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Why is it assumed that the symmetry transformations change either the state $\psi$ or the operator $O$ but not both simultaneously? For example, if you assume $\psi\to T\psi$, you take $O\to O$ and if you assume $\psi\to \psi$, you take $O\to T^{-1}OT$. What is the logic here? This bit is not explained in the textbooks I am using (Griffiths and Shankar).

Qmechanic
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Solidification
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  • Does this answer your question? Symmetry transformations of states and operators – Tobias Fünke Jan 23 '23 at 07:40
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    I don't think it does. I think matrix elements as well as expectation values do change under a transformation. For example, under parity, the expectation value of the position operator changes as $\langle X\rangle\to \langle X^\prime\rangle=-\langle X\rangle$. So any argument based on the "expectation values don't change under a transformation" is not clear to me. – Solidification Jan 23 '23 at 07:49
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    I don't understand. Can you please give a definition of symmetry you have in mind? – Tobias Fünke Jan 23 '23 at 07:50
  • If you mean the notion of symmetry in Wigner's theorem, then we require $|\langle \psi,\phi\rangle|=|\langle \psi^\prime,\phi^\prime\rangle|$ and (roughly) find that all such transformations are either unitary or anti-unitary. Say in the unitary case, e.g. translations, we thus have $\psi^\prime =U \psi$. And yes, obviously the expectation value of $X$ will change in this new states compared to $\psi$- but this is nothing surprising, I'd say. If you shift your wave function which is centered at $0$ to $+a$, then the translated wavefunction is centered at $a$. – Tobias Fünke Jan 23 '23 at 07:53
  • Under translation. the expectation value of position operator changes as $\langle X\rangle\to \langle X^\prime\rangle=\langle X\rangle+a$. Therefore, $ \langle X^\prime\rangle\neq \langle X\rangle$. You definition of symmetry does not say that expectation value of X or any other operator O does not change. It only says that the symmetry operators are unitary or antiunitary. This was my point in the answer provided by ɪdɪət strəʊlə – Solidification Jan 23 '23 at 08:15
  • I mean, I disagree with the answer provided by ɪdɪət strəʊlə – Solidification Jan 23 '23 at 08:19
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    Okay, I see. But why you are asking about a concept without a more or less clear definition of the concept? Here on PSE are many questions regarding the various definitions of symmetry etc. in quantum mechanics. Put differently, for you and the people who might answer your question, it might help a lot to cite a clear definition of symmetry you mean and a passage where it is said that either only states or only operators are transformed. – Tobias Fünke Jan 23 '23 at 08:51
  • @TobiasFünke Instead of defining the concept of a transformation, allow me to put it like this. Suppose I make a transformation (e.g., translation, rotation, parity, etc). I want to know why should it affect either the states or the operators but not both simultaneously. Why? The existing answer argues as follows: Since expectation values are unchanged it suffices to consider that only states change or only operators change but not both. But I have clearly two clear examples where expectation value does change under a transformation. (References: R. Shankar, J.J. Sakurai). – Solidification Jan 23 '23 at 11:16
  • Sorry, I really don't understand. If you would read my comments and the answer I have linked carefully, I guess, it would answer all your questions. I cannot say really more or any better than the answers in the link. – Tobias Fünke Jan 23 '23 at 11:22
  • @TobiasFünke Which answer, in particular, do you think is correct and want me to look? – Solidification Jan 23 '23 at 11:27
  • The one with the most upvotes (at this time). Additionally, as I've said, there are several posts here on PSE discussing the different notions of symmetry in quantum mechanics. And again: It would make everything much easier if you would give a definition of the kind of symmetry you are considering. Or, at least, edit the question and tell us why you think one should transform both the operators and states. Your question is not really clear at all. – Tobias Fünke Jan 23 '23 at 11:30

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You have exactly the same situation with time evolution. In Schrödinger picture of quantum mechanics, the state of the system is assumed to evolve according to $$i\hbar{d\over dt}|\psi(t)\rangle=\hat H|\psi(t)\rangle \ \Leftrightarrow\ |\psi(t)\rangle=e^{-i\hat Ht/\hbar}|\psi(0)\rangle$$ while operators $\hat O$ are time-independent. Averages evolve according to $$\langle O(t)\rangle=\langle\psi(t)|\hat O|\psi(t)\rangle =\langle\psi(0)|e^{i\hat Ht/\hbar}\hat Oe^{-i\hat Ht/\hbar}|\psi(0)\rangle$$ You see that you could obtain the same expression of $\langle O(t)\rangle$ by assuming that the state is time-independent, i.e. $|\psi(t)\rangle=|\psi(0)\rangle$, but the operators evolve according to $$\hat O(t)=e^{i\hat Ht/\hbar}\hat Oe^{-i\hat Ht/\hbar} \ \Leftrightarrow\ i\hbar{d\over dt}\hat O(t)=[\hat H,\hat O(t)]$$ We have recovered Heisenberg picture of QM.

Time evolution is a special case of a transformation that can be applied to a quantum state. The general case is obtained by replacing $e^{-i\hat Ht/\hbar}$ by $\hat T$. Either you consider that the state is affected by the transformation as $|{\psi'}\rangle=\hat T|\psi\rangle$ and you assume that operators are not or you consider that the state is invariant but then operators should transform as $\hat O'=\hat T^+\hat O\hat T$ in such a way that the expression of averages $$\langle O'\rangle=\langle\psi'|\hat O|\psi'\rangle =\langle\psi|\hat O'|\psi\rangle =\langle\psi|\hat T^+\hat O\hat T|\psi\rangle$$ is recovered. Note that the average $\langle O'\rangle$ after transformation is not equal to $\langle O\rangle$. We just assumed that the two pictures give the same answer.

Christophe
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