Angular momentum-Eigenvalues & Eigenfunctions

Why angular momentum?

• Doesn't show up in 1-d. It's a 3-d phenomenon.
• I claimed a connection between quantum number $l$ and angular momentum, but it's not at all obvious from what we've done so far.
• Classical mechanics: angular momentum of a particle in a central potential is conserved.
• Quantum: Any observable $\hat{Q}$ that commutes with the Hamiltonian $\hat{H}$ is a constant in time. (Also means that they have a common set of eigenfunctions & eigenvalues).

Strategy:

1. Work out the commutation relations between the Cartesian components of angular momentum and a total angular momentum operator, $\hat L^2$, where classically: $$\myv L=\myv r \times \myv p$$.
2. Define and use some ladder operators which raise and lower the eigenfunctions of the $\hat L_z$ operator.
3. This will allow us to find the eigenvalues of the $\hat L_z$ and $\hat L^2$ operators without ever finding the eigenfunctions.
4. Finally, we'll find the eigenfunctions themselves. Spoiler: They turn out to be the spherical harmonics, $Y_l^m(\theta, \phi)$.

Commutation relations in 3-d

Problem 4.1 - Work out all the other canonical commutation relationships between the component of $\myv r \equiv \{x, y, z\}$ and $\myv p\equiv\{p_x, p_y, p_z\}$

We had the operators $\hat x = x$ and $\hat p_x=\frac \hbar i\frac{\del}{\del x}$ operate on a test function $f(x)$ to show that $[x,p]=i\hbar$.

So, lets now have operators act on a test function $f(x,y,z)$ For example: $$xyf(x,y,z)=yxf(x,y,z) \Rightarrow xy-yx=[x,y]=0$$

When operators $\hat x$ and $\hat y$ commute, that means that we can switch their order without changing the result.

It's more compact to write these commutation relations in this notation: With $x\equiv r_1$, $y\equiv r_2$, $z\equiv r_3$ and $p_x\equiv p_1$, $p_y\equiv p_2$, $p_z\equiv r_3$ we find: $$[r_i,r_j]=0$$

You'll show in homework that: $$[p_i,p_j]=0$$ and $$[r_i,p_j]=i\hbar\delta_{ij}=-[p_j,r_i].$$

Use the results above to work out $$[yp_z,zp_x]=?$$

You found $$[yp_z,zp_x]=yp_zzp_x-zp_xyp_z = yp_x[p_z,z]=-i\hbar yp_x.$$

Angular momentum-Cartesian components

I'm violating my earlier convention to write hats on the top of all operators....

In classical physics, angular momentum $\myv L$ is $$\myv{ L} = \myv{r}\times\myv{p}.$$

In terms of Cartesian components... $$L_x=y p_z-zp_y\ ; \ L_y=z p_x-xp_z\ ; \ L_z=x p_y-yp_x.$$

Do the quantum mechanical operators corresponding to the different components of $\myv L$ commute? Consider: $$\begineq [\hat L_x,\hat L_y] &=&[yp_z-zp_y,zp_x-xp_z]\\ &=&[yp_z,zp_x]-[yp_z,xp_z]-[zp_y,zp_x]+[zp_y,xp_z]\\ &=& yp_x[p_z,z] -0 -0 +p_yx[z,p_z]\\ &=& -i\hbar yp_x +i\hbar p_yx = i\hbar (xp_y-yp_x) &=&i\hbar \hat L_z \endeq $$

Cyclically permuting through the other possibilities: $$[\hat L_x,\hat L_y] = i\hbar \hat L_z;\ [\hat L_y,\hat L_z] = i\hbar \hat L_x;\ [\hat L_z,\hat L_x] = i\hbar \hat L_y.$$

Notice that when we take the cross product of two identical vectors $\myv{A} \times \myv{A}=\myv{C}$, that $C_z=A_x A_y - A_y A_x=[A_x,A_y]$, so we could compactly write the three expressions above as...

$$\hat{\myv L} \times \hat{\myv L} = i\hbar \hat{\myv L} \label{cross} $$

Some consequences...

• The usual rule for classical vectors that $\myv{A}\times\myv{A} = 0$ doesn't hold for $\hat{\myv L}$ with its non-commuting components.
• You can't simultaneously measure $\hat L_x$ and $\hat L_y$ unless the system is in a state with $\hat L_z=0$.
• It's impossible to find states that are simultaneous eigenstates of $\hat L_x$ and $\hat L_y$.

Total angular momentum operator, $\hat J^2$

Angular momentum, $\myv L=\myv r \times \myv p$ should have the units of position (m) times momentum (kg$\cdot$m/sec).

