Angular Momentum

The angular momentum operator $L$ is obtained by substituting $r$ and $p$ by their corresponding quantum mechanical operators $\hat{r}$ and $-i \hbar \nabla_r$ in the classical expression of the angular momentum$L=r\times p$. The Cartesian components of $L$ are:

$$L_x = -i \hbar (y\frac{\partial}{\partial z}- z \frac{\partial}{\partial y}) = y p_z - z p_y,$$

$$L_y = -i \hbar (z\frac{\partial}{\partial x}- x \frac{\partial}{\partial z}) = z p_x - x p_z,$$

$$L_z = -i \hbar (x\frac{\partial}{\partial y}- y \frac{\partial}{\partial x}) = x p_y - y p_x.$$

These components satisfy the following commutation relations:

$$\begin{aligned}\ [L_x, L_y] &= [yp_z - zp_y, zp_x -xp_z],\\ &= ... ...] p_y,\\ &=-i\hbar(yp_x - xp_y),\\ &=i\hbar L_z. \\ \end{aligned}$$

Exercise 20:

Show that,

$$\boxed{L\times L= i \hbar L}.$$

Hint: Show that, $i \hbar L_x=[L_y,L_z]$. Note, that this expression corresponds to the cyclic permutation where $y$ is substituted by $z$, $x$ by $y$, and $z$ by $x$, in the commutation relation $i \hbar L_x=[L_y,L_z]$. Cyclic permutations can be represented by the following diagram:

Having obtained the commutation relations we can show that $L^2$ commutes with the Cartesian components of $L$, e.g.,

$$[L^2, L_x]=0.$$

We consider that,

$[L^2, L_x]=[L_x^2 +L_y^2 +L_z^2, L_x],$

$[L^2, L_x]=[L_y^2, L_x] + [L_z^2, L_x],$

$[L^2, L_x]=L_y[L_y, L_x] + [L_y, L_x]L_y +L_z[L_z, L_x] + [L_z, L_x] L_z,$   and

since$\ [L_y, L_x]=-i\hbar L_z, \ [L_y, L_x]=-i\hbar L_z, \ [L_z, L_x] = i\hbar L_y, \$   then$$,

$$[L^2, L_x]=0.$$

Due to the cyclic permutations we can also conclude that,

$$[L^2, L_y]=0,$$   and$$\qquad [L^2, L_z]=0.$$

According to these equations both the magnitude of the angular momentum and one (any) of its components can be simultaneously determined, since there is always a set of eigenfunctions that is common to $L^2$ and any of the three Cartesian components. Remember, however, that none of the individual components commute with each other. Therefore, if one component is determined the other two are completely undetermined.

Eigenvalues of $L^2$ and $L_z$: Ladder Operators

In order to find eigenfunctions $Y$ that are common to $L^2$ and $L_z$,

$$L^2Y=aY,$$ (29)

and

$$L_zY=bY,$$ (30)

we define the ladder operators,

$L_+ = L_x + i L_y,$

$L_- = L_x - i L_y,$

where $L_+$ is the raising operator, and $L_-$ is the lowering operator.

In order to show the origin of these names, we operate Eq. (30) with $L_+$ and we obtain,

$L_+ L_z Y=b L_+Y.$

Then, we substitute $L_+L_z$ by $[L_+, L_z] + L_z L_+$, where

$[L_+, L_z]=[L_x+iL_y,L_z]=[L_x,L_z]+i[L_y,L_z]$.

Since,$\hspace{.5cm} [L_x,L_z]=-i\hbar L_y,$   and$\qquad [L_y,L_z]=i \hbar L_x, \$   then$$

$L_+L_z-L_zL_+= -i \hbar(L_y-iL_x) = -\hbar L_+.$

Consequently,

$(-\hbar L_+ + L_z L_+)Y = b L_+ Y,$

and,

$L_z(L_+ Y) = (b + \hbar)(L_+ Y).$

Thus the ladder operator $L_+$ generates a new eigenfunction of $L_z$ (e.g., $L_+ Y$) with eigenvalue $(b+\hbar)$ when such operator is applied to the eigenfunction of $L_z$ with eigenvalue $b$ (e.g., Y). The operator $L_+$ is therefore called the raising operator.

