- #1
LAHLH
- 409
- 1
Hi,
In QM symmetries can be represented by unitary operators. For example for rotations: [tex] \hat{U}_{R}\psi(\vec{x})=\psi(R^{-1}\vec{x}) [/tex], which is simple enough, as it just says that the vale of the rotated wavefunction at some point is the value of the old wavefunction at the pre-rotated point. Similary for a translation we could have [tex] \hat{T}(a)\psi(x)=\psi(x-a) [/tex] and so on. This much I understand.
Also we have similarity that is to say any representation D of a group is equivalent to another representation [tex] D'=SDS^{-1} [/tex] where S is just a non singular matrix that relates the basis vectors [tex] \vec{e}_i=S_{ji}\vec{f}_j [/tex] say. So this kind of equivalence is just looking at the operator in another basis effectively.
Now when we come to QFT, obviously the main symmetry of interest in Lorentz transformations and it is quite common to see written, [tex] U(\Lambda)^{-1}\Psi(x)U(\Lambda)=D(\Lambda)\Psi(\Lambda^{-1}x)[/tex]. I think what may be confusing me is that now [tex] \Psi[/tex] is not a state, it is an operator. If I'm not mistaken then, we are in the Heisenburg picture, so the states are staying fixed instead of transforming as in my first paragraph results (i.e. [tex] \hat{U}_{R}\psi(\vec{x})=\psi(R^{-1}\vec{x}) [/tex] etc), and instead these unitary operators are acting to evolve the operators themselves, as usual in Heisenburg. So this relation isn't a similarity transformation after all?
I don't understand however, why we need the D on the RHS, and why is this D not unitary?
just not quite 100% clear on what's going on here.
In QM symmetries can be represented by unitary operators. For example for rotations: [tex] \hat{U}_{R}\psi(\vec{x})=\psi(R^{-1}\vec{x}) [/tex], which is simple enough, as it just says that the vale of the rotated wavefunction at some point is the value of the old wavefunction at the pre-rotated point. Similary for a translation we could have [tex] \hat{T}(a)\psi(x)=\psi(x-a) [/tex] and so on. This much I understand.
Also we have similarity that is to say any representation D of a group is equivalent to another representation [tex] D'=SDS^{-1} [/tex] where S is just a non singular matrix that relates the basis vectors [tex] \vec{e}_i=S_{ji}\vec{f}_j [/tex] say. So this kind of equivalence is just looking at the operator in another basis effectively.
Now when we come to QFT, obviously the main symmetry of interest in Lorentz transformations and it is quite common to see written, [tex] U(\Lambda)^{-1}\Psi(x)U(\Lambda)=D(\Lambda)\Psi(\Lambda^{-1}x)[/tex]. I think what may be confusing me is that now [tex] \Psi[/tex] is not a state, it is an operator. If I'm not mistaken then, we are in the Heisenburg picture, so the states are staying fixed instead of transforming as in my first paragraph results (i.e. [tex] \hat{U}_{R}\psi(\vec{x})=\psi(R^{-1}\vec{x}) [/tex] etc), and instead these unitary operators are acting to evolve the operators themselves, as usual in Heisenburg. So this relation isn't a similarity transformation after all?
I don't understand however, why we need the D on the RHS, and why is this D not unitary?
just not quite 100% clear on what's going on here.