# Dirac notation

## Main Question or Discussion Point

hey guys just a quick question , within the Dirac notation I we have bras and kets.Is it allowable to simply hermitianly conjugate everything , e.g:

<w|c> = <b|c> - <d|c>
Can we then:
<c|w> = <c|b> -<c|d>

Or is there some subtly hidden rule.

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Try expanding (<w|c> = <b|c> - <d|c>)*

stevendaryl
Staff Emeritus
hey guys just a quick question , within the Dirac notation I we have bras and kets.Is it allowable to simply hermitianly conjugate everything , e.g:

<w|c> = <b|c> - <d|c>
Can we then:
<c|w> = <c|b> -<c|d>

Or is there some subtly hidden rule.
The quantity $\langle w | c \rangle$ is just a complex number, and it has the property that $(\langle w | c \rangle^* = \langle c | w \rangle$. So it's perfectly fine to apply the $^*$ operation to both sides of an equality.

The quantity $\langle w | c \rangle$ is just a complex number, and it has the property that $(\langle w | c \rangle^* = \langle c | w \rangle$. So it's perfectly fine to apply the $^*$ operation to both sides of an equality.
Okay well that leads to my real conundrum:

<w|c><c|w> = α = P
conjugation of both sides
(<w|c><c|w>)* = α* = P*
<c|w><w|c> =α*
<c|w><w|c><w|c><c|w> = α2
=(<w|c><c|w>)2 = <w|c><c|w><w|c><c|w>

but does not this imply

<c|w><w|c> = <w|c><c|w> which means <w|c><c|w> real?

i dont understand why that would be the case as the operator P should act differently when conjugated.

Last edited:
yes the number alpha is real
what is the définition of your operator P?

Okay well that leads to my real conundrum:

but does not this imply

<c|w><w|c> = <w|c><c|w> which means <w|c><c|w> real?
Yes, <c|w><w|c> = |<c|w>|^2 is a real number.