# Angular momentum – raising and lowering operators

Required math: calculus

Required physics: 3-d Schrödinger equation

References: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education Section 4.3.1 & Problem 4.18.

The angular momentum operator ${\mathbf{L}}$ in quantum mechanics has three components that are not mutually observable. In the calculation of the eigenvalues of ${L^{2}}$ and ${L_{z}}$, we made use of the raising and lowering operators ${L_{\pm}}$, defined as follows:

$\displaystyle L_{\pm}\equiv L_{x}\pm iL_{y} \ \ \ \ \ (1)$

We showed that the effect of these operators on an eigenfunction ${f_{l}^{m}}$ of ${L^{2}}$ and ${L_{z}}$ is to generate a new eigenfunction with the properties

 $\displaystyle L^{2}(L_{\pm}f_{l}^{m})$ $\displaystyle =$ $\displaystyle \hbar^{2}l(l+1)(L_{\pm}f_{l}^{m})\ \ \ \ \ (2)$ $\displaystyle L_{z}(L_{\pm}f_{l}^{m})$ $\displaystyle =$ $\displaystyle \hbar(m\pm1)(L_{\pm}f_{l}^{m}) \ \ \ \ \ (3)$

That is, the new eigenfunction has the same eigenvalue as the original eigenfunction with respect to the operator ${L^{2}}$ and an eigenvalue raised or lowered by ${\hbar}$ with respect to ${L_{z}}$.

It is useful to derive a formula which gives the exact operation of the raising and lowering operators on the eigenfunctions. That is, we seek a formula of form

$\displaystyle L_{\pm}f_{l}^{m}=A_{l}^{m}f_{l}^{m\pm1} \ \ \ \ \ (4)$

for some constant ${A_{l}^{m}}$. It turns out we can find the value of this constant without knowing anything more about the eigenfunctions than that they are normalized.

To do this, we need to show that ${L_{\mp}}$ is the hermitian conjugate of ${L_{\pm}}$. To be strict about this, we need first to show that ${L_{x}}$ and ${L_{y}}$ are hermitian. Since they are observables, they have to be, but we can show it explicitly anyway.

To prove that the components of angular momentum are Hermitian, we can use the same technique as that used for proving that linear momentum is Hermitian. The difference here is that matrix elements are integrated over 3 dimensions rather than one. We can see how the proof goes for the case of ${L_{x}=yp_{z}-zp_{y}=-i\hbar y\partial/\partial z+i\hbar z\partial/\partial y}$.

 $\displaystyle \langle g|L_{x}f\rangle$ $\displaystyle =$ $\displaystyle -i\hbar\int\int\int g^*\left(y\frac{\partial f}{\partial z}-z\frac{\partial f}{\partial y}\right)dxdydz \ \ \ \ \ (5)$

We can consider the first term in this integral, since all other terms in this and the other integrals for the other components work in the same way. Using integration by parts over ${z}$ we get

 $\displaystyle \int\int\int g^*y\frac{\partial f}{\partial z}dzdydz$ $\displaystyle =$ $\displaystyle \int\int g^*y(f|_{z=-\infty}^{z=\infty})dydx-\int\int\int y\frac{\partial g^*}{\partial z}fdzdydx \ \ \ \ \ (6)$

The first term is zero for the usual reason that any function ${f}$ that is physically meaningful must be normalizable and so must go to zero at infinity. The second term uses the fact that ${y}$ is a constant when taking the derivative with respect to ${z}$. If we plug this result and the similar one for the second term in the integral back into the original equation, we get

 $\displaystyle \langle g|L_{x}f\rangle$ $\displaystyle =$ $\displaystyle -i\hbar\int\int\int g^*\left(y\frac{\partial f}{\partial z}-z\frac{\partial f}{\partial y}\right)dxdydz\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle i\hbar\int\int\int f\left(y\frac{\partial g^*}{\partial z}-z\frac{\partial g^*}{\partial y}\right)dxdydz\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \langle L_{x}g|f\rangle \ \ \ \ \ (9)$

Once we know that ${L_{x}}$ and ${L_{y}}$ are Hermitian, then it follows that ${L_{\mp}^{\dagger}=(L_{x}\mp i\hbar L_{y})^{\dagger}=L_{x}\pm i\hbar L_{y}=L_{\pm}}$.

Now to get the constant ${A_{l}^{m}}$, we have

 $\displaystyle \langle L_{\pm}f_{l}^{m}|L_{\pm}f_{l}^{m}\rangle$ $\displaystyle =$ $\displaystyle \langle f_{l}^{m}|L_{\mp}L_{\pm}f_{l}^{m}\rangle\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \langle f_{l}^{m}|(L^{2}-L_{z}^{2}\mp\hbar L_{z})f_{l}^{m}\rangle\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \hbar^{2}(l(l+1)-m^{2}\mp m)\langle f_{l}^{m}|f_{l}^{m}\rangle\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle (A_{l}^{m})^{2}\langle f_{l}^{m\pm1}|f_{l}^{m\pm1}\rangle \ \ \ \ \ (13)$

The second line uses the equation derived earlier:

$\displaystyle L^{2}=L_{\pm}L_{\mp}+L_{z}^{2}\mp\hbar L_{z} \ \ \ \ \ (14)$

and the third line uses the eigenvalues of ${L^{2}}$ and ${L_{z}}$.

Assuming all the ${f_{l}^{m}}$ are normalized,

 $\displaystyle A_{l}^{m}$ $\displaystyle =$ $\displaystyle \hbar\sqrt{l(l+1)-m^{2}\mp m}\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \hbar\sqrt{(l\mp m)(l\pm m+1)} \ \ \ \ \ (16)$

Applying ${L_{+}}$ to ${f_{l}^{l}}$ or ${L_{-}}$ to ${f_{l}^{-l}}$ results in ${A_{l}^{m}}$ being zero, as required.

## 10 thoughts on “Angular momentum – raising and lowering operators”

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2. B.D.

In line (10), why did you take the inner product in order to find A^m_l? I understand the math and the logic afterward, but I’m not quite sure why you’ve taken the inner product of the LHS of the given equation with itself, and I would like to understand.

Also, thanks so much for posting these solutions! They’ve been a great help in understanding material and studying for tests!

1. gwrowe Post author

Taking the inner product allows us to eliminate the wave functions using the normalization condition ${\langle f_{l}^{m}|f_{l}^{m}\rangle=1}$, thus leaving an equation for ${A_{\ell}^{m}}$ without any wave functions getting in the way.

3. Luan Nico

I believe there is an error in equations (11) and (12), the plus or minus sign should be a $\pm$, not a $\mp$, because you just replaced equation (14). This error propagates to equation (15), where there should be a $\pm$ too, but it’s fixed in the last equation (16).

1. gwrowe Post author

The equations are actually correct as they stand. Equation 14 has the term ${L_{\pm}L_{\mp}}$ but equation 10 has ${L_{\mp}L_{\pm}}$ so we need to swap ${\mp}$ with ${\pm}$ in equation 14:

$\displaystyle L^{2}=L_{\mp}L_{\pm}+L_{z}^{2}\pm\hbar L_{z}$