Dirac notation

  • Thread starter Niles
  • Start date
  • #1
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Homework Statement


Hi

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

[tex]
\left| \mu \right\rangle = \sum\limits_v {\left| v \right\rangle \left\langle {v}
\mathrel{\left | {\vphantom {v \mu }}
\right. \kern-\nulldelimiterspace}
{\mu } \right\rangle } = \sum\limits_v {\left\langle {\mu }
\mathrel{\left | {\vphantom {\mu v}}
\right. \kern-\nulldelimiterspace}
{v} \right\rangle ^* \left| v \right\rangle }
[/tex]

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

[tex]
\sum\limits_v {\left\langle {\mu }
\mathrel{\left | {\vphantom {\mu v}}
\right. \kern-\nulldelimiterspace}
{v} \right\rangle ^* \left| v \right\rangle } = \sum\limits_v {\left\langle {v}
\mathrel{\left | {\vphantom {v \mu }}
\right. \kern-\nulldelimiterspace}
{\mu } \right\rangle \left| v \right\rangle } \ne \sum\limits_v {\left| v \right\rangle \left\langle {v}
\mathrel{\left | {\vphantom {v \mu }}
\right. \kern-\nulldelimiterspace}
{\mu } \right\rangle }
[/tex]

What am I missing here?

Best,
Niles.
 

Answers and Replies

  • #2
vela
Staff Emeritus
Science Advisor
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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.
 
  • #3
1,868
0
Yeah, you are right. Thanks.
 

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