- #1

Gabriel Maia

- 72

- 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.

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.

Last edited: