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I was surprised to learn that not every $|jm\rangle$ can be rotated into a given $|jm'\rangle$ with any finite rotation $R(\alpha, \beta, \gamma)$, and I am curious whether or not there is any physical intuition that would let us tell when this can and can't be done. I think I understand mathematically why this is the case. Rotations about a given axis are generated by the angular momenta operators, and we can get the matrix corresponding to a finite rotation by exponentiating. That is, we can say:

$$ R(\alpha, \beta, \gamma) = e^{-i\alpha L_x/\hbar}e^{-i\beta L_y/\hbar}e^{-i\gamma L_z/\hbar} $$

For a given subspace corresponding to a fixed $j$, only the first $2j$ terms in the series expansion of each exponential will be linearly independent and we can can actually write down a compact expression for each of this operators. For simplicity, I will take $j=1$. Then:

$$ e^{-i\theta L_i/\hbar} = \mathbb{1} - \frac{i}{\hbar}\sin (\theta) L_i + \frac{(\cos(\theta) -1)}{\hbar^2} L_i^2 $$

If you turn plug in the forms for $L_x$ and $L_y$ and turn the crank, the above rotation becomes:

$$ \begin{pmatrix} \cos^2(\alpha/2) & -i\sin\alpha / \sqrt{2} & -\sin^2(\alpha/2) \\ -i\sin\alpha / \sqrt{2} & \cos \alpha & -i\sin\alpha / \sqrt{2} \\ -\sin^2(\alpha/2) & -i\sin\alpha / \sqrt{2} & \cos^2(\alpha/2) \end{pmatrix} \times \begin{pmatrix} 1-\sin^2(\beta/2) & -\sin\beta / \sqrt{2} & \sin^2(\beta/2) \\ \sin\beta / \sqrt{2} & \cos \beta & -\sin\beta / \sqrt{2} \\ \sin^2(\beta/2) & \sin\beta / \sqrt{2} & 1-\sin^2(\beta/2) \end{pmatrix} \\ \times \begin{pmatrix} e^{-im\gamma} & 0 & 0 \\ 0 & e^{-im\gamma} & 0 \\ 0 & 0 & e^{-im\gamma} \end{pmatrix} $$

It's messy, but you can show using this for example that if you start at $|1, 1 \rangle$ you can't rotate into $|1, 0 \rangle$, but you can rotate into $|1,-1\rangle$. This is super odd to me, as I can't see much that would distinguish these two cases. Both $|1,0\rangle$ and $|1,-1\rangle$ are in the same subspace as $|1,1\rangle$, both are basis vectors, and classically if you had an angular momentum such that its projection along $z$ was $1$, you could rotate it both so that its projection was $0$ (rotate around $x$ by $\pi/2$) or $-1$ (rotate around $x$ by $\pi$). So what is different between these two states that makes one rotation possible and the other not?

Are there any arguments on physical grounds we could have made to arrive at the above conclusion without going through the math? Is there a general way to classify which states you can reach from a given starting state via finite rotations?

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    $\begingroup$ You might inspect the structure of Wigner d matrices, nullifying the ss element.. $\endgroup$ – Cosmas Zachos Feb 4 at 23:03
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    $\begingroup$ I'm very confused by your rotation matrices. The spin 1 representation is not projective so $\alpha=2\pi$ should be trivial. Where is the factor of $1/2$ coming from? i.e., why is it $\sin(\alpha/2)$ instead of just $\sin(\alpha)$? Also, the spin 1 representation is the fundamental representation of $SO(3)$, i.e., $U(R)=R$. Your rotation matrices should be the standard Euler matrices $$\begin{pmatrix}\cos\alpha&\sin\alpha&0\\-\sin\alpha&\cos\alpha&0\\0&0&1\end{pmatrix}$$which is definitely not what you wrote. Or am I talking nonsense? $\endgroup$ – AccidentalFourierTransform Feb 6 at 22:41
  • $\begingroup$ @ AccidentalFourierTransform Indeed, his paradigm is quite flawed. Since he started working in the spherical basis, he should stick to Wigner's matrices and simply observe, as I invited him to, that to rotate away from the highest m state, he needs π for all spins, which cannot possibly take him to all states, manifestly, except for spin 1/2, which is inevitable, from unitarity. Why he could imagine he could, I have no clue. $\endgroup$ – Cosmas Zachos Feb 6 at 22:59
  • $\begingroup$ @AccidentalFourierTransform It's possible I made a mistake, I used the identity above for j=1 rotation matrices and plugged stuff in. Not sure what to tell you other than those matrices are the result, or I made a mistake. $\endgroup$ – gabe Feb 7 at 19:40
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Are there any arguments on physical grounds we could have made to arrive at the above conclusion without going through the math?

Yes. This answer uses some mathematial notation, but it doesn't use any matrices. It uses only geometric intuition and the most basic general principles of quantum physics.

Notation

Consider an irreducible representation of the spin group (the double cover of the rotation group), and use this notation:

  • $\hat u$ is a unit vector in 3d space.

