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Hermitian conjugate of operators

  • Thread starter v_pino
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  • #1
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Homework Statement



Find the hermitian conjugates, where A and B are operators.

a.) AB-BA
b.) AB+BA
c.) i(AB+BA)
d.) [itex] A^\dagger A [/itex]


Homework Equations



[itex] (AB)^\dagger =B^\dagger A^\dagger [/itex]



The Attempt at a Solution



Are they correct and can I simplify them more?

a.) [itex] (AB-BA)^\dagger = (AB)^\dagger - (BA)^\dagger= B^\dagger A^\dagger - A^\dagger B^\dagger = BA-AB = -(AB-BA) [/itex]

b.) [itex] (AB+BA)^\dagger = (AB)^\dagger + (BA)^\dagger= B^\dagger A^\dagger + A^\dagger B^\dagger = BA+AB [/itex]

c.) [itex] [i(AB+BA)]^\dagger = -i[(AB)^\dagger + (BA)^\dagger]= -i(B^\dagger A^\dagger + A^\dagger B^\dagger) = BA+AB [/itex]

d.) [itex] [A^\dagger A]^\dagger = A^\dagger (A^\dagger)^\dagger=A^\dagger A [/itex]
 

Answers and Replies

  • #2
dextercioby
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You might review point c.)
 
  • #3
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For point C.) [itex] -i(B^\dagger A^\dagger + A^\dagger B^\dagger) [/itex]
 
  • #4
dextercioby
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Alright. The last equality is not true. That's what I meant.
 
  • #5
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Is it ok until [itex] -i(B^\dagger A^\dagger + A^\dagger B^\dagger) [/itex]?
 
  • #6
dextercioby
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Yes, it is.
 

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