Spin matrices for particle of spin 1

By the way I'm trying to generate things like spin operators for large spins in hopes of using it in "spin networks" and "spin foams".Also, note that Ix and Iy are not uniquely defined. There are infinitely many operators that satisfy the commutation relation [Ix,Iy] = i Iz. This is why we usually just stick with Iz, I+ and I-.There are many ways to define Ix and Iy. One way is to use the general theory of angular momentum, as suggested by George Jones above. Another way, as I mentioned, is to use the spinor representation. If you really want to use the lowering and raising operators, you can use the following relations:Ix =
  • #1
genloz
76
1

Homework Statement


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

Homework Equations


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

The Attempt at a Solution


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

So I understand where [tex]S_{z}[/tex] 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 [tex]S_{x}[/tex] and [tex]S_{y}[/tex] 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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  • #2
genloz said:
So I understand where [tex]S_{z}[/tex] comes from...

From where does [itex]S_z[/itex] come?

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

Depending on how you found [itex]S_z[/itex], you might be able to use the same technique to find [itex]S_x[/itex] and [itex]S_y[/itex] first, or to find [itex]S_+[/itex] and [itex]S_-[/itex] 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 [itex]S_z[/itex]? General states?

Is this question from a course? From a book? If so which course and book?
 
  • #3
Thanks very much for your reply...

But I'm very confused now... This exact question "Construct the spin matrices ([tex]S_{x}[/tex], [tex]S_{y}[/tex], and [tex]S_{z}[/tex]) for a particle of spin 1. Determine the action of [tex]S_{z}[/tex], [tex]S_{+}[/tex], and [tex]S_{-}[/tex] 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 [tex]S_{z}[/tex].

How would I find eigenstates of [tex]S_{z}[/tex]? 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...
 
  • #4
Have you seen the general theory of angular momentum in your quantum course?

Stuff like ([itex]\hbar = 1[/itex]):

[tex]J_z \left| jm \right> = m \left| jm \right>;[/tex]

[tex]J_+ \left| jm \right> = \sqrt{j \left( j + 1 \right) - m \left(m + 1 \right)} \left| j,m+1 \right>;[/tex]

[tex]J_- \left| jm \right> = \sqrt{j \left( j + 1 \right) - m \left(m - 1 \right)} \left| j,m-1 \right>.[/tex]

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

[tex]\left< jm \right| J_\pm \left| jm \right>[/tex]

for [itex]j=1[/itex] and [itex]m = -1, 0, 1.[/itex]
 
Last edited:
  • #5
George Jones, you mean

[tex]J_- \left| jm \right> = \sqrt{j \left( j + 1 \right) - m \left(m - 1 \right)} \left| j,m-1 \right>.[/tex]

?
 
  • #6
malawi_glenn said:
George Jones, you mean

[tex]J_- \left| jm \right> = \sqrt{j \left( j + 1 \right) - m \left(m - 1 \right)} \left| j,m-1 \right>.[/tex]

?

Yes, thanks. I have corrected my previous post.
 
  • #7
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 [tex]S[/tex], [tex]S_z[/tex] comes from demanding [tex]S_z|S,m> = m |S,m>[/tex] and is thus a diagonal (2S+1)x(2S+1) matrix with elements [tex]S, S-1,...,-S[/tex]
(Note: I assume [tex]|S,S> = (1,0,...,0)[/tex] and [tex]|S,-S> = (0,0,...,1)[/tex] )
The [tex]S_x[/tex] and [tex]S_y[/tex] matrices - and consequently [tex]S_+[/tex] and [tex]S_-[/tex] - come from rotations of Sz about the y and z axes, [tex]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}[/tex],
where the Uy matrix is comprised by the elements ([tex]m, m' = S, S-1,..., -S[/tex] and [tex]\theta\in [0,\pi][/tex]):

[tex]U^{y}_{m,m'}(\theta) = \sqrt{(S-m)! (S+m)! (S-m')! (S+m')!}\sum_{x=Max(0,m'-m)}^{Min(S+m',S-m)}\frac{(-1)^x cos(-\theta/2)^{2 S + m' - m - 2 x} sin(-\theta/2)^{2 x + m - m'}}{(S + m' - x)! (S - m - x)! x! (x + m - m')!}[/tex]

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

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

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

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

In the former case, [tex]k[/tex] refer to matrix index (e.g. [tex]U_{1,1}[/tex] is the upper left element of the matrix), while in the latter case [tex]m[/tex] refer to magnetic number indexing (e.g. [tex]U_{S,S}[/tex] is the upper left element of the matrix). I use the former formula and it gives the expected results.
 
  • #8
How do you get the rotation matrix about the y-axis when you don't know what the Sy generator is?
 
