Hello, hello guys welcome back to quantum mechanics. I mean this video we're actually tackling operators, commuting and uncommuting operators. I'm actually doing a follow-up example from the previous video that I did where we're able to show two operators that actually commuted. You know I'm saying. And in this video the question is can more momentum and position be determined simultaneously? You know what I'm saying? So basically what is being asked is do these two operators which is the position operator and the momentum operator? Do they actually commute? You know what I'm saying? And if this equals to zero then yes a momentum and position can be determined simultaneously. If you can satisfy this condition then momentum and position can be determined simultaneously. Okay let's just dive right into it. Position operator, the momentum operator, it applies to G of X minus momentum operator. Okay the question is what does this equals to? Okay the position operator is just an X. Right? Then the momentum operator is minus I, H, T, over T of X being applied to G of X minus the momentum operator minus I, H, prime, T, over T of X. Right? And then the position operator is just an X. I just X multiplied by G of X. Okay cool. Continuing from here. Okay there's nothing I can really simplify further on on this side. I can just multiply everything out. So it's going to be minus I, H multiplied by X, T over T of X, in the derivative, the derivative applies to G of X minus. Okay negative multiplied by negative is a positive, right? So it's going to be a positive I, H. Then the derivative of X, Pi by G of X. Okay. Okay. This remains the same. Right? Plus I, H. Okay so looking here, this is a different differentiation in terms of X being applied to X multiplied by G of X. So we are going to use what I think you guys would recall from high school. The product rule. You know, we're going to use the product rule. So the product rule is going to be X multiplied by D over D of X multiplied by G of X plus G of X. Then the derivative of X is going to be A1, right? And then we'll continue minus I. X, D derivative, we're applying to G of X plus I, H, Pi, X. Okay. Plus I, H from G of X. Okay. Can you guys see that this and this are the same? Right? So this and this, the addition will equal us to zero. But we are not left with a zero. You know what I'm saying? We are not left with a zero. The answer that we remain with is I, try to be the Planck's constraint, but maybe G of X. So what does this say? This is not equals to zero. Hence, or therefore momentum and position. Can not be determined simultaneously. Right? So this is just an example that I wanted to just do, you know, for you guys to see that, you know, it's not always zero, right? The arc is where it's not zero, where the commutator is not zero, that is being applied. See if that is not zero. And this is one of those cases when momentum and position cannot be determined simultaneously because they are operators do not commute. You know, all right guys, stay cool, enjoy the rest of your day.