What property of the spectrum of the oscillator follows from the commutator $[H,a^\dagger]=\hbar\omega a^\dagger$? I was thinking first about that you can't know both values simultaneously, but $a^\dagger$ is not an observable. My next guess was that it has something to do with the fact that you can only move up and down the ladder of the energies once at a time, but I don't know what specifically.


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$a^\dagger$ is the creation operator in this picture. What that means is that it takes a particular eigenstate of the Hamiltonian(with particular $E$) and then increases it's energy by an amount $1/2$ $\hbar \omega$. as,

$$a^\dagger \left \lvert n \right \rangle = (n+1)^{1/2} \left \vert n+1 \right \rangle \, .$$

Obviously then the commutator cannot be zero because if it commuted with the Hamiltonian it cannot change the eigenstates.

Also, $a^\dagger$ is not Hermitian, as you pointed out, So it being an observable doesn't make sense.

  • $\begingroup$ Could you explain a bit as to why if it commuted with the Hamiltonian it couldn't change the eigenstates? I'm still a bit shaky on that $\endgroup$ Aug 24, 2016 at 4:28
  • 1
    $\begingroup$ Sure. See when two operators commute it means they have the same eigenstates, i.e. if |n> is an eigenstate of Hamiltonian, it's also an eigenstate of $a^\dagger$. Thus $a^\dagger$ |n> will be some scalar(eigenvalue) multiplied by |n> itself. Thus it won't be possible for this operator to change the eigenstate. $\endgroup$
    – Ari
    Aug 24, 2016 at 4:41

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