# Tag Info

## New answers tagged trace

3

Every Hermitian traceless matrix $H$ is in $\mathfrak{su}(N)$ since $\mathrm{Tr}(H) = 0$ and so $$\exp(\mathrm{tr}(\mathrm{i}H)) = \det(\exp(\mathrm{i}H)) = 1$$ so $\exp(\mathrm{i}H)$ is unitary with determinant $1$, hence in $\mathrm{SU}(N)$. The gauge field is always in the Lie algebra of the gauge group since it is introduced to cancel terms that are ...

0

I think it's easier to see this if you start from the matter representations rather than the vector field side. For example, imagine that you have some matter field $\psi$ that transforms under some simple Lie group $G$ representation according $$\psi \to g \psi$$ where $g \in G$. Now, the derivative term is not invariant if $g=g(x)$ as one has ...

1

To take the partial trace you need to build the sum over the matrix elements w.r.t. the same input and output basis, as you probably already used to calculate the partial traces you gave. In Dirac notation this is often written as: $$tr_A(L_{AB}) =\sum_i \langle i|_A L_{AB} |i\rangle_A=\langle0|0\rangle\langle 0|0\rangle ... 1 Let H_A \otimes H_B be your Hilbert space, and O be an operator acting on this composite space. Then O can be written has$$ O = \sum_{i,j} c_{ij} M_i \otimes N_j$$where the M_i's and N_j's act on H_A and H_B respectively. Then the partial trace over H_A defined as$$tr_{H_A}(O) = \sum_{i,j} c_{ij} tr(M_i) N_j , and similarly for $H_B$.

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