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How does one prove the anti-commutation of the operators $$e^{\hat{y}} , \hat{P}_y^2 $$ where $\hat{y}$ and $\hat{P}_y$ are the standard position operator and translation generator operator in quantum mechanics, respectively. $\hbar=1$

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  • $\begingroup$ What have you tried? $\endgroup$ Mar 27, 2021 at 13:42
  • $\begingroup$ Use definition of exponential of operator, employ the 'uncertainty relation' for each term such that I get an error term for each exponential term. E.g, for the cube term of the exponential, $$\hat{P}_y^2\hat{y}^3 = \hat{y}^3\hat{P}_y^2-3i[\hat{y}^2\hat{P}_y+\hat{P}_y\hat{y}^2]$$ We know that $$[\hat{y}^2\hat{P}_y+\hat{P}_y\hat{y}^2]=2\hat{P}_y\hat{y}^2-2i\hat{y}$$ and thus bracket term in first equation is non zero, this is true for all terms in the series for the exponential. $\endgroup$
    – TheDawg
    Mar 27, 2021 at 13:55
  • $\begingroup$ Do you know what the commutator of $p$ with a function of $x$ is? Maybe this could help you. $\endgroup$ Mar 27, 2021 at 15:23
  • $\begingroup$ I know that expression, but that is commutation involving operator $\hat{p}$, while I have $\hat{p}^2$ $\endgroup$
    – TheDawg
    Mar 27, 2021 at 16:08
  • $\begingroup$ Well, there are some other useful relations regarding cases like $[AB,C]= \ldots$ Write down the commutator and then you should see how to rewrite $[p^2, f(x)]$ in terms of $[p,f(x)]$. $\endgroup$ Mar 27, 2021 at 16:15

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The exponential is the Lagrange shift operator for momenta, i.e. $$ e^{-ia \hat y} f(\hat p) e^{ia \hat y}= e^{ a \partial_p} f(\hat p) e^{-a \partial_p} = f(\hat p + a). $$ If you wish to focus on your stated particular case, $f(x)= x^2$, $a=i$, $$ e^{ \hat y} \hat p ^2 e^{ -\hat y}= (\hat p +i)^2,~~~\leadsto \\ e^{ \hat y} \hat p ^2 + \hat p ^2e^{ \hat y} =( (\hat p+i)^2+ \hat p^2 )e^{\hat y} . $$ It's hard to see why you'd need this partial result.

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  • $\begingroup$ The two operators are part of some Hamiltonian which, I want to prove, can not be diagonalized, that is, the time propagator being the exponential of the hamiltonian, sandwiched between position bra and momentum ket, can not be written as the exponential of some scalar value multiplied by the position bra momentum ket. $\endgroup$
    – TheDawg
    Mar 28, 2021 at 9:32
  • $\begingroup$ So, is this relevant? $\endgroup$ Mar 31, 2021 at 16:06

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