Angular momentum operators on matrix form

In summary, the conversation discussed a 3-dimensional subspace H and its basis states. The usual operator relations for L_{z} and L^{2} were stated, along with the expressions for L_{+} and L_{-}. The solution for L_{+} was given as a matrix with specific entries, and the method for finding these entries was discussed. The conversation concluded with a request for clarification on the conceptual understanding of the problem.
  • #1
Denver Dang
148
1

Homework Statement


Hi.

I'm given a 3-dimensional subspace [itex]H[/itex] that is made up of the states [itex]|1,-1\rangle[/itex], [itex]|1,0\rangle[/itex] and [itex]|1,1\rangle[/itex] with the states defined as [itex]|l,m\rangle[/itex] and [itex]l=1[/itex] as you can see.

The usual operator relations for [itex]L_{z}[/itex] and [itex]L^{2}[/itex] applies, and also:
[tex]L_{+} = L_{x}+iL_{y}[/tex]
[tex]L_{-} = L_{x}-iL_{y}[/tex]

Then I'm told to express the operators [itex]L_{+}[/itex] and [itex]L_{-}[/itex] in [itex]H[/itex].

The answer for [itex]L_{+}[/itex] is supposed to be:
[tex]{{L}_{+}}=\sqrt{2}\hbar \left[ \begin{matrix}
0 & 0 & 0 \\
1 & 0 & 0 \\
0 & 1 & 0 \\
\end{matrix} \right]
[/tex]
And the transposed for [itex]L_{-}[/itex]

But I'm really not sure how that is found.

Homework Equations

The Attempt at a Solution


The [itex]\sqrt{2}\hbar[/itex] probably comes from the fact that:
[tex]{{L}_{\pm }}\left| l,m \right\rangle =\sqrt{l\left( l+1 \right)-m\left( m\pm 1 \right)}\hbar \left| l,m \right\rangle[/tex]

But I can't figure out why the entries in the matrix is like that.
My first thought was that it should be diagonal, as it was made up of the 3 bases, but as you can see, it is not diagonal.So I was hoping someone could explain what I'm missing out ?Thanks in advance.
 
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  • #2
Denver Dang said:
The [itex]\sqrt{2}\hbar[/itex] probably comes from the fact that:
[tex]{{L}_{\pm }}\left| l,m \right\rangle =\sqrt{l\left( l+1 \right)-m\left( m\pm 1 \right)}\hbar \left| l,m \right\rangle[/tex]

I don't think that is right; it should change the value of m, and also m may not be greater than [itex]\ell[/itex].

Denver Dang said:
The answer for [itex]L_{+}[/itex] is supposed to be:
[tex]{{L}_{+}}=\sqrt{2}\hbar \left[ \begin{matrix}
0 & 0 & 0 \\
1 & 0 & 0 \\
0 & 1 & 0 \\
\end{matrix} \right][/tex]

This acts on column vector that has the first row for m = -1, the second row for m =0, and the third row for m= 1.

The first row of the matrix is all zero because no matter what, when [itex]L_{+}[/itex] acts on a state, there is never anything remaining in the m = -1 position. The second row takes whatever was in the m = -1 position and moves it to the m=0 position (multiplied by a constant). The third row of the the matrix takes whatever is in the m=0 and moves it to m =1 position, multiplied by a constant.

This was a conceptual question and not a homework exercise, right?
 

FAQ: Angular momentum operators on matrix form

What is the matrix form of the angular momentum operator?

The matrix form of the angular momentum operator is a 3x3 matrix that represents the three components of the operator: Lx, Ly, and Lz. It is typically denoted as [Lx, Ly, Lz] and can be used to calculate the total angular momentum of a system.

How is the matrix form of the angular momentum operator derived?

The matrix form of the angular momentum operator is derived from the mathematical definition of the operator, which is based on the cross product of position and momentum vectors. By applying the rules of matrix multiplication, the components of the operator can be represented in a matrix form.

What is the significance of the matrix form of the angular momentum operator?

The matrix form of the angular momentum operator is significant because it allows us to mathematically describe and calculate the angular momentum of a system. It is a crucial tool in quantum mechanics and is used in many applications, such as determining the angular momentum states of particles.

Can the matrix form of the angular momentum operator be used to calculate other properties?

Yes, the matrix form of the angular momentum operator can be used to calculate other properties, such as the total energy and position of a particle. This is because angular momentum is a fundamental property of particles and is related to other physical quantities.

How does the matrix form of the angular momentum operator relate to the Heisenberg uncertainty principle?

The matrix form of the angular momentum operator is related to the Heisenberg uncertainty principle through the commutation relations between the components of the operator. These relations, also known as the uncertainty relations, state that the product of uncertainties in position and momentum, as well as angular momentum, cannot be simultaneously minimized. This is a fundamental principle in quantum mechanics that has significant implications in understanding the behavior of particles.

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