# Anti-commutation of operators proof

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. $$\hbar=1$$

• What have you tried? Commented Mar 27, 2021 at 13:42
• 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. Commented Mar 27, 2021 at 13:55
• Do you know what the commutator of $p$ with a function of $x$ is? Maybe this could help you. Commented Mar 27, 2021 at 15:23
• I know that expression, but that is commutation involving operator $\hat{p}$, while I have $\hat{p}^2$ Commented Mar 27, 2021 at 16:08
• 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 $[p,f(x)]$. Commented Mar 27, 2021 at 16:15

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.