Calculating Commutator of Position and Momentum: Troubleshooting Tips

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In summary, the problem is that the commutator is wrong because anti-hermitian matrices are not the case here.
  • #1
antibrane
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I am attempting to calculate the commutator [itex][\hat{X}^2,\hat{P}^2][/itex] where [itex]\hat{X}[/itex] is position and [itex]\hat{P}[/itex] is momentum and am running into the following problem. The calculation goes as follows,

[tex]
[\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)
[/tex]

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

[tex]
[\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
[/tex]

which is wrong because I know from a theorem that if [itex]\hat{A}[/itex] is Hermitian and [itex]\hat{B}[/itex] is Hermitian then [itex][\hat{A},\hat{B}][/itex] 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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  • #2
16180339887 said:
[tex]
[\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
[/tex]

which is wrong because I know from a theorem that if [itex]\hat{A}[/itex] is Hermitian and [itex]\hat{B}[/itex] is Hermitian then [itex][\hat{A},\hat{B}][/itex] 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 ?
 
  • #3
Doesn't [itex]2\hbar^2[/itex] destroy the anti-hermicity, since

[tex]\left(2\hbar^2\right)^{\dagger}\neq -2\hbar^2[/tex]
 
  • #4
16180339887 said:
Doesn't [itex]2\hbar^2[/itex] destroy the anti-hermicity, since

[tex]\left(2\hbar^2\right)^{\dagger}\neq -2\hbar^2[/tex]

[tex]
(4i\hbar XP + 2\hbar^2)^\dagger ~=~ -4i\hbar PX + 2\hbar^2
~=~ -4i\hbar(XP - i\hbar) + 2\hbar^2 ~=~ -4i\hbar XP - 4\hbar^2 + 2\hbar^2
~=~ -(4i\hbar XP + 2\hbar^2)
[/tex]
 
  • #5
Use the rule:

[tex]
[\hat{A}^{2}, \hat{B}^{2}] = [\hat{A}^{2}, \hat{B}] \, \hat{B} + \hat{B} \, [\hat{A}^{2} \, \hat{B}]
[/tex]

and

[tex]
[\hat{A}^{2}, \hat{B}] = \hat{A} \, [\hat{A}, \hat{B}] + [\hat{A}, \hat{B}] \, \hat{A}
[/tex]
 
  • #6
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.
 

FAQ: Calculating Commutator of Position and Momentum: Troubleshooting Tips

What is the commutator of [x^2,p^2]?

The commutator of [x^2,p^2] is a mathematical expression that describes the relationship between the operators x^2 and p^2. It is written as [x^2,p^2] = x^2p^2 - p^2x^2. This commutator is used in quantum mechanics to calculate the uncertainty in the position and momentum of a particle.

How do you calculate the commutator of [x^2,p^2]?

The commutator of [x^2,p^2] is calculated by subtracting the product of the operators x^2 and p^2 in one order from the product in the opposite order. This can be written as [x^2,p^2] = x^2p^2 - p^2x^2.

What does the commutator of [x^2,p^2] tell us about a particle's position and momentum?

The commutator of [x^2,p^2] is related to the Heisenberg uncertainty principle, which states that the more precisely we know a particle's position, the less precisely we know its momentum, and vice versa. The commutator helps us calculate the minimum uncertainty in the position and momentum of a particle.

Is the commutator of [x^2,p^2] always zero?

No, the commutator of [x^2,p^2] is not always zero. In fact, it is only equal to zero when the operators x^2 and p^2 commute, meaning they can be rearranged without changing the result. In general, the commutator of two non-commuting operators will be non-zero.

How is the commutator of [x^2,p^2] related to the energy of a particle?

The commutator of [x^2,p^2] is related to the energy of a particle through the Heisenberg uncertainty principle. As the uncertainty in the position and momentum of a particle decreases, the uncertainty in its energy increases. This is because the operators x^2 and p^2 are related to the energy operator, and the commutator helps us calculate the minimum uncertainty in the energy of a particle.

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