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Norm of a vector related to coherent states

  1. May 1, 2015 #1
    Hi,

    For the past couple of days I've been attempting to derive the equality (for any normalised ##\varphi##)
    [tex]||(a+za^\dagger+ \lambda)\varphi ||^2 = ||(a^\dagger+\bar{z}a+ \bar{\lambda} )\varphi ||^2 + (|z|^2-1)||\varphi ||^2[/tex]
    First of the summation seemed like a typo. So I first tried to prove this but got nowhere. As such I tried to change it to an equality. ( I entertained this idea because I found, and verified, some other typos in previous equations )


    Below is an outline of my attempt to derive the second equality (assuming a typo)
    Here we consider a system of bosons with only one mode hence ##[ a, a^\dagger ] = 1##.
    I expanded the norm as ##||(a+za^\dagger+ \lambda)\varphi ||^2 = \langle (a^\dagger+\bar{z}a+ \bar{\lambda} )\varphi , (a+za^\dagger+ \lambda)\varphi \rangle##

    I just cannot get a factor ##|z|^2## out of there. Because the inner product is defined as ##\langle\alpha \psi,\phi\rangle = \bar{\alpha}\langle \psi, \phi \rangle## while ##\langle \psi,\alpha\phi\rangle = \alpha\langle \psi, \phi \rangle##.

    So the straightforward way doesn't work. Next I tried to use the first equality.
    Lets call the two norms in this equality A resp. B, could I maybe use A-B was my idea.
    Because in the text the commutation relations were mentioned which (could) show up directly this way.
    I'm not going to write everything I tried (a lot) down here but I tried various tricks etc.

    Thanks

    Joris
     
  2. jcsd
  3. May 1, 2015 #2

    fzero

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    It's a little clearer if you write it in the notation

    $$||(a+za^\dagger+ \lambda)\varphi ||^2 = \langle \varphi| (a^\dagger+\bar{z}a+ \bar{\lambda} )(a+za^\dagger+ \lambda)|\varphi \rangle.$$

    Then you will see that

    $$||(a+za^\dagger+ \lambda)\varphi ||^2 - || (a^\dagger+\bar{z}a+ \bar{\lambda} )\varphi ||^2 = \langle \varphi| [a^\dagger+\bar{z}a+ \bar{\lambda} ,a+za^\dagger+ \lambda]|\varphi \rangle$$

    and a bit of algebra on the commutator should lead you to the result.
     
  4. May 1, 2015 #3
    Thanks I will check this tomorrow since it's late already.
    Unbelievable I didn't notice that, guess its one of those case where prolonged staring makes it worse :)

    Joris
     
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