00:01
For the given problem, we are given a quantum mechanical operators a digger.
00:06
These are ladder operators.
00:09
We are asked to find what will be the commutation relation.
00:12
We have to prove that x, the position and momentum operators, they commute to form to get i times h cut.
00:25
So we have to prove this.
00:27
In order to do that, we start by adding operating.
00:32
Operator a, a dagger.
00:36
When we add them, we get the result that is a two times of square root of mass times omega divided by two h cut times position operator.
00:52
Then if we subtract these two operators a minus a dagger, we get a 2 i divided by 2m omega h cut operator p for the momentum.
01:09
We simply wrote the position and momentum operator in terms of a ladder operators.
01:15
So we'll write x position operator will be then a plus a dagger divided by two times of m mass omega divided by two each cut where our position momentum operator will be a minus a dagger divided by two i times square a root of 2m omega h cut.
01:45
The next step we will use these position and momentum operators to find a commutation relation that is if we take the commutation of position and momentum we know this is equal to x operator times a position operator minus position operator times momentum operators times position operator sorry so we'll find and this part and this part, then we'll subtract them to see if we are getting i times h cut.
02:22
So first we will find h x operator times a p operator.
02:30
We see the result we get here is a square root of 2m omega h cut divided by 4 pi, 4 i sorry, times a square root of m times omega divided by two edge cut and there is two position operators so the letter operators they multiply a plus a dagger multiplies with a minus a dagger they multiply to form a square so let's call this let's keep it this one here let's multiply these two letter operators uh a square minus a a dagger plus a dagger a minus a dagger square.
03:23
We know a square, operator square, is equal to its conjugate, a dagger square, which is equal to 1.
03:34
And we are also given a property that says the commutation of a and a dagger is equal to 1...