Eigenstates and eigenvalues of angular momentum

(Part of the Wolverhampton Lectures of Physics's Quantum Physics Course)

We now solve the equations with which we departed last time:

\begin{align} L_z\psi&=\mu\psi\\ L^2\psi&=\lambda\psi\,. \end{align}

That is to say, we need to find out the $\mu$, $\lambda$ and $\psi$ that are compatible with the $L_z$ and $L^2$ quantum algebra that we introduced in the previous lecture. This will be achieved purely by algebra. Like in the harmonic oscillator case, we first introduce the ladder operators:

$$L_\pm\equiv L_x\pm i L_y\,.$$

It is easy to see that $L_\pm$ commutes with $L^2$ since the latter commutes with $L_x$ and $L_y$ and commutation is linear. The algebra of this (these 2) operator(s) with $L_z$ is interesting:

$$[L_z,L_\pm]=[L_z,L_x]\pm i[L_z,L_y]=i\hbar(L_y\mp iL_x)=\pm\hbar(L_x\pm iL_y)=\pm\hbar L_\pm$$

We now turn to the beautiful quantum algebra of angular momentum. The trick will be to show that if $\psi$ is an eigenfunction of $L^2$ and $L_z$, as stated above, then so is also $L_\pm\psi$. We proceed to prove this and to find the corresponding eigenvalues.

The $L^2$ case is the simplest. That $L_\pm\psi$ is an eigenfunction of $L^2$ means that $L^2(L_\pm\psi)$ is proportional to $L_\pm\psi$, but indeed, since $L^2$ and $L_\pm$ commute:

$$L^2 L_\pm\psi=L_\pm L^2\psi=L_\pm\lambda\psi=\lambda L_\pm\psi_L$$

so this is true and, furthermore, they have the same eigenvalue $\lambda$. The projection operator is slightly more interesting:

$$L_z L_\pm\psi=(L_\pm L_z\pm\hbar L_\pm)\psi=(L_\pm\lambda+\hbar L_\pm)\psi=(\lambda\pm\hbar)L_\pm\psi$$

meaning this time that if $\psi$ is an eigenfunction of $L_z$ with eigenvalue $\lambda$, so is $L_\pm\psi$ with eigenvalue $\lambda\pm\hbar$, i.e., we add or subtract one quantum of angular momentum. Whence the name:

  • $L_+$ adds one quantum of angular momentum, and is called the creation operator.
  • $L_-$ removes one quantum of angular momentum, and is called the destruction operator.

This is pretty similar to the ladder operators of the harmonic oscillator, except that will find here that the states are bounded on both sides. Indeed, adding one quantum forever to the $z$-component, we would get to the point where the $z$-component overcomes the total $L^2$ since $L^2$ retains the same eigenvalue. This cannot be, so, like for the harmonic oscillator, one must hit a top-rung where the addition is forbidden. There must exist, therefore, a "top" function $\psi_T$ (T for "top") such that $L_+\psi_T=0$. The same applies to the annihilation $L_-$ with a function $\psi_B$ which touches the bottom rung, i.e., such that $L_-\psi_B=0$.

Now one can check by direct calculations that:

$$L^2=L_\pm L_\mp+L_z^2\mp\hbar L_z\,.$$

This is straightforward (quantum) algebra: Since $L_\pm L_\mp=(L_x\pm iL_y)(L_x\mp iL_y)=L_X^2+L_y^2\mp i L_xL_y\pm iL_yL_x$ (do not forget quantum refers to noncommutative) and since $L_yL_x=L_xL_y+i\hbar L_z$

Applying the bottom part of this equation to $\psi_T$, we find

\begin{align} L^2\psi_T&=L_-L_+\psi_T+L_z^2\psi_T+\hbar L_z\psi_T\\ &=0+\mu_T^2\psi_T+\hbar\mu_T\psi_T \end{align}

and since $L^2\psi_T=\lambda\psi_T$, then:


Doing the same with top part of the $L^2$ equation applied to $\psi_B$, to find that


Equating the right-hand-side of the two previous equations, both equal to $\lambda$, we find:


We are looking for $\mu_T$ and $\mu_B$ and can tidy up a bit the notation by assuming that they are defined as $\hbar\tilde\mu_{B,T}$ and solve for corresponding equation for the $\tilde\mu$: 


This is an odd-looking equation (for two unknown) that, if you think about it, is quadratic, so there should be two solutions. If we don't see the solution right away, one can use the machinery of algebra for one of the unknown:


Does this help? We also need to see that the term below the square root is a perfect square, $(2\tilde\mu_B-1)^2$, so the two solutions are:

\begin{align} \tilde\mu_T&=\tilde\mu_B-1\,,\\ \tilde\mu_T&=-\tilde\mu_B\,. \end{align}

Now, the first solution does not make sense, because it says the top $z$-component, the highest one we can get, is smaller than the bottom $z$-component of $L_z$. The second solution, if $\tilde\mu_B$ is negative, makes sense: the top and bottom rungs have opposite values. It's also a beautiful, symmetrical, sort of expected result. So we have this structure for the eigenvalues:

$$\tilde\mu_B=-\hbar\mu\quad-\hbar(\mu-1)\quad-\hbar(\mu-2)\quad\cdots\quad \hbar(\mu-2)\quad \hbar(\mu-1)\quad \hbar\mu=\tilde\mu_T$$

where we jump by one quantum of $\hbar$ each time. Let's consider we have $N$ such steps, for integer $N$, meaning we reach the top state from the bottom one after adding $N\hbar$. Then:


which constrains the possible values of $\mu$ to be either integers or half-integers:

$$\mu=N/2\ \hbox{for integer $N$.}$$

The possible eigenvalues for the $z$-component of angular momentum can therefore be of the type:

