I How to define expectation value in relativistic quantum mechanics?

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In relativistic quantum mechanics, the expectation value of an operator is defined differently than in non-relativistic quantum mechanics due to the altered scalar product. The standard expectation value formula is modified to account for the time dependence of the wave function. For operators that do not depend on position, the expectation value can be expressed as an integral involving the time derivative of the wave function. In more general cases, the expectation value incorporates the spatial dependence of the operator. This approach ensures consistency with the principles of relativistic quantum theory.
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How to define expectation value in relativistic quantum mechanics?
In non relativistic quantum mechanics, the expectation value of an operator ##\hat{O}## in state ##\psi## is defined as $$<\psi |\hat{O}|\psi>=\int\psi^* \hat{O} \psi dx$$.
Since the scalar product in relativistic quantum has been altered into $$|\psi|^2=i\int\left(\psi^*\frac{\partial \psi}{\partial t}-\frac{\partial \psi^*}{\partial t}\psi\right)dx$$
how do we define expectation value of an operator ##\hat{O}## in state ##\psi##?
 
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Go to the momentum space (via Fourier transform) and then define scalar product, probability and expectation value as in "ordinary" QM.
 
Foracle said:
Since the scalar product in relativistic quantum has been altered into $$|\psi|^2=i\int\left(\psi^*\frac{\partial \psi}{\partial t}-\frac{\partial \psi^*}{\partial t}\psi\right)dx$$
how do we define expectation value of an operator ##\hat{O}## in state ##\psi##?
$$\langle \hat O\rangle=i\int\left(\psi^*\hat O\frac{\partial \psi}{\partial t}-\frac{\partial \psi^*}{\partial t}\hat O\psi\right)dx.$$
works if ##O## does not depend on ##x##. In general,
$$\langle \hat O(x)\rangle=i\int\left(\psi^*\frac{\partial \hat O(x)\psi}{\partial t}-\frac{\partial \psi^*}{\partial t}\hat O(x)\psi\right)dx.$$
 
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