Calculating Commutator of Position and Momentum: Troubleshooting Tips

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The discussion focuses on calculating the commutator [\hat{X}^2,\hat{P}^2] for position and momentum operators, revealing an error in the initial calculation. The user initially arrives at a result that contradicts the theorem stating that the commutator of two Hermitian operators should be anti-Hermitian. A suggestion is made to check the order of the operators, as this affects the anti-Hermitian property. The user acknowledges the oversight regarding the order of X and P, indicating a resolution to the problem. The conversation emphasizes the importance of operator order in quantum mechanics calculations.
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I am attempting to calculate the commutator [\hat{X}^2,\hat{P}^2] where \hat{X} is position and \hat{P} is momentum and am running into the following problem. The calculation goes as follows,

<br /> [\hat{X}^2,\hat{P}^2]=-\left(\underbrace{[\hat{P}^2,\hat{X}]}_{-2i\hbar\hat{P}}\hat{X}+\hat{X}\underbrace{[\hat{P}^2,\hat{X}]}_{-2i\hbar\hat{P}}\right)=2i\hbar\left(\hat{P}\hat{X}+\hat{X}\hat{P}\right)<br />

and using that [\hat{X},\hat{P}]=i\hbar we find that

<br /> [\hat{X}^2,\hat{P}^2]=2i\hbar\left[\left(\hat{X}\hat{P}-i\hbar\right)+\hat{X}\hat{P}\right]=4i\hbar\hat{X}\hat{P}+2\hbar^2<br />

which is wrong because I know from a theorem that if \hat{A} is Hermitian and \hat{B} is Hermitian then [\hat{A},\hat{B}] is anti-Hermitian, which is definitely not the case here. What am I doing wrong?

Thanks in advance for any help.
 
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16180339887 said:
<br /> [\hat{X}^2,\hat{P}^2]=2i\hbar\left[\left(\hat{X}\hat{P}-i\hbar\right)+\hat{X}\hat{P}\right]=4i\hbar\hat{X}\hat{P}+2\hbar^2<br />

which is wrong because I know from a theorem that if \hat{A} is Hermitian and \hat{B} is Hermitian then [\hat{A},\hat{B}] is anti-Hermitian, which is definitely not the case here. What am I doing wrong?

Maybe, in the last term, when checking whether it's anti-Hermitian, are you
forgetting to swap P and X ?
 
Doesn't 2\hbar^2 destroy the anti-hermicity, since

\left(2\hbar^2\right)^{\dagger}\neq -2\hbar^2
 
16180339887 said:
Doesn't 2\hbar^2 destroy the anti-hermicity, since

\left(2\hbar^2\right)^{\dagger}\neq -2\hbar^2

<br /> (4i\hbar XP + 2\hbar^2)^\dagger ~=~ -4i\hbar PX + 2\hbar^2<br /> ~=~ -4i\hbar(XP - i\hbar) + 2\hbar^2 ~=~ -4i\hbar XP - 4\hbar^2 + 2\hbar^2<br /> ~=~ -(4i\hbar XP + 2\hbar^2)<br />
 
Use the rule:

<br /> [\hat{A}^{2}, \hat{B}^{2}] = [\hat{A}^{2}, \hat{B}] \, \hat{B} + \hat{B} \, [\hat{A}^{2} \, \hat{B}]<br />

and

<br /> [\hat{A}^{2}, \hat{B}] = \hat{A} \, [\hat{A}, \hat{B}] + [\hat{A}, \hat{B}] \, \hat{A}<br />
 
oh you are right i was forgetting to change the order of X and P. Thanks. Thanks for your comment as well Dickfore I will use that.
 
Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?

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