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Cheers to everyone. I' ve got a serious doubt about the following: consider the annihilation operator $\hat a$. For practical reasons, I sometimes find useful redefining it in the following way : $\hat a' =\hat a e^{i \phi}$, with $\phi \in \mathbb R$. If I add a new global phase to each eigenstate of $\hat a^\dagger \hat a$, $| 1 \rangle \rightarrow | 1 \rangle e^{i \phi}, \quad | 2 \rangle \rightarrow | 2 \rangle e^{2 i \phi} \,\dots$, I have a new annihilation operator $\hat a'$ and a new equivalent Hilbert space.

Is this $\hat a'$ physically reliable? Consider the time evolution of a state with Hamiltonian $\mathcal H = \alpha \hat a + \alpha^* \hat a^\dagger$, with $\alpha \in \mathbb C$. With the transformation described above $\alpha$ can be considered to be real without loss of generality. Is this correct?

Qmechanic
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SoterX
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1 Answers1

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$$\mathcal H = \alpha \hat a + \alpha^* \hat a^\dagger\\ =\alpha e^{-i\phi}~\hat a' + (\alpha e^{-i\phi})^* \hat {a'}~^\dagger\\ \equiv \alpha' \hat a' + (\alpha ')^* \hat {a'}~^\dagger.$$

Names, by themselves, cannot affect physical relevance. Arbitrary complex number coefficients present differently in the unprimed and primed representations, which amounts to a complex rotation. For given coefficients, fixed, there is a complex rotation to make them real. What's your point?

Cosmas Zachos
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  • Thanks for the reply. I should explain myself. The hamiltonian on top can be used to perform various rotations over the bloch sphere of a two-level system, but by applying this transformation you reduce the rotations to reflection over the x axis. This means that to fully describe any rotation can be reduced to a so-called pi-shift. This is what I was wondering. – SoterX Sep 03 '20 at 13:10
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    This answer might be of use. The primed and unprimed systems are completely equivalent. There is a real α for some direction, and a real α' for another. – Cosmas Zachos Sep 03 '20 at 13:51