Angular momentum – eigenfunctions

Required math: algebra, calculus

Required physics: quantum mechanics, angular momentum

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Sec. 4.3.2 & Problem 4.21.

We have seen how to calculate the eigenvalues of the angular momentum operators {L^{2}} and {L_{z}} using only the commutation relations for the three components of {\mathbf{L}}. This doesn’t tell us the eigenfunctions, however, so we’ll consider that here. The procedure unfortunately involves a lot of calculation, so we just have to get on with it.

It is easiest to work in spherical coordinates, so we first need to express {\mathbf{L}} in spherical coordinates. We have

\displaystyle   \mathbf{L} \displaystyle  = \displaystyle  -i\hbar\mathbf{r}\times\mathbf{\nabla}
\displaystyle  \nabla \displaystyle  = \displaystyle  \hat{r}\frac{\partial}{\partial r}+\hat{\theta}\frac{1}{r}\frac{\partial}{\partial\theta}+\hat{\phi}\frac{1}{r\sin\theta}\frac{\partial}{\partial\phi}

The radius vector is {\mathbf{r}=r\hat{r}} and the cross product of any vector with itself is zero, so the angular momentum operator becomes

\displaystyle   \mathbf{L} \displaystyle  = \displaystyle  -i\hbar\left[\hat{r}\times\hat{\theta}\frac{\partial}{\partial\theta}+\hat{r}\times\hat{\phi}\frac{1}{\sin\theta}\frac{\partial}{\partial\phi}\right]
\displaystyle  \displaystyle  = \displaystyle  -i\hbar\left[\hat{\phi}\frac{\partial}{\partial\theta}-\hat{\theta}\frac{1}{\sin\theta}\frac{\partial}{\partial\phi}\right]

using the cross product relations for the spherical unit vectors {\hat{r}\times\hat{\theta}=\hat{\phi}} and {\hat{r}\times\hat{\phi}=-\hat{\theta}}.

We need to extract from this formula the values for the three components of {\mathbf{L}}, so we need to convert this equation so that the vectors are the cartesian unit vectors. To do this, we can use the formulas

\displaystyle   \hat{\theta} \displaystyle  = \displaystyle  \cos\theta\cos\phi\hat{i}+\cos\theta\sin\phi\hat{j}-\sin\theta\hat{k}
\displaystyle  \hat{\phi} \displaystyle  = \displaystyle  -\sin\phi\hat{i}+\cos\phi\hat{j}

Plugging these into the equation for {\mathbf{L}} and collecting terms, we get

\displaystyle   \mathbf{L} \displaystyle  = \displaystyle  -i\hbar\left[(-\sin\phi\hat{i}+\cos\phi\hat{j})\frac{\partial}{\partial\theta}-(\cos\theta\cos\phi\hat{i}+\cos\theta\sin\phi\hat{j}-\sin\theta\hat{k})\frac{1}{\sin\theta}\frac{\partial}{\partial\phi}\right]
\displaystyle  L_{x} \displaystyle  = \displaystyle  -i\hbar\left[-\sin\phi\frac{\partial}{\partial\theta}-\cot\theta\cos\phi\frac{\partial}{\partial\phi}\right]
\displaystyle  L_{y} \displaystyle  = \displaystyle  -i\hbar\left[\cos\phi\frac{\partial}{\partial\theta}-\cot\theta\sin\phi\frac{\partial}{\partial\phi}\right]
\displaystyle  L_{z} \displaystyle  = \displaystyle  -i\hbar\frac{\partial}{\partial\phi}

The raising and lowering operators can be worked out from these formulas, and we get

\displaystyle   L_{\pm} \displaystyle  = \displaystyle  L_{x}\pm iL_{y}
\displaystyle  \displaystyle  = \displaystyle  -i\hbar\left[(-\sin\phi\pm i\cos\phi)\frac{\partial}{\partial\theta}-\cot\theta(\cos\phi\pm i\sin\phi)\frac{\partial}{\partial\phi}\right]
\displaystyle  \displaystyle  = \displaystyle  \hbar\left[\pm(\cos\phi\pm i\sin\theta)\frac{\partial}{\partial\theta}+i\cot\theta(\cos\phi\pm i\sin\phi)\frac{\partial}{\partial\phi}\right]
\displaystyle  \displaystyle  = \displaystyle  \hbar\left[\pm e^{\pm i\phi}\frac{\partial}{\partial\theta}+i\cot\phi e^{\pm\phi}\frac{\partial}{\partial\phi}\right]
\displaystyle  \displaystyle  = \displaystyle  \pm\hbar e^{\pm i\phi}\left[\frac{\partial}{\partial\theta}\pm i\cot\phi\frac{\partial}{\partial\phi}\right]

We now have all the bits we need to calculate {L^{2}}. We use the formula

\displaystyle  L^{2}=L_{+}L_{-}+L_{z}^{2}-\hbar L_{z}

We work out the first term by applying the two operators to a test function {f}:

