High School Conjugation , involving operators in Dirac Notation.

[Somali_Physicist]

In a PDF i was looking through i came about a question
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.

 

[stevendaryl]

In a PDF i was looking through i came about a question
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.

'

I'm not sure what your point of confusion is.

1. $\langle v|P^\dagger|u\rangle = (\langle u|P|v\rangle)^*$: That's just the definition of $P^\dagger$ (I think $P^\dagger$ is used more often than $P^x$)
2. $= (\langle u|a\rangle \langle b|v\rangle)^*$: That's just using the definition of $P$.
3. $= (\langle u|a\rangle)^* (\langle b|v\rangle)^*$: That's just using the fact that the complex conjugate of a product is just the product of the complex conjugates.
4. $= \langle a|u\rangle \langle v|b\rangle$: That's just using the fact that for matrix elements, $\langle A|B\rangle^* = \langle B | A \rangle$.
5. $= \langle v | b \rangle \langle a | u \rangle$: Since matrix elements are just numbers, you can change the order in a product.

So we've shown that $\langle v | P^\dagger | u \rangle = \langle v | b \rangle \langle a | u \rangle$

 

[Somali_Physicist]

'

I'm not sure what your point of confusion is.

1. $\langle v|P^\dagger|u\rangle = (\langle u|P|v\rangle)^*$: That's just the definition of $P^\dagger$ (I think $P^\dagger$ is used more often than $P^x$)
2. $= (\langle u|a\rangle \langle b|v\rangle)^*$: That's just using the definition of $P$.
3. $= (\langle u|a\rangle)^* (\langle b|v\rangle)^*$: That's just using the fact that the complex conjugate of a product is just the product of the complex conjugates.
4. $= \langle a|u\rangle \langle v|b\rangle$: That's just using the fact that for matrix elements, $\langle A|B\rangle^* = \langle B | A \rangle$.
5. $= \langle v | b \rangle \langle a | u \rangle$: Since matrix elements are just numbers, you can change the order in a product.

So we've shown that $\langle v | P^\dagger | u \rangle = \langle v | b \rangle \langle a | u \rangle$

Hmm step 4 to 5 got me
for $\langle a|u\rangle \langle v|b\rangle$:
couldnt $\langle a|u\rangle$ give you a matrix and hence you can't simply change the product around?

 

[Nugatory]

Hmm step 4 to 5 got me
for $\langle a|u\rangle \langle v|b\rangle$:
couldnt $\langle a|u\rangle$ give you a matrix and hence you can't simply change the product around?

$\langle a|u\rangle$ is a number, one that is equal to $\langle u|a \rangle^*$

 

[Somali_Physicist]

$\langle a|u\rangle$ is a number, one that is equal to $\langle u|a \rangle^*$

So you can never get a 2 x m bra with m x 2 row?
Implying that for a given pair there will always be a 1 row matrix with n elements along with a 1 column matrix with n elements.

If that's the case then it makes sense!