# 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 $$\widehat{a}$$$$\widehat{a}$$$$^{+}$$

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Homework Helper
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.

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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!