# Computing Inner Product of Tensor Product

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.

• The ket $|w_3w_2\rangle$ doesn't really make sense without more information and context. Dec 5, 2020 at 19:27
• 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? Dec 5, 2020 at 19:33
• 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. Mar 31, 2021 at 15:31