TITLE: The Inner Product of In and Out Spins
QUESTION [0 upvotes]: In Quantum Mechanics: The Theoretical Minimum, Hrabovsky and Susskind define the in and out vector like this:
$$
| i \rangle = \frac{1}{\sqrt{2}} | u \rangle + \frac{i}{\sqrt{2}} | d \rangle \\~\\
| o \rangle = \frac{1}{\sqrt{2}} | u \rangle - \frac{i}{\sqrt{2}} | d \rangle \\~\\
$$
We are asked to prove $ \langle i | o \rangle = 0$. Simple enough, but when we perform the multiplication:
$$
\frac{1}{2} \langle u | u \rangle - \frac{i^2}{2} \langle d | d \rangle \\~\\
\frac{1}{2} \langle u | u \rangle + \frac{1}{2} \langle d | d \rangle \\~\\
= 1 \: \text{if} \langle u | d \rangle \: \text{and} \: \langle d | d \rangle \: \text{are both positive and constant.}
$$
This certainly isn't zero! What are we missing here?

REPLY [2 votes]: Be careful in a complex vector space.
$$\langle i|=\frac{1}{\sqrt{2}}\langle u|-\frac{i}{\sqrt{2}}\langle d|$$
so you forgot to take the complex conjugate.
The result is immediate after that.
{"set_name": "stack_exchange", "score": 0, "question_id": 720748}