4.2 An­gu­lar Mo­men­tum

Be­fore a so­lu­tion can be found for the im­por­tant elec­tronic struc­ture of the hy­dro­gen atom, the ba­sis for the de­scrip­tion of all the other el­e­ments and chem­i­cal bonds, first an­gu­lar mo­men­tum must be dis­cussed. Like in the clas­si­cal New­ton­ian case, an­gu­lar mo­men­tum is es­sen­tial for the analy­sis, and in quan­tum me­chan­ics, an­gu­lar mo­men­tum is also es­sen­tial for de­scrib­ing the fi­nal so­lu­tion. More­over, the quan­tum prop­er­ties of an­gu­lar mo­men­tum turn out to be quite un­ex­pected and im­por­tant for prac­ti­cal ap­pli­ca­tions.

4.2.1 De­f­i­n­i­tion of an­gu­lar mo­men­tum

The old New­ton­ian physics de­fines an­gu­lar mo­men­tum $\vec{L}$ as the vec­to­r­ial prod­uct ${\skew0\vec r}$ $\times$ ${\skew0\vec p}$, where ${\skew0\vec r}$ is the po­si­tion of the par­ti­cle in ques­tion and ${\skew0\vec p}$ is its lin­ear mo­men­tum.

Fol­low­ing the New­ton­ian anal­ogy, quan­tum me­chan­ics sub­sti­tutes the gra­di­ent op­er­a­tor $\hbar\nabla$$\raisebox{.5pt}{$/$}$​${\rm i}$ for the lin­ear mo­men­tum, so the an­gu­lar mo­men­tum op­er­a­tor be­comes:

$${\skew 4\widehat{\vec L}}= \frac{\hbar}{{\rm i}} {\skew 2\w... ...\partial}{\partial y}, \frac{\partial}{\partial z} \right) %$$ (4.18)

Un­like the Hamil­ton­ian, the an­gu­lar mo­men­tum op­er­a­tor is not spe­cific to a given sys­tem. All ob­ser­va­tions about an­gu­lar mo­men­tum will ap­ply re­gard­less of the phys­i­cal sys­tem be­ing stud­ied.

Key Points

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The an­gu­lar mo­men­tum op­er­a­tor (4.18) has been iden­ti­fied.

4.2.2 An­gu­lar mo­men­tum in an ar­bi­trary di­rec­tion

The in­tent in this sub­sec­tion is to find the op­er­a­tor for the an­gu­lar mo­men­tum in an ar­bi­trary di­rec­tion and its eigen­func­tions and eigen­val­ues.

For con­ve­nience, the di­rec­tion in which the an­gu­lar mo­men­tum is de­sired will be taken as the $z$-​axis of the co­or­di­nate sys­tem. In fact, much of the math­e­mat­ics that you do in quan­tum me­chan­ics re­quires you to se­lect some ar­bi­trary di­rec­tion as your $z$-​axis, even if the physics it­self does not have any pre­ferred di­rec­tion. It is fur­ther con­ven­tional in the quan­tum me­chan­ics of atoms and mol­e­cules to draw the cho­sen $z$-​axis hor­i­zon­tal, (though not in [25] or [52]), and that is what will be done here.

Fig­ure 4.7: Spher­i­cal co­or­di­nates of an ar­bi­trary point P.

Things fur­ther sim­plify greatly if you switch from Carte­sian co­or­di­nates $x$, $y$, and $z$ to “spher­i­cal co­or­di­nates” $r$, $\theta$, and $\phi$, as shown in fig­ure 4.7. The co­or­di­nate $r$ is the dis­tance from the cho­sen ori­gin, $\theta$ is the an­gu­lar po­si­tion away from the cho­sen $z$-​axis, and $\phi$ is the an­gu­lar po­si­tion around the $z$-​axis, mea­sured from the cho­sen $x$-​axis.

