Chapter 4: Problem 19

(a) Demonstrate that if \(\left[j_{1 q}, j_{2 q^{\prime}}\right]=0\) for all \(q, q^{\prime},\) then (b) Go on to show that if \(j_{1} \times j_{1}=i \hbar j_{1}\) and \(j_{1} \times j_{2}=-j_{2} \times j_{1}\) \(j_{2} \times j_{2}=i \hbar j_{2},\) then \(j \times j=\) ihj where \(j=j_{1}+j_{2}\)

Short Answer

Expert verified

The procedure begins by noting that the given commutation relation implies the two operators independently. Using the properties of cross products and commutation, it is shown that \(j_{1} \times j_{2}\) equals \(j_{2} \times j_{1}\). Subsequently, these results are used to find the expression for \(j \times j\), which is shown to be \(i \hbar j\).

Step by step solution

Commutation Relation

Given the commutation relation \([j_{1 q}, j_{2 q^{\prime}}] = 0\) for all \(q, q^{\prime}\), this means these two operators commute, and thus can be known independently of each other.

Cross Product with Commutation

Because \([j_{1} \times j_{2}] = j_{1} \times j_{2} - j_{2} \times j_{1}\), applying the provided properties, we get \(i \hbar j_{1} - (-j_{2} \times j_{1}) = i \hbar j_{1} + j_{2} \times j_{1}\). As operators \(j_{1}\) and \(j_{2}\) are independent, the order doesn't matter in the equation, thus \(j_{1} \times j_{2} = j_{2} \times j_{1}\).

Derive the final cross product

The operator \(j = j_{1} + j_{2}\) is given. So \(j \times j = (j_{1} + j_{2}) \times (j_{1} + j_{2}) = j_{1} \times j_{1} + j_{1} \times j_{2} + j_{2} \times j_{1} + j_{2} \times j_{2}\). Applying the given properties, it simplifies to \(i \hbar j_{1} + 0 + 0 + i \hbar j_{2} = i \hbar (j_{1} + j_{2}) = i \hbar j\).

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(a) Demonstrate that if \(\left[j_{1 q}, j_{2 q^{\prime}}\right]=0\) for all \(q, q^{\prime},\) then (b) Go on to show that if \(j_{1} \times j_{1}=i \hbar j_{1}\) and \(j_{1} \times j_{2}=-j_{2} \times j_{1}\) \(j_{2} \times j_{2}=i \hbar j_{2},\) then \(j \times j=\) ihj where \(j=j_{1}+j_{2}\)

Short Answer

Step by step solution

Commutation Relation

Cross Product with Commutation

Derive the final cross product

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