How to Derive the 3x3 Spin Matrices S(+) and S(-)?

[philip041]

For spin 1, the states are |1>, |-1> and |0>

These are written as:

|1> = column matrix[1 0 0]

|0> = column matrix [0 1 0]

|-1> = column matrix [0 0 1]

I need to find the 3 x 3 matrices for S(+) and S(-) which operate on these kets to give the correct answers eg.

S(-)|1> = sqrt(2)*h(bar)*|0>

I had tried getting linear eqns from the following:

[a b c] * [1] = h(bar)*sqrt(2)* [0]
[d e f] [0] [1]
[g h i] [0] [0]


but i just get

a + 0 + 0 = 0
d + 0 + 0 = 1* h(bar) * sqrt(2)
g + 0 + 0 = 0

This doesn't look remotely useful... I know the lowering operator is:

[ 0 0 0 ]
[sqrt(2) 0 0 ]
[ 0 sqrt(2) 0 ]

But how do I get there?

Cheers for any help

Philip

 

[philip041]

it took out my carefully placed spacing in the eqns, hope this is still legible...

 

[dextercioby]

Apply the operator S_{-} both on |1>, |0> and |-1> and see what you get.

 

[philip041]

But I'm meant to be working out S_{1}..

I have checked the answer given and it works, by trying S_{-}|1> i get the right thing but i don't understand how you would work out S_{-} if you weren't given the answer?

 

[buffordboy23]

But I'm meant to be working out S_{1}..

I have checked the answer given and it works, by trying S_{-}|1> i get the right thing but i don't understand how you would work out S_{-} if you weren't given the answer?

Your column matrices for the spin states only have one nonzero term. Let the lowering operator matrix have the general form:

$$\textbf{S}_{-} = \begin{bmatrix} A & B & C \\ D & E & F \\ G & H & I \end{bmatrix}$$

Now,

$$\textbf{S}_{-}\left|1\right> =$$ ?

What is the question mark equal to in terms of the generalized matrix elements? Do the same for the other two spin-vectors, and it should be evident how you can construct the matrix.

 

[philip041]

I think you get a 1 x 3 matrix with h(bar)*sqrt(2) in front of it, but only because I know those answers from a previous question I did.

$$\textbf{S}_{-} = \begin{bmatrix} A \\ D \\ G \end{bmatrix}$$The others, would be:

$$\textbf{S}_{-} = \begin{bmatrix} B \\ E \\ H \end{bmatrix}$$

$$\textbf{S}_{-} = \begin{bmatrix} C \\ F \\ I \end{bmatrix}$$

Is this what you meant by a generalised matrix?

 

[philip041]

argh, obviously ignore the S(-) i was using your code cause I don't know how to do it myself. The S(-)s are meant to be ?s

 

[philip041]

ohhhhhhhhhhhhhhhhhhhhh, i just got it cheers.

once you put the column vectors i just posted equal to the states you are expecting ie. B= 1 and H = 1

cheers

 

[buffordboy23]

Exactly right. And what exactly are these generalized 1x3 matrices? What is the purpose of the lowering operator?

EDIT: I think your coefficient values B=1 and H=1 may be off. You know what the matrix is supposed to be, so you can verify.

Actually, to be more clear,

$$<br /> \textbf{S}_{-}\left|1\right> = \begin{bmatrix} A \\ D \\ G \end{bmatrix} =\sqrt2\hbar \left|0\right><br />$$

 

[philip041]

sorry was being lazy by 1 i meant $$sqrt(2)\hbar [\tex]<br /> <br /> they are the eigenvalues?$$

 

[buffordboy23]

Be careful using the word "eigenvalue" here, since the lowering operator does not return an eigenvector (the same vector that it was acting on). I almost made the same mistake when thinking about how to explain it to you. Now, if you have to construct the Pauli Spin matrices, then these operators do return eigenvectors with the appropriate eigenvalue.