Why does ⟨μ|v⟩* = ⟨v|μ⟩ in the completeness relation?

[Niles]

Homework Statement

Hi

Please take a look at the following equality found in my book:

$$\left| \mu \right\rangle = \sum\limits_v {\left| v \right\rangle \left\langle {v}<br /> \mathrel{\left | {\vphantom {v \mu }}<br /> \right. \kern-\nulldelimiterspace}<br /> {\mu } \right\rangle } = \sum\limits_v {\left\langle {\mu }<br /> \mathrel{\left | {\vphantom {\mu v}}<br /> \right. \kern-\nulldelimiterspace}<br /> {v} \right\rangle ^* \left| v \right\rangle }$$

The asterix denotes complex conjugation. I cannot see why the second equality holds, since

$$\sum\limits_v {\left\langle {\mu }<br /> \mathrel{\left | {\vphantom {\mu v}}<br /> \right. \kern-\nulldelimiterspace}<br /> {v} \right\rangle ^* \left| v \right\rangle } = \sum\limits_v {\left\langle {v}<br /> \mathrel{\left | {\vphantom {v \mu }}<br /> \right. \kern-\nulldelimiterspace}<br /> {\mu } \right\rangle \left| v \right\rangle } \ne \sum\limits_v {\left| v \right\rangle \left\langle {v}<br /> \mathrel{\left | {\vphantom {v \mu }}<br /> \right. \kern-\nulldelimiterspace}<br /> {\mu } \right\rangle }$$

What am I missing here?


Niles.

 

[vela]

Why don't you think the last equality holds? You're just writing <v|u>, which is a number, behind |v> instead of in front of it.

 

[Niles]

Yeah, you are right. Thanks.