# Applying $J_+$ to spin

I tried to check that $$|1,1\rangle = |\uparrow \uparrow \rangle$$ Which it is, according to my lecture notes. But I'm going wrong somewhere. I did it by applying $J_+$ to both sides of $|1,0\rangle = \frac{1}{\sqrt{2}}|\uparrow \downarrow + \downarrow \uparrow\rangle$:

$$|1,0\rangle = |1,1\rangle$$ And from the RHS: $$J_+ \frac{1}{\sqrt{2}}|\uparrow \downarrow + \downarrow \uparrow\rangle = \frac{1}{\sqrt{2}}J_+|\uparrow \downarrow + \downarrow \uparrow\rangle$$ $$=\frac{1}{\sqrt{2}}(J_+|\uparrow \downarrow \rangle + J_+|\downarrow \uparrow\rangle)$$ Considering just the first term, $$J_+ |\uparrow \downarrow \rangle = J_+|\uparrow\rangle \otimes |\downarrow \rangle + |\uparrow\rangle \otimes J_+ |\downarrow\rangle$$ $$=(|\uparrow\rangle\otimes |\uparrow\rangle) = |\uparrow \uparrow\rangle$$ Repeating the process with the other term, my final answer actually turns out to be $$|1,1\rangle = \frac{2}{\sqrt{2}} |\uparrow \uparrow\rangle$$ Where have I gone wrong though?

PHYS366 is hard right?

$J_{+}\vert j, m\rangle = \sqrt{j-m}\sqrt{j+m+1} \vert j, m+1 \rangle$

So when you apply the operator to the LHS you get:

$J_{+}\vert 1, 0\rangle = \sqrt{1-0}\sqrt{1+0+1} \vert 1, 0+1 \rangle = \sqrt{2}\vert1, 1 \rangle = \frac{2}{\sqrt{2}}\vert1, 1 \rangle$

So by comparing your answer for the RHS you get to $\vert \uparrow\uparrow \rangle= \vert 1, 1 \rangle$.

I can't believe he gave us 9 questions! Mental...

• It is hard! That was a silly mistake though. Thanks! Feb 8, 2017 at 22:48