How Does the Transformation in Sakurai's Quantum Mechanics Work?

[omoplata]

The following is from 'Modern Quantum Mechanics' by J.J. Sakurai, page 159.

\left( \frac{ \hbar }{ 2 } \right) \exp \left( \frac{i S_{z} \phi}{\hbar} \right) \left{ left( \mid + \rangle \langle - \mid \right) + \left( \mid - \rangle \langle + \mid \right) right} \exp \left( \frac{- i S_{z} \phi}{\hbar}

= \left( \frac{\hbar}{2} \right) \left( e^{i \phi / 2} \mid + \rangle \langle - \mid e^{i \phi / 2} + e^{- i \phi / 2} \mid - \rangle \langle + \mid e^{- i \phi / 2} \right)

How do I get from the first to the second?

 

[omoplata]

Oops. Posted before I wanted to. How do I delete this?

 

[dextercioby]

You can't delete the thread, just individual posts, as long as nobody answered to you already. Just click the edit button on your first post and add the full question.

 

[Fredrik]

You can edit your post for 12 hours after you posted it.

Keep in mind that you need to refresh and resend after each preview to see the right LaTeX images.

 

[omoplata]

I ran into trouble with LaTeX.

This is what I wanted to post. What am I doing wrong?

for the first expression,

\left( \frac{ \hbar }{ 2 } \right) \exp \left( \frac{i S_{z} \phi}{\hbar} \right) \left{ left( \mid + \rangle \langle - \mid \right) + \left( \mid - \rangle \langle + \mid \right) right} \exp \left( \frac{- i S_{z} \phi}{\hbar}

for the second expression,

= \left( \frac{\hbar}{2} \right) \left( e^{i \phi / 2} \mid + \rangle \langle - \mid e^{i \phi / 2} + e^{- i \phi / 2} \mid - \rangle \langle + \mid e^{- i \phi / 2} \right)

Looks like I got the second one right. What's wrong with the first one?

 

[Fredrik]

\left( \frac{ \hbar }{ 2 } \right) \exp \left( \frac{i S_{z} \phi}{\hbar} \right) \left{ left( \mid + \rangle \langle - \mid \right) + \left( \mid - \rangle \langle + \mid \right) right} \exp \left( \frac{- i S_{z} \phi}{\hbar}
...
What's wrong with the first one?

No \right) at the end. A couple of missing \ symbols. { isn't displayed unless you type it as \{.

 

[dextercioby]

OK, the operators act on the |\pm\rangle vectors, according to their definitions. These vectors are eigenvectors of the S_z spin operator, so making the 2 calculations shouln't be difficult. The exponential of a self-adjoint operator is a unitary operator. If you know the eigenvalue of the s-a operator, it's easy to find the eigenvalues of the unitary operator which is associated to it.

 

[omoplata]

Thanks bigubau.

I think there's something wrong with my browser or something ( firefox 3.6.13 in ubuntu ). I can see the first LaTeX expression where Fredrik has quoted me. But I can't see it on my own post.

 

[Fredrik]

Your post doesn't contain any tex tags. I typed them around the math expression I quoted.

 

[omoplata]

Your post doesn't contain any tex tags. I typed them around the math expression I quoted.

Oh, no. I mean my very first post in this tread. I can't see the first LaTeX expression.

 

[omoplata]

OK, I can see it now.

But I couldn't see it before.

My conclusion is one of two things.
1. I am going crazy.
2. My browser was displaying an incorrect LaTeX image and something (refreshing the browser? ) happened which caused the LaTeX image to be displayed correctly.

Either way, thanks for the help everyone.

 

[Fredrik]

OK, I can see it now.

But I couldn't see it before.

My conclusion is one of two things.
1. I am going crazy.
2. My browser was displaying an incorrect LaTeX image and something (refreshing the browser? ) happened which caused the LaTeX image to be displayed correctly.

Either way, thanks for the help everyone.

Definitely 2, although that doesn't rule out that 1 is true as well. You need to refresh and resend after each preview, and if you didn't do that after the last change you made before you saved the changes, you need to refresh the page one more time after it shows up the first time.

Edit: I noticed today that you may have to refresh the page after saving the changes, even if you did refresh and resend after each preview. I also found out that the time you're allowed to edit a post is 11 hours and 40 minutes, not 12 hours as I said above.

 

[Dyson]

I tend to calculate it in the representation of S_{z}'s. So,
S_{z}\mid +>=\frac{\hbar}{2}\mid +>
S_{z}\mid - >=-\frac{\hbar}{2}\mid ->
Then, according to the definition of exp(\hat{F}),
exp(\frac{iS_{z} \phi}{\hbar})\mid + >=exp(\frac{i\phi}{2})\mid + >
exp(\frac{iS_{z} \phi}{\hbar})\mid - >=exp(\frac{-i\phi}{2})\mid - >
Then, operating the equations above with dagger, you can get,
< + \mid exp(\frac{-iS_{z} \phi}{\hbar})=< + \mid exp(\frac{-i\phi}{2})
< - \mid exp(\frac{-iS_{z} \phi}{\hbar})=< - \mid exp(\frac{i\phi}{2})
You can work the second now.