Really quick question: It feels somehow wrong to write it out, so just correct me: $$\langle x\rangle\langle p\rangle=\langle\psi|x|\psi\rangle\langle\psi|p|\psi\rangle =\langle\psi|xp|\psi\rangle=\langle xp\rangle.$$ The critical point probably consist of taking use of $|\psi\rangle\langle \psi| = 1$, which might not be valid.
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5$\begingroup$ Why do you think that $|\psi\rangle\langle \psi|=1$ (where I assume that with $1$ you mean the identity operator)? $\endgroup$– Tobias FünkeCommented Feb 10, 2023 at 22:02
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3$\begingroup$ This is not valid. $\endgroup$– don't train ai on meCommented Feb 10, 2023 at 22:14
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$\begingroup$ But what about the completeness relation? $\endgroup$– LeonCommented Feb 11, 2023 at 11:48
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1$\begingroup$ @Leon you should use the @ username function to notify another user. The completeness relation is something different in general and the difference is obvious if you compare both expressions... $\endgroup$– Tobias FünkeCommented Feb 11, 2023 at 15:04
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1 Answer
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Firstly, $$|\psi \rangle \langle \psi |\neq 1 \tag{1}$$ in general. The operator (1) will project a state onto the $|\psi\rangle$ state.
Another way that you can see that your manipulation is wrong is it implies that $\hat{x}$ and $\hat{p}$ commute, which we know to not be true (to see this consider the fact that $\langle x \rangle \langle p \rangle = \langle p \rangle \langle x \rangle$ and then do the same manipulations you did above).
I don't think you could simplify this general form simpler than is useful.
answered Feb 10, 2023 at 22:14