Calculating the Commutator of R_1,R_2 in Coordinates

In summary, the conversation discusses the calculation of the commutator [R_1,R_2] and the confusion arising from evaluating differentials. The conversation also explores the meaning of expressions such as \partial_yy\partial_x and the use of the product rule in computing commutators. Ultimately, it is shown that the correct result for [R_1,R_2] is -R_3, with similar results for [R_2,R_3] and [R_3,R_1].
  • #1
Oxymoron
870
0
Could someone check if I have done this right.

[tex]R_1 = x^2\partial_3 - x^3\partial_2[/tex]
[tex]R_2 = x^3\partial_1 - x^1\partial_3[/tex]
[tex]R_3 = x^1\partial_2 - x^2\partial_1[/tex]

Where [itex]x^i[/itex] are coordinates.

I need to calculate the commutator [itex][R_1,R_2][/itex].

[tex][R_1,R_2] = x^2\partial_3x^3\partial_1 - x^3\partial_1x^2\partial_3 + x^3\partial_2x^1\partial_3 - x^1\partial_3x^3\partial_2
- (x^3\partial_2x^3\partial_1-x^3\partial_1x^3\partial_2) - (x^2\partial_3x^1\partial_3-x^1\partial_3x^2\partial_3)
[/tex]
[tex]
.\quad\quad\quad = x^2x^3\partial_3\partial_1 - x^2x^3\partial_1\partial_3 + x^1x^3\partial_2\partial_3-x^1x^3\partial_2\partial_3
- x^3x^3\partial_2\partial_1 + x^3x^3\partial_1\partial_2-x^1x^2\partial_3\partial_3+x^1x^2\partial_3\partial_3
[/tex]
[tex].\quad\quad\quad = x^2\partial_1-x^2\partial_1+x^1\partial_2-x^1\partial_2 -x^3x^3\partial_2\partial_1 + x^3x^3\partial_1\partial_2 - x^1x^2\partial_3\partial_3 + x^1x^2\partial_3\partial_3
[/tex]
[tex]
.\quad\quad\quad = 0
[/tex]

And as a result,

[tex][R_1,R_2] = [R_2,R_3] = [R_3,R_1] = 0[/tex]

by cyclically permuting the indices.
 
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  • #2
I think you're wrong...Check the angular momentum so(3) algebra...

Daniel.
 
  • #3
You're right. But I don't know where I went wrong. I would have expected

[tex][R_1,R_2] = R_3[/itex]
 
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  • #4
I think my problem is evaluating the differentials. Here is an example of what I am doing in a simpler problem.

let [itex]A=\partial_x[/itex], [itex]B=\partial_y[/itex], and [itex]C=x\partial_y-y\partial_x[/itex].

Im going to compute [itex][A,C][/itex] and let me know if I've done something wrong...

[tex][A,C] = [\partial_x,x\partial_y-y\partial_x][/tex]
[tex]= \partial_x(x\partial_y - y\partial_x)-(x\partial_y-y\partial_x)\partial_x[/tex]
[tex]= \partial_xx\partial_y-\partial_xy\partial_x - x\partial_y\partial_x+y\partial_x\partial_x[/tex]

Now I am pretty sure [itex]\partial_xx = 1[/itex] so the first term is [itex]\partial_y[/itex].

The second term [itex]\partial_xy\partial_x = \partial_x\partial_xy[/itex] since the partial derivatives commute. The second term equals the fourth term so they cancel.

This leaves the third term. [itex]x\partial_y\partial_x=\partial_y(\partial_xx) = \partial_y(1) = 0[/itex]

So I would say the answer is

[tex][A,C] = \partial_y = B[/itex]

how does this look?
 
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  • #5
To be honest I don't think I've done this right at all.

My basic question is what does

[tex]\partial_yy\partial_x[/tex]

equal? And is it the same thing as

[tex]y\partial_x\partial_y[/tex]

EDIT: Could

[tex]\partial_yy\partial_x = \partial_y(y)\partial_x + y\partial_x\partial_y[/tex]?

If this calculation is right could someone tell me how.
 
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  • #6
Oxymoron said:
To be honest I don't think I've done this right at all.

My basic question is what does

[tex]\partial_yy\partial_x[/tex]

equal? And is it the same thing as

[tex]y\partial_x\partial_y[/tex]

EDIT: Could

[tex]\partial_yy\partial_x = \partial_y(y)\partial_x + y\partial_x\partial_y[/tex]?

If this calculation is right could someone tell me how.

What [tex]\partial_yy\partial_x[/tex] means depends on how you are applying the [tex]\partial_y[/tex]. From a commutator, for example:
[tex] [\partial_y, y\partial_x] = \partial_y(y\partial_x)-y\partial_x(\partial_y)[/tex]
In this case the y operator acts on the whole expression so you use the product rule.

-Dan
 
  • #7
Ah, of course! So in fact

[tex][R_1,R_2] = -R_3[/tex]

and...

[tex][R_2,R_3] = -R_1[/tex]
[tex][R_3,R_1] = -R_2[/tex]
 
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1. What is the commutator of R1 and R2 in coordinates?

The commutator of two operators R1 and R2 in coordinates is a mathematical operation that measures how much the order of the operators affects the final result when applied to a function. It is defined as [R1, R2] = R1R2 - R2R1.

2. How do you calculate the commutator of R1 and R2 in coordinates?

To calculate the commutator of R1 and R2 in coordinates, you can use the above formula [R1, R2] = R1R2 - R2R1. You will need to substitute the coordinates for R1 and R2 into the formula and perform the necessary mathematical operations to find the commutator.

3. What is the significance of calculating the commutator of R1 and R2 in coordinates?

The commutator of R1 and R2 in coordinates is important because it helps to determine the compatibility of two operators. If the commutator is zero, then the operators are said to commute and can be applied in any order. If the commutator is non-zero, then the operators do not commute and the order in which they are applied matters.

4. Can the commutator of R1 and R2 in coordinates be negative?

Yes, the commutator of R1 and R2 in coordinates can be negative. The commutator can take on any real value, including negative values, depending on the specific operators and coordinates being used.

5. How is the commutator of R1 and R2 in coordinates used in quantum mechanics?

In quantum mechanics, the commutator of R1 and R2 in coordinates is used to determine the uncertainty in the measurement of two observables. The uncertainty principle states that the product of the uncertainties of two non-commuting observables must be greater than or equal to the magnitude of their commutator. This means that the commutator plays a crucial role in understanding the fundamental limits of measurement in quantum mechanics.

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