  • $N=0$, we only have zero.
  • $N=1$, one can go from $-{\hbar\over 2}$ to ${\hbar\over 2}$ in a step of $\hbar$ indeed.
  • $N=2$, we go from $-\hbar$ to $0$ to $+\hbar$, the whole sequence in 2 steps.
  • $N=3$, we go from $-{3\hbar\over2}$ to $-{\hbar\over2}$ to ${3\hbar\over2}$ and then to $+{3\hbar\over2}$, in 3 steps.
  • $N=4$, that's $-2\hbar, -\hbar, 0, \hbar, 2\hbar$, etc.

We have found all the $\mu$, one per wavefunction. We still have to find the one $\lambda$ that corresponds to all of them. Coming back to the link between $\lambda$ and $\mu_T$, we see that the eigenvalue of $L^2$ for a given $N$ is of the type

$$\displaystyle\hbar^2\left({N\over 2}\left({N\over 2}+1\right)\right)$$

for integer $N$, and the eigenvalues of $L_z$ are $\hbar m$ for $m=-{N\over2}$ to $+{N\over2}$. If we call $l\equiv N/2$ as is the usual convention, we can denote the joint eigenfunctions of $L_z$ and $L^2$ as $\psi_l^m$.

We can now work out what these wavefunctions are in real-space by going back to the differential operators. That is, we want to know the functional form of $\psi_l^m(x, y, z)$ or, since we are dealing about rotation, in the polar form $\psi_l^m(r,\theta,\phi)$. For this, we need to express our differential operators in terms of the polar coordinates, which we already did to some extent in earlier lectures, and that we will now bring this to its conclusion. We write $\vec L=\vec r\times\vec p=-i\hbar \vec r\times\nabla$ in spherical coordinates, with

$$\nabla=\hat r\partial_r+\hat\theta{1\over r}\partial_\theta+\hat\phi{1\over r\sin\theta}\partial_\phi\,,$$

so that

$$\displaystyle\vec L=-i\hbar\left( \vec r\times \hat r\partial_r+\vec r\times \hat\theta{1\over r}\partial_\theta+\vec r\times\hat\phi{1\over r\sin\theta}\partial_\phi \right)\,.$$

Since $\vec r\times\hat r=0$, $\hat r\times\hat\theta=\hat\phi$ and $\hat r\times\hat\phi=-\hat\phi$, we find:

$$\vec L=-i\hbar(\hat\phi\partial_\theta-\hat\theta{1\over\sin\theta}\partial_\phi)\,.$$

Since the creation and destruction operators are in terms of $L_x$ and $L_y$, we still need to go back to Cartesian coordinates. We do this from the relations we know from 3D vector calculus:

\begin{align} \hat\theta&=(\cos\theta\cos\phi)\hat i+(\cos\theta\sin\phi)\hat j-(\sin\theta)\hat k\,,\\ \hat\phi&=-(\sin\phi)\hat i+(\cos\phi)\hat j\,. \end{align}

From this, one can see, by substitution and proper collecting, that the all-important $z$-component takes a particularly simple (and now familiar) form:


This is consistent with the results we got in 2D with the harmonic oscillator. We leave it as an exercise to similarly exprss $L_x$ and $L_y$ (and justify $L_z$ above) and check for yourself that, eventually:

$$L_\pm=\pm\hbar e^{\pm i\phi}\left(\partial_\theta\pm i\cot\theta\partial_\phi\right)\,.$$

This is a cotangent there. How often do we see them? Note that the radial dependence has disappeared. Which is appropriate: rotation is about the angles! With these expressions, we can now compute $\psi_l^m$, starting with:

$$L_z\psi_l^m=\hbar m\psi_l^m=-i\hbar\partial_\phi\psi_l^m$$

so, assuming separation of variable (which is fine since we are after a basis), this one at least is easy to solve:


where we don't need to retain the sign as $m$ is a signed integer anyway. The $\Theta(\theta)$ part is more troublesome, but we actually already passed through such an ordeal. We also let you check by yourself that, using again $L^2=L_+L_-+L_z^2-\hbar L_z$ but this time with differential operators in Cartesian space, that one arrives at the following differential equation:

$$\displaystyle\sin\theta{d\over d\theta}\left(\sin\theta{d\Theta\over d\theta}\right)+[l(l+1)\sin^2\theta-m^2]\Theta=0$$

which we have already encountered when solving the angular part of the Schrödinger equation. The solution was given there and so, the eigenstates of angular momentum are none less than the Spherical Harmonics $Y_l^m$.

The only thing is that, of course, solutions are for integer $l$. The algebraic method also opened the door for solutions with half-integer angular momentum. But these have no solutions in the $\theta, \phi$ (real, physical, geometric... how to call it?) space. We will see next lectures that there are such rotations, however, but that they take place elsewhere!