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Spin-1 particles' spin operator

  • #1
The S[itex]_{z}[/itex] operator for a spin-1 particle is

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

I'm given the particle state

|[itex]\phi[/itex]>=[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[itex]_{1}[/itex] = [1 // 0 // 0]

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

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

I 've normalized the given state. It gives me

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

which can be written as

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

I would expect to write this state to be a combination of v[itex]_{1}[/itex], v[itex]_{0}[/itex] and v[itex]_{-1}[/itex] 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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Answers and Replies

  • #2
TSny
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v[itex]_{1}[/itex] = [1 // 0 // 0]

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

v[itex]_{-1}[/itex] = [0 // 0 // -1]
Your v[itex]_{0}[/itex] vector is written incorrectly.
 
  • #3
I'm sorry... what's wrong with it?
 
  • #4
TSny
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An eigenvector (autovector) is never the zero vector .
 
  • #5
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[itex]_{z}[/itex]-[itex]\omega[/itex]I)=0

where [itex]\omega[/itex] is the eigenvalue and I is the identity matrix. This procedure gave me

([itex]\hbar^{2}[/itex]-[itex]\omega^{2}[/itex])[itex]\omega[/itex]=0

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

(S[itex]_{z}[/itex]-[itex]\omega[/itex]I)v=0

and for [itex]\omega[/itex]=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 [itex]\omega[/itex]=0 then?
 
  • #6
TSny
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I then used these results (one at a time) in

(S[itex]_{z}[/itex]-[itex]\omega[/itex]I)v=0

and for [itex]\omega[/itex]=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.
 
  • #7
Beginning from the eigenvalue equation

S[itex]_{z}[/itex][itex]\stackrel{\rightarrow}{v}[/itex]=[itex]\omega[/itex][itex]\stackrel{\rightarrow}{v}[/itex]

we have

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


where v[itex]_{a}[/itex], v[itex]_{b}[/itex] e v[itex]_{c}[/itex] are the components of the vector [itex]\stackrel{\rightarrow}{v}[/itex].

Applying the matrix operator S[itex]_{z}[/itex] we have the equations

[itex]\hbar[/itex]v[itex]_{a}[/itex]=[itex]\omega[/itex]v[itex]_{a}[/itex]

0*v[itex]_{b}[/itex]=[itex]\omega[/itex]v[itex]_{b}[/itex]

-[itex]\hbar[/itex]v[itex]_{c}[/itex]=[itex]\omega[/itex]v[itex]_{c}[/itex]

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

0*v[itex]_{b}[/itex]=0*v[itex]_{b}[/itex]

It could be anything. How can I find it?
 
  • #8
TSny
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if [itex]\omega[/itex]=0 I see v[itex]_{a}[/itex] and v[itex]_{c}[/itex] are zero too but I'm not sure about v[itex]_{b}[/itex] because

0*v[itex]_{b}[/itex]=0*v[itex]_{b}[/itex]

It could be anything. How can I find it?
Right, v[itex]_{b}[/itex] could be any number. But you want your eigenstates to be normalized.
 
  • #9
Right right... in this case I have the condition v[itex]^{2}[/itex][itex]_{b}[/itex]=1. And then v[itex]_{b}[/itex]=[itex]{\pm}[/itex]1.

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

Am I free to choose the sign? I think I am. I don't see any constraint to my choice.
 
  • #10
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!
 
  • #11
In this case I will choose v[itex]_{b}[/itex]=1.

Thank you!
 
  • #12
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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