1
$\begingroup$

Suppose we are in a total spin $J=j$ vector space and there is the angular momentum operator $\boldsymbol{J}$. The Hilbert space then has $2j+1$ states:

$$ |m=-j\rangle, |m=-j+1\rangle, \ldots, |m=j-1\rangle, |m=j\rangle $$

Any state $|\psi\rangle$ can be expressed as

$$ |\psi\rangle = \sum_{m=-j}^j c_{m} |m\rangle $$

Thus the state is fully determined by the $2j+1$ coefficients $c_m$. Note that the states $|m\rangle$ are defined with respect to a certain quantization axis which can be chosen to point in any direction we like. Let's denote this original quantization axis direction as the $\hat{\boldsymbol{z}}$ direction. If we have $|\psi\rangle = |m=1 \rangle$ then this tells us that

$$ \langle\boldsymbol{J}\cdot\hat{\boldsymbol{z}}\rangle = \langle J_z\rangle = \langle\psi|J_z|\psi\rangle = 1\times \hbar $$

That is, this state has a an expected projection of angular momentum onto the $\hat{\boldsymbol{z}}$ axis of exactly $\hbar$. (I include this to drive home the point that while we can choose any quantization axis the choice of quantization axis does have implications for the physical interpretation of states).

We can perform a change of quantization axis by performing a rotation of our coordinate system $R(\boldsymbol{\zeta})$ where $\boldsymbol{\zeta}$ represents an axis and angle of rotation. This unitary transformation gives us new basis states $|m'\rangle$ defined by

$$ R(\boldsymbol{\zeta})|m\rangle_{\boldsymbol{z}} = \sum_{m'=-j}^j d_{m'm}^{(j)}(\boldsymbol{\zeta})|m'\rangle_{\boldsymbol{n}} $$

Where the $d_{m'm}^{(j)}(\boldsymbol{\zeta})$ are the Wigner rotation matrices. I am following quantum and atom optics by Steck. I've added a subscript to the kets to indicate that they are basis states with respect to a certain quantization axis. Here $\hat{\boldsymbol{n}}$ represents the new quantization axis found by rotation the old quantization axis $\hat{\boldsymbol{z}}$ through the rotation associated with $\boldsymbol{\zeta}$.

I actually have 2 questions.

  1. Is it possible to fully determine a general state $|\psi\rangle$ (which is restricted to the total $J=j$ subspace*) if $\langle J_n \rangle = \langle \boldsymbol{J}\cdot\hat{\boldsymbol{n}}\rangle$ is known for $|\psi\rangle$ for all directions $\hat{\boldsymbol{n}}$? For example from that information could we determine all of the $c_m$ with respect to a particular quantization axis?

  2. What if instead of knowing $\langle J_n \rangle$ we only the magnitude of the coefficient of the stretch state with respect to each direction $\hat{\boldsymbol{n}}$? Would this be enough information?

*What if we do not restrict to a particular total spin $J=j$ subspace? What can be learned from these expectations?

$\endgroup$
0
$\begingroup$

I'm probably misunderstanding your question screened behind a wall of formalism. Looks like you are really asking a question about the spherical basis. Rather than a thrust-and-parry avalanche of comments, I'm reviewing the obvious.

So, let's illustrate it out for elementary QM, spin 1/2, so J=σ/2, hence, for $\hat n= (n_1,n_2,n_3)$, $n_1^2+n_2^2+n_3^2=1$, $$ \tfrac{1}{2}\hat n \cdot \vec \sigma = \tfrac{1}{2}\begin{pmatrix} n_3 & n_1 - in_2 \\ n_1 + in_2 & -n_3 \end{pmatrix} ~. $$ You may trivially compute its eigenvectors, $$ \psi_+ = \begin{pmatrix} n_3 + 1 \\ n_1 + in_2 \end{pmatrix}/ \sqrt{2(1+n_3)} ; \qquad \psi_- = \begin{pmatrix} in_2 - n_1 \\ n_3 + 1 \end{pmatrix}/\sqrt{2(1+n_3)} ~, $$ but you don't need to.

You already know its maximal eigenvalue (1/2) eigenvector is (1,0) for $\hat n=\hat z$, and so, for any $\hat n$, for this vector, $$ \langle \hat n\cdot \vec J\rangle= n_3/2 \leq 1/2 ~. $$ Conversely, if you are given the (real) r.h.side, you've virtually specified $\hat n$, up to an immaterial complex phase rotating $n_1$ to $n_2$.

Varying this $n_3$ from 1 to 0 to -1 will give you all rotations of the quantization axis, and hence its eigenvectors, from the north to the south pole.

In reverse, for arbitrary states, the maximum of such expectation values specifies that state's North pole, and hence the rotation from the nominal North pole of your chosen coordinate system, as evident upon contemplation of the evident notional diagram involved.

Does the paradigm deconstruct your problem? There are, of course, formal analogs to this, for each and every spin matrix.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.