Undergrad Matrix Representation of the Angular Momentum Raising Operator

Click For Summary
The discussion focuses on the calculation of the matrix elements for the angular momentum raising operator L(+) with l = 1 and m values of -1, 0, and 1. The user encounters a problem with the element L(+)|0,-1>, expecting a value of √2ħ but instead calculating it as 0ħ|0,0>. The confusion arises from incorrectly applying the general formula for the raising operator, specifically using the wrong value for l. After clarification, it is confirmed that l should be 1, leading to the correct application of the formula and resolving the issue. The user acknowledges the mistake and expresses gratitude for the assistance.
hnicholls
Messages
49
Reaction score
1
TL;DR
Matrix Representation of the Angular Momentum Raising Operator. Calculating L(+).
In calculating the matrix elements for the raising operator L(+) with l = 1 and m = -1, 0, 1 each of my elements conforms to a diagonal shifted over one column with values [(2)^1/2]hbar on that diagonal, except for the element, L(+)|0,-1>, where I have a problem.

This should be value [(2)^1/2]hbar; however, I get L(+)|0,-1> = (0(0+1)-(-1)((-1)+1))^1/2|0,-1+1> = 0hbar|0,0>. This would be a 0 value not [(2)^1/2]hbar. Not sure where I making my mistake.
 
Physics news on Phys.org
Please use ## ( beginning and end) for Latex rendeting to make post easier to read.
 
L+ with l = 1 and m = -1, 0, 1

L+|0,-1> = (0(0+1)-(-1)((-1)+1))^1/2|0,-1+1> = 0ħ|0,0>. This would be a 0 value not √2ħ. Not sure why this wrong.
 
L+ with l = 1 and m = -1, 0, 1

L+|0,-1> = √[0(0+1)-(-1)((-1)+1)]|0,-1+1> = 0ħ|0,0>. This would be a 0 value not √2ħ. Not sure why this wrong.
 
In the following I use ##\hbar=1## and write ##|m \rangle## (we are in the subspace with ##l=1## anyway). Then the "raising operator" acts on these basis states by
$$\hat{L}_+|1 \rangle=0, \quad \hat{L}_+ |0 \rangle=\sqrt{2}|1 \rangle, \quad \hat{L}_{+} |-1 \rangle=\sqrt{2} |0 \rangle.$$
Here I used the general formula
$$|\hat{L}_+ |l,m \rangle=\sqrt{(l-m)(l+m+1)} |l,m+1 \rangle.$$
The matrix representation with the basis ##|1 \rangle##, ##0 \rangle##, ##|-1 \rangle## (in that order) thus reads
$$\hat{L}_+=\begin{pmatrix} 0 & \sqrt{2} & 0 \\ 0 & 0 & \sqrt{2} \\ 0 & 0 &0 \end{pmatrix} $$
 
  • Like
Likes Mentz114
Applying the general formula

$$|\hat{L}_+ |l,m \rangle=\sqrt{(l-m)(l+m+1)} |l,m+1 \rangle.$$

$$|\hat{L}_+ |l,m \rangle=\sqrt{(0-(-1))(0+(-1)+1)} |l,m+1 \rangle.$$

$$|\hat{L}_+ |l,m \rangle=\sqrt{(+1)(0)} |0,0 \rangle.$$

$$|\hat{L}_+ |l,m \rangle=\sqrt{0} |0,0 \rangle.$$

I don't see how matrix element column 3 row 2 can be$$\sqrt{2} $$I still get 0. Thank you for the response.
 
hnicholls said:
Applying the general formula

You calculated this wrong. These states all have ##l = 1##, not ##l = 0##. Try the formula with ##l = 1##.
 
  • Like
Likes vanhees71
Got it! Thanks! Did not correctly apply l = 1 as the subspace and the m values (1,0,-1) as the basis states.
 
  • Like
Likes vanhees71

Similar threads

  • · Replies 1 ·
Replies
1
Views
936
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 10 ·
Replies
10
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 2 ·
Replies
2
Views
920
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 0 ·
Replies
0
Views
1K