Electron Spin Introduction Our solution of the TISE in three dimensions for one-electron atoms resulted in quantum states that are uniquely specified by the values of the three quantum numbers n, l, and ml. This picture was very successful in explaining many characteristics of the hydrogen atom, even more so than the Bohr model. However, experiments showed that this picture was incomplete. This gap was filled by the introduction of the concept of intrinsic angular momentum or spin angular momentum of the electron. Much like an orbiting planet, an electron in an atom has two independent angular momenta: (i) the orbital angular momentum (due to orbital motion) and (ii) an intrinsic or spin angular momentum, which is analogous to a planet spinning about its axis. To understand the consequences of spin, we must first revisit the idea of magnetic dipole moment. To treat the topic of atomic magnetic dipole moment, we shall use classical and semi-classical ideas because they are much simpler to work with. The final results are, however, in agreement with the full quantum mechanical treatment. Orbital Magnetic Dipole Moments Consider an electron of mass m, charge –e and velocity magnitude v executing a circular orbit of radius r in a Bohr atom. The circular motion of the electron constitutes a current loop. The current is given by µ q e ev I I = = = . (12.1) T 2πr v 2πr ( ) r –e Recall that the conventional current is directed opposite to the v direction of the electron’s motion. This current loop constitutes a magnetic dipole, with a magnetic dipole moment (also called magnetic moment) µ given by µ = IAnˆ, (12.2) where A is the area of the loop and nˆ is a unit vector whose direction is perpendicular to the plane of the loop and is given by the right hand rule (fingers in the direction of the current, outstretched thumb points in direction of nˆ ). The magnitude of µ is given by ⎛ ev ⎞ 2 evr µ = IA = ⎜ ⎟(π r )= . (12.3) ⎝ 2π r ⎠ 2 The magnitude L of the orbital angular momentum is given by L = mvr. Thus, eL µ = . (12.4) 2m Since L is quantized in units of , we can write 1 ⎛ e ⎞⎛ 1 ⎞ µ = ⎜ ⎟⎜ ⎟ L, (12.5) ⎝ 2m ⎠⎝ ⎠ where e 2m has units of magnetic moment. We define the Bohr magneton e µ ≡ . (12.6) B 2m −24 −5 Note that µB = 9.27×10 J/T = 5.79×10 eV/T. Rewriting Eq. (12.5) gives µ L µ = B . (12.7) In vector form, we obtain µB L µl = − , (12.8) where the minus sign appears because the electron has a negative charge, so its velocity is opposite in direction to the conventional current. The subscript “l” indicates that the magnetic moment is associated with the orbital angular momentum. Eq. (12.8) is the same result obtained using quantum mechanics. Since L and Lz are quantized according to L = l(l +1) (12.9) and Lz = ml , (12.10) respectively, we have µl = µB l(l +1) (12.11) and m . (12.12) µlz = −µB l Hence, µl is quantized, and the vector µl , like L , can only point in certain directions. Now if a magnetic dipole µ is immersed in a magnetic field B , it will experience a torque given by τ = µ × B, (12.13) 2 which tends to align the dipole with the field (just like a magnet suspended in the Earth’s field). There is an orientational potential energy associated with this torque (external work is needed to twist the dipole moment out of alignment with the field, which increases the potential energy). The potential energy is given by Um = −µ ⋅ B. (12.14) Thus, the potential energy is lowest when µ is parallel to B and highest when µ is antiparallel to B . Stern-Gerlach Experiment In a uniform B field, no net force is exerted on a magnetic dipole. However, in a non-uniform magnetic field, a force is exerted on a magnetic dipole. [Show diagram of apparatus.] Consider an atomic dipole in a quantum state whose magnetic moment depends only on the orbital angular momentum. If the field varies along the z-axis, then one can show that ∂B F = µ z . (12.15) z lz ∂z Note that is quantized according to Eq. (12.12): m , with m l, l 1,...,l 1,l. µlz µlz = −µB l l = − − + − Thus, if a beam of atoms is in a state with orbital angular momentum quantum number l, then the force exerted on an atom due to the non-uniform field will depend on the value of ml. If the force is transverse to the motion of the atoms, it follows that the deflected beam will be split into 2l+1 discrete segments. Note that 