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In a QM text I am using (Sakurai 2nd edition 'Modern Quantum Mechanics'), he describes the classical Euler rotations by the rotation $$R_{z'}(\gamma)R_{y'}(\beta)R_{z}(\alpha) = R_{z}(\alpha)R_{y}(\beta)R_{z}(\gamma)$$ where $z'$ and $y'$ subscripts on the left is rotation with respect to the rotated axis. He then defines the corresponding rotations in quantum mechanics by $\mathcal{D}_{z}(\alpha)\mathcal{D}_{y}(\beta)\mathcal{D}_z(\gamma) = \text{exp}(\frac{-i \sigma_3 \alpha}{2})\text{exp}(\frac{-i \sigma_2 \beta}{2})\text{exp}(\frac{-i \sigma_3 \gamma}{2}) = \left( {\begin{array}{cc} e^{\frac{-i(\alpha + \gamma)}{2}}\cos(\frac{\beta}{2}) & -e^{\frac{-i(\alpha - \gamma)}{2}}\sin(\frac{\beta}{2}) \\ e^{\frac{-i(\alpha - \gamma)}{2}}\sin(\frac{\beta}{2}) & e^{\frac{i(\alpha + \gamma)}{2}}\cos(\frac{\beta}{2}) \\ \end{array} } \right)$

He then states "This matrix is clearly of the unitary umimodular form. Conversely, the most general 2x2 unitary unimodular matrix can be written in this Euler angle form."

Why is this particular sequence which defines the euler rotation important? As I understand the QM matrix is given a special name "$j = 1/2$ irreducible representation of the rotation operation $\mathcal{D}$" (not very catchy but...).

Lastly, why is it important that the QM representation is a unitary unimodular matrix and conversely that any unitary unimodular matrix can be written in the Euler angle form, Is it simply important to show that the classical notion carries over to QM?

Thanks.

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  • The form is convenient because it closely matches the sequence used to go from body-fixed to space-fixed frame in classical mechanics (note that in CM the middle rotation is usually about the $x$-axis, not $y$). The same reasoning for the sequence - and therefore the same physical interpretation to every term in the sequence - holds in both CM and QM.
  • It is also convenient in QM because the standard basis states are eigenstates of $L_z$ so that \begin{align} \langle LM'\vert R_z(\alpha)R_y(\beta)R_z(\gamma)\vert LM\rangle &=e^{i\alpha M'}e^{i\gamma M}\langle LM'\vert R_y(\beta)\vert LM\rangle\\ &=e^{i\alpha M'}e^{i\gamma M}d^L_{M'M}(\beta) \end{align} where one can show that $\langle LM'\vert R_y(\beta)\vert LM\rangle=d^L_{M'M}(\beta)$ is real. The choice of a rotation about $x$ would produce a matrix $\langle LM'\vert R_x(\beta)\vert LM\rangle$ that is complex.
  • The key point is for the transformation to be unitary, not necessarily unitary and unimodular. The reason is quite simple: if $U$ is unitary and $\vert\psi'\rangle:= U\vert\psi \rangle$ then the transformation $U$ preserves the inner product $$ \langle \psi'\vert\psi'\rangle = \langle \psi\vert U^\dagger U\vert \psi\rangle=\langle \psi\vert\psi\rangle\, . $$ In general det$(U)=e^{i\zeta}$ and it is always possible to write $U$ as a diagonal matrix $D$ with phases only such that det$(D)=e^{i\zeta}$ and multiplied by a unitary matrix of determinant $+1$, i.e. a unimodular unitary matrix.

Edit: Quoting Goldstein's Classical Mechanics

... the sequence of rotations used to define the final orientation of the coordinate system is to some extent arbitrary. ... A total of twelve convention is therefore possible...

From a group theory perspective a factorization where the first and last axes are the same is really convenient since $$ R_z(\alpha)R_y(\beta)R_z(\gamma)=R_z(\alpha+\gamma) \left[R_z(-\gamma)R_y(\beta)R_z(\gamma)\right] $$ and $\left[R_z(-\gamma)R_y(\beta)R_z(\gamma)\right]$ rotates the $y$ axis about the $z$ axis (i.e. a conjugation) so that $$ \left[R_z(-\gamma)R_y(\beta)R_z(\gamma)\right]=R_{y'}(\beta) $$ with $$ \hat y'=\hat y\cos\gamma+\hat x\sin\gamma $$ is a rotation by an angle $\beta$ about an axis $\hat y'$ in the $xy$ plane.

Conversely, in QM, the expression $$ R_z(\alpha)R_y(\beta)R_z(\gamma)= \left[R_z(\alpha)R_y(\beta)R_z(-\alpha)\right]R_z(\alpha+\gamma) $$ rotates (up to a phase) the angular momentum state $\vert LM\rangle$ around the axis $\hat y''= \hat y\cos\alpha-\hat x\sin\alpha$.

In general, any factorization where the first and last operations are rotations (or generalized rotations) about the same axes can thus be given a reasonably intuitive interpretation of the type given above.

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  • $\begingroup$ Interesting answer. I think the op is interested in why the particular sequence of rotations is chosen in classical mechanics to describe the euler rotations? I also don't see why this particular sequence is particularly important but it is quoted in Sakurai as stated. $\endgroup$ – Alex Apr 28 '17 at 13:01
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    $\begingroup$ @Alex I've edited my answer to partially answer your query and add some details. $\endgroup$ – ZeroTheHero Apr 28 '17 at 13:37
  • $\begingroup$ So the euler rotations don't really describe arbitrary sequence of rotations, it deals more with rotations where as you state the first and last rotations are about the same spatial axis? $\endgroup$ – Alex Apr 28 '17 at 13:56
  • $\begingroup$ This is nomenclature but the most common use of Euler angles is with the first and last rotations the same. $\endgroup$ – ZeroTheHero Apr 28 '17 at 17:13

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