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Basic Linear Alegbra for Quantum Mechanics

  1. Apr 21, 2015 #1
    I do not understand a work example in the book: ‘Quantum Mechanics DeMystified”.

    On page 212, part of Example 7-5:

    Given: Let { |a> |b> } be an orthonormal two-dimensional basis

    Let Operator A be given by:

    A = |a>< a |- i| a><b |+ i| b><a |- |b><b |

    Then: (The following part I do not understand)

    A squared = (|a>< a|- i| a><b |+ i| b><a |- |b><b |) (|a>< a |- i| a><b |+ i| b><a |- |b><b |)

    =|a>< a|(|a>< a |) + |a>< a|(- i| a><b |)- i| a><b |( i| b><a |) - i| a><b |(- |b><b |)

    + i| b><a |(|a>< a|) + i| b><a |(- i| a><b |)- |b><b |( i| b><a |)- |b><b |(- |b><b |)

    = |a>< a|- i| a><b |+|a>< a|+ i| a><b |+|b><b |+ i| b><a |- i| b><a |+|b><b |

    = 2|a>< a| + 2|b><b |

    Most grateful if someone could help!
     
  2. jcsd
  3. Apr 21, 2015 #2

    Quantum Defect

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    Homework Helper
    Gold Member

    Which step is confusing you?

    You might try to think about the "bra" and "ket" as vectors that make sense to you, for example 2-D unit vectors

    I.e. |a> = column vector (1,0); <a| = row vector (1,0); |b> = column vector (0,1); <b| = row vector (0,1)

    Things like <a|a> = 1 = <b|b> -- you can see this easily with the vectors above. Also, <a|b> = <b|a> = 0
     
  4. Apr 21, 2015 #3
    Thank you for your response.

    I am confused by the first step, the every basic Ket - Bra alegbra operation.

    For example:

    How to do the multiplication of:

    |a>< a|(|a>< a |) = ?

    |a>< a|( i| b><a |) = ?
     
  5. Apr 21, 2015 #4
    I assume i is just √-1? Then:
    |a>< a|(|a>< a |) =|a>< a|a>< a | = |a>< a |

    |a>< a|( i| b><a |) = i|a>< a| b><a | = 0

    i is just a number and numbers be moved to the front, back where ever you find convenient.
     
  6. Apr 21, 2015 #5
    Hi Qiao,
    Many Many Thanks! I got it now!
     
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