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The position shift operator $T^{a}$ (where $a \in \mathbb{R}$ ) takes a real valued wavefunction $\psi$ on $\mathbb{R}$ to its translation $\psi_{a}$, $T^{a} \psi(x)=\psi_{a}(x)=\psi(x+a)$. A practical operational calculus representation of the linear operator $T^{a}$ in terms of the plain derivative $\frac{d}{d x}$ was introduced by Lagrange, $$T^{a}=e^{a\frac{d}{d x}}. $$

Of course this is the well known unitary translation operator in quantum mechanics and the exponent can be written to express it in terms of the momentum operator $$T^{a}=e^{a\frac{i}{\hbar}(\frac{\hbar}{i}\frac{d}{d x})} =e^{a\frac{i}{\hbar}\hat{P_x}}.$$

But what about $$T^{ia}=e^{ia\frac{d}{d x}}=e^{\frac{-a}{\hbar}(\frac{\hbar}{i}\frac{d}{d x})}=e^{\frac{-a}{\hbar}\hat{P_x}}$$ where $\psi$ is a real valued wavefunction of a real variable, $i$ the imaginary unit and $a\in \mathbb{R}$. It's meaningful to consider its action on a real-valued function? $$T^{ia}\psi(x)=e^{ia\frac{d}{d x}}\psi(x)=???$$ What would be the expected result?

There's any connection to the Wick rotation? It's possible to perform a Wick rotation in the variable to get a meaningful result in some way?

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Beforehand, you may check my answer to your related question at math.stackexchange.

I doubt there is a deep connection with Wick's rotation. The main purpose behind the imaginary prefactor is mathematical convenience. Indeed, the standard shift operator $e^{a\partial_x}$ is anti-unitary (for a real $a$) because the derivative $\partial_x$ itself is skew-hermitian, while $i\partial_x$ is a hermitian operator, which makes $e^{ia\partial_x}$ unitary.

Of course, the initially real-valued wavefunction becomes complex-valued, because $e^{ia\partial}\psi(x) = \psi(x+ia)$, but it is not a problem in quantum mechanics, since the only meaningful quantity $-$ because it embodies the measured physical observables of the system $-$ turns out to be the probability distribution represented by $|\psi(x+ia)|^2$ after translation, which is indeed real thanks to the modulus.

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