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Using ladder operators, I can find eigenstates $\psi_n$ with eigenenergies $$E_n=\hbar\omega\left(n+\frac{1}{2}\right). $$ In my textbook, ladder operators work like $$ a\psi_n = c_n \psi_{n-1}$$ $$ a^\dagger \psi_n = d_n \psi_{n+1} $$ where $c_n$ and $d_n$ are propotional constants.

But, how do I assure that the difference between eigenstates is just $\hbar\omega$ and there is no other eigenstates between them? In other words, how do I assure that ladder operators make the right next eigenstate?

Qmechanic
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Jungmin Kim
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    Related: http://physics.stackexchange.com/q/23028/2451 , http://physics.stackexchange.com/q/54691/2451 and links therein. – Qmechanic Aug 10 '15 at 00:10

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Observe that the number operator $N := a^\dagger a$ is positive-semidefinite because $$ \langle \psi \vert N \psi \rangle = \lvert a \lvert \psi \rangle \rvert^2 \ge 0$$ and hence has no negative eigenvalues.

Since $a\lvert n \rangle = \sqrt{n}\lvert n - 1 \rangle$ for a normalized eigenstate of $N$ with eigenvalue $n$ (which one can derive from the commutation relation $[a,a^\dagger] = 1$), non-integer eigenvalues $q$ would make it possible to create a state with negative eigenvalue by applying $a^{\lceil q \rceil}$ to it, where $\lceil q \rceil$ is the smallest integer larger than $q$.

But such negative eigenvalues cannot exist, so there are no non-integer eigenstates.

ACuriousMind
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