Applying $p$ times the raising operation to Y, we obtain:

$$L_Z L_+^p Y = (b + \hbar p)L_+^p Y.$$

Exercise 21: Show that:

$$L_Z L_-^p Y = (b - \hbar p)L_-^p Y.$$

Therefore $L_+$ and $L_-$ generate the following ladder of eigenvalues:

... $b-3\hbar$         $b-2\hbar$         $b-\hbar$         $b$         $b+\hbar$         $b+2\hbar$         $b+3\hbar$ ...

Note that all functions $L_\pm^p Y$ generated by the ladder operators are eigenfunctions of $L^2$ with eigenvalue equal to $a$ (see Eq. (29)).

Proof:

$L^2 L_\pm^p Y = L_\pm^p L^2 Y = L_\pm^p a Y,$

since $[L^2, L_x] =[L^2, L_y] = [L^2, L_\pm] = 0$, and therefore, $[L^2, L_\pm^p] = 0$.

Note that the ladder of eigenvalues must be bounded:

$$L_zY_k= b_kY_k,$$

with $Y_k= L_{\pm}^k Y$, and $b_k=b\pm k \hbar$.

Therefore,

$L_z^2 Y_k = b_k^2 Y_k$,

$L^2 Y_k = a Y_k$,

$\underbrace{(L_x^2+L_y^2)Y_k} =(a-b_k^2)Y_k$.

non-negative physical quantity $\Longrightarrow$ $(a-b_k^2)$ has to be positive:

$$a \geq b_k^2, \Longrightarrow a^{\frac{1}{2}} \geq \vert b_k\vert,$$

$$\boxed{a^{\frac{1}{2}} \geq b_k \geq -a^{\frac{1}{2}}}$$

In order to avoid contradictions,

$$L_+ Y_{max} = 0,$$   and$$\qquad L_-Y_{min}=0.$$

$L_+L_-Y_{min} = 0,$

$L_+L_-=(L_x+iL_y)(L_x-iL_y),$

$L_+L_-=L_x^2 -i(\underbrace{L_xL_y - L_yL_x}) + L_y^2,$

$\qquad \qquad \qquad \qquad i\hbar L_z$

$L_+L_-=L_x^2+L_y^2 +\hbar L_z = L^2 - L_z^2 + \hbar L_z.$

Therefore,

$$a - b_{min}^2 + \hbar b_{min} =0,$$ (31)

because,

$$L_z^2 Y_{min}=b_{min}^2 Y_{min}, \ L^2 Y_{min} = a Y_{min}, \ L_z Y_{min}=b_{min} Y_{min}.$$

Analogously,

$L_-L_+Y_{max} = 0.$

$\qquad \qquad \Downarrow$

$(L^2 - L_z^2 - \hbar L_z) Y_{max} =0,$ and

$$a - b_{max}^2 - \hbar b_{max} =0.$$ (32)

Eqs. (31) and (32) provide the following result:

$$(b_{min}^2-b_{max}^2) - \hbar(b_{min} + b_{max}) =0 \ \Rightarrow \ \boxed{b_{min}= - b_{max}}.$$

Furthermore, we know that $b_{max} = b_{min} + n \hbar$, because all eigenvalues of $L_z$ are separated by units of $\hbar$. Therefore,

$2 b_{max}=n\hbar \Longrightarrow b_{max}= \frac{n}{2} \hbar = j \hbar$, where$\hspace{.20cm} j = \frac{n}{2},$

$a = b_{min}^2 - \hbar b_{min} = j^2 \hbar^2 + \hbar^2 j= \hbar ^2 j(j +1)$, and $b=-j\hbar, (-j +1)\hbar, (-j +2)\hbar, ..., j\hbar.$

Note that these quantization rules do not rule out the possibility that $j$ might have half-integer values. In the next section we will see that such possibility is, however, ruled out by the requirement that the eigenfunctions of $L^2$ must be 2$\pi$-periodic.

