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Let $|w_2 \rangle = | + \rangle |-\rangle$ and $|w_3\rangle = |-\rangle |+\rangle$. Show that $\langle w_3 | w_3 w_2 \rangle = 0$.

I get $\langle w_3 | w_3 w_2 \rangle = (\langle - | \langle + | ) (| - \rangle |+\rangle |+\rangle |-\rangle$) but I'm not sure how to combine this into inner products.

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  • $\begingroup$ The ket $|w_3w_2\rangle$ doesn't really make sense without more information and context. $\endgroup$
    – Arthur
    Dec 5, 2020 at 19:27
  • $\begingroup$ This book has lots of typos and no online errata. I think you're right, the ket doesn't make sense. $|w_3\rangle$ is in $C^4$ so $|w_3 w_2\rangle$ would be in $C^8$ which wouldn't make sense to do an inner product with $|w_3\rangle \in C^4$. Right? $\endgroup$
    – jmacuna
    Dec 5, 2020 at 19:33
  • $\begingroup$ I think the way of understanding this is as follows: $\left<{w_{3},w_{3}w_{2}}\right>= \left<{w_{3},w_{3}}\right> \left<{w_{3},w_{2}}\right>= 1\cdot{ \left<{w_{3},w_{2}}\right> }= 1\cdot{0}=0 $ , I mean, I don't see other manner of interpreting that computation. $\endgroup$
    – ferolimen
    Mar 31, 2021 at 15:31

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