Why do we conjugate operators in QFT?

[gentsagree]

Why do we multiply some operator A both on the left and on the right with, say, A and A^(-1) in order to perform some kind of conjugation?

If it helps, the example I'm thinking of is the relationship between Schrodinger and Heisenberg operators in QFT.

Thanks.

 

[Joel Martis]

Hi. I'm sorry but I don't understand what you are saying. Could you elaborate on the example?

 

[The_Duck]

Suppose we have some states $| \psi \rangle$, $| \phi \rangle$ and an operator $A$. We can compute the matrix element $\langle \phi | A | \psi \rangle$. Then we can apply some transformation operator $T$ to the states to get new states $T | \psi \rangle$, $T | \phi \rangle$, and we can compute the matrix element of A between the new states, namely $\langle \phi | T^\dagger A T | \psi \rangle$. We can see that this is equivalent to computing the matrix element of the transformed operator $T^\dagger A T$ between the original states $| \psi \rangle$ and $| \phi \rangle$. So performing this sort of conjugation on operators is equivalent to performing a certain transformation on states. That's why this sort of conjugation appears so much.

 

[gentsagree]

The_Duck, thanks, great answer.