Spherical Coordinates

Spherical coordinates are defined as follows,

$z=r$   Cos$\theta$,

$y=r$   Sin$\theta$   Sin$\phi$,

$x=r$   Sin$\theta$   Cos$\phi$,

where $\theta$, and $\phi$ are defined by the following diagram,

Exercise 22: Write the Cartesian components of the linear momentum operator $\hat{p}$: $\hat{p}_x$, $\hat{p}_y$ and $\hat{p}_z$ in spherical coordinates.

Hint:

$$\Bigg (\frac{\partial g}{\partial x} \Bigg )_{y,z} = \Bigg (\frac... ...al x} \Bigg )_{y,z} \Bigg (\frac{\partial f}{\partial r} \Bigg )_{\theta,\phi},$$

where $g=g(x, y, z)$, and $f=f(r(x, y, z), \theta(x, y, z), \phi(x, y, z))$.

$r = \sqrt{(x^2 + y^2 + z^2)}$,

$\frac{y}{x} =$   tan$\phi$,

Cos$\theta = \frac{z}{r} = \frac{z}{(x^2 + y^2 + z^2)^{\frac{1}{2}}}$.

$$\Bigg (\frac{\partial \text{Cos} \theta}{\partial x }\Bigg)_{y,z}... ...\frac{r^2 \text{Cos} \theta \text{Sin} \theta \text{Cos} \phi}{r^3 Sin \theta},$$

$$\Bigg (\frac{\partial \text{tan} \theta}{\partial x }\Bigg)_{y,z}... ... \text{Sin} \phi \text{Cos}^2 \phi}{r^2 \text{Sin}^2 \theta \text{Cos}^2 \phi},$$

$$\Bigg (\frac{\partial r}{\partial x}\Bigg)_{y,z}= \frac{1}{2} \fr... ...al r}{\partial x}\Bigg )_{y,z} = \frac{r \text{Sin} \theta \text{Cos} \phi}{r}.$$

Exercise 23: Show that,

$$L_x = i\hbar\left(\text{Sin} \phi \frac{\partial}{\partial \theta... ...theta}{\text{Sin} \theta}\text{Cos} \phi \frac{\partial}{\partial \phi}\right),$$

$$L_y = - i\hbar\left(\text{Cos} \phi \frac{\partial}{\partial \the... ...eta}{\text{Sin} \theta}\text{Sin} \phi \frac{\partial}{\partial \theta}\right),$$

and

$$L_z=-i \hbar \frac{\partial}{\partial \phi}.$$

Squaring $L_x$, $L_y$ and $L_z$ we obtain,

$$L^2= -\hbar^2\left (\frac{\partial ^2}{\partial \theta ^2} + \fra... ...a} + \frac{1}{\text{Sin}^2 \theta} \frac{\partial ^2}{\partial \phi^2} \right).$$

Eigenfunctions of $L^2$

Since $L^2$ does not depend on r, $\Rightarrow$ $Y=Y(\theta, \phi)$. Furthermore, if $Y$ is an eigenfunction of $L_z$ then,

$$L_zY=bY.$$

$$-i \hbar \frac{\partial Y}{\partial \phi} = bY \qquad \Rightarrow... ...rtial \phi} = \frac{1}{Y} \frac{\partial Y}{\partial \phi}= -\frac{b}{i \hbar}.$$

                         $\boxed{Y=A \hspace{.1cm} \text{exp} \left(\frac{ib\phi}{\hbar} \right)}.$

Since $Y(\phi + 2\pi) = Y(\phi),$ we must have

$$e^{i\frac{2\pi b}{\hbar}}= 1, \qquad \Rightarrow \qquad 2\pi \frac{b}{\hbar} = 2 \pi m,$$   with$$\hspace{0.5 cm} \qquad m=0, \pm 1, \pm 2, ...$$

Therefore, $\boxed{b=m\hbar}$, where $m$ is an integer.

