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I want to know how to get the following result:

$$ e^{-i2\pi J_y / \hbar}|j, m\rangle = (-1)^{2j}|j, m\rangle $$

for an arbitrary spin state $|j, m \rangle$.

What I've tried is to expand the exponential in a power series and using the fact that $\hat{J_y}^2 = 1$ and $\hat{J_y}^3 = \hat{J_y}$ but I'm only getting the positive value (which according to the formula above would be for integer states).

Where does the spin being either integer or half integer come into this?

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  • $\begingroup$ First, shouldn't the operator $e^{-i2\pi J_y / \hbar}$ be acting on the ket $|j,m\rangle$? And are you sure it's $J_y$ you want? Because $|j,m\rangle$ is an eigenket of $J_z$ $\endgroup$
    – ErickShock
    Commented Dec 13, 2018 at 2:19
  • $\begingroup$ @ErickShock whoops! You're right, I'll edit the question $\endgroup$
    – Feng
    Commented Dec 13, 2018 at 2:19
  • $\begingroup$ @ErickShock also, I believe it is $J_y$ that we want, that's how it's written in the question $\endgroup$
    – Feng
    Commented Dec 13, 2018 at 2:24

1 Answer 1

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$$J_z|J,m\rangle=m|J,m\rangle\quad\rightarrow \quad e^{-2\pi i J_z}|J,m\rangle=e^{-2\pi i m}|J,m\rangle$$ Now if $J$ is a half-integer, so is $m$. So $e^{-2\pi i m}=e^{-\pi i}=-1$ regardless of which $m$ it is. Similarly if $J$ is an integer then $e^{-2\pi i m}=+1$.

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    $\begingroup$ If it really is $J_y$ then that doesn't change the argument. Since $|J,m\rangle$ is a superposition of y-eigenstates, all of which have either half-integer or integer $m_y$ depending on $J$ $\endgroup$
    – octonion
    Commented Dec 13, 2018 at 2:26
  • $\begingroup$ Hi thanks for your answer, could you explain what you mean that is $|J, m \rangle$ is a superposition of y-eigenstates and why this ensures that our answer doesn't change when we swap out $J_z$ for $J_y$ $\endgroup$
    – Feng
    Commented Dec 13, 2018 at 2:35
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    $\begingroup$ @David Feng he means that $|j,m_z\rangle = \sum_{m_y} C_{m_y} |j, m_y\rangle$. You can check that $j$ must be the same by writing $J_y$ in terms of the ladder operators $J_\pm = J_x \pm i J_y$ and remembering that $J_\pm$ only increases/decreases the values of $m_z$. $\endgroup$
    – ErickShock
    Commented Dec 13, 2018 at 3:11
  • $\begingroup$ @DavidFeng, Yes ErickShock explained what I meant. If $j$ is half integer, $m_y$ is also a half integer for each state in that sum, and so acting with $\exp(-2\pi i J_y)$ you get a minus sign for every single term. $\endgroup$
    – octonion
    Commented Dec 13, 2018 at 22:41

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