Harmonic oscillators and commutators.

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In the discussion on harmonic oscillators and commutators, it is established that the creation and annihilation operators for independent oscillators do not affect each other, leading to the conclusion that [a_{+}, a'_{-}] = 0. This indicates that operators from one oscillator commute with those from another, as they operate on separate states. The reasoning behind this is that the operators for one oscillator have no influence on the states of another. Additionally, it is confirmed that other combinations of these operators also commute, such as [a'_{-}a_{+}, a'_{+}, a_{+}] = 0. Overall, the commutation relations hold true across independent oscillators.
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Homework Statement



If we have a harmonic oscillator with creation and annhilation operators a_{-} a_{+}, respectively. The commutation relation is well known:

[a_{+},a_{-}] = I

However, if we have two independent oscillators with operators a'_{-} a'_{+}

As the operators are the same function would they commute if mixed.

i..e is [a_{+},a'_{-}] = I

Still valid?
 
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When you say two operators A and B commute, it means AB=BA so that [A,B]=AB-BA=0.

As to your question about the commutation relation, the answer is no. The creation and annihilation operators for the one oscillator commute with the operators for the other oscillator.
 
Last edited:
So [a_{+},a'_{-}] = 0

Is there any reason for this beside the operators for one oscillator have no effect on the states of another?

Would assume that any other mix of these operators would commute?

e.g [a'_{-}a_{+},a'_{+},a_{+}] = 0
 
The position and momentum operators, x1 and p1, for one oscillator commute with the position and momentum operators, x2 and p2 for the second oscillator:
\begin{align*}
[x_i, x_j] &= 0 \\
[p_i, p_j] &= 0 \\
[x_i, p_j] &= i\hbar\delta_{ij}
\end{align*}It follows from these that a primed and unprimed operator commute.
 

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