Solving Commutator Problem: Find [a,a†]

[nmsurobert]

Homework Statement

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.

Homework Equations

[a,b] = ab -ba

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?

 

[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?

 

[nmsurobert]

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?

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.

 

[TSny]

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

 

[Sugyned]

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

 

[TSny]

Hello Sugyned, welcome to PF!

Please show your work.

 

[Sugyned]

Hello Sugyned, welcome to PF!

Please show your work.

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)I don't know much about this, maybe I'm completely wrong.
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.

 

[strangerep]

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

Maybe you should start a new thread, and state what your problem you're trying to answer?

 

[Sugyned]

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

Maybe you should start a new thread, and state what your problem you're trying to answer?

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.

 

[strangerep]

It's actualy this exact same problem shown up here.

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.