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Homework Help: Annihilation creation operator

  1. Oct 27, 2006 #1
    Hey,

    I am calculating the expectation value of the position x. I have the wave function
    psi(x) = |1> + |2>

    so I use the equation <x> = <psi|(a+a^+|psi> to calculate the mean value. So I get

    (<1| + <2|)(a+a^+)(|1> + |2>)

    which I reduce to

    <1|a|1> + <1|a|2> + <2|a|1> + <2|a|2> + <1|a+|1> + <1|a+|2> + <2|a+|1> + <2|a+|2>

    if I use the properties of annihilation and creation I get some strange results such fx
    sqrt(1) <1|0> or sqrt(3)<2|3> which are totally wrong. What have I done wrong?

    Thanks in advance
     
  2. jcsd
  3. Oct 28, 2006 #2

    OlderDan

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    I am not following the notation. If a is an operator and a^ is an operator, then why is there a + after a^ and why are there no a^ in your expansion?
     
    Last edited: Oct 29, 2006
  4. Oct 28, 2006 #3
    Maybe I have expressed the problem badly. a is the annihilation operator and a+ is the creation operator. I have a two state system |1>, |2> with a wavefunction (psi) = |1> + |2>. My problem is to perform the multiplication in order to find the expectation value:

    (<1| + <2|)(a + a+)(|1> + |2>)

    if I expand this calculation I get some strange results. How to multiple this? Any help appreciated - thanks in advance
     
  5. Oct 28, 2006 #4
    <1|0>=<2|3>=0

    In the above sum, only two terms are nonzero.
     
  6. Oct 28, 2006 #5

    OlderDan

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    So a|n> = |n-1> and a+|n> = |n+1>? and the states are orthogonal? Is that right? Euclid seems to know what you are doing, and it seems to be consistent with this interpretation. What are <1|1> and <2|2>? Are the states orthonormal? This is not a hint. I am asking because I want to know.
     
  7. Oct 29, 2006 #6
    Yes the two kets |1> and |2> are orthonormale. My problem in the multiplication was the <2|3>, <1|0> which are zero.
     
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