In terms of these spher­i­cal co­or­di­nates, the $z$-​com­po­nent of an­gu­lar mo­men­tum sim­pli­fies to:

$$\fbox{$\displaystyle \L _z \equiv \frac{\hbar}{{\rm i}} \frac{\partial}{\partial\phi} $} %$$ (4.19)

This can be ver­i­fied by look­ing up the gra­di­ent op­er­a­tor $\nabla$ in spher­i­cal co­or­di­nates in [41, pp. 124-126] and then tak­ing the com­po­nent of ${\skew0\vec r}\times\nabla$ in the $z$-​di­rec­tion.

In any case, with a bit of thought, it clearly makes sense: the $z$-​com­po­nent of lin­ear mo­men­tum clas­si­cally de­scribes the mo­tion in the di­rec­tion of the $z$-​axis, while the $z$-​com­po­nent of an­gu­lar mo­men­tum de­scribes the mo­tion around the $z$-​axis. So if in quan­tum me­chan­ics the $z$ lin­ear mo­men­tum is $\hbar$$\raisebox{.5pt}{$/$}$​${\rm i}$ times the de­riv­a­tive with re­spect the co­or­di­nate $z$ along the $z$-​axis, then surely the log­i­cal equiv­a­lent for $z$ an­gu­lar mo­men­tum is $\hbar$$\raisebox{.5pt}{$/$}$​${\rm i}$ times the de­riv­a­tive with re­spect to the an­gle $\phi$ around the $z$-​axis?

Any­way, the eigen­func­tions of the op­er­a­tor $\L _z$ above turn out to be ex­po­nen­tials in $\phi$. More pre­cisely, the eigen­func­tions are of the form

$$C(r,\theta) e^{{\rm i}m \phi} %$$ (4.20)

Here $m$ is a con­stant and $C(r,\theta)$ can be any ar­bi­trary func­tion of $r$ and $\theta$. For his­tor­i­cal rea­sons, the num­ber $m$ is called the “mag­netic quan­tum num­ber”. His­tor­i­cally, physi­cists have never seen a need to get rid of ob­so­lete and con­fus­ing terms. The mag­netic quan­tum num­ber must be an in­te­ger, one of $\ldots,-2,-1,0,1,2,3,\ldots$ The rea­son is that if you in­crease the an­gle $\phi$ by $2\pi$, you make a com­plete cir­cle around the $z$-​axis and re­turn to the same point. Then the eigen­func­tion (4.20) must again be the same, but that is only the case if $m$ is an in­te­ger. To ver­ify this, use the Euler for­mula (2.5).

Note fur­ther that the or­bital mo­men­tum is as­so­ci­ated with a par­ti­cle whose mass is also in­di­cated by $m$. This book will more specif­i­cally in­di­cate the mag­netic quan­tum num­ber as $m_l$ if con­fu­sion be­tween the two is likely.

The above so­lu­tion is eas­ily ver­i­fied di­rectly, and the eigen­value $L_z$ iden­ti­fied, by sub­sti­tu­tion into the eigen­value prob­lem $\L _zCe^{{{\rm i}}m\phi}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $L_zCe^{{{\rm i}}m\phi}$ us­ing the ex­pres­sion for $\L _z$ above:

$$\frac{\hbar}{{\rm i}} \frac{\partial C e^{{\rm i}m\phi}}{\p... ...{\rm i}} {\rm i}m C e^{{\rm i}m\phi} = L_z C e^{{\rm i}m\phi}$$

It fol­lows that every eigen­value is of the form:

$$\fbox{$\displaystyle L_z = m_l\hbar \mbox{ for $m_l$ an integer} $} %$$ (4.21)

So the an­gu­lar mo­men­tum in a given di­rec­tion can­not just take on any value: it must be a whole mul­ti­ple $m_l$, (pos­si­bly neg­a­tive), of Planck's con­stant $\hbar$.

