- #36
Rorshach
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- 0
Then it is [itex]\hat{x}[/itex], but I don't know its formula, or if it even has one besides [itex]\hat{x}[/itex]
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Unfortunately, you haven't understood your mistake. You're still doing the same thing you did before. It doesn't work at all like (A+B)CD = (AD + BD)CD.Rorshach said:Gosh, I finally understood my mistake:P I just treated the whole thing as simple standard multiplication, just like I should do in normal mathematical case, however in case of operators I should just multiply every operator by the variable on its right side:D In normal situation it seems illogical, looking like this: ##(A+B)CD=(AD+BD)CD##, but in case of operators it should look like this ##(-i\hbar\frac{\partial }{\partial x}x+i\hbar\frac{\partial}{\partial x})(-i\hbar\frac{\partial }{\partial x})\psi=(-i\hbar(x\frac{\partial \psi}{\partial x}+\psi\frac{\partial x}{\partial x})+i\hbar\frac{\partial \psi}{\partial x})(-i\hbar\frac{\partial \psi}{\partial x})##
That's not correct. Where's you get ##\frac{\partial x}{\partial x}## from?Rorshach said:##-i\hbar x\frac{\partial x}{\partial x}##
Now use the fact that ##-\hbar^2 = (-ih)^2## and rewrite the final result in terms of ##\hat{p}_x## to show that ##[\hat{p}_x^2,\hat{x}] = (-i\hbar)2\hat{p}_x##.Rorshach said:I think I finally did it. Here it goes:
$$-i\hbar\frac{\partial }{\partial x}(-i\hbar\frac{\partial x\psi}{\partial x}+i\hbar x\frac{\partial \psi}{\partial x})+(-i\hbar\frac{\partial xf}{\partial x}+i\hbar x\frac{\partial f}{\partial x}),$$ where we let ##f=-i\hbar\frac{\partial \psi}{\partial x}##. Then it goes on:
$$-i\hbar\frac{\partial }{\partial x}(-i\hbar(x\frac{\partial \psi}{\partial x}+\psi\frac{\partial x}{\partial x})+i\hbar x\frac{\partial \psi}{\partial x})+(-i\hbar(x\frac{\partial f}{\partial x}+f\frac{\partial x}{\partial x})+i\hbar x\frac{\partial f}{\partial x})$$
$$-i\hbar\frac{\partial }{\partial x}(-i\hbar x\frac{\partial \psi}{\partial x}-i\hbar\psi+i\hbar x\frac{\partial \psi}{\partial x})+({\color{red}-}i\hbar x\frac{\partial f}{\partial x}-i\hbar f+i\hbar x\frac{\partial f}{\partial x})$$
$$-\hbar^2\frac{\partial \psi}{\partial x}+(-i\hbar f)
= \hbar^2\frac{\partial \psi}{\partial x}-{\color{red} {(i\hbar)}}(-i\hbar\frac{\partial \psi}{\partial x})
= -\hbar^2\frac{\partial \psi}{\partial x}-\hbar^2\frac{\partial \psi}{\partial x}
= -2\hbar^{\color{red}2} \frac{\partial \psi}{\partial x}$$.
Please tell me that this time I am right.
A commutator is a mathematical operation that involves taking the difference between two quantities and multiplying it by a third quantity. In the context of solving with Jacobian, commutators are used to calculate the rate of change of a function with respect to its variables.
Commutators are useful in solving with Jacobian because they allow us to express the rate of change of a function in terms of its partial derivatives. This makes it easier to solve systems of equations involving multiple variables.
Yes, commutators can be used to solve a wide range of problems involving Jacobian, including optimization, differential equations, and physics problems.
One limitation of using commutators is that they can only be used for differentiable functions. Additionally, they may not always provide the most efficient or accurate solution to a problem.
There are many resources available online that provide detailed explanations and examples of using commutators in solving with Jacobian. You can also consult textbooks or seek guidance from a mathematics tutor or professor.