# QM having difficulty on proofs of operators

1. Nov 23, 2006

### kel

I know this is a simple part of Quantum Mechanics, but I seem to be having trouble with it, I'm not sure if my math is just wrong or if I'm applying it wrong.

The questions that I have are:

Prove the following for arbitrary operators A,B and C:
(hint-no tricks, just write them out in full)

i- [A,c1B+c2C] = c1[A,B] + c2[A,C]

So far I've got

A[c1B+c2C] - [c1B+c2C]A = (Ac1B+Ac2C) - (c1BA + c2CA)

giving

Ac1B+Ac2C - c1BA + c2CA = c1[AB] + c2[AC] - c1[BA] + c2[CA]

but I don't know what to do from here - something should cancel, but I think my workings may be wrong.

ii-[A,BC] = [AB]C + B[A,C]

iii-[A,[B,C]] + [B,[C,A]] + [C,[B,A]] = 0

I may be able to do this one based on number ii above, but I need some help on that one first please.

Cheers
Kel

2. Nov 23, 2006

### neutrino

Nothing cancels, just rearrange the terms. Btw, the second 'plus' should actually be a 'minus'.

3. Nov 23, 2006

### kel

ok, I've got the first one.

i- [A,c1B+c2C] = c1[A,B] + c2[A,C]
= (Ac1B+Ac2C) - (c1BA + c2CA)
= Ac1B + Ac2C - c1BA - c2CA
= c1AB - c1BA + c2AC - c2CA
= c1[AB-BA] + c2[AC-CA]
= c1[A,B] + c2[A,C]

but how, do I go about the second one??
[A,BC] = [AB]C + B[A,C]
is it?
[A,BC] = A[BC] - [BC]A

and if so (or not) where do I go from here??

Kel

4. Nov 24, 2006

### dextercioby

$$[A,BC]=ABC-BCA=ABC-BAC+BAC-BCA = \ ... \$$

Daniel.

5. Nov 24, 2006

### dextercioby

iii is written incorrectly. The third double commutator should be [C,[A,B]].

Daniel.