Spin matrices for particle of spin 1

genloz
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Homework Statement


Construct the spin matrices (S_{x}, S_{y}, and S_{z}) for a particle of spin 1. Determine the action of S_{z}, S_{+}, and S_{-} on each of these
states.

Homework Equations


S=\sqrt{1(1+1)}\hbar
m=-s,-s+1,...,s-1,s

The Attempt at a Solution


m=-1,0,0,0,0,0,0,0,-1
S=\sqrt{2}\hbar
S_{z}=\hbar \[ \left( \begin{array}{ccc}<br /> 1 &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; -1 \end{array} \right)\]

So I understand where S_{z} comes from... I know what the answers for the matrices are from http://en.wikipedia.org/wiki/Pauli_matrices but I don't know how to go about finding S_{x} and S_{y} even with the commutation laws etc... and I don't really understand what it means by 'the action of' in the second part despite numerous searchs on google...

Any help is much appreciated, thankyou!
 
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genloz said:
So I understand where S_{z} comes from...

From where does S_z come?

I don't know how to go about finding S_{x} and S_{y}

Depending on how you found S_z, you might be able to use the same technique to find S_x and S_y first, or to find S_+ and S_- first.

I don't really understand what it means by 'the action of' in the second part

Neither do I. What states?

The eigenstates of S_z? General states?

Is this question from a course? From a book? If so which course and book?
 
Thanks very much for your reply...

But I'm very confused now... This exact question "Construct the spin matrices (S_{x}, S_{y}, and S_{z}) for a particle of spin 1. Determine the action of S_{z}, S_{+}, and S_{-} on each of these states." is from a quantum mechanics course. The relevant equations were equations I thought were relevant... and I thought that the equation listed for m gave the matrix S_{z}.

How would I find eigenstates of S_{z}? And what is the usual way for determining spin matrices? I can't find this info out anywhere on the internet and am getting more and more confused by the second...
 
Have you seen the general theory of angular momentum in your quantum course?

Stuff like (\hbar = 1):

J_z \left| jm \right&gt; = m \left| jm \right&gt;;

J_+ \left| jm \right&gt; = \sqrt{j \left( j + 1 \right) - m \left(m + 1 \right)} \left| j,m+1 \right&gt;;

J_- \left| jm \right&gt; = \sqrt{j \left( j + 1 \right) - m \left(m - 1 \right)} \left| j,m-1 \right&gt;.

If so, you can find the matices by considering stuff like

\left&lt; jm \right| J_\pm \left| jm \right&gt;

for j=1 and m = -1, 0, 1.
 
Last edited:
George Jones, you mean

J_- \left| jm \right&gt; = \sqrt{j \left( j + 1 \right) - m \left(m - 1 \right)} \left| j,m-1 \right&gt;.

?
 
malawi_glenn said:
George Jones, you mean

J_- \left| jm \right&gt; = \sqrt{j \left( j + 1 \right) - m \left(m - 1 \right)} \left| j,m-1 \right&gt;.

?

Yes, thanks. I have corrected my previous post.
 
This is an old thread, but people are bound to come back looking for these answers, so here's my 2 cents.
For a given S, S_z comes from demanding S_z|S,m&gt; = m |S,m&gt; and is thus a diagonal (2S+1)x(2S+1) matrix with elements S, S-1,...,-S
(Note: I assume |S,S&gt; = (1,0,...,0) and |S,-S&gt; = (0,0,...,1) )
The S_x and S_y matrices - and consequently S_+ and S_- - come from rotations of Sz about the y and z axes, S_x = U_y(\pi/2).S_z.U^{\dagger}_y(\pi/2) \text{ and } S_y = U_z(\pi/2).U_y(\pi/2).S_z.\left(U_z(\pi/2).U_y(\pi/2)\right)^{\dagger},
where the Uy matrix is comprised by the elements (m, m&#039; = S, S-1,..., -S and \theta\in [0,\pi]):

U^{y}_{m,m&#039;}(\theta) = \sqrt{(S-m)! (S+m)! (S-m&#039;)! (S+m&#039;)!}\sum_{x=Max(0,m&#039;-m)}^{Min(S+m&#039;,S-m)}\frac{(-1)^x cos(-\theta/2)^{2 S + m&#039; - m - 2 x} sin(-\theta/2)^{2 x + m - m&#039;}}{(S + m&#039; - x)! (S - m - x)! x! (x + m - m&#039;)!}

and Uz is diagonal with (\phi\in [0,2 \pi]):

U^z_{k,k}(\phi) = e^{i (k-1) \phi}, \text{ } k = 1,2,...,2 S+1

There is some ambiguity here however, since another Uz is quoted:

U^z_{m,m}(\phi) = e^{-i m \phi}, \text{ } m = S,...,-S

In the former case, k refer to matrix index (e.g. U_{1,1} is the upper left element of the matrix), while in the latter case m refer to magnetic number indexing (e.g. U_{S,S} is the upper left element of the matrix). I use the former formula and it gives the expected results.
 