Com­pare that with the lin­ear mo­men­tum com­po­nent $p_z$ which can take on any value, within the ac­cu­racy that the un­cer­tainty prin­ci­ple al­lows. $L_z$ can only take dis­crete val­ues, but they will be pre­cise. And since the $z$-​axis was ar­bi­trary, this is true in any di­rec­tion you choose.

It is im­por­tant to keep in mind that if the sur­round­ings of the par­ti­cle has no pre­ferred di­rec­tion, the an­gu­lar mo­men­tum in the ar­bi­trar­ily cho­sen $z$-​di­rec­tion is phys­i­cally ir­rel­e­vant. For ex­am­ple, for the mo­tion of the elec­tron in an iso­lated hy­dro­gen atom, no pre­ferred di­rec­tion of space can be iden­ti­fied. There­fore, the en­ergy of the elec­tron will only de­pend on its to­tal an­gu­lar mo­men­tum, not on the an­gu­lar mo­men­tum in what­ever is com­pletely ar­bi­trar­ily cho­sen to be the $z$-​di­rec­tion. In terms of quan­tum me­chan­ics, that means that the value of $m$ does not af­fect the en­ergy. (Ac­tu­ally, this is not ex­actly true, al­though it is true to very high ac­cu­racy. The elec­tron and nu­cleus have mag­netic fields that give them in­her­ent di­rec­tion­al­ity. It re­mains true that the $z$-​com­po­nent of net an­gu­lar mo­men­tum of the com­plete atom is not rel­e­vant. How­ever, the space in which the elec­tron moves has a pre­ferred di­rec­tion due to the mag­netic field of the nu­cleus and vice-versa. It af­fects en­ergy very slightly. There­fore the elec­tron and nu­cleus must co­or­di­nate their an­gu­lar mo­men­tum com­po­nents, ad­den­dum {A.39}.)

Key Points

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Even if the physics that you want to de­scribe has no pre­ferred di­rec­tion, you usu­ally need to se­lect some ar­bi­trary $z$-​axis to do the math­e­mat­ics of quan­tum me­chan­ics.

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Spher­i­cal co­or­di­nates based on the cho­sen $z$-​axis are needed in this and sub­se­quent analy­sis. They are de­fined in fig­ure 4.7.

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The op­er­a­tor for the $z$-​com­po­nent of an­gu­lar mo­men­tum is (4.19), where $\phi$ is the an­gle around the $z$-​axis.

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The eigen­val­ues, or mea­sur­able val­ues, of an­gu­lar mo­men­tum in any ar­bi­trary di­rec­tion are whole mul­ti­ples $m$, pos­si­bly neg­a­tive, of $\hbar$.

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The whole mul­ti­ple $m$ is called the mag­netic quan­tum num­ber.

4.2.2 Re­view Ques­tions

1.

If the an­gu­lar mo­men­tum in a given di­rec­tion is a mul­ti­ple of $\hbar$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1.054 57 10$\POW9,{-34}$ J s, then $\hbar$ should have units of an­gu­lar mo­men­tum. Ver­ify that.

So­lu­tion an­gub-a

2.

What is the mag­netic quan­tum num­ber of a macro­scopic, 1 kg, par­ti­cle that is en­cir­cling the $z$-​axis at a dis­tance of 1 m at a speed of 1 m/s? Write out as an in­te­ger, and show dig­its you are not sure about as a ques­tion mark.

So­lu­tion an­gub-b

3.

Ac­tu­ally, based on the de­rived eigen­func­tion, $C(r,\theta)e^{{\rm i}{m}\phi}$, would any macro­scopic par­ti­cle ever be at a sin­gle mag­netic quan­tum num­ber in the first place? In par­tic­u­lar, what can you say about where the par­ti­cle can be found in an eigen­state?

