(Solved): quantum mechanics 3. Sketch the splitting of s, p and d levels u ...

quantum mechanics

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3. Sketch the splitting of s, p and d levels under a magnetic field. Draw the possible transitions in the cases \( s \leftrightarrow p \) and \( p \leftrightarrow d \) and estimate the relative intensities of the different lines. 4. 12 Motion in a Uniform Magnetic Field. Although this chapter has been devoted so far only to central forces, it is a convenient place to examine briefly the effects which occur when an external magnetic-field is present. Assuming a hydrogen-like atom placed in a magnetic field B, we can write the classical Hamiltonian of the electron as \[ H=(\mathrm{p}+(e / c) \mathrm{A})^{2} / 2 m_{m}+V(x) \] where the vector potential A is related to B through \[ B=\nabla \times A \] and is chosen in such a way as to satisfy the Lorentz gauge condition, (?) \[ \nabla \cdot \mathrm{A}=0 . \] We assume that both \( \mathrm{A} \) and \( \mathrm{B} \), as well as \( V \) (r), a re time-independent. According to our ear lierprescription, the quantal Hamiltonian operator in the \( x \)-representation is given by \( (4.12-1) \) with \( p=2 \) t . Now using the relations \[ \mathbf{W}(\mathbf{r}) \cdot \mathrm{p}-\mathrm{p} \cdot \mathrm{W}(\boldsymbol{r})=\ell \wedge \nabla \cdot \mathbf{W}(\boldsymbol{r}), \quad(4,12=4) \] (see Problem 3.7-2) and \( (4.12-3) \), we find that \[ \mathbf{A} \cdot \mathbf{p}-\mathbf{p} \cdot \mathbf{A}=0 \] so that we may expand \( (4,12-1) \) and replace the term \( p \cdot A+A \cdot F \) by \( 2 \mathrm{~A} \cdot \) p. This procedure gives \[ H=\mathrm{p}^{2} / 2 m_{e}+\left(e / m_{e} c\right) \mathrm{A} \cdot \mathrm{p}+\left(e^{2} / 2 m_{e} e^{2}\right) \mathrm{A}^{8}+V(\mathrm{r}) \quad(4,12-6) \] as the quantal Hamiltonian for a (spinless) particle moving in a magnetic field. The second and third terms depend generally on \( \theta \), \( \phi \) in addition to \( r_{,} \), and \( L^{2} \) and \( L_{x} \) are no longer expected to commute with the Hamiltonian \( (4,12-6) \) or to represent constants of the motion. Since the size of an atomic system is usually small in comparison to the region over which an external magnetic field mar change appreciably, it is not unreasonable to treat the magnetic field as uniform. Then we can take the vector potential to be \[ A=\frac{1}{2} B \times r \] as can be seen by substitution into \( (4,12-2) \). In this case, the second term in \( (4,12-6) \) becomes \[ \left(e / m_{e} c\right) \mathrm{A} \cdot \mathrm{p}=\left(\varepsilon / 2 m_{e} c\right)(\mathbf{B} \times \mathbf{r}) \cdot \mathrm{p}=\left(e h / 2 m_{e} c\right) \mathrm{B} \cdot \mathrm{L} \quad(4.12-8) \] since in our units \( A L=\mathbf{r} \times \mathbf{p} \). This permits us to write \( (4.12-6) \) in the form \[ H=\mathrm{p}^{2} / 2 m_{e}+\left(e \hbar / 2 m_{e} e\right) \mathrm{B} \cdot \mathrm{L}+\left(e^{2} / 2 m_{n} 0^{2}\right) \mathrm{A}^{2}+V(r) . \quad(4,12-9) \] Recalling that the interaction energy between a magnetic dipole of moment \( M \) and a magnetic field is \( -M \). \( B \), we may interpret the second term in \( (4.12-9) \) as the energy of interaction of the external field with the magnetic dipole