# Change of basis from $J_y$'s eigenbasis to $J_z$'s eigenbasis for arbitrary $j$

I have been trying to compute the inner product ($$j$$ being fixed) $$\langle m_y|a_z\rangle \tag{1}$$ where $$J_y|m_y\rangle = m |m_y\rangle$$ and $$J_z|a_z\rangle = a |a_z\rangle.$$ I tried writing $$|m_y\rangle = \sum_{n=-j}^j c_n |n_z\rangle \tag{2}$$ and then used $$J_y = (J_+ - J_-)/(2i)$$ to get the recurrence relation $$c_n + c_{n+1} \sqrt{(j-n)(j+n+1)}/m = c_{n-1}\sqrt{(j+n)(j-n+1)}/m \tag{3}$$ where coefficients for $$n<-j$$ and $$n>j$$ are zero.

In response to the comment, I'm outlining how I got this equation. $$J_y|m_y\rangle = \frac{1}{2i} \sum_{n} (J_+ - J_-) c_n |n_z\rangle = m\sum_n c_n |n_z\rangle$$

Using $$J_{\pm}|m_z\rangle = \sqrt{(j\mp m)(j\pm m + 1)}|(m\pm 1)_z\rangle$$, you just have to redefine the indices and then compare term by term (since these vectors are linearly independent) and you get this equation.

Now, I am thinking about how to solve this relation. I tried writing the matrix equation for it but that looks a bit difficult. Once I get it, the answer is, of course, $$c_a$$.

My question is if there is a faster, nicer way to compute this. If not, then please suggest some simple method to solve this equation.

Second (and more important) question is that I actually need this inner product for computing $$e^{-iJ_y\lambda} \Pi_a e^{i J_y \lambda} \tag{4}$$ where $$\Pi_a$$ is the projector to $$J_z$$'s eigenspace with eigenvalue $$a$$ (I'm using BCH.) Can you suggest a simpler alternative for computing this?

• Is this the representation transformation between $\{J^2, J_y\}$ and $\{J^2, J_z\}$? Commented Jan 2, 2019 at 11:43
• How can you get the equation $c_n + c_{n+1} \sqrt{(j-n)(j+n+1)}/m = c_{n-1}\sqrt{(j+n)(j-n+1)}/m$ ? Commented Jan 2, 2019 at 11:51
• I am not sure what you mean by representation transformation. I don't know group theory. I have outlined how I got this equation and is probably right (at worst it could have a silly mistake) Commented Jan 2, 2019 at 12:30
• Have you tried spin 1/2? With Wigner’s d-matrices?? Commented Jan 2, 2019 at 12:45
• I don't want it for spin 1/2, but even if I did, I'd directly exponentiate the exponentials. That might be easier. Commented Jan 2, 2019 at 14:33

I am not sure if there is a simpler answer, but this is an alternative. When computing the representations of SO$$(3)$$, one often finds the need for the matrix element $$$$d^j(\beta)^{m'}_{\ \ \ m} = \langle j,m'|\mathrm{e}^{-i \, \frac{\beta}{\hbar} \, J_y}|j,m \rangle \, ,$$$$ where $$|j,m\rangle$$ is the standard eigenstate of $$J^2$$ and $$J_z$$. There are expressions for this matrix element such as Wigner's formula, which you can find in, for instance, eq. 3.8.33, at the end of section 3.8 in Sakurai's Modern Quantum Mechanics, revised edition.

Now you see that you can compute an arbitrary matrix element of the operator you mention using this result: $$$$\langle j_2 , m_2 | \mathrm{e}^{- i \, J_y \, \lambda} |j_1 , m_1 \rangle\langle j_1 , m_1 |\mathrm{e}^{ i \, J_y \, \lambda} |j_3 , m_3 \rangle \, = \, \delta_{j_1,j_2} \, \delta_{j_1,j_3} \, d^{j_1}( \hbar \, \lambda)^{m_2}_{\ \ \ m_1} \, d^{j_1}(- \, \hbar \, \lambda)^{m_1}_{\ \ \ m_3} \, .$$$$

• I'll try it. I've always tried to run away from these d matrices. Commented Jan 2, 2019 at 14:31
• @physicophilic The problem you're solving is the very definition of the Wigner D matrices. You run away from them to your own detriment. Commented Jan 2, 2019 at 14:36
• No problem OP. In my answer I picked the convention with $\hbar$ which can be removed if you are not using that normalization. Also, I picked a specific form for $\Pi_a$. According to what I understand from your question, $\Pi_a$ is a projector of the form $\Pi_a = \Pi_{j,m}$ which acts as $|j,m \rangle \langle j,m|$. Commented Jan 2, 2019 at 14:41

The best method to deal with $$e^{-i\lambda J_y}$$ is to use the Wigner d-matrices, which are defined by $$d^j_{mm'}(\lambda)=\langle jm\vert e^{-i\lambda J_y} \vert jm'\rangle\, .$$ Thus for instance \begin{align} \langle jM\vert e^{-i \lambda J_y}\Pi_a e^{i \lambda J_y} \vert jM'\rangle&= \sum_{mm'} \langle jM\vert e^{-i \lambda J_y}\vert jm\rangle \langle jm\vert \Pi_a \vert jm'\rangle\langle jm'\vert e^{i \lambda J_y} \vert jM'\rangle\, ,\\ &=\sum_{mm'} d^j_{Mm}(\lambda) \langle jm\vert \Pi_a \vert jm'\rangle d^j_{m'M'}(-\lambda) \end{align} Tables of these $$d$$-functions can be found in multiple sources, including this wiki entry. Moreover, various limits of these are well studied.

Mathematica has a built-in command called WignerD but beware that their definition of angles is not standard. In your specific case, you would need WignerD[{j,m1,m2},0,$$-\lambda$$,0] with the negative argument to make the Mathematica function agree with the list of wiki entries linked above.

• Thanks! I didn't know about Mathematica's inbuilt function! Might be just what I need. Commented Jan 2, 2019 at 14:28
• You want to check beforehand to make sure of the sign convention. For some very strange reason it's not standard in the Wolfram package. Commented Jan 2, 2019 at 14:39