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Chapter 4: Problem 11

Calculate the matrix elements (a) \(\left\langle 0,0\left|l_{z}\right| 0,0\right\rangle\) (b) \(\left\langle 2,1\left|l_{+}\right| 2,0\right\rangle\) (c) \(\left\langle 2,2\left|l_{+}^{2}\right| 2,0\right\rangle,\) (d) \(\left\langle 2,0\left|l_{+} l_{-}\right| 2,0\right\rangle\) (e) \(\left\langle 2,0\left|l \downarrow_{+}\right| 2,0\right\rangle,\) and (f) \(\left\langle 2,0\left|l^{2} l_{z} l^{2}+\right| 2,0\right\rangle\)

Short Answer

Expert verified
\(\left\langle 0,0\left|l_{z}\right| 0,0\right\rangle = 0\), \(\left\langle 2,1\left|l_{+}\right| 2,0\right\rangle = \sqrt{6}\hbar\), \(\left\langle 2,2\left|l_{+}^{2}\right| 2,0\right\rangle = 0\)

Step by step solution

01

Calculate \(\left\langle 0,0\left|l_{z}\right| 0,0\right\rangle\)

The given matrix element represents the application of the angular momentum operator along the z-direction (l_{z}) on a quantum state. This quantum state, \(|0,0\rangle\), has quantum numbers l=0 and m=0. Since l_{z} behaves as an eigenoperator when acting upon these quantum states, it multiplies the state by an eigenvalue given by \(m\hbar\). That gives the value \(0* \hbar\). Therefore, \(\left\langle 0,0\left|l_{z}\right| 0,0\right\rangle = 0\).
02

Calculate \(\left\langle 2,1\left|l_{+}\right| 2,0\right\rangle\)

Here, the operator \(l_{+}\) (the raising operator) is acting on the quantum state \(|2,0\rangle\). By the definition of \(l_{+}\), it raises the magnetic quantum number (m) by 1. Therefore, the initial state becomes \(|2,1\rangle\). And the multiplicative constant accompanying the change is given by \(\sqrt{l(l+1)-m(m+1)}\hbar = \sqrt{2*3-0*1}\hbar = \sqrt{6}\hbar\). Therefore, \(\left\langle 2,1\left|l_{+}\right| 2,0\right\rangle = \sqrt{6}\hbar\).
03

Calculate \(\left\langle 2,2\left|l_{+}^{2}\right| 2,0\right\rangle\)

Here, the operator \(l_{+}^2\) is acting on the quantum state \(|2,0\rangle\). This operator raises the magnetic quantum number by 2. However, when it tries to raise the state \(|2,2\rangle\) to \(|2,4\rangle\), the resulting state is not valid, because m cannot exceed l, and thus the resulting operator when acted on \(|2,2\rangle\) will give 0. Therefore, \(\left\langle 2,2\left|l_{+}^{2}\right| 2,0\right\rangle = 0\).

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Most popular questions from this chapter

What are the possible outcomes of a single measurement of the \(z\) -component of spin angular momentum of an electron in the spin state \((1 / 4)^{1 / 2} \alpha-(3 / 4)^{1 / 2} \beta ?\)

Suppose that in place of the actual angular momentum commutation rules, the operators obeyed \(\left[l_{x}, l_{y}\right]=-\mathrm{i} \hbar l_{z^{*}}\) What would be the roles of \(l_{2}=l_{x} \pm l_{y} ?\)

Use the vector model of angular momentum to derive the value of the angle between the vectors representing (a) two \(\alpha\) spins, (b) an \(\alpha\) and a \(\beta\) spin in a state with \(S=1\) and \(M_{\mathrm{S}}=+1\) and \(M_{\mathrm{S}}=0,\) respectively.

Using the Pauli matrix representation, reduce each of the operators (a) \(s_{x} s_{y}\) (b) \(s_{x} s_{y}^{2} s_{z}^{2},\) and (c) \(s_{x}^{2} s_{y}^{2} s_{z}^{2},\) to a single spin operator.

In some cases \(m_{11}\) and \(m_{12}\) may be specified at the same time as \(j\) because although \(\left[f^{2}, j_{1 z}\right]\) is non-zero, the effect of \(\left[j^{2}, j_{12}\right]\) on the state with \(m_{i 1}=j_{1}, m_{i 2}=j_{2}\) is zero. Confirm that \(\left[j^{2}, j_{1 z}\right]\left|j_{1} j_{1} ; j_{2} / j_{2}\right\rangle=0\) and \(\left[j^{2}, j_{1 z}\right]\left|j_{1},-j_{1} ; j_{2},-j_{2}\right\rangle=0\).
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