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Obtain the matrix representations of J-hat (subscript y) for the state with j=1 in te

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  • #1
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Homework Statement



The raising and lowering angular momentum operators, J-hat(subscript +), J-hat(subscript -) are defined in terms of the Cartesian components J-hat(subscript x), J-hat(subscript y), J-hat(subscript z) of angular momentum J-hat by J-hat(+)=J-hat(x)+iJ-hat(y) and J-hat(-)=J-hat(x)-iJ-hat(y).

Obtain the matrix representation of J(subscript y) for the state with j=1 in terms of the set of eigenstates of J-hat(subscript z).

The Attempt at a Solution



J(subscript y)=(-i/2) (0 sqrt 2 0)
(-sqrt 2 0 sqrt 2)
(0 -sqrt 2 0)

I don't know why though. And what does it mean why 'in terms of the set of eigenstates J-hat(z)?
 

Answers and Replies

  • #2
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I had trouble typing out the matrix properly

it is supposed to be:
(0 sqrt2 0)
(-sqrt2 0 sqrt2)
(0 -sqrt2 0)
 
  • #3
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its the 'in terms of the set of eigenstates of J-z' that confuse me. If it wasn't for those words at the end I might almost be able to do the question. So what do I do? Find the matrix representation of J (y) normally, and then operate with J(y) onto J(z) and then find its eigenvalue, and those eigenvalues are the eigenstates? I tried to operate with J(y) onto J(z) but got nowhere. I feel like I have no idea what I am doing. Please help.
 
  • #4
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I copied this somewhere from the internet:

the ordered basis is:
|1 1>, |1 0>,|1,-1>

The matrix representation of the operator J(z) in the ordered basis is:

J(z)=
<1,1|J z|1,1> <1,1|J z|1,0> <1,1|J z|1,-1>
<1,0|J z|1,1> <1,0|J z|1,0> <1,0|J z|1,-1>
<1,-1|J z|1,1> <1,-1|J z|1,0> <1,-1|J z|1,-1>

since the basis sets are eigenstates of J z
J z=
1 0 0
0 0 0
0 0 -1

but how did they get:
1 0 0
0 0 0
0 0 -1??

please help
 
  • #5
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The basis (the possible values for m) for j=1 are m=-1,0,1. Write the corresponding states in vector representation as
[itex]
m=-1: \hspace{2cm} \begin{pmatrix}1 \\ 0 \\ 0 \end{pmatrix} \\
m=0: \hspace{2cm} \begin{pmatrix}0 \\ 1 \\ 0 \end{pmatrix} \\
m=1: \hspace{2cm} \begin{pmatrix}0 \\ 0 \\ 1 \end{pmatrix}
[/itex]
The Jz operator is diagonal in this representation and has these as eigenvectors. The diagonal elements will simply be the possible m-values (that is, m=-1,0,1), giving the matrix you asked about.

In general, for spin j, the diagonal elements of Jz are the possible m-vales, going from m=j to m=-j in integer steps.
 

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