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Bra-Ket operation

  1. May 29, 2010 #1
    I have a basic question that I have overlooked in the past, given that you have
    <psi2|A = lamda2<psi2|, where <| is a bra and lamda2 is the eigenvalue. If you were to multiply the equation by |psi1>, why do you get <psi2|A|psi1> = lamda2<psi2|psi1> and not |psi1><psi2|A = lamda2|psi1><psi2| ? Wouldn't the former defeat the purpose of operator algebra?
     
  2. jcsd
  3. May 29, 2010 #2
    It depends on from what side you multiplied |psi1> on your expression.
    This is similar to multiplying of columns and rows in matrix algebra :smile:

    For example,
    |psi1> *| <psi2|A = lamda2<psi2| => |psi1><psi2|A = lamda2|psi1><psi2| ;

    <psi2|A = lamda2<psi2| | *|psi1> => <psi2|A|psi1> = lamda2<psi2|psi1> ;
     
  4. May 29, 2010 #3
    Okay, I didn't know you can multiply it from both sides. I know that A is a matrix, but the bra's and ket's are just vectors.. and i know that the matrix A can be found using the orthonormal basis, but i can't seem to make sense of it.
     
  5. May 29, 2010 #4
    Do you know of a website or (intuitive) method i can use to learn the properties of bra-ket algebra? How did you learn it, for instance?
     
  6. May 29, 2010 #5
    To put it formally:
    Every isolated physical system has associated with it a complex inner product
    space or Hilbert Space. This is known as its state space. A
    complete description of the system is given by its state vector, |ψ>, a
    unit-normalised vector in the state space.

    The bra has two equivalent definitions, as (rather fearsome sounding)
    linear functionals and as elements of a dual vector space.

    Try googling some of the above (and below) key words for more info. Hope that helps :D

    Linear, Hermitian and Unitary Operators
    Closure Relation and matrix representation of operators
    Projectors
    Basis sets, norms and 'complete set of commuting operators' (CSCOs)
     
  7. May 29, 2010 #6

    Fredrik

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    I learned some of it from Sakurai, but I didn't like his presentation, so I spent a lot of time thinking about it until I had figured it all out (and then checked with other sources so I knew that I was right). This is how I would have wanted it presented to me.
     
  8. May 29, 2010 #7
    Hi,ian2012.
    Uning identity relation I = Σn|n><n| where |n> are eigenketvectors of some Observable, A=IAI= Σn,m |n><n|A|m><m| = Σn,m |n> Anm <m|. It shows that something appears |><| is matrix or operator.
    Regards.
     
    Last edited: May 29, 2010
  9. May 30, 2010 #8
    Thanks for all of your posts. I have, since, extensively been reading about Dirac notation, dual space, projectors, etc.. but I still have one query regarding an operation with commutators.
    How do you go from: <x'|x(hat)A(hat) - A(hat)x(hat)|x> to x'<x'|A(hat)|x> - <x'|A(hat)|x>x ?
    (You may recognize the commutation relation [x(hat), A(hat)]).
    From what I understand x(hat)|x> = x|x> and <x'|x(hat) = <x'|x' , which is halfway there. How do you multiply the rest of the equation?
     
  10. May 30, 2010 #9
    Due to x(hat) and A(hat) being linear operators you can split
    <x'|x(hat)A(hat) - A(hat)x(hat)|x> into <x'|x(hat)A(hat)|x> - <x'|A(hat)x(hat)|x>, then simply use the eigenfunction relations you stated, namely:
    x(hat)|x> = x|x> and <x'|x(hat) = <x'|x' then you've done it. :D
     
  11. May 30, 2010 #10
    What linearity condition do you use ?
     
  12. May 30, 2010 #11

    Fredrik

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    DJsTeLF is right that you need to split the expression into those two parts before you use the eigenvalue equations, but it's not the linearity of the operators that allows you to do that, it's that the inner product is linear in the second variable.
     
  13. May 30, 2010 #12
    Oh sorry, I understand what i did wrong.
    Is it right to say: (A(hat) + B(hat)).|psi> = A(hat)|psi> + B(hat)|psi> ?
    If so, then I believe that is the condition I was looking for... Well.. it has to be correct as it works for replacing the operator with real numbers.
     
    Last edited: May 30, 2010
  14. May 30, 2010 #13

    Fredrik

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    Yeah, I should have mentioned that too. The first step is to do what you just did. That's just the definition of the sum of two operators. The second step is to use either that the inner product is linear in the second variable or that a bra is a linear functional.
     
  15. May 31, 2010 #14
    Yeah, the property is overlooked on many sites I have been on. However, I looked in Dirac's Principles of QM and he explains it there, which is fantastic. But it's a quite basic property of linear operators you can infer from simple derivatives.
     
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