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Harmonic oscillator in 2D - applying operators

  1. Feb 23, 2015 #1
    Hello, I juste don't know how this was done it is on the solutionnary of a very long exercise and i am not getting this calculation

    1. The problem statement, all variables and given/known data


    <1,0| ax+ay++ax+ay+axay++axay|0,1> = <1,0|1,0>
    2. Relevant equations
    3. The attempt at a solution
    We have that |0,1> = ay+ |0,0>
    I don't understand how they did the simplification to get to the <1,0|1,0>, i am missing something
    Thanks
     
  2. jcsd
  3. Feb 23, 2015 #2

    DrClaude

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    Staff: Mentor

    Can you figure out what
    (ax+ay++ax+ay+axay++axay)|0,1>
    is?

    (Hint: try it term by term.)
     
  4. Feb 23, 2015 #3
    ax+ay+|1,0 > =ax+ay+ay+|0,0 >
    ax+ay|1,0 >=ax+ayay+|0,0 >
    axay+|1,0 >=axay+ay+|0,0 >
    axay|1,0 >=axayay+|0,0 >

    But then I don't know how to simplify all that
     
  5. Feb 23, 2015 #4

    DrClaude

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    Staff: Mentor

    You're going backwards: reduce the number of operators. The same way that ay+|0,0 > = |0,1 >, what is ay|1,0 > ?
     
  6. Feb 23, 2015 #5
    I am so confused, are you saying that I should not replace |1,0> by ay+| 0 0 > ?
    ay|1,0 > = 0 ?
     
  7. Feb 23, 2015 #6

    DrClaude

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    Staff: Mentor

    Indeed. The final expression, <1,0|1,0> has no more operators. What you need to do is to apply the operators successively to get the resulting state vector (ket).

    That's not correct. You need to revise creation and annihilation operators.
     
  8. Feb 23, 2015 #7
    I edited my answer
    ay|1,0> = sqrt(0)|1, 0-1> = 0
     
  9. Feb 23, 2015 #8

    DrClaude

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    Staff: Mentor

    I don't like the intermediate step, as there is no such thing as |1, -1>. The result is simply ay|1,0> = 0.

    So now go back to my question in post #2.
     
  10. Feb 23, 2015 #9
    ax+ay++ax+ay+axay++axay|0,1> = ax+ay++axay+|1,0>
    ax+ay+|1,0>=|2,1>
    axay+|1,0>=|0,1>
     
  11. Feb 23, 2015 #10
    There is a mistake somewhere it is
    ax+ay+|0,1>=|1,2>
    axay+|0,1>=0

    So I get to |1,2> and not |1,0>
     
  12. Feb 23, 2015 #11
    Oh my god I have to see this again I applied ay to |1,0> and not |0,1> so i cancelled them
    I am left with

    <1,0|1,2>=0 but why ? and <1,0|1,0>=1
     
  13. Feb 23, 2015 #12

    DrClaude

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    Staff: Mentor

    Because eigenstates of the harmonic oscillator are orthogonal. It is the y part, <0|2>, that is zero.

    For normalized states, yes.
     
  14. Feb 23, 2015 #13
    Thank you very much !
     
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