Two component field theory and unitary transformation

[zakk87]

I have a theory described by a 2-component field [itex]\psi_i[/itex] (i'm working with BCS in Nambu-Gorkov representation, but any other field theory would be ok, that's why I'm posting in this subforum), and the lagrangian it's defined in the following way:

[tex]\mathscr{L}=\psi^\dagger \Gamma \psi[/tex]

where [itex]\Gamma[/itex] is an opportune 2x2 matrix. Now, I know from theory that applying a unitary transformation won't change the physics of the system, but I can't always see how it's so. The simplest counter-example I came up with is this one:

[tex]\Gamma=\left( \begin{array}{cc} \partial_\mu & 0 \\ 0 & \partial_\mu \end{array} \right) [/tex]

and by imposing [itex]\det \left( \Gamma \right) = 0 [/itex] and taking the Fourier transform I can find the dispersion relation: [itex]k_\mu k^\mu = 0 \Longrightarrow \omega = \left| \mathbf{k} \right| [/itex]. But if I apply the following unitary transformation [itex]U=\left( \begin{array}{cc} e^{- \mathrm{i} \alpha \left( x \right)} & 0 \\ 0 & e^{\mathrm{i} \alpha \left( x \right)} \end{array} \right) [/itex] i get:

[tex]\Gamma' \equiv U^{-1} \Gamma U =\left( \begin{array}{cc} \partial_\mu - \mathrm{i} \partial_\mu \alpha \left( x \right) & 0 \\ 0 & \partial_\mu + \mathrm{i} \partial_\mu \alpha \left( x \right) \end{array} \right) [/tex]

and now the dispersion relation reads: [itex] k_\mu k^\mu + k^2 \alpha \left( x \right)^2 = 0 [/itex] which is quite different. Now, the dispersion relation contains lots of physical information about the system and it shouldn't have changed.

My guess is that those two dispersion relations are essentially the same one. The change is related to [itex]U[/itex] being the infinitesimal generator of translation in momenta space, and that the second dispersion relation is just the first one "Doppler shifted", i.e. written in another intertial frame of reference. Am I right? And if so, is there anyway I can show that the second dispersion relation is just the first one written in a different frame of reference?

 

[eljose79]

Thank you for sharing your theory and question with us. I can confirm that your understanding of the unitary transformation and its effect on the dispersion relation is correct. Let me explain in more detail.

In quantum field theory, a unitary transformation is a mathematical operation that changes the basis in which the fields are written. This means that the fields themselves do not change, but their representation in terms of a new basis does. In your example, the unitary transformation U changes the basis of the fields \psi_i from the original basis to a new one, which is given by the eigenvectors of U. This transformation does not affect the physics of the system, as you correctly state, since the fields themselves do not change.

However, when we apply this unitary transformation to the Lagrangian, we see a change in the dispersion relation. This is because the dispersion relation is a physical quantity that depends on the basis chosen for the fields. In your case, the second dispersion relation is indeed the first one "Doppler shifted" to a different frame of reference, as you correctly pointed out. This is because the transformation U is related to translation in momentum space, which changes the frame of reference for the system. Therefore, the two dispersion relations are essentially the same, just written in different frames of reference.

To show this mathematically, we can use the fact that the Lagrangian is invariant under unitary transformations. This means that the equations of motion derived from the Lagrangian will also be invariant. Therefore, the physical predictions of the system will not change, regardless of the chosen basis for the fields. In your case, this means that the two dispersion relations will give the same physical predictions for the system, despite looking different.

I hope this explanation helps to clarify your question. Keep up the good work in exploring and understanding quantum field theory!