Matrix Representation of the Angular Momentum Raising Operator

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Discussion Overview

The discussion revolves around the matrix representation of the angular momentum raising operator \( L_+ \) for the case where \( l = 1 \) and \( m = -1, 0, 1 \). Participants explore the calculation of matrix elements and the application of the general formula for the raising operator.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant calculates the matrix elements for \( L_+ \) and finds that all elements conform to a diagonal shifted over one column, except for \( L_+ |0,-1\rangle \), which they believe should yield \( \sqrt{2}\hbar \) but instead results in 0.
  • Several participants reiterate the calculation for \( L_+ |0,-1\rangle \) and arrive at the same conclusion of 0, questioning the correctness of the expected value of \( \sqrt{2}\hbar \).
  • Another participant provides the general formula for the raising operator and applies it, confirming that \( L_+ |0,-1\rangle \) results in 0, while also presenting the matrix representation of \( L_+ \) based on the correct basis states.
  • A later reply points out that the calculations were incorrectly based on \( l = 0 \) instead of \( l = 1 \), suggesting that the participants should apply the formula with \( l = 1 \) to resolve the confusion.
  • One participant acknowledges the mistake in applying the correct value of \( l \) and expresses gratitude for the clarification.

Areas of Agreement / Disagreement

Participants generally disagree on the value of the matrix element \( L_+ |0,-1\rangle \), with some asserting it should be \( \sqrt{2}\hbar \) while others calculate it as 0. The discussion remains unresolved regarding the correct application of the raising operator formula.

Contextual Notes

There is a noted confusion regarding the application of the raising operator formula, particularly the dependence on the quantum number \( l \) and its implications for the calculations. Some assumptions about the basis states and their corresponding values may not have been clearly defined.

hnicholls
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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.
 
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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
$$\hat{L}_+=\begin{pmatrix} 0 & \sqrt{2} & 0 \\ 0 & 0 & \sqrt{2} \\ 0 & 0 &0 \end{pmatrix} $$
 
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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##.
 
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Got it! Thanks! Did not correctly apply l = 1 as the subspace and the m values (1,0,-1) as the basis states.
 
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