moment a rising from the electron's orbital angular momentum. The latter is thus given by \[ \mathrm{M}=-\left(e \hbar / 2 m_{0} 0\right) \mathrm{L}=-\mu_{3} \mathrm{~L}, \] where the constant \( (4.12-10) \) \( \mu_{0}=e k /\left(2 m_{e} o\right)=0.9273 \times 10^{-20} \mathrm{erg} / g \) auss \( =5.788 \times 10^{-9} \mathrm{eV} / \mathrm{gauss} \) is called \( B o x^{\prime}{ }^{\prime} s \) magneton, and is the fundamental unit for atomic magnetic moments, * We shall now consider two important problems involving a uniform magnetic field, (a) Electron moving under a central force plus a weak magnetic field. In the presence of a weak magnetic field, the third term in \( (4,12-9) \) can be neglected, and we may write \[ H=\mathrm{p}^{2} / 2 m_{0}+\mu_{0} B L_{\tau}+V(r)=H_{0}+\mu_{0} B L_{x} \] (4. 12-12) where \( H_{0}=\mathrm{p}^{2} / 2 \mathrm{~m}_{0}+V(\boldsymbol{r}) \) represents the Hamiltonian in the absence of the magnetic field, and we have taken \( \mathrm{B} \) to be along the \( \mathrm{Z} \) axis. Since the potential \( V(r) \) is central, it follows that \( L^{2} \) and \( L_{2} \) will both commute with the Hamiltonian \( (4.12-12) \) and will be constants of the motion. In such a case, the eigenfunctions of \( H \) are of the form \( \psi_{n 2 n}(r)=R_{n 1}(r) Y_{1 s} \), Operating on \( \phi_{010} \) with \( (4.12-12) \), we have \( =t_{n 1}=E \psi_{2 \ln } \), that is \[ \begin{array}{l} \left(H_{0}+H_{0} B m\right) \psi_{a l n}=E \psi_{\mathbf{n} n^{\prime}} \text { or }\left(E_{\mathrm{at}}+\mu_{0} B m\right) \psi_{a t e}=E \psi_{a 12} \quad(4,12-13) \\ \text { re } m \text { is the eigenvalue of } L_{2} \text { and } E_{0} \text { is the eigenvalue of } \end{array} \] where \( m \) is the eigenvalue of \( L_{2} \) and \( E_{01} \) is the eigenvalue of \( H_{0} \) i. e. the energy in the absence of the external ficld. Then If instead of a negative electron we have a positively charged particle (suchas a positronor a proton) we get \( \mathrm{M}=+\left(e^{\circ} / 2 m_{0}\right. \) c) \( \mathrm{L} \). \[ E=E_{\mathrm{nt}}+\mu_{8} B m_{2} . \] Thus we see that the magnetic field introduces an additional dependence on the quantum number \( m \) and splits each of the original central force levels \( B_{n 1} \) into \( 2 l+1 \) sublevels having a spacing \( \mu_{0} B \). In other words the magnetic field lifts the essential degeneracy with respect to \( m \) that exists when only central forces are present. The eigenfunctions, however, remain the same in the weak-field as in the no-field case, In a transition \( n, l, m \rightarrow n^{\prime}, l^{\prime}, m^{\prime} \), the frequency \( v \) is given by \( v=\left(E-E^{\prime}\right) / h=v_{0}+\left(\mu_{8} B / h\right) \Delta m \) (4. \( 12-15) \) where \( v_{0} \) is the frequency for zero magnetic field. Since the selection rules for \( m \) are \( \Delta m=0, \pm 1 \), we conclude that each spectral line will be split into three lines offrequencies, \( v_{0}, v_{0}+\mu_{0} B / h \) and \( \nu_{0}-\mu_{8} B / h \), respectively, This situation correspond 5 to the socalled normal \( Z \) eman effeet. (b) Free electron moving in a magnetic field of arbitrary strength. In this case, \( V(r) \equiv 0 \) and \( (4.12-9) \) reduces to \[ H=\mathrm{p}^{2} / 