Now, $\hbar=1.054572 \times 10^{-34}$ J s, and J s = (N m) s = (kg m^2 / s^2) s= (kg m / s) m. So $\hbar$ already has units of angular momentum. So, let's define a unit-less angular momentum operator, $$\hat J\equiv \hat L/\hbar$$ so as not to have to keep writing out the $\hbar$ all the time. With this definition, the commutation relations in Eq. $\ref{cross}$ can be written: $$\hat{\myv J}\times\hat{\myv J} = i\hat{\myv J}.$$

The (total angular momentum)^2 operator is $$\hat J^2\equiv\hat{\myv J}\cdot\hat{\myv J}=\hat J_x^2+\hat J_y^2+\hat J_z^2.$$ (Usually we'll just call this "the total angular momentum operator", without mentioning that it's *really* angular momentum *squared*).

Commutator $[ \hat J^2, \hat J_x]=0$

Next, we'll show that $\hat J^2$ commutes with $J_x$, one of the components of angular momentum: $$\begineq [\hat J^2,\hat J_x] &=& [(\hat J_x^2+\hat J_y^2+\hat J_z^2),\hat J_x]\\ &=&\color{red}{[\hat J_x^2,\hat J_x]} +\color{blue}{[\hat J_y^2,\hat J_x]}+\color{green}{[\hat J_z^2,\hat J_x]}\\ &=&\color{red}{0} + \color{blue}{\hat J_y\hat J_y\hat J_x-\hat J_x\hat J_y\hat J_y} +\color{green}{\hat J_z\hat J_z\hat J_x-\hat J_x\hat J_z\hat J_z}\\ &=&\color{blue}{\hat J_y\hat J_y\hat J_x-\hat J_x\hat J_y\hat J_y} +\color{green}{\hat J_z\hat J_z\hat J_x-\hat J_x\hat J_z\hat J_z}\\ &&-\hat J_y\hat J_x\hat J_y +\hat J_y\hat J_x\hat J_y\ -\hat J_z\hat J_x\hat J_z +\hat J_z\hat J_x\hat J_z\\ \endeq $$ Where the black terms introduced on the last line add up to zero.

Now, we can group each of the four colored terms with the black term just below it to write the equation above as $$\begineq [\hat J^2,\hat J_x] &=& \hat J_y[\hat J_y,\hat J_x] +[\hat J_y,\hat J_x]\hat J_y+\hat J_z[\hat J_z,\hat J_x]+[\hat J_z,\hat J_x]\hat J_z\\ &=&\hat J_y(-i\hat J_z) +(-i\hat J_z)\hat J_y+\hat J_z(+i\hat J_y)+(+i\hat J_y)\hat J_z\\ &=&0\\ \endeq $$

And for the other components as well... $$[\hat J^2,\hat J_x]=0;\ [\hat J^2,\hat J_y]=0;\ [\hat J^2,\hat J_z]=0.$$

Simultaneous eigenstates

Anytime two operators commute, it's possible that they have the same eigenvectors (or "eigenstates"). So, taking the last of those last 3 commutation relations: $$[\hat J^2,\hat J_z]=0$$ This would mean that if there is an eigenstate $f$ of the $\hat J_z$ operator, with eigenvalue $\mu$: $$\hat J_z f = \mu f,$$ Then this same eigenstate $f$ should also be an eigenstate of the $\hat J^2$ operator, albeit with a different eigenvalue $\lambda$: $$\hat J^2 f= \lambda f.$$

Ladder operators

Let us consider the intriguing properties of the two ladder operators $\hat J_+$ and $\hat J_-$, compactly written in one expression as:

$$\hat J_{\pm} \equiv \hat J_x \pm i\hat J_y.$$

Note that $\hat J_+$ is the Hermitian conjugate of $\hat J_-$, and vice versa.

Show this...

$[\hat J^2,\hat J_\pm]=0$ and consequences

1. $\hat J^2$ commutes with all the Cartesian components of angular momentum $\hat J_x$, $\hat J_y$, or $\hat J_z$
2. Therefore, $\hat J^2$ should commute with any linear combination of $\hat J_x$, $\hat J_y$, and $\hat J_z$.
3. The ladder operators: $$\hat J_{\pm} \equiv \hat J_x \pm i\hat J_y,\nonumber$$ are linear combinations of $\hat J_x$ and $\hat J_y$.

