- #1
spaghetti3451
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- 33
Let the generators of the SU(2) algebra be ##\tau_{1}##, ##\tau_{2}## and ##\tau_{3}##.
Consider an ##N## dimensional representation, which means that the ##\tau_{i}## are ##N \times N## matrices which act on some ##N##-dimensional vector space.
Consider the ladder operators ##\tau_{\pm}=\tau_{1}\pm i\tau_{2}##. Let's work in a basis where ##\tau_{3}## is diagonal, and let ##|m\rangle## denote a unit normalized eigenstate of ##\tau_{3}## with eigenvalue ##m##.
Now, I can show that ##\tau_{\pm}## can raise/lower the eigenvalue of ##|m\rangle## by 1, since ##\tau_{3}(\tau_{\pm}|m\rangle)=(m\pm 1)(\tau_{\pm}|m\rangle)##.
I can also show that ##\tau_{-}## can annihilate the state ##|1\rangle##, since ##\tau_{3}(\tau_{-}|1\rangle)=(1-1)(\tau_{-}|1\rangle)=0##
How do I show that ##\tau_{+}## also annihilates some state?
Consider an ##N## dimensional representation, which means that the ##\tau_{i}## are ##N \times N## matrices which act on some ##N##-dimensional vector space.
Consider the ladder operators ##\tau_{\pm}=\tau_{1}\pm i\tau_{2}##. Let's work in a basis where ##\tau_{3}## is diagonal, and let ##|m\rangle## denote a unit normalized eigenstate of ##\tau_{3}## with eigenvalue ##m##.
Now, I can show that ##\tau_{\pm}## can raise/lower the eigenvalue of ##|m\rangle## by 1, since ##\tau_{3}(\tau_{\pm}|m\rangle)=(m\pm 1)(\tau_{\pm}|m\rangle)##.
I can also show that ##\tau_{-}## can annihilate the state ##|1\rangle##, since ##\tau_{3}(\tau_{-}|1\rangle)=(1-1)(\tau_{-}|1\rangle)=0##
How do I show that ##\tau_{+}## also annihilates some state?