Proving commutator relation between H and raising operator

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The discussion focuses on proving the commutator relation [H, a†] = hωa†, with a participant encountering a negative sign due to mistakenly using [a†, a] instead of [a, a†]. They realize that the error arises from altering the order of operations in their calculations. A hint suggests calculating [H, a†] |1> using the properties of the Hamiltonian and the raising operator. Another participant points out the mistake in the steps taken and advises maintaining the order of operators. Ultimately, the original poster corrects their approach and successfully derives the correct answer.
guyvsdcsniper
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Homework Statement
Prove the commutator relation [H,a*]=hwa*
Relevant Equations
[H,a*]=hwa*
I am going through my class notes and trying to prove the middle commutator relation,
Screen Shot 2022-08-25 at 10.06.11 PM.png


I am ending up with a negative sign in my work. It comes from [a,a] being invoked during the commutation. I obviously need [a,a] to appear instead.

Why am I getting [a,a] instead of [a,a]?

IMG_1106.JPG
 
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quittingthecult said:
Homework Statement:: Prove the commutator relation [H,a*]=hwa*
Relevant Equations:: [H,a*]=hwa*

I am going through my class notes and trying to prove the middle commutator relation, View attachment 313257

I am ending up with a negative sign in my work. It comes from [a,a] being invoked during the commutation. I obviously need [a,a] to appear instead.

Why am I getting [a,a] instead of [a,a]?

View attachment 313258
Hint: Calculate ##[H, a^{\dagger} ] |1>## using ##H|n> = (n + 1/2) \hbar \omega |n>## and ##a^{\dagger} |1> = c |2>##. What happens?

-Dan
 
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Seems to me the step (2) is wrong, you are changing the order of operation there

In step (1) you have ## (a^\dagger a + \frac{1}{2})a^\dagger - a^\dagger(a^\dagger a + \frac{1}{2}) ##
But in step (2) you have ## a^\dagger (a^\dagger a + \frac{1}{2} - a^\dagger a - \frac{1}{2})##

Redo step (1) to (2), keep the order of operators unaltered.
 
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It seems to me step (1) is wrong. The commutator disappeared…

Too early in the morning, you just expanded the commutator. I would not do this, I would apply commutator rules for ##[AB,C] = A[B,C]+[A,C]B##.
 
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Thanks to all, I have seen the trivial mistake I made. I was able to get the correct answer now.
 
So is there some elegant way to do this or am I just supposed to follow my nose and sub the Taylor expansions for terms in the two boost matrices under the assumption ##v,w\ll 1##, then do three ugly matrix multiplications and get some horrifying kludge for ##R## and show that the product of ##R## and its transpose is the identity matrix with det(R)=1? Without loss of generality I made ##\mathbf{v}## point along the x-axis and since ##\mathbf{v}\cdot\mathbf{w} = 0## I set ##w_1 = 0## to...

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