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

  • #1

Main Question or Discussion Point

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
[tex]\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, [/tex] 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
[tex]\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. [/tex]
Why have these operators been assigned seemingly at random to one of the states?

My intuition suggests that
[tex]\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*}[/tex]
[tex]\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*}.[/tex]
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:

Answers and Replies

  • #2
George Jones
Staff Emeritus
Science Advisor
Gold Member
7,273
808
Some steps have been left out.

What is ##L_+ Y_{11}##?

What is ##L_- Y_{10}##?
 
  • #3
What is ##L_+ Y_{11}##?
So since [tex]L_\pm Y_{\ell m} = \hbar \sqrt{\ell (\ell + 1) - m (m \pm 1)} Y_{\ell (m\pm1)},[/tex] we find that ##L_+ Y_{11} = 0##.

What is ##L_- Y_{10}##?
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
George Jones
Staff Emeritus
Science Advisor
Gold Member
7,273
808
So since [tex]L_\pm Y_{\ell m} = \hbar \sqrt{\ell (\ell + 1) - m (m \pm 1)} Y_{\ell (m\pm1)},[/tex] we find that ##L_+ Y_{11} = 0##.
Yes, so this term contributes nothing.

Using the same logic as before, ##L_- Y_{10} = 2\hbar Y_{1,-1}##, right?
Yes. Even though this term is non-zero, it doesn't contribute anything to the integral (inner product). Why?
 
  • #5
Yes. Even though this term is non-zero, it doesn't contribute anything to the integral (inner product). Why?
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
George Jones
Staff Emeritus
Science Advisor
Gold Member
7,273
808
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?
Yes.
 
  • #7

Related Threads on Why does a sum of operators act on the state like this?

Replies
3
Views
2K
Replies
3
Views
2K
Replies
6
Views
2K
Replies
3
Views
733
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
10
Views
6K
Replies
15
Views
1K
Replies
16
Views
950
Replies
15
Views
3K
Top