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Showing raising operator raising!

  1. Jan 16, 2013 #1
    Hey,

    I have a question on showing how the raising operator in QM raises a particular eigenstate by 1 unit, the question is showed below:

    Raising_Operator_Q.png

    I think I know how to do this but not sure if what I'm doing is sufficient:

    [tex]\hat{N}a^{\dagger}|n>=([\hat{N},a^{\dagger}]+a^{\dagger}\hat{N})|n>[/tex]

    So I considered the N operator acting on the raising operator, we know the commutation relation stated on the RHS, so this simplifies to:

    [tex]\hat{N}a^{\dagger}|n>=(a^{\dagger}+a^{\dagger}\hat{N})|n>[/tex]

    Letting N act on the state N we attain:

    [tex]\hat{N}a^{\dagger}|n>=(1+n)a^{\dagger}|n>[/tex]

    I'm not sure if it's enough to say now that by definition of the raising operator 'a-dagger', it raises the state n to n+1 and we conclude that 'a-dagger' acting on a state n is equal to some constant multiplied by (1+n) and the state 1+n.

    Cheers,
    SK
     
  2. jcsd
  3. Jan 16, 2013 #2

    Simon Bridge

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    I'd be more explicit than that:

    from (given) ##[\hat{N},a^\dagger]|n\rangle = a^\dagger |n\rangle## ... expand the commutator

    ##\Leftrightarrow \hat{N}[a^\dagger|n\rangle ] - a^\dagger [\hat{N}|n\rangle ] = a^\dagger |n\rangle## ... since: ##\hat{N}|n\rangle = n|n\rangle## (given)

    ##\Leftrightarrow \hat{N}[a^\dagger|n\rangle] - n[a^\dagger|n\rangle] = a^\dagger |n\rangle##

    ##\Rightarrow \hat{N}[a^\dagger|n\rangle] = a^\dagger |n\rangle+na^\dagger|n\rangle = (n+1)[a^\dagger |n\rangle]##

    i.e. ##a^\dagger |n\rangle## is an eigenstate of ##\hat{N}## with eigenvalue n+1 ...

    ##\Rightarrow a^\dagger |n\rangle=|n+1\rangle##

    ... is |n+1> normalized already?
     
    Last edited: Jan 16, 2013
  4. Jan 16, 2013 #3

    Simon Bridge

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    BTW: It can be useful, if you are writing a LOT of fancy notation, to define the most common combinations ... eg: ##\renewcommand{\bra}[1]{\langle {#1} |}
    \renewcommand{\ket}[1]{|{#1}\rangle}
    \renewcommand{\braket}[1]{ \langle #1 \rangle }##

    \renewcommand{\bra}[1]{\langle {#1} |}
    \renewcommand{\ket}[1]{|{#1}\rangle}
    \renewcommand{\braket}[1]{ \langle #1 \rangle }

    so that \bra{n}\; \ket{n,l,m,s}\; \braket{\psi}\; \bra{\psi}H\ket{\psi} gives you: $$\bra{n}\; \ket{n,l,m,s}\; \braket{\psi}\; \bra{\psi}H\ket{\psi}$$
     
  5. Jan 17, 2013 #4
    Ahh yes, that's better. I think the ket 'n+1' is not normalised already but that's only due to the normalisation factor 'c' they've put in in the question.

    I think what you have written is sufficient, they do not ask you to determine 'c' - so I presume we don't have to in this case.

    Also your second post coding hasn't all come up properly on my screen!
     
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