So­lu­tion an­gub-c

4.2.3 Square an­gu­lar mo­men­tum

Be­sides the an­gu­lar mo­men­tum in an ar­bi­trary di­rec­tion, the other quan­tity of pri­mary im­por­tance is the mag­ni­tude of the an­gu­lar mo­men­tum. This is the length of the an­gu­lar mo­men­tum vec­tor, $\sqrt{\vec{L}\cdot\vec{L}}$. The square root is awk­ward, though; it is eas­ier to work with the square an­gu­lar mo­men­tum:

$$L^2 \equiv \vec L \cdot \vec L$$

This sub­sec­tion dis­cusses the $\L ^2$ op­er­a­tor and its eigen­val­ues.

Like the $\L _z$ op­er­a­tor of the pre­vi­ous sub­sec­tion, $\L ^2$ can be writ­ten in terms of spher­i­cal co­or­di­nates. To do so, note first that

$${\skew 4\widehat{\vec L}}\cdot {\skew 4\widehat{\vec L}}= ... ...kew0\vec r}\cdot (\nabla \times ({\skew0\vec r}\times \nabla))$$

(That is the ba­sic vec­tor iden­tity (D.2) with vec­tors ${\skew0\vec r}$, $\nabla$, and ${\skew0\vec r}\times\nabla$.) Next look up the gra­di­ent and the curl in [41, pp. 124-126]. The re­sult is:

$$\L ^2 \equiv -\frac{\hbar^2}{\sin\theta} \frac{\partial}{... ...c{\hbar^2}{\sin^2\theta} \frac{\partial^2}{\partial \phi^2} %$$ (4.22)

Ob­vi­ously, this re­sult is not as in­tu­itive as the $\L _z$ op­er­a­tor of the pre­vi­ous sub­sec­tion, but once again, it only in­volves the spher­i­cal co­or­di­nate an­gles. The mea­sur­able val­ues of square an­gu­lar mo­men­tum will be the eigen­val­ues of this op­er­a­tor. How­ever, that eigen­value prob­lem is not easy to solve. In fact the so­lu­tion is not even unique.

Ta­ble 4.2: The first few spher­i­cal har­mon­ics.

The so­lu­tion to the prob­lem may be sum­ma­rized as fol­lows. First, the nonunique­ness is re­moved by de­mand­ing that the eigen­func­tions are also eigen­func­tions of $\L _z$, the op­er­a­tor of an­gu­lar mo­men­tum in the $z$-​di­rec­tion. This makes the prob­lem solv­able, {D.14}, and the re­sult­ing eigen­func­tions are called the spher­i­cal har­mon­ics $Y_l^m(\theta,\phi)$. The first few are given ex­plic­itly in ta­ble 4.2. In case you need more of them for some rea­son, there is a generic ex­pres­sion (D.5) in de­riva­tion {D.14}.

These eigen­func­tions can ad­di­tion­ally be mul­ti­plied by any ar­bi­trary func­tion of the dis­tance from the ori­gin $r$. They are nor­mal­ized to be or­tho­nor­mal in­te­grated over the sur­face of the unit sphere:

$$\int_{\theta=0}^{\pi}\int_{\phi=0}^{2\pi} Y_l^m(\theta,\ph... ...underline m}$} \\ 0 & \mbox{otherwise} \end{array}\right. %$$ (4.23)

The spher­i­cal har­mon­ics $Y_l^m$ are some­times sym­bol­i­cally writ­ten in “ket no­ta­tion” as ${\left\vert l\:m\right\rangle}$.

What to say about them, ex­cept that they are in gen­eral a mess? Well, at least every one is pro­por­tional to $e^{{{\rm i}}m\phi}$, as an eigen­func­tion of $\L _z$ should be. More im­por­tantly, the very first one, $Y^0_0$ is in­de­pen­dent of an­gu­lar po­si­tion com­pared to the ori­gin (it is the same for all $\theta$ and $\phi$ an­gu­lar po­si­tions.) This eigen­func­tion cor­re­sponds to the state in which there is no an­gu­lar mo­men­tum around the ori­gin at all. If a par­ti­cle has no an­gu­lar mo­men­tum around the ori­gin, it can be found at all an­gu­lar lo­ca­tions rel­a­tive to it with equal prob­a­bil­ity.