2 m_{0}+\mu_{9} B L_{n}+\left(e^{2} / 2 m_{0} c^{2}\right) \mathrm{A}^{2} . \] When the \( Z \) axis is chosen parallel to the magnetic field, \( (4,12-7) \) becomes \[ A=\frac{b}{2} B(-i y+j x) \text {, } \] so that \( \mathrm{A}^{2}=\frac{1}{4} B^{2}\left(x^{8}+y^{2}\right) \). Hence \[ \begin{array}{l} H=\mathrm{p}^{2} / 2 m_{e}+\mu_{a} B L_{\mathrm{a}}+\left(e^{2} B^{2} / 8 m_{0} o^{2}\right)\left(x^{2}+y^{2}\right)=H_{x y}+H_{2}+H_{0}, \\ H_{x y}=\left(p_{x}^{3}+p_{y}^{2}\right) / 2 m_{e}+\left(\epsilon^{2} B^{2} / 8 m_{a} c^{2}\right)\left(x^{8}+y^{2}\right), \quad(4,12-16) \\ H_{x}=p_{z}^{2} / 2 m_{t} \\ (4.12-17) \\ \text { and } \\ H_{\mathrm{a}}=H_{8} B L_{2} . \\ \end{array} \] The three terms in the Hamiltonian commute among themselves and are subject to very simple interpretations, \( H_{\pi y} \) describes the motion of a two-dimensional harmonic oscillator in the XY plane having an angular frequency \( w=e B / 2 m_{e} c, H_{x} \) describes the motion of a free particle along the \( \mathrm{Z} \) axis, and \( H_{a} \) represents the magnetic Hamilonian interaction due to the orbital angular momentum of the electron, We now proceed to determine the eigenfunctions and eigenvalues of the system. We already know that \( H_{\tau} \) has an eigen- function \( e^{i k z} \), with an eigenvalue \( E_{2}=A^{a} p^{3} / 2 m_{*} \), representing the energy of a free electron moving along the \( Z \) axis with momentum 4i. Moreover, we recognize that the eigenfunction of \( H_{\mathrm{a}} \) is \( e^{t=\phi} \) =here \( \pi=0, \pm 1, \pm 2, \ldots \). The corresponding eigenvalue \( E_{\mathrm{s}}=\mathrm{H}_{8} \mathrm{Pm}_{4} \), is that of an electron having a \( \mathrm{Z} \) component of the orbital angular momentum equal to \( m h \). The character of these solutions and the separability of the Hamiltonian \( H \) suggests that we introduce cylindrical coordinates \( \rho, \phi, z \), and a product wave function \[ \psi(\mathbf{r})=R(p) e^{1 n \phi_{e}} e^{i k z} . \] If we introduce the cyclotron frequency (the angular frequency of an electron in a magnetic field) \[ v_{0}=e B / m_{e} o \] \[ \begin{array}{l} \text { and set } \\ \left.\qquad 5=(e B / 2 A c) \rho^{2}, \quad E=B_{x y} / \hbar \omega_{a}\right) \end{array} \] (4. 12-21) then \( (4.12-20) \) can be reduced to \( 5 \mathrm{~d}^{8} R / \mathrm{d} \xi^{2}+\mathrm{d} R / \mathrm{d} 5+\left[\varepsilon-m^{3} / 4 \xi-\frac{2}{4} \xi\right] R=0 \). It is clear that as \( \xi \rightarrow \infty \), we arrive at anasymptotic value for \( R(\xi) \) \( (4.12-22) \) of \( e^{=5 / 2} \). Similarly, for \( \xi \rightarrow 0 \), we obtain \( A(\xi)=5^{a / a} \). This suggests that we write \[ F(\xi)=\xi \mathrm{s} \mid / e^{-\xi} / \mathrm{a} F(\xi) \] \( (4,12-24) \) where we use \( |\mathrm{m}| \) to assure the proper behavior of \( f(\xi) \) at \( \xi=0 \). Substitution of \( (4,12-24) \) into \( (4,12-23) \) yields \[ \left.