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In the Sakurai book "Modern quantum mechanics" (pg. 263) an operator $S$ is said to be invariant under a unitary transformation $T$ if:

$$T^\dagger S T = S.$$

Where that definition come from?

My guess is that it comes from the similarity transformation of matrix in liner algebra:

$$M^{\prime}=T^{-1} M T.$$

So if $M= T^{-1}M T $ then I know that the matrix doesn't change when I change the basis of the vector space. If $T$ is unitary then $T^{-1} = T^{\dagger}$

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    $\begingroup$ You seem to already have answered your own question. What exactly do you expect answerers to add here? $\endgroup$
    – ACuriousMind
    Nov 17, 2021 at 12:09
  • $\begingroup$ I am self-studying so looking to confirmation/denial, more general definition or some physics concept underlying that definition. $\endgroup$
    – Andrea
    Nov 17, 2021 at 12:16

1 Answer 1

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You could also derive that expression in the following way. Let us consider a state $\rvert \Psi \rangle$. The expectation value of the operator $S$ in that state is \begin{equation} \langle\Psi\lvert S\rvert\Psi \rangle \end{equation} Now let us suppose we act with the operator $T$ on the state $\rvert \Psi \rangle$. The new expectation value is \begin{equation} \langle\Psi\lvert T^\dagger S T\rvert \Psi \rangle \end{equation} If we require that a measurement of the observable $S$ be independent of wether we perform it on the state $\rvert\Psi \rangle$ or $T\rvert\Psi \rangle$, then the two expectation values must be the same, i.e. \begin{equation} \langle\Psi\lvert S\rvert\Psi \rangle=\langle\Psi\lvert T^\dagger S T\rvert \Psi \rangle \end{equation} From which \begin{equation} T^\dagger S T = S \end{equation} If $T$ is unitary, then $T^\dagger=T^{-1}$.

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