I have a general question related to rotations of wave functions. I have never really come across this in any of the core QM books, and was curious to know this.

Consider, then, a wave function that consists of an angular part and a spin part of a spin-1/2 particle given by, say: $$ |\psi \rangle = (z) ⊗ \begin{pmatrix} 1 \\ -i \end{pmatrix} . $$

What will happen if I rotate this state, say, around the y-axis by $\frac{\pi}{2}$ radians?

My approach is this, which I am sure is quite naive and possibly wrong. I will greatly appreciate any help.

My approach to this problem:

Write the angular part in terms of spherical harmonics (not caring too much about normalization for now): $$ |\psi \rangle = | l = 1, m = 0 \rangle ⊗ \begin{pmatrix} 1 \\ -i \end{pmatrix} . $$

Now, when we rotate it, does the rotation matrix act independently on the angular part and the spin part? So, the angular states and spin states rotate like: \begin{align} $ R|\psi \rangle &= \sum_{m'}^{} d_{m'm}^{l} \left(\frac{\pi}{2} \right) | l = 1, m' \rangle ⊗ d_{m'm}^{l} \left(\frac{\pi}{2} \right) \begin{pmatrix} 1 \\ -i \end{pmatrix} \\ R|\psi \rangle & = \left( \frac{-1}{\sqrt{2}} |l = 1, m = 1 \rangle + \frac{1}{\sqrt{2}} |l = 1, m = -1 \rangle \right) ⊗ \begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix} \begin{pmatrix} 1 \\ -i \end{pmatrix} \\ R|\psi \rangle & = \left( \frac{-1}{\sqrt{2}} |l = 1, m = 1 \rangle + \frac{1}{\sqrt{2}} |l = 1, m = -1 \rangle \right) ⊗ \frac{1}{\sqrt{2}} \begin{pmatrix} 1 - i \\ 1 - i \end{pmatrix}$ \end{align} Hopefully the question is understood clearly. So, is this above calculation correct? Or is there something wrong with it.

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    $\begingroup$ “ Now, when we rotate it, does the rotation matrix act independently on the angular part and the spin part?“ Yes. $\endgroup$
    – knzhou
    Dec 10, 2019 at 7:29

2 Answers 2


Just use the properties of the direct product. For any Kronecker product of two representation vectors, not just space and spin, $\Psi \otimes \psi$, so using capitals for the left space and lower case for the right, you learned that the corresponding rotation generating operator acting on them is the coproduct $$ \Delta (J) = J\otimes 1\!\! 1 + 1\!\!\! 1 \otimes j , $$ seen to satisfy the same Lie algebra as $J$ and $j$.

But the two terms commute with each other, so the full rotation trivially splits, $$ e^{i\theta \Delta(J)}=\exp ({i\theta ( J\otimes 1\!\! 1 + 1\!\!\! 1 \otimes j) }) =e^{i\theta ~ J\otimes 1\!\! 1 } e^{i\theta ~ 1\!\!\! 1 \otimes j} = e^{i\theta ~ J }\otimes e^{i\theta ~ j} , $$ so your reducible representation transforms as $$ ( e^{i\theta ~ J }\otimes e^{i\theta ~ j}) (\Psi \otimes \psi)= (e^{i\theta ~ J } \Psi )\otimes (e^{i\theta ~ j}\psi). $$ It's up to you to choose your angle to be a right one.

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    $\begingroup$ Thank you. This is amazing. $\endgroup$
    – user244805
    Dec 10, 2019 at 23:49

Here's a slightly different take, if only in formalism. Several Translation Operators can be created as the complex exponential of the conjugate observable operator. For example, the complex exponential of the Hamiltonian divided by $i\hbar$ is the unitary time translation operator. In the case of rotations, construct the angular momentum operator with respect to the desired axis of rotation: Rotations in Quantum Mechanics

As I recall they go into this in some detail in Bransden and Joachain. Griffiths leaves it as an exercise.



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