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I have asked a question at math.stackexchange that have a physical meaning.

My assumption: Suppose $a$ and $a^\dagger$ is Hermitian adjoint operators and $[a,a^\dagger]=1$. I want to prove that there are no inverse operators for $a$ and $a^\dagger$.

I thought that this assumption purely mathematical, but I have no answers there. Maybe I am missing something?

I will clarify that $a$ and $a^\dagger$ are just Creation and annihilation operators for quantum harmonic oscillator.

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up vote 11 down vote accepted

The ground state of the harmonic oscillator $|0\rangle$ obeys $$a|0\rangle = 0$$ which means that the action of $a$ can't be undone: once you act with it on a state, you set to zero the coefficient in front of $|0\rangle$ in the decomposition into occupation eigenstates. Any candidate inverse operator $a^{-1}$ acting on zero will give you zero again; you will never get $0$ back so it implies that there's no operator $a^{-1}$ that would satisfy $$ a^{-1}a = {\bf 1}.$$ On the other hand, there is an inverse from the opposite side that obeys $$aa^{-1} = {\bf 1}.$$ The action of this $a^{-1}$ on $|n\rangle$ is simply defined as $|n+1\rangle/\sqrt{n+1}$ or whatever coefficient is needed for it to be inverse. I can write this one-sided inverse operator as $a^\dagger (aa^\dagger)^{-1}$ which is well-defined because $aa^\dagger$ only has nonzero eigenvalues.

For $a^\dagger$, the claims are reverted, of course. There exists an inverse that obeys $$ (a^{\dagger})^{-1} a^\dagger = {\bf 1}$$ but not the other one.

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Thank you a lot! – Oiale Apr 14 '13 at 19:28

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