# I Why does a sum of operators act on the state like this?

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1. May 11, 2016

### Harry Smith

I'm reading through my quantum physics lecture notes (see page 216 of the lecture notes for more details) and under the ladder operators section there is a discussion of the expectation value of $L_x$ for a state $\psi = R(r) \left( \sqrt{ \frac{2}{3}} Y_{11} - \sqrt{ \frac{1}{3}} Y_{10} \right)$ such that
$$\langle \psi | L_x | \psi \rangle = \frac{1}{2} \bigg\langle \sqrt{ \frac{2}{3}} Y_{11} - \sqrt{ \frac{1}{3}} Y_{10} \bigg| L_+ + L_- \bigg| \sqrt{ \frac{2}{3}} Y_{11} - \sqrt{ \frac{1}{3}} Y_{10} \bigg\rangle,$$ where $L_\pm$ are the angular momentum ladder operators and $Y_{\ell m}$ are the spherical harmonics.

Now, this all makes sense to me, however the next step states that
$$\langle \psi | L_x | \psi \rangle = \frac{1}{2} \bigg\langle \sqrt{ \frac{2}{3}} Y_{11} - \sqrt{ \frac{1}{3}} Y_{10} \bigg| \sqrt{ \frac{2}{3}} L_- Y_{11} - \sqrt{ \frac{1}{3}} L_+ Y_{10} \bigg\rangle.$$
Why have these operators been assigned seemingly at random to one of the states?

My intuition suggests that
\begin{align*} (A+B) |u_1+u_2 \rangle &= A |u_1+u_2 \rangle + B |u_1+u_2 \rangle \\&= | A u_1 + A u_2 \rangle + | B u_1 + B u_2 \rangle\end{align*}
\begin{align*}\implies \langle u_1+u_2 | (A+B) |u_1+u_2 \rangle &= \langle u_1+u_2 | A u_1 + A u_2 \rangle + \langle u_1+u_2 | B u_1 + B u_2 \rangle \\&= \langle u_1 | A u_1 \rangle + \langle u_2 | A u_1 \rangle + \langle u_1 | A u_2 \rangle + \langle u_2 | A u_2 \rangle \\&\ \ \ \ \ \ \ \ + \langle u_1 | B u_1 \rangle + \langle u_2 | B u_1 \rangle + \langle u_1 | B u_2 \rangle + \langle u_2 | B u_2 \rangle \end{align*}.
Where am I going wrong?

Please note that this is not a homework problem, so full solutions are welcome. I may need each tiny step written out to understand why this is happening.

Last edited: May 11, 2016
2. May 11, 2016

### George Jones

Staff Emeritus
Some steps have been left out.

What is $L_+ Y_{11}$?

What is $L_- Y_{10}$?

3. May 11, 2016

### Harry Smith

So since $$L_\pm Y_{\ell m} = \hbar \sqrt{\ell (\ell + 1) - m (m \pm 1)} Y_{\ell (m\pm1)},$$ we find that $L_+ Y_{11} = 0$.

Using the same logic as before, $L_- Y_{10} = 2\hbar Y_{1,-1}$, right?

Here, my first inference was that since $\ell$ is only valid for $\ell = 0,1,2,...$ then $Y_{1,-1} = 0$. However, other problems (such as 13.5.1 in the notes) successfully use $Y_{1,-1}$.

4. May 11, 2016

### George Jones

Staff Emeritus
Yes, so this term contributes nothing.

Yes. Even though this term is non-zero, it doesn't contribute anything to the integral (inner product). Why?

5. May 11, 2016

### Harry Smith

Ah, because the spherical harmonics are orthonormal, so we need only consider the states that are already present because the others will have an inner product of zero?

6. May 11, 2016

### George Jones

Staff Emeritus
Yes.

7. May 11, 2016

### Harry Smith

Thank you so much for your help, George!