \frac{i}{\sqrt{2}} (\Psi_{2,1, + 1} - \Psi_{2,1, - 1} ) &= \frac{1}{8\sqrt{2\pi}a_0^{3/2}}\frac{r}{a_0}e^{-\frac{r}{2a_0}}\sin\theta \sin\phi\\ \end{align}

These functions are no longer the eigenfunctions of the operator $\hat L_z$ and describe states with an indefinite projection of the angular momentum. But what exactly do they describe? If there is no dedicated direction, then all three functions $p_x, p_y, p_z$ must be equiprobable, that is, the state of the electron must be a linear combination of these three functions, and the shape of the cloud must therefore be spherically symmetric. How, then, can the directional valence be explained to form a chemical bond?

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    $\begingroup$ When there is some other atom nearby, the spherical symmetry is broken, and $L_z$ is not a good quantum number. There will be a difference in energy for different directions of the $p$ orbitals. $\endgroup$ – Pieter Feb 21 '18 at 7:57
  • $\begingroup$ @Pieter But if some other atom nearby, then there is some prefered direction, and $L_z$ should be good quantum number. Also, the good wave function should be not $p$'s, but $\psi$ themselves. Am I right? Why then chemistі used $p$'s-functions for bond explanation? $\endgroup$ – Sergio Feb 21 '18 at 8:12
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Just to be clear, even without a magnetic field, the eigenstates of the hydrogen atom are simply not all spherically symmetric. The space of eigenstates of a given energy is spherically symmetric. Spherical symmetry means that if we transform an eigenstate by rotating it, it will give another eigenstate, not that each eigenstate is spherically symmetric.

Now, if we take a hydrogen atom at a particular energy, and then let it sit in a spherically symmetric environment with which it somehow interacts weakly for a long time, it should end up in a spherically symmetric superposition of eigenstates of the energy we started with, i.e. it should end up looking like a sphere. If we then measure the angular momentum we are equally likely to find it pointing in any direction.

The wavefunctions you mentioned as used in chemistry -- let's focus on the two ''new'' ones $p_x$ and $p_y$ -- describe energy eigenstates with total angular momentum $1$ and moreover angular momentum $0$ in the $x$ and $y$ direction respectively (you missed a factor of $i$ in the definition of $p_y$ btw). The simplest way to see this is to consider the space of states of total angular momentum $1$ as a three dimensional vector space (this makes sense as long as you only look at the angular momentum operators). In this vector space, the angular momentum operators act like matrices of the form $$L_x \sim \begin{pmatrix} 0 & 0 & 0 \\0 & 0 & i \\ 0 & -i & 0 \end{pmatrix},\quad L_y \sim \begin{pmatrix} 0 & 0 & -i \\0 & 0 & 0 \\ i & 0 & 0 \end{pmatrix},\quad L_z \sim \begin{pmatrix} 0 & i & 0 \\-i & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} $$ and the $\Psi_{2,1,x}$ are represented by $$ \Psi_{2,1,0} \sim \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}, \quad \Psi_{2,1,1} \sim \begin{pmatrix} 1 \\ -i \\ 0 \end{pmatrix}, \quad \Psi_{2,1,-1} \sim \begin{pmatrix} 1 \\ i \\ 0 \end{pmatrix}.$$ Now if you take the linear combinations of $\Psi_{2,1,\pm1}$ that you write, you can easily see that they are in the null space of $L_x$ and $L_y$ respectively.

As soon as you put more than one hydrogen atom in your universe, there can be a preferred direction (as there definitely is in molecules), and this will put your hydrogen atom in a particular not-spherically-symmetric eigenstate, which allows for the typical picture of a valence bond.

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    $\begingroup$ If $p_z$ has $L_z = 0$, then $p_y$ --- $L_y = 0$ and $p_x$ --- $L_x = 0$. Am I right? $\endgroup$ – Sergio Feb 21 '18 at 21:44
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    $\begingroup$ You are completely right, thanks! I was too quick, made typos, and was thinking of the spin 1/2 representation, not the spin 1 case we are discussing here. I will edit my answer. $\endgroup$ – Stijn Feb 22 '18 at 9:09

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