In order to find eigenfunctions that are common to $L_z$ and $L^2$ we assume $A$ to be a function of theta, $A = A(\theta)$:

$$L^2Y=-\hbar^2 \left( \frac{\partial^2A}{\partial \theta ^2} + \fr... ...r} \right ) = a A (\theta) \text{exp} \left ( \frac{i b \phi}{\hbar} \right ),$$

$$-\hbar^2 \left ( Sin^2 \theta \frac{\partial^2 A}{\partial \theta... ...l \theta} - \frac{b^2}{\hbar ^2} A \right ) = a A (\theta) \text{Sin}^2 \theta.$$ (33)

Making the substitution $x=$Cos$\theta$ we obtain,

$$(1-x^2) \frac{d^2 A}{d x^2} - 2 x \frac{d A}{dx} + \left( \frac{a}{\hbar^2} -\frac{m^2}{1-x^2}\right) A =0.$$ (34)

Exercise 24: Obtain Eq. (34) from Eq. (33).

Eq. (34) is the associated Legendre equation, whose solutions exist only for $a= \hbar^2l(l+1)$, and $b=-l\hbar, (-l+1)\hbar, ..., l \hbar$ (i.e., the quantum number $l$ is an integer greater or equal to zero, with $\vert m\vert \leq l$). The solutions of the associated The solutions of the associated Legendre polynomials,  $A (l, m) = P_l^{\vert m\vert}($Cos$\theta)$,

For example, the normalized polynomials for various values of $l$ and $m$ are:

$A(0, 0) = 1/ \sqrt{2}$,

$A(1, 0) = \sqrt{3/2}$   Cos$\theta$,

$A(1, \pm1) = \sqrt{3/4}$   Sin$\theta$,

...

The eigenstates that are common to $L^2$ and $L_z$ are called spherical harmonics and are defined as follows,

$\boxed{Y_l^m(\theta, \phi) = P_l^{\vert m\vert}(\text{Cos} \theta) e^{i m \phi}}.$

The spherical harmonics are normalized as follows,

$$\int_0^{2\pi}d\phi \int _{-1}^1 d$$Cos$$\theta \ {Y_{l'}^{m'}}^*(\theta, \phi) Y_l^m(\theta, \phi)= \delta _{l l'}\delta_{mm'}.$$

Rotations and Angular Momentum

A coordinate transformation that corresponds to a rotation can be represented by the following diagram:

This diagram shows that vector $\vec{r}$ can be specified either relative to the axes (x, y, z), or relative to the axes (x', y', z'), where these two sets of coordinates are defined relative to each other as follows,

$$\boxed{\bar{r}' = R(\alpha, z) \bar{r}},$$ (35)

where, $\bar{r}'$ is the same vector $\bar{r}$ but with components expressed in the primed coordinate system.

$$\alpha: \$$   Angle$$, \qquad z:$$   Rotation axis

$x=r$   Cos$\phi,$

$y=r$   Sin$\phi,$

$x'= r$   Cos$(\phi - \alpha) = r($Cos$\phi$   Cos$\alpha +$   Sin$\phi$   Sin$\alpha),$

$y'= r$   Sin$(\phi - \alpha) = r($Sin$\phi$   Cos$\alpha -$   Cos$\phi$   Sin$\alpha),$

$z' = z,$

$x' = x$   Cos$\alpha + y$   Sin$\alpha,$

$y' = y$   Cos$\alpha - x$   Sin$\alpha.$

Therefore, the coordinate transformation can be written in matrix representation as follows,

$$\left ( \begin{array}[c]{c} x'\\ y'\\ z'\end{array} \right ) = \b... ...0 & 1\\ \end{pmatrix}\left ( \begin{array}[c]{c} x\\ y\\ z\end{array} \right ).$$

