# Commutator problem

1. Oct 10, 2016

### nmsurobert

1. The problem statement, all variables and given/known data
a = √(mω/2ħ)x + i√(1/2ħmω), a = √(mω/2ħ)x - i√(1/2ħmω),

find [a,a]

the solution is given. it should be 1.
2. Relevant equations
[a,b] = ab -ba

3. The attempt at a solution
im guessing there is something I'm missing or I'm not doing something somewhere.
I'm just doing what the formula says.

(√(mω/2ħ)x + i√(1/2ħmω))(√(mω/2ħ)x - i√(1/2ħmω)) -(√(mω/2ħ)x - i√(1/2ħmω))(√(mω/2ħ)x + i√(1/2ħmω))

and from that i get zero. what i am forgetting to do?

2. Oct 10, 2016

### strangerep

I think you mis-typed the definitions of $a$ and $a^\dagger$. Shouldn't there be a $p$ (momentum) in the 2nd term of each?

3. Oct 10, 2016

### nmsurobert

i just noticed that. i copied it straight from the homework so I'm guessing the teacher made a mistake.

here i am banging my head on the table.

4. Oct 10, 2016

### TSny

Yes, the teacher left out the p in the second term of each expression. x and p are operators here. Do they commute?

5. May 15, 2017

### Sugyned

It still seems like zero to me. How can I commute those terms?

6. May 15, 2017

### TSny

Hello Sugyned, welcome to PF!

7. May 16, 2017

### Sugyned

Shouldnt poisson parenthesis be:

(dA/dQ)(dA†/dP) - (dA/dP)(dA†/dQ)
?
That would result in:
-aib-iba = -2iab

if a=√(mω/2ħ)
and b = √(1/2ħmω)

My university is protesting because the teachers weren't payed some salaries that the government owes to them, so the quantum mechanics teacher isn't giving classes, thus, we have a test coming up next tuesday.

8. May 16, 2017

### strangerep

The original question is not about Poisson brackets -- it's about quantum operators.

9. May 16, 2017

### Sugyned

It's actualy this exact same problem shown up here. The notes I have are quite confusing, I can't just simply read it and understand it.

10. May 16, 2017

### strangerep

Note that the original problem as stated in post #1 is wrong. See my post #2 where I corrected it.

With this correction,... are you familiar with the canonical commutation relations between $x$ and $p$?

From these, you should be able to calculate $[a, a^\dagger]$, using the linearity property of commutators. It's little more than a 1-liner.