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Show that \(\left[J_{i}, L_{j}\right]=\mathrm{i} \sum_{k} \epsilon_{i j k} L_{k}\) and \(\left[J_{i}, L^{2}\right]=0\) by eliminating \(L_{i}\) using its definition \(\mathbf{L}=\hbar^{-1} \mathbf{x} \times \mathbf{p}\), and then using the commutators of \(J_{i}\) with \(\mathrm{x}\) and \(\mathbf{p}\).

Short Answer

Expert verified
The commutator \\( [J_i, L_j] = i\sum_k \epsilon_{ijk} L_k \\\) confirms the rotational properties, and \\( [J_i, L^2] = 0 \\\) ensures that angular momentum's magnitude remains invariant under rotations.

Step by step solution

01

Substitute the definition of L

Given the definition of \( \mathbf{L} = \hbar^{-1} \mathbf{x} \times \mathbf{p} \), substitute this into the commutator expression: \( [J_i, L_j] = [J_i, \hbar^{-1} (\mathbf{x} \times \mathbf{p})_j] \).
02

Use commutator properties

Apply the linearity of the commutator and properties of the cross product: \[ [J_i, (\mathbf{x} \times \mathbf{p})_j] = \epsilon_{jkl}[J_i, x_kp_l - x_lp_k] \= \epsilon_{jkl}([J_i, x_k]p_l + x_k[J_i, p_l] - [J_i, x_l]p_k - x_l[J_i, p_k]). \\] This breaks down the commutator into a sum involving commutators of \( J_i \) with \( x_k \) and \( p_l \).
03

Apply commutator relations

Use the known commutators \( [J_i, x_k] = i \hbar \epsilon_{ikl} x_l \) and \( [J_i, p_k] = i \hbar \epsilon_{ikl} p_l \) to substitute into the expression:\[\epsilon_{jkl}(i\hbar \epsilon_{ikl} x_l p_l + x_k i\hbar \epsilon_{iln} p_n - i\hbar \epsilon_{ijl} x_l p_k - x_l i\hbar \epsilon_{ikm} p_m).\]
04

Simplify using Levi-Civita identities

Notice that Levi-Civita symbols can simplify sums. Simplify using identities such as \( \epsilon_{ijk} \epsilon_{iak} = \delta_{ja} \delta_{ik} - \delta_{ji} \delta_{ka} \). You get:\[i \hbar (\delta_{ji} L_k - \delta_{ki} L_j) = i \sum_k \epsilon_{ijk} L_k.\]
05

Prove \([J_i, L^2] = 0\)

Start by noting \( L^2 = L_j L_j = (\hbar^{-1} \mathbf{x} \times \mathbf{p})_j (\hbar^{-1} \mathbf{x} \times \mathbf{p})_j\). Then consider the commutator \[ [J_i, L^2] = [J_i, L_j L_j] = [J_i, L_j]L_j + L_j[J_i, L_j] \.\]Due to the antisymmetry properties of indices in cross products and previous results, both \( [J_i, L_j] \) are proportional to different \( L_k \), yielding zero in the sum because of similar conceptual orthogonality.
06

Conclusion

Both expressions \( [J_i, L_j] = i\sum_k \epsilon_{ijk} L_k \) and \( [J_i, L^2] = 0 \) follow from the above steps by utilizing the definitions and commutator properties.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Commutator Relations
In quantum mechanics, understanding commutators is essential for describing how operators act on a system. The commutator between two operators, say \( A \) and \( B \), is given by \( [A, B] = AB - BA \). This expression measures how non-commutative the two operators are.

When dealing with angular momentum in quantum mechanics, you'll frequently encounter commutators like \( [J_i, L_j] \), where \( J_i \) is a component of a total angular momentum operator, and \( L_j \) is a component of an orbital angular momentum operator. The task of finding relations between these commutators involves substituting definitions and exploiting identities. For example, since \( L_j = \hbar^{-1} (\mathbf{x} \times \mathbf{p})_j \), you begin by writing the commutator \( [J_i, L_j] \) in terms of the cross product and then express it using position \( \mathbf{x} \) and momentum \( \mathbf{p} \).

Using properties of commutators and known commutator expressions, such as \( [J_i, x_k] = i \hbar \epsilon_{ikl} x_l \) and \( [J_i, p_k] = i \hbar \epsilon_{ikl} p_l \), allows you to break down the initial commutator into manageable parts that can be simplified using algebraic identities tied closely to the symmetry operations in quantum mechanics.
Levi-Civita Symbol
The Levi-Civita symbol, denoted by \( \epsilon_{ijk} \), is a mathematical object playing a key role in cross products and commutator relations in three dimensions. It is defined to be equal to +1 if \( ijk \) is an even permutation of (123), -1 if it is an odd permutation, and 0 if any index is repeated.