The operator associated with the coordinate transformation is $P_R(\alpha)$, defined as follows:

$$\hat{P}_R(\alpha ,z) f(\bar{r}) = f[R^{-1}(\alpha ,z) \bar{r}],$$

where $R^{-1}$ is the transpose of R, i.e., $R^{-1}= \begin{pmatrix}\text{Cos} \alpha & -\text{Sin}\alpha & 0\\ \text{Sin} \alpha& \text{Cos} \alpha & 0 \\ 0 &0 & 1\\ \end{pmatrix}.$

Therefore, $\hat{P}_R(\alpha,z) f(\bar{r}) = f(x$   Cos$\alpha - y$   Sin$\alpha, x$   Sin$\alpha + y$   Cos$\alpha, z)$.

An infinitesimal rotation is defined as follows,

$\hat{P}_R(\delta, z) f({\bar r}) = f(x-y \delta, x\delta+y, z)$,

$\hat{P}_R(\delta, z) f({\bar r}) = f(x,y, z) - y \delta \frac{\partial f}{\partial x}+ x\delta \frac {\partial f}{\partial y}$,

$\hat{P}_R(\delta, z) f({\bar r}) = f(x,y, z) + \delta (x \frac{\partial}{\partial y} - y \frac {\partial}{\partial x})f(x,y,z)$

recall that, $-i \hbar(x \frac {\partial}{\partial y} - y \frac {\partial}{\partial x}) = L_z$, therefore,

$\hat{P}_R (\delta, z) f({\bar r}) = (1+ \frac{i}{\hbar} \delta L_z) f({\bar r})$.

A finite rotation through an angle $\alpha$ can be defined according to $n$ infinitesimal rotations, after subdividing $\alpha$ into $n$ angle increments, $\alpha = n \delta$, and taking the limit $n \rightarrow \infty$, and $\delta \rightarrow 0$.

$$\hat{P}_R (\alpha, z) = \lim \begin{sb} { n \rightarrow \infty ... ...\left (1+ i \frac{\delta}{\hbar}L_z \right )^n = e^{\frac{i}{\hbar}\alpha L_z}.$$

In general, a finite rotation through an angle $\alpha$ around an arbitrary axis specified by a unit vector $\hat{n}$ is defined as follows,

$$\boxed{\hat{P}_R(\alpha, \hat{n})= e^{\frac{i}{\hbar}\alpha \hat{n} \cdot L} }.$$

This equation establishes the connection between the operator associated with a coordinate transformation and the angular momentum operator.

Note:

It is important to note that if coordinates are transformed according to $\bar{r}' = {\bf R} \bar{r}$, the Hamiltonian is transformed according to a similarity transformation, which is defined as follows:

$$\hat{H}'= \hat{P}_R \hat{H} \hat{P}_R^{-1}.$$

Proof:

Consider, $f(r) \equiv \hat{H}(r) \phi(r) = E \phi (r)$,

$\hat{P}_R f(r)= \hat{P}_R H(r) \hat{P}_R^{-1}\hat{P}_R \phi(r) = E \phi (R^{-1}r)$,

$\hat{P}_R H(r) \hat{P}_R^{-1} \phi(R^{-1}r) = E \phi (R^{-1}r)= H(R^{-1}r) \phi(R^{-1}r)$.

Therefore, $H(R^{-1}r) = \hat{P}_R H(r) \hat{P}_R^{-1}$.

It is, therefore, evident that the Hamiltonian is an invariant operator (i.e., $H(r) = H(R^{-1}r)$) under a coordinate transformation, $\bar{r}' = R \bar{r}$, whenever the operator associated with the coordinate transformation commutes with the Hamiltonian, $[\hat{P}_R, H] = 0$.