How to Find the Trace of ABA^-1 in Component Form?

sarrfriend
Messages
6
Reaction score
0
Let A, B be matrices with components Aμν , Bμν such that μ, ν = 0, 1, 2, 3. Indices are lowered and raised with the metric gμν and its inverse gμν. Find the trace of ABA-1 in component form?
Since A and B are generalized versions of tensors, finding their inverse becomes very tedious if we try to solve this by brute force, isn't it? Is there an easier way to find the solution?
 
Physics news on Phys.org
welcome to pf!

hi sarrfriend! welcome to pf! :smile:

hint: tr(ABC) - tr(CAB) = … ? :wink:
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top