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5_angular_momentum

# 5_angular_momentum - V Angular Momentum Grifﬁths Chapter...

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Unformatted text preview: V. Angular Momentum Grifﬁths Chapter 4 1 A. Orbital Angular Momentum The principal quantum number n of the hydrogen atom gives the energy of the state. The azimuthal and magnetic quantum numbers l and m are related to the orbital angular momentum. 2 A. Orbital Angular Momentum In classical physics, the orbital angular momentum is given by L = r × p. Lx = ypz − zpy Ly = zpx − xpz In cartesian coordinates Lz = xpy − ypx We use the same deﬁnition in quantum physics, but these are now operators. Lx = Y Pz − ZPy L=R×P Ly = ZPx − XPz Lz = XPy − Y Px 3 A. Orbital Angular Momentum 1. Some properties [Lx , Ly ] = i￿Lz The orbital angular momentum operators don’t commute. L × L = i￿L It’s useful to deﬁne the square of the magnitude of the orbital angular momentum. L2 ≡ L2 + L2 + L2 x y z L2 DOES commute with the orbital angular momentum operator components. [Ly , Lz ] = i￿Lx [Lz , Lx ] = i￿Ly [L2 , Lx ] = 0 [L2 , L] = 0 [L2 , Ly ] = 0 [L2 , Lz ] = 0 See board V.A.1 4 A. Orbital Angular Momentum 2. Eigenvalues Operators that don’t commute obey uncertainty relations. Operators that DO commute share eigenfunctions. Since L2 commutes with each angular momentum component, we can ﬁnd simultaneous eigenfunctions with one of them. Let’s choose Lz. We can also deﬁne ladder operators based on our choice of Lz as a reference. L± ≡ Lx ± iLy From this we get the eigenvalue equations Lz flm = ￿mflm L2 flm = ￿2 l(l + 1)flm 5 A. Orbital Angular Momentum integer STEPS Lz flm = ￿mflm L2 flm = ￿2 l(l + 1)flm some, as of yet, unknown f unction of position positive integer and HALF integer values l = 0, 1/2, 1, 3/2, ... See board V.A.2 6 m = -l, -l+1, ..., l-1, l A. Orbital Angular Momentum 3. Eigenfunctions Lz flm = ￿mflm L2 flm = ￿2 l(l + 1)flm We have found the eigenvalues for the orbital angular momentum operator, but what are these eigenfunctions, flm ? We shall see that they are the spherical harmonics, so that the quantum numbers l and m of the hydrogen atom correspond to angular momentum values. BUT... what about the half integer values? 7 A. Orbital Angular Momentum We have found the eigenfunctions or the orbital angular momentum operator, but what are these eigenfunctions, flm ? The angular momentum operator is L = (ħ/i)(r×∇). Write this in spherical coordinates and we get ￿ Lx = i ￿ ￿ ∂ ∂ − sin φ − cos φ cot θ ∂θ ∂φ ￿ ￿ ￿ ∂ ∂ Ly = + cos φ − sin φ cot θ i ∂θ ∂φ ￿∂ Lz = i ∂φ See board V.A.3.a 8 A. Orbital Angular Momentum ￿∂ Lz = i ∂φ The raising and lowering operators in spherical coordinates are L± = ±￿e ±iφ ￿ ∂ ∂ ± i cot θ ∂θ ∂φ ￿ These can be used to ﬁnd L2. L = −￿ 2 2 ￿ 1∂ sin θ ∂θ ￿ ∂ sin θ ∂θ ￿ 1∂ + sin2 θ ∂φ2 2 ￿ See Grifﬁths problem 4.21 9 A. Orbital Angular Momentum ￿∂ Lz = i ∂φ ￿ ￿ ￿ ￿ 2 1∂ ∂ 1∂ 2 2 L = −￿ sin θ + sin θ ∂θ ∂θ sin2 θ ∂φ2 flm is an eigenfunction of Lz and L2. m Lz fL = ￿∂ m fl = ￿mflm i ∂φ Equivalent to the azimuthal equation, Griffiths 4.21 L2 flm = −￿ 2 ￿ 1∂ sin θ ∂θ ￿ ∂ sin θ ∂θ ￿ ￿ 1∂ + flm = ￿2 l(l + 1)flm sin2 θ ∂φ2 2 Angular equation, Griffiths 4.18 Solution: spherical harmonics 10 A. Orbital Angular Momentum So the eigenfunctions of Lz and L2 are the spherical harmonics. The quantum numbers l and m enumerate the values of the orbital angular momentum. But what about the half integer values we found using the ladder operators? Why did separation of variables not ﬁnd them? Separation of variables looks for solutions as functions of r, θ and φ. The half integer solutions are apparently not functions of position. They, therefore, are not related to ORBITAL angular momentum, but are solutions corresponding to a type of angular momentum intrinsic to a quantum mechanical particle. The classical analogue to such a quantity is SPIN. Grifﬁths warns us not to take this analogy too far. 11 B. Spin To write the mathematical properties of this “spin” type of angular momentum, we just copy the mathematical properties of the orbital angular momentum, which is summarized by the commutation relations. [Sx , Sy ] = i￿Sz S h as the cartesian c omponents Sx, Sy a nd Sz. [Sy , Sz ] = i￿Sx [Sz , Sx ] = i￿Sy Therefore, many of our previous derivations for orbital angular momentum apply also for spin, just replacing L with S. 12 B. Spin Therefore, many of our previous derivations for orbital angular momentum apply also for spin, just replacing L with S. Lz flm = ￿mflm L2 flm = ￿2 l(l + 1)flm whereas flm was a function of position, fsm i s n ot, so it’s more convenient just to write it in D irac notation. flm c ould have been written t his way too -> |lm>. S 2 |sm￿ = ￿2 s(s + 1)|sm￿ L± = Lx ± iLy S± = Sx ± iSy Sz |sm￿ = ￿m|sm￿ Griffiths Problem 4.18 ￿ S± |sm￿ = ￿ s(s + 1) − m(m ± 1)|s(m ± 1)￿ l = 0, 1, 2, · · · m = −l, −l + 1, · · · , l − 1, l 1 3 s = 0, , 1, , · · · m = −s, −s + 1, · · · , s − 1, s 2 2 13 B. Spin 1 3 s = 0, , 1, , · · · m = −s, −s + 1, · · · , s − 1, s 2 2 Spin can have both integer and half integer values since there are position independent solutions to the angular momentum operators for both these cases. The hydrogen atom electron, for example, can take on different l orbital angular momentum values since orbital angular momentum is a function of position. Since spin angular momentum is not a function of position, a quantum mechanical particle can only ever have ONE value s for spin. Spin is therefore an intrinsic property of a quantum mechanical particle, like its mass or charge. An electron has spin 1/2, for example, or a photon has spin 1. 14 B. Spin 1. Spin 1/2 Spin 1/2 systems are important because protons, neutrons, electrons, and other particles in quantum physics all have spin 1/2. It is also one of the simplest quantum systems to study. If s=1/2, then m can only have two values: 1/2 or -1/2. There are therefore only two spin states, |½,½> or |½,-½>. “spin up” ↑,+ “spin down” ↓, - We can write these states as orthonormal column vectors 15 B. Spin We can write these states as orthonormal column vectors ￿ ￿ ￿ ￿ 1 0 χ+ = χ− = 0 1 “spin up” ↑,+ “spin down” ↓,- Any general spin state of a spin 1/2 particle is then a superposition of these. ￿ ￿ a χ= = aχ+ + bχ− b Spin operators can also be written in terms of this basis. ￿ ￿ ￿ Sx = σ x 32 1 0 2 S2 = ￿ 01 4 ￿ spin operator components Sy = σ y 2 total spin (squared ) ￿ Sz = σ z 2 16 B. Spin Spin operators can also be written in terms of this basis. ￿ ￿ 32 1 0 2 S= ￿ 01 4 ￿ ￿ 01 ￿ σx ≡ Sx = σ x 10 2 ￿ ￿ 0 −i ￿ Pauli Spin Matrices σy ≡ Sy = σ y i0 2 ￿ ￿ ￿ 10 Sz = σ z σz ≡ 2 0 −1 See board V.B.1 From this, we see that χ+ and χ- are eigenfunctions (eigenspinors) of S2 and Sz. Eigenvalue 3/4 ħ2 for S2 for both χ+ and χ-. Eigenvalues +ħ/2 and -ħ/2 for Sz for χ+ and χ-, respectively. 17 B. Spin The total wave function of a quantum mechanical particle might depend on, for example, the coordinates x, y, z as well as s and m. Unlike x, y and z, s and m are not continuous. The spinor space is a complex, two dimensional space, unlike the continuous Hilbert spaces that we’ve discussed so far. We will ignore these complications and not look, in this course, at complete wave functions including spin. For spin problems, the wave function is a two component spinor, as in Grifﬁths example of Larmor precession. See Grifﬁths Example 4.3 B. Spin 2. Addition of Angular Momentum What if the electron in the hydrogen atom is not in the ground state and has some orbital angular momentum? The means there is both spin and orbital angular momentum, so we have to add the angular momentum. The TOTAL ANGULAR MOMENTUM of a system is denoted by J=L+S total a ngular momentum orbital a ngular momentum spin a ngular momentum The total angular momentum obeys the same form of commutation relations as the orbital and spin angular momenta. B. Spin J=L+S The total angular momentum obeys the same form of commutation relations as the orbital and spin angular momenta. [Jx , Jy ] = i￿Jz J h as the cartesian c omponents Jx, Jy a nd Jz. [Jy , Jz ] = i￿Jx [Jz , Jx ] = i￿Jy To add the momenta, one uses PRODUCT STATES and CLEBSCH-GORDAN coefﬁcients. C oefficients of the tensor p roduct expansion Tensor products: ⊗ The same is true if we add spin to spin – for example proton spin to electron spin to get the total spin angular momentum of the hydrogen atom. We will skip studying the mathematics of all this. ...
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