# QM commutators

1. May 12, 2004

### rick1138

Does anyone know of any tables that show the commutation relations of all QM opeartors? Any information would be appreciated.

2. May 13, 2004

### slyboy

You don't need a table, they can all be derived simply from

$$[x,p]=i\hbar$$

That's the beauty of physics and the thing that distinguishes it from Botany or stamp collecting.

You may need a couple more for relativistic quantum physics and spin, but since you didn't specify I am assuming you mean non-relativistic QM.

3. May 16, 2004

### Janitor

Yes indeed. There are some important commutation relations involving rotations and boosts.

4. May 16, 2004

### rick1138

I was looking for non-relativistic, just wondered if such a thing existed. The matrical rep of the Serret-Frenet formulae is very neat because it makes them very easy to remember. Now that I've looked at all the classical QM relations together at once I see that, as Slyboy stated a table is not really necessary, because they are all so similar. Thanks everyone.

5. May 16, 2004

### Janitor

But if you had been looking for relativistic commutation relations, one way to sum up the relations for generators of rotation J and generators of boost K are that they form the algebra SO(3,1), or equivalently SU(2) x SU(2).

Tell that to your friends to impress them.

6. May 17, 2004

### slyboy

but perhaps you should read a book on Lie algebras first, just in case any of them know what you are talking about.

7. May 18, 2004

### rick1138

I've read about 20 books on Lie Algebras and Superalgebras - I know more math than physics.

8. May 18, 2004

### Tom Mattson

Staff Emeritus
Well, you'd also need to know that $[x_i,x_j]=0$ and $[p_i,p_j]=0$.

Yes, and you need the commutators for spin even in nonrelativistic QM, since spin is not derived from x and p.

9. May 19, 2004

### slyboy

Yes, but you can derive the relations for orbital angular momentum and they are essentially the same.

10. May 19, 2004

### Tom Mattson

Staff Emeritus
They're exactly the same, but that's not the point. The point is that spin isn't a function of x and p, and so the commutators for the spin operators must be added to the set of "fundamental" commutators.