# What does m_l represent and why does it correspond to each of the orbitals for a given subshell?

Feb 3, 2016

For the angular distribution function ${Y}_{l}^{{m}_{l}} \left(\theta , \phi\right)$, ${m}_{l}$ stands for the projection of the angular momentum $l$. As it turns out, it is basically the shape of the orbitals because it corresponds to the $z$ angular momentum of the orbital.

THE MAGNETIC QUANTUM NUMBER, FOR THE 4F ORBITAL

${m}_{l}$, the magnetic quantum number, takes on the values $0 , \pm 1 , \pm 2 , . . . , \pm l$, where $l$ is an integer. For each $l$, there corresponds an orbital name, so $l = 0 \to s$, $l = 1 \to p$, $l = 2 \to d$, $l = 3 \to f$, $l = 4 \to g$, etc.

For the $4 f$ orbital, $n = 4$ and $l = 3$, meaning that ${m}_{l} = 0 , \pm 1 , \pm 2 , \pm 3$.

Therefore, there are $7$ ${m}_{l}$ values for the $4 f$ orbital, and there happens to be 7 different $4 f$ orbitals:

• $4 {f}_{y \left(3 {x}^{2} - {y}^{2}\right)} : {m}_{l} = - 3$
• $4 {f}_{z \left({x}^{2} - {y}^{2}\right)} : {m}_{l} = - 2$
• $4 {f}_{y {z}^{2}} : {m}_{l} = - 1$
• $4 {f}_{{z}^{3}} : {m}_{l} = 0$
• $4 {f}_{x {z}^{2}} : {m}_{l} = + 1$
• $4 {f}_{x y z} : {m}_{l} = + 2$
• $4 {f}_{x \left({x}^{2} - 3 {y}^{2}\right)} : {m}_{l} = + 3$

But why are there $7$? Yes, ${m}_{l}$ correlates with how there are $7$, but there are other reasons why there are $7$ unique $f$ orbitals.

MORE-OR-LESS WHY THERE ARE $\setminus m a t h b f \left(2 l + 1\right)$ ORBITALS POSSIBLE

Consider the $2 p$ orbitals as a simpler example.

When you have ${m}_{l} = - 1 , 0 , + 1$, you know you get three orbitals: $2 {p}_{x}$, $2 {p}_{y}$, and $2 {p}_{z}$, but ${m}_{l}$ only correlates with three orbitals without further explanation.

You could also take the $2 {p}_{z}$ orbital and think of it another way. If you want the angular momentum in the $z$ direction, you can examine the following angular part of the Schrodinger equation in spherical coordinates:

hatL_zY_l^(m_l)(theta,phi) = m_lℏY_l^(m_l)(theta,phi)

resembling $\hat{H} \psi = E \psi$.

Without getting into too many complicated details (you don't have to know the three expressions for the ${Y}_{l}^{{m}_{l}} \left(\theta , \phi\right)$), you can assert that the $z$ angular momentum ${L}_{z}$ can be expressed for three different ${m}_{l}$ values, which trace out the $2 {p}_{z}$ orbital in units of ℏ, where ℏ = h/(2pi) and $h$ is Planck's constant.

You have three relevant ${m}_{l}$ projections depicted (ignore the pm2ℏ; that's because the diagram is of a $d$ orbital, but we can apply it to the $2 {p}_{z}$ orbital):

• m_l = -1 -> L_z => -ℏ (forming the bottom lobe, labeled -ℏ)
• ${m}_{l} = 0 \to {L}_{z} \implies 0$ (forming the central dot, or node, labeled $0$)
• m_l = +1 -> L_z => ℏ (forming the top lobe, labeled ℏ)

Naturally, the top lobe is symmetrically-shaped in relation to the bottom lobe.

THE ORTHOGONALITY CONDITION

Now, a quantum mechanical postulate/requirement for an orbital is that it be orthogonal (perpendicular) to all the other possible orbitals.

Two mathematical ways of conveying that are the dot product and the cross product:

$\hat{x} \cdot \hat{y} = \left\langle1 , 0 , 0\right\rangle \cdot \left\langle0 , 1 , 0\right\rangle = 0$
$\hat{x} \cdot \hat{z} = \left\langle1 , 0 , 0\right\rangle \cdot \left\langle0 , 0 , 1\right\rangle = 0$
$\hat{y} \cdot \hat{z} = \left\langle0 , 1 , 0\right\rangle \cdot \left\langle0 , 0 , 1\right\rangle = 0$
$\hat{x} \times \hat{y} = \left\langle1 , 0 , 0\right\rangle \times \left\langle0 , 1 , 0\right\rangle = \left\langle0 , 0 , 1\right\rangle = \hat{z}$
$\hat{y} \times \hat{z} = \left\langle0 , 1 , 0\right\rangle \times \left\langle0 , 0 , 1\right\rangle = \left\langle1 , 0 , 0\right\rangle = \hat{x}$
$\hat{z} \times \hat{x} = \left\langle0 , 0 , 1\right\rangle \times \left\langle1 , 0 , 0\right\rangle = \left\langle0 , 1 , 0\right\rangle = \hat{y}$

Since the $2 {p}_{z}$ lies along the $z$ axis and since the $z$ axis is the $2 {p}_{z}$ orbital's axis of symmetry, the only way the other orbitals can be orthogonal is that they lie along the $\setminus m a t h b f \left(x\right)$ and $m a t h b f \left(y\right)$ axes (as you may have recognized in the pattern from the above dot and cross products).

Clearly, the $x$, $y$, and $z$ axes, as you've seen many times in $x y z$ coordinate systems, are perpendicular.

As a result of ${m}_{l} = - 1 , 0 , + 1$ as well as the orthogonality condition, there are three orbitals possible: $2 {p}_{x}$, $2 {p}_{y}$, and $2 {p}_{z}$.

Similarly, because there are $7$ values of ${m}_{l}$ for the $4 f$ orbitals, they are constructed to be particularly shaped according to the above angular momentum Schrodinger equation, in such a way that they are all orthogonal to each other.

There just happens to be $7$ different orbital shapes that are all orthogonal to each other, just in more complex ways than the $2 p$ example.