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

In summary, acting the raising operator on an l=1, m=1 quantum state for a hydrogen wavefunction would result in zero, as M_max is equal to L and cannot be larger than that. Similarly, acting the lowering operator on the lowest possible state would also result in zero, as M_min is equal to -L. This is due to the fact that m cannot be larger than l.
  • #1
neish88
2
0
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!
 
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  • #2
it will be zero

L+ on |L,L> = 0

since M_max = L

Same with L- on |L,-L>

since M_min = -L
 
  • #3
Much appreciated thank you!
 

1. What is a raising operator in the context of l=1 m=1 state?

In quantum mechanics, a raising operator is a mathematical operator that increases the quantum number associated with a specific quantum state. In the l=1 m=1 state, the raising operator increases the magnetic quantum number by one unit, resulting in a new state with l=1 and m=2.

2. How does the raising operator affect the energy of an l=1 m=1 state?

The raising operator does not affect the energy of an l=1 m=1 state. It only changes the quantum number associated with the state, which can affect other properties such as the angular momentum.

3. Can the raising operator be applied to any state with l=1 m=1?

No, the raising operator can only be applied to a state with specific quantum numbers. In the case of l=1 m=1, the raising operator can only be applied to the state with l=1 and m=1.

4. What happens if the raising operator is applied multiple times to an l=1 m=1 state?

Each time the raising operator is applied, the magnetic quantum number increases by one unit. Therefore, applying the operator multiple times will result in a state with a higher magnetic quantum number, such as l=1 m=3 or l=1 m=4.

5. How does the raising operator affect the probability of finding an electron in an l=1 m=1 state?

The raising operator does not directly affect the probability of finding an electron in a specific state. However, by changing the quantum number associated with the state, it can indirectly affect the probability through other properties such as the angular momentum.

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