Quantum Mechanics Operator Commutation Relations

[rick1138]

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

 

[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.

 

[Janitor]

You may need a couple more for relativistic quantum physics and spin- Slyboy

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

 

[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.

 

[Janitor]

I can't resist adding-

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.

 

[slyboy]

Tell that to your friends to impress them.

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

 

[rick1138]

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

 

[quantumdude]

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

[x,p]=i\hbar

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

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.

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

 

[slyboy]

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

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

 

[quantumdude]

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

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.