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In my book it states the magnetic quantum number “$m_\ell$" is the spatial orientation of the orbital with respect to a standard set of coordinates. I get what $n$ and $\ell$ is, but what does it mean visually or conceptually, if $m_\ell$ is $1$ vs $-1$? For instance if $n = 4$, $\ell = 1$ and $m_\ell = -1$ vs. $n = 4$, $\ell = 1$ and $m_\ell = 1$?

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    $\begingroup$ Related question - quoting: "Magnetic quantum number (ml) - indicates the orientation of a particular orbital in space. ... While each orbital is indeed a p-orbital, we can describe each orbital uniquely by assigning this third quantum number to indicate its position in space. " Does that answer your question? $\endgroup$ – Gaurang Tandon Feb 17 '18 at 3:05
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The magnetic quantum number can be best explained if it's viewed through quantum mechanics.

The orbitals you have read are nothing but the solutions of Schrodinger's equation $\hat{H}\Psi = E\Psi$. We can express the Laplacian $\nabla^2$ in spherical coordinates as such:

$$ -\frac{\hbar^2}{2m}\nabla^2=-\frac{-h^2}{2m}\frac{1}{r^2}\left\{\frac{\partial}{\partial r}\left(r^2\frac{\partial}{\partial r}\right) + \frac{1}{\sin{\theta}}\left[\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial}{\partial\theta}\right)\right]+\frac{1}{\sin^2{\theta}}\frac{\partial^2}{\partial\phi^2}\right\}$$

or alternatively as:

$$ -\frac{\hbar^2}{2m}\nabla^2=-\frac{-h^2}{2m}\frac{1}{r^2}\left[\frac{\partial}{\partial r}\left(r^2\frac{\partial}{\partial r}\right)\right] + {L^2}/{2mr^2}$$

where $L^2$ is the square of the angular momentum operator.

So, for solving Schrodinger's equation, one can independently solve the radial part and the angular part as the variables are independent. So, if you want to solve the angular part i.e $L^2 \Psi = E_1 \Psi$. You will find the solution as spherical harmonics ($Y_l^m (\theta,\phi)$) and the eigen values related to the azimuthal quantum numbers as $l(l+1)\hbar^2$. The $\phi$ part depends on eigenvalues as $m_l\hbar$ where $m_l$ is your magnetic quantum number which can take value from $-l$ to $+l$.

This calculation also yields another special result. If the electron is introduced in an external magnetic field, the angular momentum vector can only make certain definite angles with the magnetic field which can be calculated as $$\cos\theta = \frac{\langle \psi{\left|L_z\right|\psi\rangle}}{\left\langle\psi|L|\psi\right\rangle} = \frac{m_l}{\sqrt{l\left(l+1\right)}}$$

So $l=1$ and $m=1$ means when applied to external magnetic field the angular momentum vector will orient at angle $\arccos(1/\sqrt2)$ or $45^\circ$. But $m=-1$ means that the angular momentum vector will orient with a magnetic field at angle $\arccos(-1/\sqrt2)$ or at $135^\circ$. Thus $m_l$ tells about the possible spatial orientation of the angular momentum vector in the presence of a magnetic field and that's why they are called magnetic quantum numbers.

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Just like $n$ and $l$, the quantum number $m_l$ is an parameter of the wave function, which describes the orbital. The Hydrogen orbitals can be split into a radial part (parametrized by $n$ and $l$) and an angular part (parametrized by $l$ and $m_l$):

\begin{equation} \Psi(r,\theta,\phi) = R_{nl}(r) Y_l^{m_l} (\theta,\phi) = R_{nl}(r) f_l^{m_l} (\theta) g_{m_l} (\phi) \end{equation}

The angular part is given as \begin{align} f_l^{m_l}(\theta) &= (-1)^{m_l} \sqrt{\frac{(2l+1)(l-{m_l})!}{4\pi (l+{m_l})!}} P_l^{m_l}(\cos \theta) \\ g_{m_l}(\phi) &= e^{i {m_l} \phi} \end{align} with $P_l^{m_l}(\cos \theta)$ being the associated Legendre polynomials. As you can see for ${m_l}\neq0$ the angular part gets complex valued.

What does this mean conceptually?

For each possible value of $l$ there are $2l+1$ possible values of $m_l$. This is why $s$ orbitals always come alone, $p$ orbitals in groups of three, $d$ orbitals in groups of five and so on. For example we have $p_z, p_y, d_{z^2}, d_{xy}$. They have different shapes (spatial orientations), but are degenerate in energy (in case of the hydrogen atom). Due to electron-electron interactions in many electron system, these energy levels can split.

The value of $m_l$ further specifies the orientation of the orbital angular momentum, while $l(l+1)$ is related to the magnitude of the orbital angular momentum. The name magenetic quantum number refers to the splitting of the energy eigenvalues for different $m_l$ when an external magnetic field is present (Zeeman effect).

What does this mean visually?

If we plot for example for $l=1$ the angular part $Y_l^{m_l}$, we get $2l+1=3$ times the typical $p$ orbital shape, each one aligned along one cartesian coordinate ($x$,$y$ and $z$).

Note however, that we have a complex part $g_{m_l}(\phi)$ for $m_l\neq0$. Plotting complex valued 3-dimensional functions is not trivial. On Wolfram I found a plot showing this thing.

complex representation of $Y_1^1$

Note that it looks like a donut, instead of the $p$ shape typically found in text books. To get there we can make a real superposition by employing Euler's formula

\begin{align} \cos(\phi) &= \frac{1}{\sqrt{2}}(e^{i\phi} + e^{-i\phi}) \\ \sin(\phi) &= \frac{1}{\sqrt{2}i}(e^{i\phi} - e^{-i\phi}) \\ \end{align}

So we combine two complex $Y_l^{m_l}$ with same $|m_l|$ to get two real valued new functions, one with $\cos(\phi)$ and one with $\sin(\phi)$. The real functions represent $p_x$ and $p_y$ orbitals, while the $p_z$ orbital is given by $Y_1^0$.

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  • $\begingroup$ I would add in your conceptual section that they can split with a magnetic field (the Zeeman effect) as well, thus explaining why it is called the magnetic quantum number. $\endgroup$ – Tyberius Feb 21 '18 at 23:12
  • $\begingroup$ I added 2 sentences about that, but it also covered in the answer by Soumik Das. $\endgroup$ – Feodoran Feb 22 '18 at 12:41

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