- #1
VantagePoint72
- 821
- 34
There's something I don't think I quite understand about spin and how it acts a generator of rotations. I'll start with quickly going over what I do understand. Suppose you want to do an infinitesimal rotation around the z-axis on some state:
[itex]
\def\ket#1{\left | #1 \right \rangle}
\ket{\psi '} = (1 - \frac{i\epsilon _z J_z}{\hbar})\ket{\psi}
[/itex]
Now, if the wavefunction is just a scalar, then projecting on the coordinate basis gives us:
[itex]
\psi '(x,y,z) = (1 - \frac{i\epsilon _z J_z}{\hbar})\psi (x,y,z)
[/itex]
and so, as usual, we can do a finite rotation of [itex]\phi[/itex] around the z-axis by breaking it down into infinitesimal rotations and composing them, giving the relation:
[itex]
\psi '(x,y,z) = e^{-i \phi L_z / \hbar}\psi (x,y,z)
[/itex]
Now, because we know what this operator does, we don't actually have to expand the exponential. We can just (passively) rotate the coordinates:
[tex]
\begin{pmatrix}
x' \\
y'
\end{pmatrix} =
\begin{pmatrix}
\cos\phi & +\sin\phi\\
-\sin\phi & \cos\phi
\end{pmatrix}
\begin{pmatrix}
x \\
y
\end{pmatrix} = R^{-1}(\phi)
\begin{pmatrix}
x \\
y
\end{pmatrix}
[/tex]
Giving:
[tex]
\psi '(x,y,z) = \psi (x\cos\phi + y\sin\phi,-x\sin\phi + y\cos\phi,z)
[/tex]
So far, so good. Now, if ##\ket{\psi}## actually requires a spinor to describe the wavefunction, [itex]\ket{\psi} \rightarrow \begin{pmatrix}
\psi _+(x,y,z) \\
\psi _-(x,y,z)
\end{pmatrix}[/itex] then ##J_z = L_z + S_z## and the act of rotating the ket has two effects: a passive rotation of the coordinate system and a shuffling together of the two wavefunction components. So, we repeat the exercise we went through for ##L_z##, derive the Pauli spin matrices from the commutators, etc., eventually finding that:
$$
e^{-i \phi S_z / \hbar} =
\begin{pmatrix}
\cos(\phi /2) - i \sin(\phi /2) & 0\\
0 & \cos(\phi /2) + i \sin(\phi /2)
\end{pmatrix}
$$
What I don't get is how this makes any sense. I mean, I can follow all the steps in the derivation, I just don't get the final result. Why isn't the overall rotation due to spin just given by the matrix ##R(\phi)## above? In other words, taking into account both ##L_z## and ##S_z##, why isn't the rotated spinor given by:
$$
\begin{align}
\begin{pmatrix}
\psi _+ '(\vec{r}) \\
\psi _- '(\vec{r})
\end{pmatrix} &= R(\phi)
\begin{pmatrix}
\psi _+ (R^{-1}(\phi)(\vec{r})) \\
\psi _- (R^{-1}(\phi)(\vec{r}))
\end{pmatrix} \\ &=
\begin{pmatrix}
(\cos\phi) \psi _+ (x\cos\phi + y\sin\phi,-x\sin\phi + y\cos\phi,z) - (\sin\phi) \psi _- (x\cos\phi + y\sin\phi,-x\sin\phi + y\cos\phi,z) \\
(\sin\phi) \psi _+ (x\cos\phi + y\sin\phi,-x\sin\phi + y\cos\phi,z) + (\cos\phi) \psi _- (x\cos\phi + y\sin\phi,-x\sin\phi + y\cos\phi,z)
\end{pmatrix}
\end{align}
$$
That's what rotating something does, so why do spinors behave differently?
[itex]
\def\ket#1{\left | #1 \right \rangle}
\ket{\psi '} = (1 - \frac{i\epsilon _z J_z}{\hbar})\ket{\psi}
[/itex]
Now, if the wavefunction is just a scalar, then projecting on the coordinate basis gives us:
[itex]
\psi '(x,y,z) = (1 - \frac{i\epsilon _z J_z}{\hbar})\psi (x,y,z)
[/itex]
and so, as usual, we can do a finite rotation of [itex]\phi[/itex] around the z-axis by breaking it down into infinitesimal rotations and composing them, giving the relation:
[itex]
\psi '(x,y,z) = e^{-i \phi L_z / \hbar}\psi (x,y,z)
[/itex]
Now, because we know what this operator does, we don't actually have to expand the exponential. We can just (passively) rotate the coordinates:
[tex]
\begin{pmatrix}
x' \\
y'
\end{pmatrix} =
\begin{pmatrix}
\cos\phi & +\sin\phi\\
-\sin\phi & \cos\phi
\end{pmatrix}
\begin{pmatrix}
x \\
y
\end{pmatrix} = R^{-1}(\phi)
\begin{pmatrix}
x \\
y
\end{pmatrix}
[/tex]
Giving:
[tex]
\psi '(x,y,z) = \psi (x\cos\phi + y\sin\phi,-x\sin\phi + y\cos\phi,z)
[/tex]
So far, so good. Now, if ##\ket{\psi}## actually requires a spinor to describe the wavefunction, [itex]\ket{\psi} \rightarrow \begin{pmatrix}
\psi _+(x,y,z) \\
\psi _-(x,y,z)
\end{pmatrix}[/itex] then ##J_z = L_z + S_z## and the act of rotating the ket has two effects: a passive rotation of the coordinate system and a shuffling together of the two wavefunction components. So, we repeat the exercise we went through for ##L_z##, derive the Pauli spin matrices from the commutators, etc., eventually finding that:
$$
e^{-i \phi S_z / \hbar} =
\begin{pmatrix}
\cos(\phi /2) - i \sin(\phi /2) & 0\\
0 & \cos(\phi /2) + i \sin(\phi /2)
\end{pmatrix}
$$
What I don't get is how this makes any sense. I mean, I can follow all the steps in the derivation, I just don't get the final result. Why isn't the overall rotation due to spin just given by the matrix ##R(\phi)## above? In other words, taking into account both ##L_z## and ##S_z##, why isn't the rotated spinor given by:
$$
\begin{align}
\begin{pmatrix}
\psi _+ '(\vec{r}) \\
\psi _- '(\vec{r})
\end{pmatrix} &= R(\phi)
\begin{pmatrix}
\psi _+ (R^{-1}(\phi)(\vec{r})) \\
\psi _- (R^{-1}(\phi)(\vec{r}))
\end{pmatrix} \\ &=
\begin{pmatrix}
(\cos\phi) \psi _+ (x\cos\phi + y\sin\phi,-x\sin\phi + y\cos\phi,z) - (\sin\phi) \psi _- (x\cos\phi + y\sin\phi,-x\sin\phi + y\cos\phi,z) \\
(\sin\phi) \psi _+ (x\cos\phi + y\sin\phi,-x\sin\phi + y\cos\phi,z) + (\cos\phi) \psi _- (x\cos\phi + y\sin\phi,-x\sin\phi + y\cos\phi,z)
\end{pmatrix}
\end{align}
$$
That's what rotating something does, so why do spinors behave differently?
Last edited:
