Effect of the raising operator on an l=1 m=1 state

  • Thread starter neish88
  • Start date
Pretty basic question but I was wondering what the result of acting the raising operator on an l=1, m=1 quantum state for a hydrogen wavefunction would be.

Specifically,

L+|1,1> = ?

I know that normally L+|l,m>=hbar(l(l+1)-m(m+1))1/2|l,m+1> but I wasn't sure if the eigenstate remained the same (since m cannot be larger than l) or if it results in zero, like if you were to act the lowering operator on the lowest possible state.

Thanks in advance!
 

malawi_glenn

Science Advisor
Homework Helper
4,782
22
it will be zero

L+ on |L,L> = 0

since M_max = L

Same with L- on |L,-L>

since M_min = -L
 
Much appreciated thank you!
 

Related Threads for: Effect of the raising operator on an l=1 m=1 state

Replies
3
Views
2K
Replies
9
Views
4K
Replies
3
Views
372
Replies
18
Views
3K
  • Posted
Replies
3
Views
1K

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving

Hot Threads

Top