Therefore $$[\hat J^2,\hat J_\pm]=0$$

Next, we can show that $\hat J_\pm \ket{jm}$ is an eigenstate of the $\hat J^2$ operator:

1. Expanding the commutator above: $$\begin{equation*}\begin{split} [\hat J^2,\hat J_\pm]f&=&0f\\ \hat J^2(\hat J_\pm f)-\hat J_\pm(\hat J^2 f)&=&0 \end{split}\end{equation*}$$
2. Re-arranging the line above: $$\begin{equation*}\begin{split} \hat J^2(\hat J_\pm f)&=&\hat J_\pm (\hat J^2 f)\\ &=&\hat J_\pm \lambda f\\ &=&\lambda(\hat J_\pm f). \end{split}\end{equation*}$$

That last line is in the form of an eigenvalue equation for the function $(\hat J_\pm f)$.

$[\hat J_z,\hat J_\pm]=\pm\hat J_\pm$ and the ladder

Using the commuting properties of the angular momentum components: $$[\hat J_x,\hat J_y] = i \hat J_z;\ [\hat J_y,\hat J_z] = i \hat J_x;\ [\hat J_z,\hat J_x] = i \hat J_y,\nonumber$$ and $$\hat J_+\equiv \hat J_x +i\hat J_y,\nonumber$$

Show that $$[\hat J_z,\hat J_+]= \hat J_+.$$

It turns out that $$[\hat J_z,\hat J_\pm]=\pm \hat J_\pm.$$

Now we can show that $\hat J_\pm$ is also an eigenstate of $\hat J_z$, but with a different eigenvalue:

1. Expanding the commutator above: $$\begin{equation*}\begin{split} [\hat J_z,\hat J_\pm]f&=&\pm \hat J_\pm f\\ \hat J_z(\hat J_\pm f)-\hat J_\pm(\hat J_z f) &=& \pm \hat J_\pm f \end{split}\end{equation*}$$
2. Re-arranging the line above: $$\begin{equation*}\begin{split} \hat J_z(\hat J_\pm f) &=&\hat J_\pm(\hat J_z f) \pm \hat J_\pm f\\ &=&\hat J_\pm(\mu f) \pm \hat J_\pm f\\ &=&\mu \hat J_\pm f \pm \hat J_\pm f\\ \hat J_z(\hat J_\pm f) &=&(\mu\pm 1) \hat J_\pm f \end{split}\end{equation*}$$

That last line is in the form of an eigenvalue equation for the function $(\hat J_\pm f)$.

It is as if the effect of the raising operator $\hat J_+$ is to kick the state into one where $\lambda$ is unchanged, but $\mu$ has been incremented (decremented) by 1 (that is, the eigenvalue with respect to $\hat L_z$ has changed by $\pm\hbar$).

What *are* the eigenvalues?

We've established that when $\hat J_\pm$ operates on some eigenstate, $f$, it leaves the angular momentum eigenvalue unchanged, but kicks $f$ into a state with a $\hat J_z$ eigenvalue greater (or lesser) by exactly 1. But where does the ladder start or end? What are the actual eigenvalues?

The norm or length of a vector, as in $|\myv V|\equiv \sqrt{\myv V\cdot\myv V}=\sqrt{V_x^2+V_y^2+V_z^2}\equiv V$, is a positive number. And this also implies that, $V_z^2 \leq V^2$, whether $V_z$ is positive or negative.

So, as we repeatedly apply the raising operator, the eigenvalue of $\hat J_z$ increases, but it cannot continue to increase forever. Eventually there must be some "top" state (indicated by $f_t$) rung on the ladder, itself a non-zero state, but such that $$\hat J_+f_t=0$$ Let's say that the eigenvalue of this state wrt the $\hat J_z$ operator is $l$, $$\hat J_zf_t=lf_t.$$

The norm^2 of the state vector $\hat J_+f\equiv\ket{\hat J_+f}$ can be calculated as follows (Remember that $(\hat J_\pm)^\dagger = \hat J_\mp$) $$\begin{equation*}\begin{split} |\ket{\hat J_+f}|^2 & = & \innerp{\hat J_+f}{\hat J_+f} = \innerp{f}{\hat J_+^\dagger\hat J_+f} \\ & = & \innerp{f}{ \hat J_-\hat J_+ f} \end{split}\end{equation*} $$