How do you get the rotation matrix about the y-axis when you don't know what the Sy generator is?
 
George Jones said:
Have you seen the general theory of angular momentum in your quantum course?

Stuff like (\hbar = 1):

J_z \left| jm \right&gt; = m \left| jm \right&gt;;

J_+ \left| jm \right&gt; = \sqrt{j \left( j + 1 \right) - m \left(m + 1 \right)} \left| j,m+1 \right&gt;;

J_- \left| jm \right&gt; = \sqrt{j \left( j + 1 \right) - m \left(m - 1 \right)} \left| j,m-1 \right&gt;.

If so, you can find the matices by considering stuff like

\left&lt; jm \right| J_\pm \left| jm \right&gt;

for j=1 and m = -1, 0, 1.
Hello,
I've to construct Ix and Iy for I=1.

so, I can construct lowering and raising operator but how do you construct cartesian operator from this equation ?

there are no definition for Ix and Iy by their action on eigenstate vector like Iz, I+ and I-...

How can I do that easily ?
 
  • #10
turin said:
How do you get the rotation matrix about the y-axis when you don't know what the Sy generator is?

By using the spinor representation.
In essence you are using combinations of spin-1/2 to represent the behaviour of arbitrarily large spins. This way you can generate operators and wavefunctions of large spins starting from the known spin-1/2 matrices.

This was shown originaly by Majorana in 1932.
I have retrieved the info from W.Thompson's Angular Momentum book.
 
  • #11
lydilmyo said:
Hello,
I've to construct Ix and Iy for I=1.

so, I can construct lowering and raising operator but how do you construct cartesian operator from this equation ?

there are no definition for Ix and Iy by their action on eigenstate vector like Iz, I+ and I-...

How can I do that easily ?

If you know what the 3x3 ladder operators look like, start with the definitions

S_{+}=S_{x}+iS_{y}

S_{-}=S_{x}-iS_{y}

and write the cartesian matrices as linear combinations of the ladder operators.

S_{x} = ...

S_{y} = ...
 
  • #12
but this relation is available only for pauli matrix (2x2) and so for spin 1/2...

please help me ! or give the matrix representation of Ix and Iy for I=1 if you know it...
 
  • #13
lydilmyo said:
but this relation is available only for pauli matrix (2x2) and so for spin 1/2...

The relations are good for any dimensionality.
 
  • #14
lydilmyo said:
but this relation is available only for pauli matrix (2x2) and so for spin 1/2...

please help me ! or give the matrix representation of Ix and Iy for I=1 if you know it...

The equations I have posted solve your problem for any spin, integer or half-integer.
Anyway, for spin = 1:

<br /> S_x = \left(<br /> \begin{array}{ccc}<br /> 0 &amp; \frac{1}{\sqrt{2}} &amp; 0 \\<br /> \frac{1}{\sqrt{2}} &amp; 0 &amp; \frac{1}{\sqrt{2}} \\<br /> 0 &amp; \frac{1}{\sqrt{2}} &amp; 0<br /> \end{array}<br /> \right)<br />

<br /> S_y = \left(<br /> \begin{array}{ccc}<br /> 0 &amp; -\frac{i}{\sqrt{2}} &amp; 0 \\<br /> \frac{i}{\sqrt{2}} &amp; 0 &amp; -\frac{i}{\sqrt{2}} \\<br /> 0 &amp; \frac{i}{\sqrt{2}} &amp; 0<br /> \end{array}<br /> \right)<br />

<br /> S_z = \left(<br /> \begin{array}{ccc}<br /> 1 &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; -1<br /> \end{array}<br /> \right)<br />
 
  • #15
so, why kaltsoplyn talk about a really hard way to determine Ix and Iy if it is so easy to construct its by this relations ?
 
  • #16
lydilmyo said:
so, why kaltsoplyn talk about a really hard way to determine Ix and Iy if it is so easy to construct its by this relations ?

Synthesizing high order spins starting from spin-1/2's is hard, although the procedure is straightforward.

The formulas I give are general that's why they look so complex. However, it's not very hard to produce any such matrix if you plug these formulas in a symbolic mathematical software package like, say, Mathematica or Maple.
 
