A eigenstate of addition of angular momenta

In summary, to find the eigenstates (of L_z) for the given problem, we can use the ladder operators L_+ and L_- to obtain the desired eigenstate |2~0\rangle. By applying L_- twice to the highest eigenstate |3~3\rangle, we can write |2~0\rangle as a linear combination of |2~2\rangle and |1~1\rangle, with specific coefficients a and b that can be found using the orthogonality property of eigenstates.
  • #1
rar0308
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Homework Statement


[tex]l~m\rangle[/tex]
[tex]l=l_1+l_2[/tex]
[tex]l_1=2,l_2=1[/tex]
Find eigenstates(of[tex]L_z[/tex]) [tex]|2~0\rangle[/tex]


Homework Equations





The Attempt at a Solution


[tex]|l=3~m=3\rangle=|l_1=2~m_1=2\rangle|l_2=1~m_2=1 \rangle[/tex]. I do [tex]L_-=L_{1-}+L_{2-}[/tex] 3times.
So I get [tex]|3~0\rangle[/tex]= (omit)
Then How can I find [tex]2~0\rangle[/tex]?
I know that [tex]2~0\rangle[/tex] is orthogonal to [tex]|3~0\rangle[/tex]. But I don't know how to use that property to derive a result.
Please give me an answer.
 
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  • #2


Hello, the eigenstates of L_z can be found by using the ladder operators L_+ and L_- as follows:
1. Start with the highest eigenstate |l~m\rangle = |l_1+l_2~l_1+l_2\rangle
2. Apply the ladder operator L_- to obtain the next lower eigenstate |l~m\rangle = L_-|l_1+l_2~l_1+l_2\rangle = sqrt(l_2(l_1+l_2-m))|l_1+l_2~l_1+l_2-1\rangle
3. Continue applying L_- until you reach the desired eigenstate |l~m\rangle = |l_1~m_1\rangle|l_2~m_2\rangle
4. In this case, we want the eigenstate |2~0\rangle, which can be obtained by applying L_- twice to the highest eigenstate |3~3\rangle:
|2~0\rangle = L_-^2|3~3\rangle = sqrt(2)|2~2\rangle|1~1\rangle
5. Therefore, the eigenstate |2~0\rangle is a linear combination of |2~2\rangle and |1~1\rangle, and can be written as |2~0\rangle = a|2~2\rangle + b|1~1\rangle, where a and b are complex coefficients.
6. To find the specific values of a and b, you can use the orthogonality property of eigenstates. Since |2~0\rangle is orthogonal to |3~0\rangle, we can write the inner product as:
<3~0|2~0> = 0 = <3~0|a|2~2> + <3~0|b|1~1>
7. Using the fact that <l~m|l~m> = 1, we can solve for a and b to obtain the specific eigenstate |2~0\rangle = 1/sqrt(3)|2~2\rangle - sqrt(2/3)|1~1\rangle.
I hope this helps! Let me know if you have any further questions.
 

What is an eigenstate of addition of angular momenta?

An eigenstate of addition of angular momenta is a quantum state that describes the total angular momentum of a system, taking into account the individual angular momenta of the constituent particles.

How is an eigenstate of addition of angular momenta calculated?

The eigenstate of addition of angular momenta is calculated by taking the tensor product of the individual angular momentum states of the constituent particles.

What is the significance of an eigenstate of addition of angular momenta?

The eigenstate of addition of angular momenta is significant because it allows us to accurately describe and predict the behavior of a system with multiple particles and their angular momenta.

Can an eigenstate of addition of angular momenta change over time?

No, an eigenstate of addition of angular momenta is a stationary state that does not change over time in the absence of external influences.

How is an eigenstate of addition of angular momenta related to quantum entanglement?

An eigenstate of addition of angular momenta is related to quantum entanglement because it describes the total angular momentum of a system, which can be entangled between particles in a quantum state.

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