Spin matrices for particle of spin 1


by genloz
Tags: matrices, particle, spin
genloz
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#1
Aug14-07, 11:57 PM
P: 76
1. The problem statement, all variables and given/known data
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.

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

3. 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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George Jones
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#2
Aug15-07, 06:59 AM
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Quote Quote by genloz View Post
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?
genloz
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#3
Aug15-07, 07:33 AM
P: 76
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...

George Jones
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#4
Aug15-07, 08:03 AM
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Spin matrices for particle of spin 1


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]
malawi_glenn
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#5
Aug15-07, 08:13 AM
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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]

?
George Jones
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Aug15-07, 08:16 AM
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Quote Quote by malawi_glenn View Post
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.
kaltsoplyn
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#7
Oct6-08, 08:22 AM
P: 5
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.
turin
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Dec4-08, 01:19 PM
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How do you get the rotation matrix about the y-axis when you don't know what the Sy generator is?
lydilmyo
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#9
Sep1-09, 05:20 AM
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Quote Quote by George Jones View Post
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 ?
kaltsoplyn
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#10
Sep1-09, 05:48 AM
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Quote Quote by turin View Post
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.
kuruman
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Sep1-09, 08:20 AM
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Quote Quote by lydilmyo View Post
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]
lydilmyo
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#12
Sep1-09, 08:49 AM
P: 9
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...
kuruman
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Sep1-09, 08:58 AM
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Quote Quote by lydilmyo View Post
but this relation is available only for pauli matrix (2x2) and so for spin 1/2....
The relations are good for any dimensionality.
kaltsoplyn
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#14
Sep1-09, 09:23 AM
P: 5
Quote Quote by lydilmyo View Post
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]
lydilmyo
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#15
Sep1-09, 09:23 AM
P: 9
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 ?
kaltsoplyn
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#16
Sep1-09, 09:27 AM
P: 5
Quote Quote by lydilmyo View Post
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.
lydilmyo
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#17
Sep1-09, 09:28 AM
P: 9
Quote Quote by kaltsoplyn View Post
[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]
lydilmyo
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#18
Sep1-09, 09:34 AM
P: 9
[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.


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