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I have a question about equivalence principle in quantum mechanics.

Consider a Schroedinger equation under gravitional field $$\left[ - \frac{1}{2m_I} \nabla^2 + m_g \Phi_{\mathrm{grav}} \right]\psi = i \partial_t \psi \tag{1} $$

where $m_I$ and $m_g$ are the inertia and gravitational masses, respectively. $\hbar=1$ unit is adopted.

To the contrary as the classical mechanics $$ m_I \frac{ d^2 x}{ dt^2} = m_g g \tag{2}$$ we can choose a transformation $x'=x-\frac{1}{2} g t^2$ to "switch off" the gravity. But it seems the transformation will not switch off gravity in quantum mechanics, Eq. (1). Does it mean quantum mechanics break the equivalence principle? (I can think about relativistic Hamiltonian, but it will not resolve the problem, as far as I can see)

I have a question about equivalence principle in quantum mechanics.

Consider a Schroedinger equation under gravitional field $$\left[ - \frac{1}{2m_I} \nabla^2 + m_g \Phi_{\mathrm{grav}} \right]\psi = i \partial_t \psi \tag{1} $$

where $m_I$ and $m_g$ are the inertia and gravitational masses, respectively. $\hbar=1$ unit is adopted.

To the contrary as the classical mechanics $$ m_I \frac{ d^2 x}{ dt^2} = m_g g \tag{2}$$ we can choose a transformation $x'=x-\frac{1}{2} g t^2$ to "switch off" the gravity. But it seems the transformation will not switch off gravity in quantum mechanics, Eq. (1). Does it mean quantum mechanics break the equivalence principle? (I can think about relativistic Hamiltonian, but it will not resolve the problem, as far as I can see.)

I have a question about equivalence principle in quantum mechanics.

Consider a Schroedinger equation under gravitional field $$\left[ - \frac{1}{2m_I} \nabla^2 + m_g \Phi_{\mathrm{grav}} \right]\psi = i \partial_t \psi \tag{1} $$

where $m_I$ and $m_g$ are the inertia and gravitational masses, respectively. $\hbar=1$ unit is adopted.

To the contrary as the classical mechanics $$ m_I \frac{ \partial^2 x}{ \partial t^2} = m_g g \tag{2}$$ we can choose a transformation $x'=x-\frac{1}{2} g t^2$ to "switch off" the gravity. But it seems the transformation will not switch off gravity in quantum mechanics, Eq. (1). Does it mean quantum mechanics break the equivalence principle? (I can think about relativistic Hamiltonian, but it will not resolve the problem, as far as I can see)

I have a question about equivalence principle in quantum mechanics.

Consider a Schroedinger equation under gravitional field $$\left[ - \frac{1}{2m_I} \nabla^2 + m_g \Phi_{\mathrm{grav}} \right]\psi = i \partial_t \psi \tag{1} $$

where $m_I$ and $m_g$ are the inertia and gravitational masses, respectively. $\hbar=1$ unit is adopted.

To the contrary as the classical mechanics $$ m_I \frac{ d^2 x}{ dt^2} = m_g g \tag{2}$$ we can choose a transformation $x'=x-\frac{1}{2} g t^2$ to "switch off" the gravity. But it seems the transformation will not switch off gravity in quantum mechanics, Eq. (1). Does it mean quantum mechanics break the equivalence principle? (I can think about relativistic Hamiltonian, but it will not resolve the problem, as far as I can see)

I have a question about equivalence principle in quantum mechanics.

Consider a Schroedinger equation under gravitional field $$\left[ - \frac{1}{2m_I} \nabla^2 + m_g \Phi_{\mathrm{grav}} \right]\psi = i \partial_t \psi \tag{1} $$

where $m_I$ and $m_g$ are the inertia and gravitational masses, respectively. $\hbar=1$ unit is adopted.

To the contrary as the classical mechanics $$ m_I \frac{ \partial^2 x}{ \partial t^2} = m_g g \tag{2}$$ we can choose a transformation $x \rightarrow x'=x-\frac{1}{2} g t^2$ to "switch off" the gravity. But it seems the transformation will not switch off gravity in quantum mechanics, Eq. (1). Does it mean quantum mechanics break the equivalence principle? (I can think about relativistic Hamiltonian, but it will not resolve the problem, as far as I can see)

I have a question about equivalence principle in quantum mechanics.

Consider a Schroedinger equation under gravitional field $$\left[ - \frac{1}{2m_I} \nabla^2 + m_g \Phi_{\mathrm{grav}} \right]\psi = i \partial_t \psi \tag{1} $$

where $m_I$ and $m_g$ are the inertia and gravitational masses, respectively. $\hbar=1$ unit is adopted.

To the contrary as the classical mechanics $$ m_I \frac{ \partial^2 x}{ \partial t^2} = m_g g \tag{2}$$ we can choose a transformation $x'=x-\frac{1}{2} g t^2$ to "switch off" the gravity. But it seems the transformation will not switch off gravity in quantum mechanics, Eq. (1). Does it mean quantum mechanics break the equivalence principle? (I can think about relativistic Hamiltonian, but it will not resolve the problem, as far as I can see)