# Rotation operator of spin

• I
• cristianbahena
When we say that Sz is the generator of rotation, we do NOT mean that applying Sz will rotate the state. We mean that the exponential of ## i S \theta /\hbar ## will rotate the state by an angle ##\theta##. This was you showed above, with an angle $\phi/2$.

#### cristianbahena

for compute:
$$e^{\frac{iS_z\phi}{\hbar}}S_x e^{\frac{-iS_z\phi}{\hbar}}$$
so, if we use $$S_x=(\frac{\hbar}{2})[(|+><-|)+(|-><+|)]$$
$$e^{\frac{iS_z\phi}{\hbar}}(\frac{\hbar}{2})[(|+><-|)+(|-><+|)] e^{\frac{-iS_z\phi}{\hbar}}$$
so, why that is equal to $$(\frac{\hbar}{2})[\frac{i\phi}{2}|+><-|\frac{i\phi}{2}+\frac{-i\phi}{2}|-><+|\frac{-i\phi}{2}]$$

??

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cristianbahena said:
for compute:
$$e^{\frac{iS_z\phi}{\hbar}}S_x e^{\frac{-iS_z\phi}{\hbar}}$$
so, if we use $$S_x=(\frac{\hbar}{2})[(|+><-|)+(|-><+|)]$$
$$e^{\frac{iS_z\phi}{\hbar}}(\frac{\hbar}{2})[(|+><-|)+(|-><+|)] e^{\frac{-iS_z\phi}{\hbar}}$$
so, why that is equal to $$(\frac{\hbar}{2})[e^{\frac{i\phi}{2}}|+><-|\frac{i\phi}{2}+\frac{-i\phi}{2}|-><+|\frac{-i\phi}{2}]$$

??
Are you sure you typed the last expression correctly?

nrqed said:
Are you sure you typed the last expression correctly?
thanks! now it is correct

Ok. But this is strange, I would expect exponentials of ##i\phi/2## instead of just factors of ## i \phi/2##. You are sure there are no exponentials in the final result??

Of course there should be exponentials since the ##|\pm \rangle## kets are eigenvectors or ##\hat{S}_z##, that's very easy to see in the notation given in the OP.

vanhees71 said:
Of course there should be exponentials since the ##|\pm \rangle## kets are eigenvectors or ##\hat{S}_z##, that's very easy to see in the notation given in the OP.
Indeed. I was trying to lead the OP to the answer without giving it away too quickly :-)

vanhees71
nrqed said:
Ok. But this is strange, I would expect exponentials of ##i\phi/2## instead of just factors of ## i \phi/2##. You are sure there are no exponentials in the final result??

sorry!

for compute:
$$e^{\frac{iS_z\phi}{\hbar}}S_ x e^{\frac{-iS_z\phi}{\hbar}}$$
so, if we use $$S_x=(\frac{\hbar}{2})[(|+><-|)+(|-><+|)]$$
$$e^{\frac{iS_z\phi}{\hbar}}(\frac{\hbar}{2})[(|+><-|)+(|-><+|)] e^{\frac{-iS_z\phi}{\hbar}}$$
so, why that is equal to $$(\frac{\hbar}{2})[e^{\frac{i\phi}{2}}|+><-|e^{\frac{i\phi}{2}}+e^{\frac{-i\phi}{2}}|-><+|e^{\frac{-i\phi}{2}}]$$

??

cristianbahena said:
sorry!

for compute:
$$e^{\frac{iS_z\phi}{\hbar}}S_ x e^{\frac{-iS_z\phi}{\hbar}}$$
so, if we use $$S_x=(\frac{\hbar}{2})[(|+><-|)+(|-><+|)]$$
$$e^{\frac{iS_z\phi}{\hbar}}(\frac{\hbar}{2})[(|+><-|)+(|-><+|)] e^{\frac{-iS_z\phi}{\hbar}}$$
so, why that is equal to $$(\frac{\hbar}{2})[e^{\frac{i\phi}{2}}|+><-|e^{\frac{i\phi}{2}}+e^{\frac{-i\phi}{2}}|-><+|e^{\frac{-i\phi}{2}}]$$

??
Good, now we are talking!

The next step is this: do you know what we get if we apply ##S_z## to ##|+ \rangle ## ?

The next question is: do you know what we get if we apply ##e^{ S_z} ## ti ## |+ \rangle ##?
(I did not include the factors of ##i\phi/\hbar## for now, we can put them back at the next step).

