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Quantum Mechanics - Unitary Operators and Spin 1/2

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Hi, I'm doing question 2/II/32D at the top of page 68 here (http://www.maths.cam.ac.uk/undergrad/pastpapers/2005/Part_2/list_II.pdf [Broken]). I have done everything except for the last sentence of the question.

This is what I have attempted so far:

[tex]|\chi\rangle=|\uparrow\rangle=\left( \begin{array}{c}
1 \\
0 \end{array} \right)[/tex]

Then [tex]U|\chi\rangle=\left( \begin{array}{c}
cos(\theta /2) \\
0 \end{array} \right)-(\boldsymbol{n}.\boldsymbol{\sigma})\left( \begin{array}{c}
isin(\theta /2) \\
0 \end{array} \right)[/tex]

Now I need to choose n. If I want the spin up state measured along the direction (sin@,0,cos@) am I correct in thinking I need the eigenvector corresponding to the +1 eigenvalue of this matrix?:

[tex]\sigma_1 sin\theta+\sigma_3 cos\theta=\left( \begin{array}{cc}
cos\theta & sin\theta \\
sin\theta & -cos\theta \end{array} \right)[/tex]

In which case this the desired state is [tex]U|\chi\rangle=\left( \begin{array}{c}
sin\theta \\
1-cos\theta \end{array} \right)[/tex]

But I don't think it's possible to choose n such that this is the case, so where have I gone wrong? Also do I need to worry about any [tex]\hbar /2[/tex] since [tex]\boldsymbol{S}=\frac{\hbar}{2}\boldsymbol{\sigma}[/tex]?

Thanks.
 
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diazona

Homework Helper
2,156
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Hi, I'm doing question 2/II/32D at the top of page 68 here (http://www.maths.cam.ac.uk/undergrad/pastpapers/2005/Part_2/list_II.pdf [Broken]). I have done everything except for the last sentence of the question.

This is what I have attempted so far:

[tex]|\chi\rangle=|\uparrow\rangle=\left( \begin{array}{c}
1 \\
0 \end{array} \right)[/tex]

Then [tex]U|\chi\rangle=\left( \begin{array}{c}
cos(\theta /2) \\
0 \end{array} \right)-(\boldsymbol{n}.\boldsymbol{\sigma})\left( \begin{array}{c}
isin(\theta /2) \\
0 \end{array} \right)[/tex]

Now I need to choose n. If I want the spin up state measured along the direction (sin@,0,cos@) am I correct in thinking I need the eigenvector corresponding to the +1 eigenvalue of this matrix?:

[tex]\sigma_1 sin\theta+\sigma_3 cos\theta=\left( \begin{array}{cc}
cos\theta & sin\theta \\
sin\theta & -cos\theta \end{array} \right)[/tex]

In which case this the desired state is [tex]U|\chi\rangle=\left( \begin{array}{c}
sin\theta \\
1-cos\theta \end{array} \right)[/tex]

But I don't think it's possible to choose n such that this is the case, so where have I gone wrong?
Right idea, wrong matrix. Think about it like this: you are trying to arrange for the component of spin along a particular axis to be [itex]+\hbar/2[/itex], so you need to find the +1 eigenvalue of the matrix that measures spin along that axis. If the axis were the z axis, you'd use [itex]S_z[/itex]. If it were the x axis, you'd use [itex]S_x[/itex]. For an arbitrary axis, though, you don't have a precomputed spin matrix, so you'll need to calculate it by applying a rotation to one of the known spin matrices.

Now, do you know how to use [itex]U[/itex] to implement a rotation? Specifically, remember what the meanings of the vector [itex]\hat{n}[/itex] and the parameter [itex]\theta[/itex] are, in the context of rotations.
Also do I need to worry about any [tex]\hbar /2[/tex] since [tex]\boldsymbol{S}=\frac{\hbar}{2}\boldsymbol{\sigma}[/tex]?
Keep in mind that [itex]\sigma[/itex] is unitless, and just make sure the units are consistent :wink:
 
Last edited by a moderator:
148
0
Right idea, wrong matrix. Think about it like this: you are trying to arrange for the component of spin along a particular axis to be [itex]+\hbar/2[/itex], so you need to find the +1 eigenvalue of the matrix that measures spin along that axis. If the axis were the z axis, you'd use [itex]S_z[/itex]. If it were the x axis, you'd use [itex]S_x[/itex]. For an arbitrary axis, though, you don't have a precomputed spin matrix, so you'll need to calculate it by applying a rotation to one of the known spin matrices.

Now, do you know how to use [itex]U[/itex] to implement a rotation? Specifically, remember what the meanings of the vector [itex]\hat{n}[/itex] and the parameter [itex]\theta[/itex] are, in the context of rotations.

Keep in mind that [itex]\sigma[/itex] is unitless, and just make sure the units are consistent :wink:
Can you tell me what I've done wrong in using the matrix [tex] \boldsymbol{m}.\boldsymbol{\sigma}=
\sigma_1 sin\theta+\sigma_3 cos\theta=\left( \begin{array}{cc}
cos\theta & sin\theta \\
sin\theta & -cos\theta \end{array} \right)
[/tex] where m=(sin@,0,cos@) (the direction for the new spin up state). Now we have to choose an n such that [tex]U=exp(-i\boldsymbol{n}.\boldsymbol{\sigma}\theta/2)[/tex] rotates our original spin up state into this new one? As you say, U here is the rotation operator which will rotate the state by an angle theta about n.
 
148
0
Actually I think I might have just figured it out. I think I've got the RIGHT matrix, just the WRONG eigenvectors. My spin up eigenvector can be simplified to (cos(@/2),sin(@/2)) and if I choose n to be n=(0,1,0) I think that will work?
 

vela

Staff Emeritus
Science Advisor
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Yes, that's correct.
 

diazona

Homework Helper
2,156
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Can you tell me what I've done wrong in using the matrix [tex] \boldsymbol{m}.\boldsymbol{\sigma}=
\sigma_1 sin\theta+\sigma_3 cos\theta=\left( \begin{array}{cc}
cos\theta & sin\theta \\
sin\theta & -cos\theta \end{array} \right)
[/tex] where m=(sin@,0,cos@) (the direction for the new spin up state). Now we have to choose an n such that [tex]U=exp(-i\boldsymbol{n}.\boldsymbol{\sigma}\theta/2)[/tex] rotates our original spin up state into this new one? As you say, U here is the rotation operator which will rotate the state by an angle theta about n.
Ah, sorry about that. I didn't go all the way to the end of the calculation so I didn't realize that the matrix you have is actually the right one for this case.
 

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