  • #9
George Jones said:
Have you seen the general theory of angular momentum in your quantum course?

Stuff like ([itex]\hbar = 1[/itex]):

[tex]J_z \left| jm \right> = m \left| jm \right>;[/tex]

[tex]J_+ \left| jm \right> = \sqrt{j \left( j + 1 \right) - m \left(m + 1 \right)} \left| j,m+1 \right>;[/tex]

[tex]J_- \left| jm \right> = \sqrt{j \left( j + 1 \right) - m \left(m - 1 \right)} \left| j,m-1 \right>.[/tex]

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

[tex]\left< jm \right| J_\pm \left| jm \right>[/tex]

for [itex]j=1[/itex] and [itex]m = -1, 0, 1.[/itex]
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

[tex]S_{+}=S_{x}+iS_{y}[/tex]

[tex]S_{-}=S_{x}-iS_{y}[/tex]

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

[tex]S_{x} = ...[/tex]

[tex]S_{y} = ...[/tex]
 
  • #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:

[tex]
S_x = \left(
\begin{array}{ccc}
0 & \frac{1}{\sqrt{2}} & 0 \\
\frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}} \\
0 & \frac{1}{\sqrt{2}} & 0
\end{array}
\right)
[/tex]

[tex]
S_y = \left(
\begin{array}{ccc}
0 & -\frac{i}{\sqrt{2}} & 0 \\
\frac{i}{\sqrt{2}} & 0 & -\frac{i}{\sqrt{2}} \\
0 & \frac{i}{\sqrt{2}} & 0
\end{array}
\right)
[/tex]

[tex]
S_z = \left(
\begin{array}{ccc}
1 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & -1
\end{array}
\right)
[/tex]
 
  • #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:
[tex]
S_x = \left(
\begin{array}{ccc}
0 & \frac{1}{\sqrt{2}} & 0 \\
\frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}} \\
0 & \frac{1}{\sqrt{2}} & 0
\end{array}
\right)
[/tex]

[tex]
S_y = \left(
\begin{array}{ccc}
0 & -\frac{i}{\sqrt{2}} & 0 \\
\frac{i}{\sqrt{2}} & 0 & -\frac{i}{\sqrt{2}} \\
0 & \frac{i}{\sqrt{2}} & 0
\end{array}
\right)
[/tex]

[tex]
S_z = \left(
\begin{array}{ccc}
1 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & -1
\end{array}
\right)
[/tex]

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

[tex]
S_y = \frac{i}{2} \left(\left(
\begin{array}{ccc}
0 & 0 & 0 \\
1 & 0 & 0 \\
0 & 1 & 0
\end{array}
\right) - \left(
\begin{array}{ccc}
0 & 1 & 0 \\
0 & 0 & 1 \\
0 & 0 & 0
\end{array}
\right)\right)
[/tex]

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

[tex]
S_y = \frac{i}{2} \left(\left(
\begin{array}{ccc}
0 & 0 & 0 \\
1 & 0 & 0 \\
0 & 1 & 0
\end{array}
\right) - \left(
\begin{array}{ccc}
0 & 1 & 0 \\
0 & 0 & 1 \\
0 & 0 & 0
\end{array}
\right)\right)
[/tex]


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:

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

[tex]
S_+ \left| 10 \right> = \sqrt{1* \left( 1 + 1 \right) - 0* \left(0 + 1 \right)} \left| 1,1 \right> = \sqrt{2}\left| 1,1 \right>;
[/tex]

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 !
 

1. What is a spin matrix and what does it represent?

A spin matrix is a mathematical tool used in quantum mechanics to describe the spin of a particle. It represents the possible spin states of a particle and how they can change over time.

2. How is a spin matrix different from a regular matrix?

A spin matrix is different from a regular matrix because it takes into account the properties of quantum particles, such as spin, which cannot be described by classical physics. It also follows different mathematical rules and can have complex numbers as elements.

3. What is the significance of the spin value 1 in a spin matrix?

The spin value of 1 in a spin matrix represents a particle with spin 1, which is a type of quantum particle that has a higher spin than a spin 1/2 particle. This means it has more possible spin states and can exhibit more complex behavior.

4. How are spin matrices used in experiments and calculations?

Spin matrices are used in experiments and calculations to predict and analyze the behavior of quantum particles with spin 1. They are also used to calculate the probabilities of different spin states and to understand how particles with spin 1 interact with other particles.

5. Can spin matrices be applied to particles with other spin values?

Yes, spin matrices can be applied to particles with other spin values, such as spin 1/2 or spin 3/2. However, the specific spin matrix used will depend on the spin value of the particle in question. Spin matrices are a flexible tool that can be adapted to different spin values in quantum mechanics.

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