"nrqed said:
Good, now we are talking!

The next step is this: do you know what we get if we apply ##S_z## to ##|+ \rangle ## ?

The next question is: do you know what we get if we apply ##e^{ S_z} ## ti ## |+ \rangle ##?
(I did not include the factors of ##i\phi/\hbar## for now, we can put them back at the next step).

i know that
$$S_z |+ >= \frac{\hbar}{2} |+ >$$

also $$D_z(\phi)= 1+i\frac{S_z}{\hbar}\phi$$ operator of infinitesimal rotations, if we apply this operator $$n$$ times:

$$(D_z(\phi))^n= (1+i\frac{S_z}{\hbar}\frac{\phi}{n})^n$$ in the limit $$lim_{n\rightarrow \infty}(D_z(\phi))^n= lim_{n\rightarrow \infty}(1+i\frac{S_z}{\hbar}\frac{\phi}{n})^n=e^{ \frac{iS_z\phi}{\hbar}}$$
im not sure why $$n$$ is pressent in $$\frac{\phi}{n}$$ when apply n-times operator D
so
i think that:
we have

$$D_z(\phi)|+ >=1+i\frac{S_z}{\hbar}\phi|+>=(1+\frac{i\phi}{2})|+>$$
then

$$(D_z(\phi))^n|+ >=(1+i\frac{S_z}{\hbar}\frac{\phi}{n})^n|+>=(1+\frac{i\phi}{2})^n|+>$$
in the limit:
$$lim_{n\rightarrow \infty}(D_z(\phi))^n|+ >=lim_{n\rightarrow \infty}(1+i\frac{S_z}{\hbar}\frac{\phi}{n})^n|+>=lim_{n\rightarrow \infty}(1+\frac{i\phi}{2})^n|+>=e^{\frac{i\phi}{2}}|+>$$

but also

$$S_z$$ is the generator of rotations so i f we apply $$S_z$$ to $$|+>$$ know the new ket was just rotated $$\phi^{'}$$, so the new ket is $$e^{i\phi^{'}}|+>$$ how i know that $$\phi^{'}= \phi$$?

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cristianbahena said:
i know that
$$S_z |+ >= \frac{\hbar}{2} |+ >$$

also $$D_z(\phi)= 1+i\frac{S_z}{\hbar}\phi$$ operator of infinitesimal rotations, if we apply this operator $$n$$ times:

$$(D_z(\phi))^n= (1+i\frac{S_z}{\hbar}\frac{\phi}{n})^n$$ in the limit $$lim_{n\rightarrow \infty}(D_z(\phi))^n= lim_{n\rightarrow \infty}(1+i\frac{S_z}{\hbar}\frac{\phi}{n})^n=e^{ \frac{iS_z\phi}{\hbar}}$$
im not sure why $$n$$ is pressent in $$\frac{\phi}{n}$$ when apply n-times operator D
so
i think that:
we have

$$D_z(\phi)|+ >=1+i\frac{S_z}{\hbar}\phi|+>=(1+\frac{i\phi}{2})|+>$$
then

$$(D_z(\phi))^n|+ >=(1+i\frac{S_z}{\hbar}\frac{\phi}{n})^n|+>=(1+\frac{i\phi}{2})^n|+>$$
in the limit:
$$lim_{n\rightarrow \infty}(D_z(\phi))^n|+ >=lim_{n\rightarrow \infty}(1+i\frac{S_z}{\hbar}\frac{\phi}{n})^n|+>=lim_{n\rightarrow \infty}(1+\frac{i\phi}{2})^n|+>=e^{\frac{i\phi}{2}}|+>$$
Good work. That's all you need to answer the problem

but also

$$S_z$$ is the generator of rotations so i f we apply $$S_z$$ to $$|+>$$ know the new ket was just rotated $$\phi^{'}$$, so the new ket is $$e^{i\phi^{'}}|+>$$ how i know that $$\phi^{'}= \phi$$?
When we say that Sz is the generator of rotation, we do NOT mean that applying Sz will rotate the state. We mean that the exponential of ## i S \theta /\hbar ## will rotate the state by an angle ##\theta##. This was you showed above, with an angle ##\phi/2##.

nrqed said:
Good work. That's all you need to answer the problem

When we say that Sz is the generator of rotation, we do NOT mean that applying Sz will rotate the state. We mean that the exponential of ## i S \theta /\hbar ## will rotate the state by an angle ##\theta##. This was you showed above, with an angle ##\phi/2##.
Good! Thanks