Conjugation , involving operators in Dirac Notation.

  • #1
In a PDF i was looking through i came about a question
for the operator P = |a><b|
find Px(adjoint)

the adjoint was defined as
<v|Px|u> = (<u|P|v>)* where u and v can be any bra and ket

now for the question:
(<u|a><b|v>)* = <v|Px|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 Px = |b><a|

I dont understand this , flipping it will confuse me for say 3 pairs.Does anyone have an explanation for this.
 

Answers and Replies

  • #2
stevendaryl
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In a PDF i was looking through i came about a question
for the operator P = |a><b|
find Px(adjoint)

the adjoint was defined as
<v|Px|u> = (<u|P|v>)* where u and v can be any bra and ket

now for the question:
(<u|a><b|v>)* = <v|Px|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 Px = |b><a|

I dont 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##
 
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  • #3
'

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 cant simply change the product around?
 
  • #4
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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 cant simply change the product around?
## \langle a|u\rangle## is a number, one that is equal to ## \langle u|a \rangle^*##
 
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  • #5
## \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 thats the case then it makes sense!
 

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