I don’t know how to apply Lorentz transformations to a wavefunction.
The wave function is a scalar in the regular QM, so it is unchanged under the Lorentz transformations. Unfortunately, the Schrodinger equation is inherently non-relativistic.
The Klein-Gordon and Dirac equations were the early attempts to “relativize” the Schrodinger equation. It didn’t work that well until the wave function was replaced with quantized fields. Those quantized fields become photons, electrons and other particles in a certain approximation. Unfortunately, the math gets quite hairy in a hurry.
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.
The wave function is a scalar in the regular QM, so it is unchanged under the Lorentz transformations. Unfortunately, the Schrodinger equation is inherently non-relativistic.
The Klein-Gordon and Dirac equations were the early attempts to “relativize” the Schrodinger equation. It didn’t work that well until the wave function was replaced with quantized fields. Those quantized fields become photons, electrons and other particles in a certain approximation. Unfortunately, the math gets quite hairy in a hurry.
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.