Ta­ble 4.3: The first few spher­i­cal har­mon­ics rewrit­ten.

There is a dif­fer­ent way of look­ing at the an­gu­lar mo­men­tum eigen­func­tions. It is shown in ta­ble 4.3. It shows that $r^lY_l^m$ is al­ways a poly­no­mial in the po­si­tion com­po­nent of de­gree $l$. Fur­ther­more, you can check that $\nabla^2r^lY_l^m$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0: the Lapla­cian of $r^lY_l^m$ is al­ways zero. This way of look­ing at the spher­i­cal har­mon­ics is of­ten very help­ful in un­der­stand­ing more ad­vanced quan­tum top­ics. These so­lu­tions may be in­di­cated as

$${\cal Y}_l^m \equiv r^l Y_l^m %$$ (4.24)

and re­ferred to as the “har­monic poly­no­mi­als.” In gen­eral the term har­monic in­di­cates a func­tion whose Lapla­cian $\nabla^2$ is zero.

Far more im­por­tant than the de­tails of the eigen­func­tions them­selves are the eigen­val­ues that come rolling out of the analy­sis. A spher­i­cal har­monic $Y_l^m$ has an an­gu­lar mo­men­tum in the $z$-​di­rec­tion

$$L_z = m\hbar$$ (4.25)

where the in­te­ger $m$ is called the mag­netic quan­tum num­ber, as noted in the pre­vi­ous sub­sec­tion. That is no sur­prise, be­cause the analy­sis de­manded that they take that form. The new re­sult is that a spher­i­cal har­monic has a square an­gu­lar mo­men­tum

$$\fbox{$\displaystyle L^2 = l (l+1) \hbar^2 $}$$ (4.26)

where $l$ is also an in­te­ger, and is called the “az­imuthal quan­tum num­ber” for rea­sons you do not want to know. It is maybe a weird re­sult, (why not sim­ply $l^2\hbar^2$?) but that is what square an­gu­lar mo­men­tum turns out to be.

The az­imuthal quan­tum num­ber is at least as large as the mag­ni­tude of the mag­netic quan­tum num­ber $m$:

$$\fbox{$\displaystyle l \mathrel{\raisebox{-1pt}{$\geqslant$}}\vert m\vert $}$$ (4.27)

The rea­son is that $\L ^2$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\L _x^2+\L _y^2+\L _z^2$ must be at least as large as $\L _z^2$; in terms of eigen­val­ues, $l(l+1)\hbar^2$ must be at least as large as $m^2\hbar^2$. As it is, with $l$ $\raisebox{-.5pt}{$\geqslant$}$ $\vert m\vert$, ei­ther the an­gu­lar mo­men­tum is com­pletely zero, for $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ $m$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0, or $L^2$ is al­ways greater than $L_z^2$.

Key Points

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The op­er­a­tor for square an­gu­lar mo­men­tum is (4.22).

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The eigen­func­tions of both square an­gu­lar mo­men­tum and an­gu­lar mo­men­tum in the cho­sen $z$-​di­rec­tion are called the spher­i­cal har­mon­ics $Y_l^m$.

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If a par­ti­cle has no an­gu­lar mo­men­tum around the ori­gin, it can be found at all an­gu­lar lo­ca­tions rel­a­tive to it with equal prob­a­bil­ity.

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The eigen­val­ues for square an­gu­lar mo­men­tum take the counter-​in­tu­itive form $L^2$ $\vphantom0\raisebox{1.5pt}{$=$}$ $l(l+1)\hbar^2$ where $l$ is a non­neg­a­tive in­te­ger, one of 0, 1, 2, 3, ..., and is called the az­imuthal quan­tum num­ber.

