# Commutator relationship

• WendysRules
Therefore, you can take out any scalar multiple of one of the arguments. In summary, by using the commutator relation ##[\hat{L}_i, \hat{L}_j] = i \hbar \epsilon_{ijk} \hat{L}_k## and the linearity of the commutator, we can show that ##[\hat{L} \cdot \vec{a}, \hat{L} \cdot \vec{b}] = i \hbar \hat{L} \cdot (\vec{a} \times \vec{b})##.

## Homework Statement

Show that ##[\hat{L} \cdot \vec{a}, \hat{L} \cdot \vec{b}] = i \hbar \hat{L} \cdot (\vec{a} \times \vec{b})##

## Homework Equations

##[\hat{L}_i, \hat{L}_j]= i \hbar \epsilon_{ijk} \hat{L}_k ##

## The Attempt at a Solution

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Maybe a naive attempt, but it has been a while. I have two ways I think this can work, but both I'm not sure. they "work", but without a more trained eye, I'm not sure if they're valid.

Start with ##[\hat{L} \cdot \vec{a}, \hat{L} \cdot \vec{b}] \rightarrow [\hat{L}_ia_i, \hat{L}_jb_j] =i \hbar \epsilon_{ijk} \hat{L}_k a_ib_i \rightarrow i \hbar\hat{L}_k \epsilon_{ijk} a_i b_i = i \hbar \hat{L} \cdot (\vec{a} \times \vec{b}) ##

The other one starts similarly, but once we put it in the ##[\hat{L}_ia_i, \hat{L}_jb_j]## form, we can take out the ##a_i## and ##b_j## giving us ##a_ib_j[\hat{L}_i,\hat{L}_j]## which from here the rest follows as above, but not sure if either approach is "valid".

How are these different ways? They look identical to me.

Orodruin said:
How are these different ways? They look identical to me.
Yes, i believe they are. Is it true that I am allowed to take out the ##a_i## from the brackets? I think that is what i conclude from above.

WendysRules said:
Is it true that I am allowed to take out the aiaia_i from the brackets?
Yes, the commutator is linear in both arguments.

WendysRules

## 1. What is a commutator relationship?

A commutator relationship is a mathematical relationship between two operators, A and B, in quantum mechanics. It is represented by [A,B] and measures the non-commutativity between the two operators.

## 2. How is the commutator relationship calculated?

The commutator relationship is calculated by taking the difference between the product of the two operators in the order AB and BA. It is represented by [A,B] = AB - BA.

## 3. What does it mean if the commutator relationship is zero?

If the commutator relationship is zero, it means that the two operators commute and can be measured simultaneously without affecting each other's values. This indicates that the physical quantities represented by the operators have a well-defined relationship.

## 4. What is the significance of the commutator relationship in quantum mechanics?

The commutator relationship is significant in quantum mechanics because it helps to determine the uncertainty in measuring two observables at the same time. If the commutator relationship is non-zero, it indicates that there is an inherent uncertainty in measuring both observables simultaneously.

## 5. Can the commutator relationship be used to determine the commutativity of any two operators?

Yes, the commutator relationship can be used to determine the commutativity of any two operators. If the commutator relationship is zero, then the two operators commute and if it is non-zero, then the two operators do not commute.