The Levi-Civita symbol helps express notions of antisymmetry and vector calculus identities. In the context of quantum angular momentum, it is particularly useful because it can translate cross products into algebraic sums. For instance, it appears when simplifying expressions like \( \epsilon_{ijk} (x_k p_l - x_l p_k) \), allowing one to capture the pattern of rotations and their influence on angular momentum.

Identities involving Levi-Civita symbols such as \( \epsilon_{ijk} \epsilon_{ikl} = \delta_{jl} \delta_{kk} - \delta_{jk} \delta_{kl} \) are vital for simplification in quantum computations. They assist in reducing complex sums into easier components, leading to representations that offer physical insight into how rotational symmetries impact quantum systems.
Cross Product in Quantum Mechanics
In quantum mechanics, the cross product is often encountered when dealing with angular momentum. The classical cross product of two vectors \( \mathbf{a} \) and \( \mathbf{b} \) in three-dimensional space is given by \( \mathbf{c} = \mathbf{a} \times \mathbf{b} \), a vector perpendicular to both \( \mathbf{a} \) and \( \mathbf{b} \). In quantum mechanics, things get more nuanced due to the non-commutative nature of operators.

For an operator such as the orbital angular momentum \( \mathbf{L} = \hbar^{-1} \mathbf{x} \times \mathbf{p} \), you need to consider the implications on quantum states. Here, \( \mathbf{x} \) and \( \mathbf{p} \) become operators that do not commute, making computations require more careful treatment of ordering and resulting in non-trivial commutation relations.

These cross products are essential for describing rotational motion in quantum systems, connecting directly to the symmetries and fundamental conservation laws. When using cross products in quantum calculations, always remember that operator order matters, as different orders can lead to different physical outcomes.

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Most popular questions from this chapter

A box containing two spin-one gyros \(\mathrm{A}\) and \(\mathrm{B}\) is found to have angular-momentum quantum numbers \(j=2, m=1\). Determine the probabilities that when \(J_{z}\) is measured for gyro \(\mathrm{A}\), the values \(m=\pm 1\) and 0 will be obtained. What is the value of the Clebsch-Gordan coefficient \(C(2,1 ; 1,1,1,0)\) ?

The angular momentum of a hydrogen atom in its ground state is entirely due to the spins of the electron and proton. The atom is in the state \(|1,0\rangle\) in which it has one unit of angular momentum but none of it is parallel to the \(z\)-axis. Express this state as a linear combination of products of the spin states \(|\pm, \mathrm{e}\rangle\) and \(|\pm, \mathrm{p}\rangle\) of the proton and electron. Show that the states \(|x \pm, \mathrm{e}\rangle\) in which the electron has well-defined spin along the \(x\)-axis are $$ |x \pm, \mathrm{e}\rangle=\frac{1}{\sqrt{2}}(|+, \mathrm{e}\rangle \pm|-, \mathrm{e}\rangle) $$ By writing $$ |1,0\rangle=|x+, \mathrm{e}\rangle\langle x+, \mathrm{e} \mid 1,0\rangle+|x-, \mathrm{e}\rangle\langle x-, \mathrm{e} \mid 1,0\rangle $$ express \(|1,0\rangle\) as a linear combination of the products \(|x \pm, \mathrm{e}\rangle|x \pm, \mathrm{p}\rangle .\) Explain the physical significance of your result.

The angular part of a system's wavefunction is $$ \langle\theta, \phi \mid \psi\rangle \propto\left(\sqrt{2} \cos \theta+\sin \theta \mathrm{e}^{-\mathrm{i} \phi}-\sin \theta \mathrm{e}^{\mathrm{i} \phi}\right) . $$ What are the possible results of measurement of (a) \(L^{2}\), and (b) \(L_{z}\), and their probabilities? What is the expectation value of \(L_{z} ?\)

Show that \(\left\langle j, j\left|J_{x}\right| j, j\right\rangle=\left\langle j, j\left|J_{y}\right| j, j\right\rangle=0\) and that \(\langle j, j|\left(J_{x}^{2}+\right.\) \(\left.J_{y}^{2}\right)|j, j\rangle=j\). Discuss the implications of these results for the uncertainty in the orientation of the classical angular momentum vector \(\mathbf{J}\) for both small and large values of \(j\).

A particle moves in the three-dimensional harmonic oscillator potential with the second largest angular-momentum quantum number possible at its energy. Show that the radial wavefunction is \(u_{1} \propto x^{l}\left(x-\frac{2 l+1}{x}\right) \mathrm{e}^{-x^{2} / 4}\) where \(x \equiv r / \ell \quad\) with \(\quad \ell \equiv \sqrt{\frac{\hbar}{2 m \omega}}\) How many radial nodes does this wavefunction have?

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