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The az­imuthal quan­tum num­ber $l$ is al­ways at least as big as the ab­solute value of the mag­netic quan­tum num­ber $m$.

4.2.3 Re­view Ques­tions

1.

The gen­eral wave func­tion of a state with az­imuthal quan­tum num­ber $l$ and mag­netic quan­tum num­ber $m$ is $\Psi$ $\vphantom0\raisebox{1.5pt}{$=$}$ $R(r)Y_l^m(\theta ,\phi)$, where $R(r)$ is some fur­ther ar­bi­trary func­tion of $r$. Show that the con­di­tion for this wave func­tion to be nor­mal­ized, so that the to­tal prob­a­bil­ity of find­ing the par­ti­cle in­te­grated over all pos­si­ble po­si­tions is one, is that

$$\int_{r=0}^\infty R(r)^* R(r) r^2 { \rm d}r = 1.$$

So­lu­tion an­guc-a

2.

Can you in­vert the state­ment about zero an­gu­lar mo­men­tum and say: if a par­ti­cle can be found at all an­gu­lar po­si­tions com­pared to the ori­gin with equal prob­a­bil­ity, it will have zero an­gu­lar mo­men­tum?

So­lu­tion an­guc-b

3.

What is the min­i­mum amount that the to­tal square an­gu­lar mo­men­tum is larger than just the square an­gu­lar mo­men­tum in the $z$-​di­rec­tion for a given value of $l$?

So­lu­tion an­guc-c

4.2.4 An­gu­lar mo­men­tum un­cer­tainty

Rephras­ing the fi­nal re­sults of the pre­vi­ous sub­sec­tion, if there is nonzero an­gu­lar mo­men­tum, the an­gu­lar mo­men­tum in the $z$-​di­rec­tion is al­ways less than the to­tal an­gu­lar mo­men­tum. There is some­thing funny go­ing on here. The $z$-​di­rec­tion can be cho­sen ar­bi­trar­ily, and if you choose it in the same di­rec­tion as the an­gu­lar mo­men­tum vec­tor, then the $z$-​com­po­nent should be the en­tire vec­tor. So, how can it al­ways be less?

The an­swer of quan­tum me­chan­ics is that the looked-for an­gu­lar mo­men­tum vec­tor does not ex­ist. No axis, how­ever ar­bi­trar­ily cho­sen, can align with a nonex­ist­ing vec­tor.

There is an un­cer­tainty prin­ci­ple here, sim­i­lar to the one of Heisen­berg for po­si­tion and lin­ear mo­men­tum. For an­gu­lar mo­men­tum, it turns out that if the com­po­nent of an­gu­lar mo­men­tum in a given di­rec­tion, here taken to be $z$, has a def­i­nite value, then the com­po­nents in both the $x$ and $y$ di­rec­tions will be un­cer­tain. (De­tails will be given in chap­ter 12.2). The wave func­tion will be in a state where $L_x$ and $L_y$ have a range of pos­si­ble val­ues $m_1\hbar$, $m_2\hbar$, ..., each with some prob­a­bil­ity. With­out def­i­nite $x$ and $y$ com­po­nents, there sim­ply is no an­gu­lar mo­men­tum vec­tor.

It is tempt­ing to think of quan­ti­ties that have not been mea­sured, such as the an­gu­lar mo­men­tum vec­tor in this ex­am­ple, as be­ing merely “hid­den.” How­ever, the im­pos­si­bil­ity for the $z$-​axis to ever align with any an­gu­lar mo­men­tum vec­tor shows that there is a fun­da­men­tal dif­fer­ence be­tween be­ing hid­den and not ex­ist­ing.

Key Points

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Ac­cord­ing to quan­tum me­chan­ics, an ex­act nonzero an­gu­lar mo­men­tum vec­tor will never ex­ist. If one com­po­nent of an­gu­lar mo­men­tum has a def­i­nite value, then the other two com­po­nents will be un­cer­tain.