The wave function is a scalar in the regular QM, so it is unchanged under the Lorentz transformations.
Eh? If I have a scalar field phi(x) in classical physics and I rotate the universe by pi/2 (an active transformation) the field not changes to phi(Mx) where M is the linear map that rotates the universe by pi/2 in the other direction. This changes phi, no? I know that if phi were a vector field then we would have the additional change that the vector rotates as well (i.e. we get M^(-1) v(Mx)), but the scalar field phi still in some sense changes.
If I wanted to check if my theory was invariant by rotations by pi/2 I would take a field that satisfied my equations, apply the above transformation to it, and see if it still satisfied my equations. What analogous transformation could I apply to a wavefunction to check if my theory was Lorentz invariant?
(Also, isn’t the wavefunction also a scalar in QFT?)
Definition of a scalar). In other words, if you change your coordinate system, a value of the scalar field at a given point in spacetime (now described by the new coordinates) is still the same number. Whereas a vector will, in general, have different components.
(Also, isn’t the wavefunction also a scalar in QFT?)
No. To quote wikipedia, “probability conservation is not a relativistically covariant concept”, because the particle number is neither conserved, nor is a covariant quantity. I.e., different observers can disagree on the number of particles, which violates the definition of a scalar. Thus the wavefunction (from which probability is derived) is not a useful concept in QFT and is replaced by fields living in the Fock space, not in the Hilbert space.
Eh? If I have a scalar field phi(x) in classical physics and I rotate the universe by pi/2 (an active transformation) the field not changes to phi(Mx) where M is the linear map that rotates the universe by pi/2 in the other direction. This changes phi, no? I know that if phi were a vector field then we would have the additional change that the vector rotates as well (i.e. we get M^(-1) v(Mx)), but the scalar field phi still in some sense changes.
If I wanted to check if my theory was invariant by rotations by pi/2 I would take a field that satisfied my equations, apply the above transformation to it, and see if it still satisfied my equations. What analogous transformation could I apply to a wavefunction to check if my theory was Lorentz invariant?
(Also, isn’t the wavefunction also a scalar in QFT?)
Definition of a scalar). In other words, if you change your coordinate system, a value of the scalar field at a given point in spacetime (now described by the new coordinates) is still the same number. Whereas a vector will, in general, have different components.
No. To quote wikipedia, “probability conservation is not a relativistically covariant concept”, because the particle number is neither conserved, nor is a covariant quantity. I.e., different observers can disagree on the number of particles, which violates the definition of a scalar. Thus the wavefunction (from which probability is derived) is not a useful concept in QFT and is replaced by fields living in the Fock space, not in the Hilbert space.