2l+1 will always be an odd number. In 1922, Stern and Gerlach performed such an experiment with a beam of silver atoms. It was found that the beam was split into two segments. Though this confirmed the idea of spatial quantization, it was inconsistent with the expected odd number of components if l = 0, 1, 2, 3,… etc. Phipps and Taylor did a similar experiment with hydrogen atoms in 1927. The H atoms were prepared so that they would be in their ground state, i.e., l = 0 and ml = 0. In this case, the atom beam should be unaffected by the field gradient. However, the beam was split into two segments. One is forced to conclude that the theory as developed up until this point is incomplete. From the splitting of the beam into two components, one infers that there is an additional dipole moment, apart from that associated with the orbital angular momentum. This additional dipole moment is due to the intrinsic angular momentum, or spin angular momentum, of the electron. The spin angular momentum is quantized in an identical fashion to its orbital counterpart: S = S = s(s +1) (12.16) and Sz = ms, (12.17) where s is the spin angular momentum quantum number and ms is the spin magnetic quantum number. Also, 3 gµB S µs = − (12.18) and g m , (12.19) µsz = − µB s where g is the electron spin g factor, and ms = –s, –s+1,…, s Now, since the hydrogen atom beam is split into two components, one finds that for a single electron (such as in a hydrogen atom) 1 1 1 s = ; m = − , + . (12.20) 2 s 2 2 (This gives the required number of components 2s+1 = 2.) By measuring the deflection of the H atoms, it was found that g = 2. (A more precise value is 2.0023.) These conclusions were confirmed by many other experiments, including the Zeeman effect, which will be studied next. 1 Since Sz = ms, it follows that the z component of S can have only two values: ± 2 , and similarly for the z component of the spin magnetic moment g m . (12.21) µsz = − µB s = µB 1 The two states corresponding to ms = ± 2 are usually called “spin up” and “spin down.” The complete specification of the quantum state of an electron in the H atom therefore requires four quantum numbers: n, l, ml, and ms. In the absence of a magnetic field, the spin up and spin down states are degenerate. Thus, the degeneracy of each quantum state must be doubled to ψ nlml account for spin. Finally, we note that both the orbital and spin angular momenta contribute to the total angular momentum: J = L + S. (12.22) Likewise, the total magnetic moment µ is given by µ µ = − B L + 2S . (12.23) ( ) It is worthwhile to note that Eq. (12.23) indicates that the total magnetic moment vector µ is not antiparallel to the total angular momentum vector J because there is no simple proportionality relationship between the two vectors. [Do PhET Stern-Gerlach simulations.] 4 Example: Obtain an expression for the transverse deflection of a beam of hydrogen atoms in the ground state in a Stern-Gerlach-type experiment. The length of the path is D and the speed of the atoms is v. Solution: Let the transverse deflection be along the x-direction. Then ∂B ∂B ∂B F = z µ = − z gµ m = z µ . z ∂z sz ∂z B s ∂z B 2 1 2 1 Fz ⎛ D ⎞ The transverse deflection distance dz = azt = ⎜ ⎟ . (We assume that any transverse 2 2 m ⎝ v ⎠ velocity is non-relativistic.) Thus, 2 ∂Bz ∂z µB ⎛ D ⎞ dz = ⎜ ⎟ . 2m ⎝ v ⎠ (For typical values, dz ≈ 1 mm.) Zeeman Effect We saw that the potential energy of a magnetic dipole in a magnetic field B is given by Um = −µ ⋅ B. (12.24) If we take the direction of the field to be the z direction, then Eq. (12.24) becomes Um = −µz B. (12.25) Consider an atom in a state in which the total spin angular momentum is zero. This occurs in atoms in which the electrons are paired, i.e., one electron is in a spin-up state and the other is in a spin-down state. This produces an effective zero-spin state. In such cases, the magnetic moment is due to orbital angular momentum alone, and the orientational PE becomes U B m B. (12.26) m = −µlz = µB l Hence, the energy of a state with orbital angular momentum quantum number l will shift in a magnetic field. The energy shift ΔE = Um depends on the value of ml. If the energy of the degenerate quantum states before the field is switched on is Enl, then after the field is switched on, the energy becomes Enl + ΔE = Enl + ml µB B.