1
$\begingroup$

In 3 dimensions, the total angular momentum (for $z$) matrix is given. It generates the rotation matrix around $z$ by $e^{-i\theta J_3/h}.$ My question is how do we actually go about doing this? I know that given the pauli matrices it would be $e^{-i\theta\sigma n/2}=\cos(\theta/2)I+i\sin(\theta/2)\sigma$ for a two dimensional rotation. But this method doesn't work for my aforementioned problem.

I.e. Where $J_z= I\hbar\begin{pmatrix}0&-1&0\\1&0&0\\0&0&0\end{pmatrix}$

I want to show that $e^{-I\theta J_3/\hbar}=\begin{pmatrix}cos(\theta)&-sin(\theta)&0\\sin(\theta)&cos(\theta)&0\\0&0&1\end{pmatrix}$

as I've seen it written in numerous texts but have no idea how it's obtained. I just want the method I don't need it worked out.

$\endgroup$
  • $\begingroup$ If there is an ambiguity I'm not aware of in my question let me know and I'll add more details $\endgroup$ – excalibirr Mar 11 at 0:58
  • 1
    $\begingroup$ This appears to be more of a physics question than a mathematics one. You should perhaps consider posting this on Physics $\endgroup$ – Brian Mar 11 at 1:01
  • $\begingroup$ @Brian theirs no underlying physics concept I'm asking about , It's a purely mathematical process, they don't like questions like that over there.... $\endgroup$ – excalibirr Mar 11 at 1:06
  • $\begingroup$ Would you mind clarifying what you mean by "the total angular momentum (for z) matrix" and "generates the rotation matrix around z"? This question seems to be about how a specific physical phenomenon may be modeled by a matrix rather than relating to matrices themselves. Forgive me if I am misunderstanding you question. Perhaps a MathJax examples of the matrices you are referring would help? $\endgroup$ – Brian Mar 11 at 1:17
  • $\begingroup$ @Brian I edited my question with the details :) $\endgroup$ – excalibirr Mar 11 at 1:43
0
$\begingroup$

To exponentiate a matrix $A$, find the eigenvalues $\lambda_i$ and eigenvectors $\xi_i$ of $A$. $e^A$ has the same eigenvectors $\xi_i$ as $A$, but with eigenvalues $e^{\lambda_i}$.

Equivalently, diagonalize $A$ by writing it as $A = P^{-1}DP$ for some diagonal matrix $D$. Then $e^A = P^{-1}e^DP$. For most applications in physics, $P$ will be a unitary matrix.

$\endgroup$
0
$\begingroup$

Hint We have $$-\frac{I}{\hbar} J_z = \pmatrix{&-1\\1\\&&0},$$ so $$\exp \left(-\frac{I}{\hbar} \theta J_z \right) = \exp \left[\theta \pmatrix{&-1\\1\\&&0} \right].$$ The standard procedure for computing an exponential matrix is to diagonalize (or more generally, decompose in Jordan normal form) the argument $A$ as $A = P D P^{-1}$ and then compute $$\exp A = P (\exp D) P^{-1} .$$ The eigenvalues of $\theta \pmatrix{&-1\\1\\&&0}$ are $\pm i \theta, 0$, so $$\exp D = \exp \pmatrix{i \theta\\&-i\theta\\&&0} = \pmatrix{\exp(i\theta)\\&\exp(-i\theta)\\&&1} .$$ All that remains is to compute the similarity matrix $P$, expand our expression for $\exp D$ using Euler's formula (namely, $\exp i\theta = \cos \theta + i \sin \theta$), and compute $P (\exp D) P^{-1}$.

$\endgroup$

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.