# Difference on the invariance of operators and their transformations under unitary operators

I am confused about, what I believe, refers to passive and active transformations in QM. What I have understood so far is that the matrix elements $$\langle \psi| \hat{H}|\phi\rangle$$ should remain unchanged under transformations; this implies that $$\hat{H} = \hat{U}^\dagger \hat{H} \hat{U}$$.

On the other hand, under certain transformation $$\hat{U}$$, an arbitrary operator $$\hat{\Omega}$$ transforms as $$\hat{U} \hat{\Omega} \hat{U}^\dagger$$.

1) Why should the matrix elements of the hamiltonian remain unchanged under a transformation? Is that the same as saying: "The evolution of the state is governed by the hamiltonian; under a transformation the system should still evolve in the same way, hence the matrix elements should be conserverd"?

2) What does $$\hat{H} = \hat{U}^\dagger \hat{H} \hat{U}$$ mean? Is that talking about how the hamiltonian transforms? Or is it only a condition on the transformation $$\hat{U}$$?

3) What is $$\hat{\Omega} \to \hat{U} \hat{\Omega} \hat{U}^\dagger$$ opposed $$\hat{U}|\phi\rangle$$. Does an active transformation mean applying $$\hat{U}$$ to every state, while a passive transformation is leaving vectors as they are and transforming the opeartors as $$\hat{\Omega} \to \hat{U} \hat{\Omega} \hat{U}^\dagger$$? Or does applying a transformation mean doing both of the above?

$$\langle \psi | \hat{H} | \phi \rangle$$ should remain unchanged under transformations.

This is certainly not true in general for an arbitrary transformation. I think perhaps what you meant is that $$\langle \psi | \hat{H} | \phi \rangle$$ should remain unchanged under a basis transformation.

This implies that $$H = U H U^\dagger$$.

This is not true. What you are asserting here is that $$[H, U]=0$$, which has no reason to be true without additional assumptions.

I think your confusion lies in what actually constitutes a unitary basis transformation. If we go to a different basis, the transformation prescription is

\begin{align*} &|\psi\rangle \mapsto U|\psi \rangle\\ &\langle \psi | \mapsto\langle \psi | U^\dagger\\ & \hat{H} \mapsto U\hat{H}U^\dagger. \end{align*}

Plug these transformations into $$\langle \psi | \hat{H} | \phi \rangle$$ and you can clearly see that that it is invariant under a basis transformation.

• > This implies $\hat{H} = U^\dagger H U$. Note the difference conjugate to the left, instead of the right. Does that change anything? From what I was told $\hat{H} = U^\dagger H U$ refers to the invariance of the Hamiltonian, not to how it transforms – Daniel Duque Mar 16 '19 at 13:21
• Would the left hand side $H$ refer to the non-transformed Hamiltonian, and the right hand side $H$ refer to the transformed Hamiltonian? In that case $H = U^\dagger U H U^\dagger U$? – Daniel Duque Mar 16 '19 at 13:26
• The left dagger is just a matter of which one you want to call the inverse, it makes no difference since they’re each daggers of one another – InertialObserver Mar 16 '19 at 17:23