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From physics courses, I'm told that (for Hydrojen atom only), for a given $\vec L$ of the electron,

$$|\vec L| * cos(\theta) = m_l,$$ where $\theta$ is the angle between the magnetic field and the orbital angular momentum $\vec L$.

Therefore, for example, if $\vec L$ lies on the $x-y$ plane, $L_z$ should be zero. However, when we see the value of $m_l$ for, say for $p$, subshell, they are $1, 0, -1$, and generally the orbitals are named as $p_x, py_, p_z$, and this confuses me. I mean if $\vec L$ is in the $z$ direction (i.e in the direction of $\vec B$), then $m_l = \pm 1$, and if it is on the $x-y$ plane, then $m_l = 0$, but naming implies as the values of $m_l$ are matched one-to-one to the names $p_x, p_y, p_z$, so am I confusing anything here, or the problem is just a silly naming of the orbitals ?

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  • $\begingroup$ I think naming them this way removes complex numbers. Otherwise it’s silly naming. In particular, how can you assign a direction to $\hat x$ and $\hat y$ in a problem with cylindrical symmetry? (Presumably the magnetic field is along $\hat z$ in your question.) $\endgroup$ – ZeroTheHero Feb 21 '18 at 4:02
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The $Y_{\ell m}$ are the spherical harmonics, with $Y_{1, -1}$ and $Y_{1, 1}$ rotating in opposite directions, eigenfunctions in spherical symmetry. Below is a figure of one of these hydrogen $2p$ orbitals, with complex phase coded as color. In an animation (when multiplied with the time dependent complex phase), the colors would circulate in one direction. For the other $Y_{\ell m}$ it would go in the other direction.

Spherical harmonic, phase as color

Now if these two spherical harmonics are added, this results in a standing wave, with static nodes and with static lobes of electron density, for example in the $x$ direction. Or when one subtracts them (adding with the opposite phase) an orbital with static lobes in the $y$ direction. The two linear combinations are appropriate to use in molecules where the spherical symmetry is broken.

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  • $\begingroup$ Peter I literally have no idea what you are trying to say. Can you explain what you mean a little bit ? $\endgroup$ – onurcanbektas Feb 21 '18 at 8:03
  • $\begingroup$ @onurcanbektas I added an image, For formulas see this question: physics.stackexchange.com/questions/387723/… $\endgroup$ – Pieter Feb 21 '18 at 10:09
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Your $p_x,p_y,p_z$ do not map to $m_l=-1,0,1$, but in stead linear combinations. See the link in Pieter's second comment.

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