Writing out the $\hat J_-\hat J_+$ operator... $$\begin{equation*}\begin{split} \hat J_-\hat J_+ & = (\hat J_x-i\hat J_y)(\hat J_x+i\hat J_y) = \hat J_x^2+\hat J_y^2+i(\hat J_x\hat J_y-\hat J_y\hat Jx)\\ & = \hat J_x^2+\hat J_y^2+i[\hat J_x,\hat J_y]\\ & = \hat J^2 - \hat J_z^2-\hat J_z \end{split}\end{equation*} $$ Since $\hat J_+ f_t=0$, the norm^2 of this state must be zero. We can calculate it in terms of $\lambda$ and $l$: $$|\ket{\hat J_+f_t}|^2 =0=\innerp{f}{\hat J^2 f_t} +\innerp{f_t}{-\hat J_z^2 f_t}+\innerp{f_t}{-\hat J_z f_t} =\lambda-l^2-l \nonumber$$ Rearranging for $\lambda$... $$\Rightarrow \lambda=l^2+l=l(l+1)\label{nobar}$$ And $\lambda$ must be a positive number since $J^2>0$.

Now we apply the same reasoning on the bottom rung: That a most negative "bottom" state $f_b$ exists, with an eigenvalue $\bar l$, fulfilling: $$\hat J_-f_b=0;\ \ \ \hat J_zf_b=\bar lf_b$$

We calculate the (vanishing) norm^2 of $\hat J_-f_b$ as: $$\begin{equation*}\begin{split} |\ket{\hat J_-f_b}|^2 &=&\innerp{f_b}{\hat J_+\hat J_- f_b}\\ &=& \innerp{f_b}{\hat J^2 f_b} +\innerp{f_b}{-\hat J_z^2 f_b}+\innerp{f_b}{+\hat J_z f_b}\\ 0&=&\lambda-\bar l^2+\bar l \end{split}\end{equation*}$$ Rearranging for $\lambda$... $$\Rightarrow \lambda=\bar l^2-\bar l\label{bar}$$ And $\lambda$ must be a positive number since $J^2>0$.

Now, we can set the two equations $\ref{nobar}$ and $\ref{bar}$ equal. There is more than one solution. But only one of them has a lower eigenvalue for the bottom rung than the top, and that is: $$\bar l=-l.$$

The raising and lowering operators increment (decrement) the eigenvalues of $\hat J_z$--let's call them $m$--by 1 between the lowest value $-l$ and the highest value $+l$. So apparently the allowed values of $m$ are $ m=-l, -l+1,....l-1, l.$ This in turn restricts the possible values of $l$ to $l=0, 1/2, 1, 3/2, 2,...$. Summarizing all of what's been figured out here, we can index the eigenstates with both an $l$ and an $m$, as in $f_l^m$ and they behave like this: $$J^2f_l^m=l(l+1)f_l^m \text{ and }\hat J_zf_l^m=mf_l^m.\nonumber$$ Or, working back to Griffith's notation, in terms of $\myv L=\hbar \myv J$:

Angular momentum eigenstates

Here's the outline of the strategy to get the eigenstates from purely angular momentum considerations:

• Write the angular momentum operator as... $$\hat{\myv L}=\myv{r}\times \hat{\myv{p}} = (\hbar / i) \myv{r}\times \myv{\nabla}$$
• Write $\myv{\nabla}$ in terms of spherical polar coordinates, (Griffiths, Eq [4.123]),
• Write the unit vectors $\hat{\theta}$ and $\hat{\phi}$ in terms of cartesian components along $\hat{x}$, $\hat{y}$, $\hat{z}$,
• Figure out $\hat L_x$, $\hat L_y$, and $\hat L_z$ in terms of spherical polar coords. For example, the simplest of these works out to $$\hat L_z = \frac{\hbar}{i}\frac{\partial}{\partial \phi}$$
• With $\hat L_x$ and $\hat L_y$ in hand, you can also work out the raising and lowering operators $\hat L_\pm=\hat L_x\pm i\hat L_y$.
• Work out $\hat L^2$.

You now need to solve two eigenvalue equations: $$\hat L^2\psi_l^m = \hbar^2 l(l+1) \psi_l^m = -\hbar^2\[ \frac{1}{\sin \theta} \frac{\del}{\del \theta}\( \sin\theta\frac{\del}{\del \theta} \) +\frac{1}{\sin^2\theta} \frac{\del^2}{\del\theta^2} \]\psi_l^m,$$

and $$\hat L_z\psi_l^m =\hbar m \psi_l^m = \frac{\hbar}{i}\frac{\del}{\del\phi}\psi_l^m.$$

These are the equations that we previously solved when trying to find a solution via the method of separation of variables. The solutions are the spherical harmonics $\psi_l^m=Y_l^m(\theta,\phi)$ for integer values of $l$.

We know that for the spherical harmonics, $l$ has to be an integer. What about those half-integer solutions $l=\hbar/2, 3\hbar/2,....$ ??