• Support PF! Buy your school textbooks, materials and every day products Here!

Transformation of electron spin vectors and operators - a problem!

  • Thread starter deneve
  • Start date
  • #1
37
0

Homework Statement


I am struggling to understand spin transformations and have used Sakurai's method of
|new basis> = U |old basis> to change basis vectors and hence should have

Sz' = Udagger Sz U

to transform the operator. I thought this should give Sz' = Sy in the workings (see attachment) but it gives Sx instead. If I use
Sz' = U Sz Udagger then I do get Sz'=Sy but according to the texts I have, the first arrangement is the correct form. So I'm really confused. Can anyone see what I'm doing wrong here.


Homework Equations



I would be really grateful for any help with this. Thank you


The Attempt at a Solution

 

Attachments

Answers and Replies

  • #2
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,617
1,252
You're not doing anything wrong. Your expectations are just not correct. Both

[tex]\frac{\hbar}{2}\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}[/tex]

and

[tex]U^\dagger \left[\frac{\hbar}{2}\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\right] U = \frac{\hbar}{2}\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}[/tex]

represent the same operator, Sz, but in two different bases.

I'm guessing what you were thinking corresponds to

[tex]U^\dagger \left[\frac{\hbar}{2}\begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}\right] U = \frac{\hbar}{2}\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}[/tex]

In the new basis, Sy is represented by the same diagonal matrix that represents Sz in the old basis.

Note that if U takes you from the z basis to the y basis, U goes the other way, from y to z. The matrix

[tex] \frac{\hbar}{2}\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}[/tex]

represents Sy in the y basis, so in the z basis, its matrix is given by

[tex]U \left[\frac{\hbar}{2}\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\right] U^\dagger = \frac{\hbar}{2}\begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}[/tex]

which is what you would expect.
 
  • #3
37
0
Vela,

Thank you so much for your kind reply. I've been going round in circles with these things all week. I need to have a think about what you kindly wrote to see if I can get my head round what is going on. I have a number of different books but they seem to conflict with one another about some other things.

I'll post these when I can get some idea what's going wrong. One thing I have struggled with is posted under "Quantum mechanics" in the physics forums and relates to states.

Some books give |+x> = 1/sqrt2(|+> + |-x>)
|-x> = 1/sqrt2(|+> - |-x>)

Others give |+x> = 1/sqrt2(|+> + |-x>)
|-x> = -1/sqrt2(|+> - |-x>)

I cannot pinpoint where the sign error is coming in from their workings except to say that I thought it would not matter but from some replys i got, I'm now not so sure? Some get these from the eigenvalue equations and some obtain them from the rotation unitary operator but I need to work at it a bit more to figure out the discrepancies.

I'm a retired mathematical physicist (teacher) that never did much quantum mechanics
and I'm trying to get to grips with it now before I get too old.
 
  • #4
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,617
1,252
Some books give |+x> = 1/sqrt2(|+> + |-x>)
|-x> = 1/sqrt2(|+> - |-x>)

Others give |+x> = 1/sqrt2(|+> + |-x>)
|-x> = -1/sqrt2(|+> - |-x>)
That's a bit strange. I've only seen the first convention, but both are correct since the two |-x> eigenstates differ only by a multiplicative constant. I'm not sure why anyone would use the second convention with its additional minus sign. It's just asking to make another sign error. :wink:

(Oh, I just noticed you wrote |-x> on the RHS. Is that supposed to be |->?)
I cannot pinpoint where the sign error is coming in from their workings except to say that I thought it would not matter but from some replys i got, I'm now not so sure? Some get these from the eigenvalue equations and some obtain them from the rotation unitary operator but I need to work at it a bit more to figure out the discrepancies.

I'm a retired mathematical physicist (teacher) that never did much quantum mechanics
and I'm trying to get to grips with it now before I get too old.
You probably can't find a sign error because there is none.
 
  • #5
37
0
That's a bit strange. I've only seen the first convention, but both are correct since the two |-x> eigenstates differ only by a multiplicative constant. I'm not sure why anyone would use the second convention with its additional minus sign. It's just asking to make another sign error. :wink:

(Oh, I just noticed you wrote |-x> on the RHS. Is that supposed to be |->?)
Oh bother! yes, sorry. It should read

Some books give |+x> = 1/sqrt2(|+> + |->)
|-x> = 1/sqrt2(|+> - |->)

Others give |+x> = 1/sqrt2(|+> + |->)
|-x> = -1/sqrt2(|+> - |->)

I have included an image of the workings I made for how the two different forms for the |-x> base vector arise. Not sure how to check which one is consistent with stuff I haven't yet got to. How do people normally construct the U matrix to rotate bases in practice - is it like I did in the attachment on my first post?

Many kind thanks for any help given
 
  • #6
37
0
Sorry but the attachment did not work the first time. Will try again
 

Attachments

Related Threads on Transformation of electron spin vectors and operators - a problem!

  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
11
Views
3K
  • Last Post
Replies
1
Views
4K
Replies
0
Views
2K
  • Last Post
Replies
3
Views
755
Replies
2
Views
2K
  • Last Post
Replies
15
Views
2K
  • Last Post
Replies
0
Views
3K
  • Last Post
Replies
7
Views
699
  • Last Post
Replies
4
Views
7K
Top