\displaystyle   L_{+}L_{-}f \displaystyle  = \displaystyle  -\hbar^{2}e^{i\phi}\left(\frac{\partial}{\partial\theta}+i\cot\theta\frac{\partial}{\partial\phi}\right)\left(e^{-i\phi}\left(\frac{\partial f}{\partial\theta}-i\cot\theta\frac{\partial f}{\partial\phi}\right)\right)
\displaystyle  \displaystyle  = \displaystyle  -\hbar^{2}(f_{\theta\theta}+i\csc^{2}\theta f_{\phi}-i\cot\theta f_{\phi\theta}+
\displaystyle  \displaystyle  \displaystyle  \,\,\,\, i\cot\theta(-i(f_{\theta}-i\cot\theta f_{\phi})+f_{\phi\theta}-i\cot\theta f_{\phi\phi}))
\displaystyle  \displaystyle  = \displaystyle  -\hbar^{2}(f_{\theta\theta}-\cot\theta f_{\theta}+\cot^{2}\theta f_{\phi\phi}+i(\csc^{2}\theta-\cot^{2}\theta)f_{\phi})

Subscripts on the test function {f} represent derivatives with respect to the subscripted variable. Using the trig identity {\csc^{2}\theta-\cot^{2}\theta=1}, we get the required result after removing the test function:

\displaystyle  L_{+}L_{-}=-\hbar^{2}\left(\frac{\partial^{2}}{\partial\theta^{2}}+\cot\theta\frac{\partial}{\partial\theta}+\cot^{2}\theta\frac{\partial^{2}}{\partial\phi^{2}}+i\frac{\partial}{\partial\phi}\right)

We now work out {L^{2}}:

\displaystyle   L^{2} \displaystyle  = \displaystyle  L_{+}L_{-}+L_{z}^{2}-\hbar L_{z}
\displaystyle  \displaystyle  = \displaystyle  -\hbar^{2}\left(\frac{\partial^{2}}{\partial\theta^{2}}+\cot\theta\frac{\partial}{\partial\theta}+\cot^{2}\theta\frac{\partial^{2}}{\partial\phi^{2}}+i\frac{\partial}{\partial\phi}\right)-\hbar^{2}\frac{\partial^{2}}{\partial\phi^{2}}+\hbar^{2}i\frac{\partial}{\partial\phi}
\displaystyle  \displaystyle  = \displaystyle  -\hbar^{2}\left[\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial}{\partial\theta}\right)+\frac{1}{\sin^{2}\theta}\frac{\partial^{2}}{\partial\phi^{2}}\right]

using the trig identity {1+\cot^{2}\theta=1/\sin^{2}\theta} to simplify the second derivative term in {\phi}.

The eigenfunction {f_{l}^{m}(\theta,\phi)} must satisfy

\displaystyle   L^{2}f_{l}^{m} \displaystyle  = \displaystyle  \hbar^{2}l(l+1)f_{l}^{m}
\displaystyle  -\hbar^{2}\left[\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial}{\partial\theta}\right)+\frac{1}{\sin^{2}\theta}\frac{\partial^{2}}{\partial\phi^{2}}\right]f_{l}^{m} \displaystyle  = \displaystyle  \hbar^{2}l(l+1)f_{l}^{m}

The eigenfunction equation for {L_{z}} is

\displaystyle  L_{z}f_{l}^{m}=-i\hbar\frac{\partial}{\partial\phi}f_{l}^{m}=\hbar mf_{l}^{m}

This equation is equivalent to

\displaystyle  \frac{1}{f_{l}^{m}}\frac{\partial^{2}f_{l}^{m}}{\partial\phi^{2}}=-m^{2}
since they both have the solution {f_{l}^{m}=g(\theta)e^{im\phi}}, where {g(\theta)} is some function of {\theta} (the precise function doesn’t matter since it cancels out on the left hand side).

Looking back at the solution to the angular part of the Schrödinger equation in three dimensions, we see that these differential equations are the same as the ones we encountered there when we used separation of variables. The solutions there were the spherical harmonics, so we see that the eigenfunctions of {L^{2}} are in fact spherical harmonics.

One final comment is worth making. When we worked out the eigenvalues of {L^{2}} we saw that they had the form {\hbar^{2}l(l+1)} where {l} could be any non-negative integer or half-integer. In the solution of the differential equations that give rise to the spherical harmonics, however, only integer values of {l} are allowed. You might think that these extra values of {l} are not physically significant, but when we come to study spin, we see that since the eigenstates of spin have no relation to spatial coordinates so that spherical harmonics are not involved, there is no reason to discard the half-integer values. In fact they turn out to be fundamental, since half-integer spins are common amongst the elementary particles (such as protons, neutrons and electrons).

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