\left.5 F^{2}+(\mid m)+1-\xi\right)^{F}\right)+\alpha{ }^{F}=0 \] where \( (4.12-25) \) \[ a=-\frac{3}{2} \mid \text { ? } \mid-\frac{a}{\varepsilon}+\epsilon . \] 4. comparison of this equation with \( (4.7-5) \) shows that \( (4,12-26) \) of \( (4.12-25) \) is the confluent hypergeometric series \( F(=\alpha,|m|+1, \xi) \). Again the normalization condition requires that the series be a olynomial, for which we must have \( a=n^{\prime} \), the degree of the polymial, a positive integer. Therefore, Eq. \( (4,12-26) \) gives \( \varepsilon= \) \( -1+\frac{1}{2}|m|+\frac{1}{2} \), or \[ E_{x y}=\left(r^{\prime}+\frac{b}{2}|m|+\frac{b}{2}\right) \hbar w_{s} \] the corresponding radial eigenfunction becomes \( (4,12-27) \) \[ \bar{A}_{n|n|}=F(=\pi,|m|+1, \xi)=L_{\mathrm{a}}^{\mathrm{a}} \mid\left(e B p^{8} / 2 \star 0\right) . \] It follows that the complete eigenfunction for the problem is where \( N \) is a suitable normalization constant. Also, noting that \( \mu_{0} \theta=\frac{1}{2} h \omega_{0} \), we can write the total energy as \[ E=\hbar^{2} k^{3} / 2 m_{0}+\left(n^{\prime}+\frac{1}{2}|m|+\frac{\hbar}{2 m}+\frac{1}{2}\right) h w_{e} . \] or, setting \( n=n^{n}+\frac{1}{2}|m|+\frac{2}{8} m \), \[ E=\hbar^{2} h^{3} / 2 m+\left(n+\frac{1}{a}\right) \hbar w_{e} \] Not surprisingly, we now see that the resultant motion of the electron is composed of a free motion along the \( Z \) axis plus a quantized circular motion about the Zaxis with a zeropointenergy \( \frac{{ }^{e}}{} \hbar \omega_{e .} \). The results we have obtained enter into the quantal description of the diamagnetism of free electrons and other phenomena such as de Haas - van Alphen effect.(9) To conclude this section, we shall presentan alternative treatment which is relatively simple and instructive. The Hamiltonian \( (4,12-1) \) becomes for a free electron \[ H=(\mathrm{p}+(b / c) \mathrm{A})^{2} / 2 m_{\mathrm{o}} \] corresponding to a classical momentum \[ \pi=\mathrm{P}+(e / c) \mathrm{A} . \] When the magnetic field is along the \( Z \) axis the vector potential is in the XY plane. Therefore \[ \pi_{x}=p_{x}+(e / c) A_{x}, \quad \pi_{y}=p_{y}+(e / c) A_{y}, \quad \pi_{z}=p_{x} \cdot(4.12-32) \] Then \[ \begin{array}{l} {\left[\pi_{x}, \pi_{y}\right]=\left[p_{x}+(e / o) A_{x}, p_{y}+(e / c) A_{y}\right]=-(e / c)\left[A_{y}, p_{x}\right]+(e / 0)\left[A_{x}, p_{y}\right]=} \\ =-(\varepsilon \hbar \epsilon / c)\left(a A_{x} / \partial x-\partial A_{x} / \partial y\right)=-(\varepsilon \hbar \epsilon / c) \theta \text {. } \\ \end{array} \] Letting \[ \pi_{x}=(e B / C)^{\frac{1}{2}} Q_{1} \quad \pi_{y}=-(e B / c)^{\frac{1}{2}} P, \] we find that \( [Q, P]=t \hbar \). Thus formally \( Q \) and \( P \) are equivalent to a coordinate and its conjugate momentum. Now we may write \( (4.12-31) \) as \[ H=\left(\pi_{x}^{3}+\pi_{y}^{z}\right) / 2 m_{0}+p_{t}^{2} / 2 m_{0}=H_{z y}^{1}+H_{s} \] so that \[ H_{x y}^{1}=\left(\pi_{x}^{2}+\pi_{y}^{2}\right) / 2 m_{n}=(e B / \theta)\left[\left(P^{2}+Q^{2}\right) / 2 m_{e}\right] \] which shows that \( H_{x,}^{\prime} \) is \( (e B / c) \) times the Hamiltonian of a harmonic oscillator having an angular frequency \( w=1 / m \). Hence the eigenvalues of \( H_{\text {ry }} \) are \[ E_{\mathrm{xy}}=(e B / c)\left\{\left(n+\frac{1}{2}\right) \hbar / m\right\}=\left(n+\frac{1}{a}\right) \hbar()_{e} \quad(4,12-34) \] where \( w_{a}=e B / \mathrm{mc} \). This result agrees with that obtained using the Schrocdinger representation if we identify \( n \) in \( (4,12-34) \) with \( n^{\prime}+ \) \( +\frac{2}{8}|m|+ \) ? m in \( \{4,12-30\} \).