  • $J(\hat u)$ is the generator of rotations about the $\hat u$-axis. For every axis $\hat u$, the generator $J(\hat u)$ is an observable represented by a self-adjoint operator on the Hilbert space.

  • $j$ is the largest eigenvalue of $J(\hat u)$ in the given irreducible representation.

  • $\big|\hat u\big\rangle$ is the eigenstate of $J(\hat u)$ with the largest eigenvalue $j$: $$ J(\hat u)\,\big|\hat u\big\rangle = j\,\big|\hat u\big\rangle. $$

We can think of $|\hat u\rangle$ as having an angular momentum with a specific orietnation in 3d space, namely along the $\hat u$-axis, but that interpretation isn't necessary for this answer to work.

Intuition

Rotations affect observables in the obvious way: applying a rotation $\hat u\to R\hat u$ transforms $J(\hat u)\to J(R\hat u)$. This immediately implies two things:

  • All of the states $|\hat u\rangle$, for all directions $\hat u$, can be obtained from each other by rotations, up to a physically irrelevant overall phase factor. (A state can be represented by a vector in the Hilbert space, but the representation is not one-to-one: vectors related to each other by an overall nonzero complex constant represent the same state.)

  • For any given direction $\hat u$, the only states that can be obtained by rotating $|\hat u\rangle$ are largest-eigenvalue eigenstates of the generators $J(R\hat u)$.

Rotating clockwise about $\hat u$ is the same as rotating counterclockwise about $-\hat u$, so the eigenstate of $J(\hat u)$ with eigenvalue $j$ corresponds to the eigenstate of $J(-\hat u)$ eigenvalue $-j$. This implies:

  • The eigenstate of $J(\hat u)$ with eigenvalue $-j$ is a rotation of the eigenstate of $J(\hat u)$ with eigenvalue $j$.

The case $j=1/2$ is special, because then $J(\hat u)$ doesn't have any other eigenstates: the only two eigenstates it has are the ones with eigenvalues $\pm j$.

Now consider a representation with $j>1/2$, and consider an eigenstate $|m\rangle$ of $J(\hat u)$ with eigenvalue $m\neq \pm j$. The operator $J(\hat u)$ is the generator of rotations about the $\hat u$ axis, so $|m\rangle$ must be invariant under rotations about the $\hat u$-axis, up to a physically meaningless overall phase. On the other hand, if $|m\rangle$ could be obtained from $|\hat u\rangle$ by a rotation, then $|m\rangle$ would be an eigenstate of $J(R\hat u)$ for some other direction $R\hat u$ with the largest eigenvalue $j$. But $J(R\hat u)$ is not invariant under rotations about the $\hat u$-axis unless $R\hat u=\pm\hat u$, so that would contradict the fact that $|m\rangle$ must be invariant under rotations about the $\hat u$-axis. Altogether, this implies

  • The eigenstate of $J(\hat u)$ with eigenvalue $-j$ is a rotation of the eigenstate of $J(\hat u)$ with eigenvalue $j$, and the other eigenstates of $J(\hat u)$ cannot be obtained by rotating the one with eigenvalue $j$.

By the way

The answer above focused on eigenstates of $J(\hat u)$, but as knzhou's answer pointed out, we can also make a more general statement: in a representation with $j>1/2$, most states are not equal (or proportional) to $|\hat u\rangle$ for any direction $\hat u$. Every state can be expressed as a superposition of these, but in most cases the superposition needs to involve more than one direction in 3d space. Furthermore, the superposition is not unique: the same state may be expressed as different superpositions, involving different sets of directions in 3d space. This follows simply from the fact that the basis $\big\{|\hat u\rangle\big\}$ is overcomplete, which in turn follows from the fact that the parameter $\hat u$ is continuous: an irreducible representation has a basis with a finite number of vectors, so the basis parameterized continuously by $\hat u$ must be overcomplete.

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    $\begingroup$ Wow, +1, I can't believe I missed how simple and elegant the criterion is! $\endgroup$ – knzhou Feb 7 at 2:57
  • $\begingroup$ Great answer, this really provides the type of insight I was looking for. @knzhou's answer was also quite good, but this one provides the type of reasoning I was after so I will award the bounty here. I am satisfied with this. $\endgroup$ – gabe Feb 7 at 19:39
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This isn’t a full answer, but an easy way to see you can’t reach every state with a rotation is that the set of rotations has three real dimensions, while the set of normalized spin $s$ states has real dimensions $2(2s+1) - 2 = 4s$ real dimensions, where the subtraction is due to tossing out a global phase and demanding normalization. This means that for spin $s=1$ and higher, almost all states can’t be reached by a rotation, just on dimensional grounds. The classical intuition doesn’t work, because a “classical” spin always has a two-dimensional state space.

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  • $\begingroup$ This is definitely an observation I hadn't made which only makes me more curious! I'm still interested in whether or not anything distinguishes the states that can be reached by rotation from a given starting state. It feels like there should be a way to tell if a certain state is reachable given a starting point, but intuition surrounding rotations in QM can quickly lead one astray. $\endgroup$ – gabe Feb 4 at 19:51

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