- #1

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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

[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

*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?
Last edited: