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Unitary Operators and symmetries in Quantum Mechanics

Physics Asked by newtothis on December 22, 2020

This is regarding symmetries and unitary transformations in quantum mechanics. Consider some infinitesimal continuous transformation given by $T$, where

$T = 1 – frac{iGepsilon}{hbar}$.

If this transformation conserves some arbitrary operator $A$, then we will need the following to hold:

$T^dagger{}A T = A$

If we choose $A$ as the Hamiltonian $H$ then, we will have:

$(1+ frac{iGepsilon}{hbar})H(1 – frac{iGepsilon}{hbar}) = H$, where I have considered $G$ to be Hermitian

Expanding this out and neglecting terms past the first order, I understand that if the transformation does indeed leave the Hamiltonian invariant, then $G$, the ‘generator’ and $H$ must commute.

My question is the following.

While writing down the form of $T$, a key factor that goes into determining this form, is that the transformation must be unitary (upto first order). Is it possible to have some transformation $S$, that is not unitary but still leaves the Hamiltonian invariant, ie

$S^dagger H S = H$

If so, will there be a conserved quantity under such transformations?

2 Answers

To my knowledge, the answer is no. The condition $$ S^dagger H S = H $$ comes from the necessity of the two operators to commute in order to have invariance under transformation. The actual condition is: $$ left[H, Sright] quad rightarrow quad H S = S H quad rightarrow quad S^{rm -1} H S = H. $$ When you write $S^dagger H S = H$, you already are assuming $S^dagger S = S^{rm -1} S = 1$, so your question should have negative answer.

Answered by SoterX on December 22, 2020

Within the context of Hamiltoniann mechanics in quantum theory, the relation that needs to hold is actually $$ T^{-1} H T = H$$ since you can see that it implies $[T,H]=0$ and by Ehrenfest's theorem $T$ remains constant in time and one can build a set of common eigenvectors to serve as a basis.

More generally there are transformations that are not unitary but still are suitable to define conserved quantities. One very famous example is the one of the Lorentz group, which is not compact, meaning that one can have transformations, e.g. boosts, which are not unitary. However they still commute with the Hamiltonian of the theory and serve to define mass shells.

Answered by ohneVal on December 22, 2020

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