Quantum Harmonic Oscillator Operator Commution

[TupoyVolk]

Quantum Harmonic Oscillator Operator Commution (solved)

EDIT
This was solved thanks to CompuChip! The entire post is also not very interesting as it was a basic mistake :P No need to waste time

This is not homework (I am not currently in college :P), but it is a mathematical question I'm stuck and I would greatly appreciate help.

The Quantum harmonic oscillator Operator method uses:

\widehat{a} = \sqrt{\frac{m\omega}{2\hbar}}(\widehat{x} + \frac{i\widehat{p}}{m\omega})
and
\widehat{a}^{+} = \sqrt{\frac{m\omega}{2\hbar}}(\widehat{x} - \frac{i\widehat{p}}{m\omega})

It also says that:
[\widehat{a},\widehat{a}^{+}] = 1

[\widehat{a},\widehat{a}^{+}] = \widehat{a}\widehat{a}^{+} - \widehat{a}^{+}\widehat{a}

I keep ending up with 2!

Here is a "proof"
http://quantummechanics.ucsd.edu/ph130a/130_notes/node169.html
But they have simply multiplied \widehat{a}\widehat{a}^{+}

I feel like I cannot continue(self-study) until I see how I'm wrong. Please help!

 

[CompuChip]

I keep ending up with 2!

Then you're doing something wrong :P
For us to see what exactly, you could post your calculation (either scanned or - preferably - nicely TeXed)

But they have simply multiplied \widehat{a}\widehat{a}^{+}

I don't see where they did that. I just see them using the bilinearity of the commutator operation, i.e.
[r x, y] = r [x, y] (when r is a real number and x, y are operators - sorry, don't feel like putting hats and stuff)
[x + y, z] = [x, z] + [y, z] (where x, y, z are operators)
together with anti-symmetry ([x, y] = -[y, x]).

The proof looks really straightforward to me, can you maybe try to explain which step exactly is giving the problem?

 

[TupoyVolk]

Thank you so much for the reply.
I believe I am wrong, because QM indeed works! I just "need" to see how.

What was written where this sentence is, had a very stupid mathematical mistake. :)

I cannot see where I am wrong.

 

[TupoyVolk]

Holy crap.
I see my giant fail.

For some reason I turned things into commutators that shouldn't be them.
Thank you so much!