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.

 

[pmacias]

Conjugation in quantum field theory (QFT) is a mathematical operation that allows us to transform operators in a way that preserves the underlying physics of the system. It is a crucial tool for understanding the dynamics of quantum systems and is used extensively in QFT calculations.

One of the main reasons for conjugation in QFT is to relate different representations of the same physical system. For example, in the case of Schrodinger and Heisenberg operators, they represent different ways of describing the same quantum state. Conjugation allows us to switch between these representations and gain a better understanding of the system.

In QFT, operators are often multiplied on both the left and right with their inverses in order to perform conjugation. This is because operators in QFT do not necessarily commute with each other, meaning that the order in which they are multiplied matters. By multiplying on both sides, we ensure that the resulting operator is properly conjugated and reflects the correct physics of the system.

Additionally, conjugation is used in QFT to transform operators into different bases, which can simplify calculations and provide a deeper understanding of the system. For example, in the case of the creation and annihilation operators in quantum field theory, conjugation allows us to switch between the position and momentum bases, which can be more convenient for certain calculations.

Overall, conjugation plays a crucial role in QFT by allowing us to relate different representations of the same physical system and transform operators into different bases. It is an essential tool for understanding the dynamics of quantum systems and is essential for performing accurate calculations in QFT.