  • #17
kaltsoplyn said:
<br /> S_x = \left(<br /> \begin{array}{ccc}<br /> 0 &amp; \frac{1}{\sqrt{2}} &amp; 0 \\<br /> \frac{1}{\sqrt{2}} &amp; 0 &amp; \frac{1}{\sqrt{2}} \\<br /> 0 &amp; \frac{1}{\sqrt{2}} &amp; 0<br /> \end{array}<br /> \right)<br />

<br /> S_y = \left(<br /> \begin{array}{ccc}<br /> 0 &amp; -\frac{i}{\sqrt{2}} &amp; 0 \\<br /> \frac{i}{\sqrt{2}} &amp; 0 &amp; -\frac{i}{\sqrt{2}} \\<br /> 0 &amp; \frac{i}{\sqrt{2}} &amp; 0<br /> \end{array}<br /> \right)<br />

<br /> S_z = \left(<br /> \begin{array}{ccc}<br /> 1 &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; -1<br /> \end{array}<br /> \right)<br />

I don't undersand why you are this square root ? It is in reality 1/2 and not \frac{1}{\sqrt{2}}
 
  • #18
<br /> S_x = \frac{1}{2} \left(\left(<br /> \begin{array}{ccc}<br /> 0 &amp; 1 &amp; 0 \\<br /> 0 &amp; 0 &amp; 1 \\<br /> 0 &amp; 0 &amp; 0<br /> \end{array}<br /> \right) + \left(<br /> \begin{array}{ccc}<br /> 0 &amp; 0 &amp; 0 \\<br /> 1 &amp; 0 &amp; 0 \\<br /> 0 &amp; 1 &amp; 0<br /> \end{array}<br /> \right)\right)<br />

<br /> S_y = \frac{i}{2} \left(\left(<br /> \begin{array}{ccc}<br /> 0 &amp; 0 &amp; 0 \\<br /> 1 &amp; 0 &amp; 0 \\<br /> 0 &amp; 1 &amp; 0<br /> \end{array}<br /> \right) - \left(<br /> \begin{array}{ccc}<br /> 0 &amp; 1 &amp; 0 \\<br /> 0 &amp; 0 &amp; 1 \\<br /> 0 &amp; 0 &amp; 0<br /> \end{array}<br /> \right)\right)<br />

this is what I find with your relations.
 
  • #19
lydilmyo said:
<br /> S_x = \frac{1}{2} \left(\left(<br /> \begin{array}{ccc}<br /> 0 &amp; 1 &amp; 0 \\<br /> 0 &amp; 0 &amp; 1 \\<br /> 0 &amp; 0 &amp; 0<br /> \end{array}<br /> \right) + \left(<br /> \begin{array}{ccc}<br /> 0 &amp; 0 &amp; 0 \\<br /> 1 &amp; 0 &amp; 0 \\<br /> 0 &amp; 1 &amp; 0<br /> \end{array}<br /> \right)\right)<br />

<br /> S_y = \frac{i}{2} \left(\left(<br /> \begin{array}{ccc}<br /> 0 &amp; 0 &amp; 0 \\<br /> 1 &amp; 0 &amp; 0 \\<br /> 0 &amp; 1 &amp; 0<br /> \end{array}<br /> \right) - \left(<br /> \begin{array}{ccc}<br /> 0 &amp; 1 &amp; 0 \\<br /> 0 &amp; 0 &amp; 1 \\<br /> 0 &amp; 0 &amp; 0<br /> \end{array}<br /> \right)\right)<br />


this is what I find with your relations.


Strange, you should get the square root and I don't see anything wrong in my eqs.
Besides your results should check against:

<br /> [S_i,S_j]=i \hbar\epsilon_{i,j,k} S_k<br />
 
  • #20
<br /> S_+ \left| s m \right&gt; = \sqrt{s \left( s + 1 \right) - m \left(m + 1 \right)} \left| s,m+1 \right&gt;;<br />

<br /> S_+ \left| 10 \right&gt; = \sqrt{1* \left( 1 + 1 \right) - 0* \left(0 + 1 \right)} \left| 1,1 \right&gt; = \sqrt{2}\left| 1,1 \right&gt;;<br />

That's where the square root comes from. When you multiply by 1/2, you get it in the denominator.
 
  • #21
of course ! sorry...

Thanks a lot...

Do you know where come from this relation between ladder and cartesian operators ?
 
  • #22
It is a definition.
 
  • #23
So, what's the physical relevance of this definition ?
 
  • #24
I assume you are unfamiliar with the Wigner-Eckart theorem, so I will wave my hands (much to the horror of some readers) and say that x + iy is proportional to e, which is like a positive rotation as the angle increases, which is like a boost in the angular momentum which is like J+. The same can be said about J- but in the opposite direction.
 
  • #25
it is like I've thought, an analogy between Ix + iIy and cosx + isinx = e(ix) that represent a rotation in complex 2D space. So It's represent the rotation of angular moment from one state to another...

And you're right, I'm not familiar with Wigne-Eckart theory ! Soon, I understand it^^
I just begin to learn quantic angular momentum theory to well understand the formalism I use to describ NMR experiment (product operator/superoperator)... Because I'm biochemist and not physician^^ but It's really interesting !

Thanks a lot !
 
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