- #1

- 117

- 12

for the operator P = |a><b|

find P

^{x}(adjoint)

the adjoint was defined as

<v|P

^{x}|u> = (<u|P|v>)* where u and v can be any bra and ket

now for the question:

(<u|a><b|v>)* = <v|P

^{x}|u>

this is the confusing step , i thought conjugated simply changed the bras and kets to the pair

e.g for (<a|c><d|b>)* = <c|a><b|d>

however the pdf states:

(<u|a><b|v>)* = <v|b><a|u>

hence the P

^{x}= |b><a|

I don't understand this , flipping it will confuse me for say 3 pairs.Does anyone have an explanation for this.