What Are the Probabilities of Measuring Each Spin State for a Spin-1 Particle?

[Gabriel Maia]

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.

 

[TSny]

v_{1} = [1 // 0 // 0]

v_{0} = [0 // 0 // 0]

v_{-1} = [0 // 0 // -1]

Your v_{0} vector is written incorrectly.

 

[Gabriel Maia]

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

 

[TSny]

An eigenvector (autovector) is never the zero vector .

 

[Gabriel Maia]

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}-\omegaI)=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}-\omegaI)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]

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

(S_{z}-\omegaI)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.

 

[Gabriel Maia]

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

\hbarv_{a}=\omegav_{a}

0*v_{b}=\omegav_{b}

-\hbarv_{c}=\omegav_{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]

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.

 

[Gabriel Maia]

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]

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!

 

[Gabriel Maia]

In this case I will choose v_{b}=1.

Thank you!

 

[TSny]

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.