# Spin-1 particles' spin operator

The S$_{z}$ operator for a spin-1 particle is

S$_{z}$=$\frac{h}{2\pi}$[1 0 0//0 0 0//0 0 -1]

I'm given the particle state

|$\phi$>=[1 // i // -2]

What are the probabilities of getting each one of the possible results?

Now... we can say the possible measure results will be 1,0,-1 and the autovectors associated to each result is

v$_{1}$ = [1 // 0 // 0]

v$_{0}$ = [0 // 0 // 0]

v$_{-1}$ = [0 // 0 // -1]

I 've normalized the given state. It gives me

|$\phi$>=$\frac{1}{\sqrt{6}}$[1 // i // -2]

which can be written as

|$\phi$>=$\frac{1}{\sqrt{6}}$(v$_{1}$ + [ 0 // i // 0] + 2v$_{-1}$)

I would expect to write this state to be a combination of v$_{1}$, v$_{0}$ and v$_{-1}$ but this vector [ 0 // i // 0 ] is tricking me. What measure is associated to it and what is the probability to measure zero?

Thank you.

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TSny
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v$_{1}$ = [1 // 0 // 0]

v$_{0}$ = [0 // 0 // 0]

v$_{-1}$ = [0 // 0 // -1]
Your v$_{0}$ vector is written incorrectly.

I'm sorry... what's wrong with it?

TSny
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An eigenvector (autovector) is never the zero vector .

So how can I find the eigenvector associated to the measure of the eigenvalue zero?

What I did was to find the eigenvalues with the condition that

det(S$_{z}$-$\omega$I)=0

where $\omega$ is the eigenvalue and I is the identity matrix. This procedure gave me

($\hbar^{2}$-$\omega^{2}$)$\omega$=0

I then used these results (one at a time) in

(S$_{z}$-$\omega$I)v=0

and for $\omega$=0 this equation gave me v=[0 // 0 // 0]

I can see why the zero vector cannot be an eigenvector but how do I find the vector associated to $\omega$=0 then?

TSny
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I then used these results (one at a time) in

(S$_{z}$-$\omega$I)v=0

and for $\omega$=0 this equation gave me v=[0 // 0 // 0]
Recheck your work for the $\omega = 0$ case. If you still get the zero vector then show your work for this case.

Beginning from the eigenvalue equation

S$_{z}$$\stackrel{\rightarrow}{v}$=$\omega$$\stackrel{\rightarrow}{v}$

we have

[$\hbar$ 0 0 // 0 0 0 // 0 0 -$\hbar$] * [v$_{a}$ // v$_{b}$ // v$_{c}$]= $\omega$[v$_{a}$ // v$_{b}$ // v$_{c}$]

where v$_{a}$, v$_{b}$ e v$_{c}$ are the components of the vector $\stackrel{\rightarrow}{v}$.

Applying the matrix operator S$_{z}$ we have the equations

$\hbar$v$_{a}$=$\omega$v$_{a}$

0*v$_{b}$=$\omega$v$_{b}$

-$\hbar$v$_{c}$=$\omega$v$_{c}$

if $\omega$=0 I see v$_{a}$ and v$_{c}$ are zero too but I'm not sure about v$_{b}$ because

0*v$_{b}$=0*v$_{b}$

It could be anything. How can I find it?

TSny
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if $\omega$=0 I see v$_{a}$ and v$_{c}$ are zero too but I'm not sure about v$_{b}$ because

0*v$_{b}$=0*v$_{b}$

It could be anything. How can I find it?
Right, v$_{b}$ could be any number. But you want your eigenstates to be normalized.

Right right... in this case I have the condition v$^{2}$$_{b}$=1. And then v$_{b}$=${\pm}$1.

I could define a normalization constant N which would be N=$\frac{1}{\sqrt{v_{b}}}$ but it is simpler to choose v$^{2}$$_{b}$=${\pm}$1.

Am I free to choose the sign? I think I am. I don't see any constraint to my choice.

TSny
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Yes, you can choose either sign (or even choose a complex number of magnitude 1). But, there is one choice that will look the best!

In this case I will choose v$_{b}$=1.

Thank you!

TSny
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Good. Note, for the vector V-1 you could have chosen [0, 0, 1] rather than [0, 0, -1].

Then your eigenvectors are [1, 0, 0], [0, 1, 0], and [0, 0, 1] which is the "standard" representation.

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