Noether's theorem in matrix form

[Heirot]

Homework Statement

Consider a quantum mechanical system described by the Lagrangian: [tex]L=Tr[\dot{U}^{\dagger}\dot{U}]=\sum_{a,b=1}^2{\dot{U}^{\dagger}_{ab}\dot{U}_{ba}}[/tex], where U is a 2x2 special unitary matrix.

Show that the Lagrangian is invariant under the following symmetry transformation: [tex]U\to LU[/tex], where L is a time independent 2x2 unitary matrix.

Use the Noether procedure to find the associated conserved quantity: [tex](J_L)_{ab}=-i(U\dot{U}^{\dagger})_{ab}[/tex].

Homework Equations

Variation of Lagrangian:

[tex]\delta L = \frac{d}{dt} (P_{ab}\delta U_{ab} + P^*_{ab}\delta U^*_{ab})[/tex]

where P_ab and P*_ab are canonical momenta for the independent variables U_ab and U*_ab, and summation over a and b is implied.

The Attempt at a Solution

From the form of the Lagrangian it is obvious that the transformation U -> LU does nothing to in. Therefore [tex]\delta L = 0[/tex] and the quantity under the derivation sign is a constant - Noether charge. I have a problem reducing it to the given form. What should I use for [tex]\delta U_{ab}[/tex] in this case?

Thanks!

 

[mark726]

Thank you for your post. Your approach to the problem is correct. To find the Noether charge associated with the given symmetry transformation, we will use the variation of the Lagrangian equation you have provided. In this case, the variation of U will be given by \delta U = L\delta U. Substituting this into the equation and simplifying, we get:

\delta L = \frac{d}{dt} \left( \sum_{a,b=1}^2{P_{ab}L_{ab}\delta U_{ab} + P^*_{ab}L^*_{ab}\delta U^*_{ab}} \right)

Since \delta L = 0, the quantity under the derivative must be a constant, which we can define as the Noether charge (J_L)_{ab}. Simplifying further, we get:

(J_L)_{ab} = P_{ab}L_{ab} + P^*_{ab}L^*_{ab}

Using the canonical momenta for U and U*, we can write this as:

(J_L)_{ab} = -i(U\dot{U}^{\dagger})_{ab}

Therefore, (J_L)_{ab} is a conserved quantity for the given Lagrangian under the symmetry transformation U -> LU. I hope this helps. Let me know if you have any further questions.