*Linear combinations are allowed by the maths of quantum mechanics.If any set of wavefunctions is a solution to the Schrödinger equation, then any set of linear combinations of these wavefunctions must also be a solution.*

An equivalent statement is that these two orbitals do not lie on the $x$- and $y$-axes, but rather bisect them.

Thus it is typical to take linear combinations of them to make the equation look prettier.

(v) The coupling of spin and orbital angular momentum.

Russell-Saunders coupling of L and S to give the total angular momentum J . (vi) Application of the results to the understanding of the periodic table.

Stack Exchange network consists of 175 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. The reason for this outcome is that the wavefunctions are usually formulated in spherical coordinates to make the maths easier, but graphs in the Cartesian coordinates make more intuitive sense for humans.

Visit Stack Exchange The form of the p orbitals that we are familiar with are the $\mathrm_x$, $\mathrm_y$, and $\mathrm_z$ orbitals: (source: Chem Tube 3D) I also know that the p subshells have the quantum number $l = 1$, meaning that the magnetic quantum number can take the three values $m_l = -1, 0, 1$. The $p_x$ and $p_y$ orbitals are constructed via a linear combination approach from radial and angular wavefunctions and converted into $xyz$.

For hydrogen and hydrogen-like atoms it is often easy to guess the starting number of the m series.

If the guess is correct a graph of line wavenumber against 1/m = 0 (m = ∞) gives the ionization energy of the fixed level.

In switching from spherical to Cartesian coordinates, we make the substitution $z=r\cos$, so: $$\Psi_=zf(r)$$ This is $\Psi_$ since the value of $\Psi$ is dependent on $z$: when $z=0;\ \Psi=0$, which is expected since $z=0$ describes the $xy$-plane.

The other two wavefunctions are unhelpfully degenerate in the $xy$-plane.

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