Orbital Angular Momentum of the Hydrogen Atom

Orbital Angular Momentum of the Hydrogen Atom

Department of Physics and Astronomy School of Arts and Sciences Rutgers, The State University of New Jersey 136 Frelinghuysen Road Piscataway, NJ 08854-8019 physics.rutgers.edu p. 848-445-9034 f. 732-445-8187 Physics and Astronomy Colloquium Physics Lecture Hall Wednesday, October 18, 2017 4:45PM Klaus Mølmer Aarhus University, Denmark Quantum Mechanics 2.0: from weird to wired Abstract: Nearly a century ago, quantum mechanics revolutionized physics and provided the theoretical basis and insights that underlie modern technologies such as lasers, integrated circuits, catalysts, nuclear power, and medical imaging. Quantum theory also came with a range of challenging paradoxes and strange phenomena that, until this day, divide physicist in fractions that hold very different beliefs about the physical interpretation of the theory and the true nature of the quantum world. We are currently witnessing a "second quantum revolution", in which the most paradoxical quantum effects are being explored for their potential applications. In the seminar, I shall review the ideas behind quantum cryptography, quantum computing and quantum metrology; I shall give examples of the experimental implementations and the progress achieved so far, and I shall discuss how new applications have led to new research questions and to new connections between different branches of physics. Coffee and Tea at 4:30 p.m. 2

Last lecture, we started with the 3D time-independent Schrodinger equation and moved into spherical coordinates (because the Coulomb potential is spherically symmetric). ~ 2 2m r2 (~x)+v (~x) (~x) =E (~x) ~ 2 2µ apple 1 r 2 @ @r r 2 @ @r + 1 r 2 sin @ @ sin @ @ + 1 r 2 sin 2 @ 2 1 e 2 @ 2 4 0 r = E 3

With some algebraic manipulation, we were able to separate the equation so that we had to solve two separate equations: apple 1 R @ @r r 2 @R @r 2µr 2 (V E) = `(` + 1) ~ 2 radial components only apple 1 Y sin @ @ sin @Y @ + 1 Y sin 2 @ 2 Y @ 2 = `(` + 1) angular components only 4

The angular differential equation has no dependence on the potential. The solution (known since 1782) is known as the spherical harmonics. apple 1 @ sin @Y 1 @ 2 Y + Y sin @ @ Y sin 2 @ 2 = `(` + 1) Let s try our last trick of separation of variable again: Y (, )= ( ) ( ) Plug that in and multiply by sin 2 (theta): apple 1 sin @ sin @ + `(` + 1) sin 2 + @ @ 1 @ 2 @ 2 =0 polar components only azimuthal components only 5

As before, we get {...} + {...} = 0, so we assign each to a constant, in this case ±m 2 1 apple sin @ @ sin @ @ 1 @ 2 + `(` + 1) sin 2 = m 2 @ 2 = m 2 Let s solve the second equation. It s pretty simple: ( )=e im Notice that the azimuthal wavefunction is complex. Also notice that there is a constant of integration, m. 6

Let s try to restrict the constant of integration by the boundary condition: This means that: ( +2 ) = ( ) e im( +2 ) = e im e im2 = e im And subsequently, e im2 =cos(m2 )+i sin(m2 ) =1 Therefore m=0, ±1, ±2, ±3, etc. As a result of the boundary condition, we have a quantized constant of integration m. This particular constant is the magnetic quantum number. 7

The full solution to the Hydrogen atom is: n`m(r,, )= s 2 na 0 3 (n ` 1)! 2n(n + `)! e r/na 0 2r na 0 ` L 2`+1 n ` 1 2r na 0 Y m ` (, ) here n is the principal quantum number, a0 is the Bohr radius, and L is the associated Laguerre polynomial... L 0 0(x) =1 L 0 1(x) = x +1 L 0 2(x) =x 2 4x +2 L 1 0(x) =1 L 1 1(x) = 2x +4 L 1 2(x) =3x 2 18x + 18 L 2 0(x) =2 L 2 1(x) = 6x + 18 8

Specifically, the spherical harmonics following solutions: Y m ` (, ) have the where m and l 0 are integers, and m has the possible values: -l, -l+1,..., -1, 0, 1,..., l-1, l. l is the orbital angular momentum quantum number. 9

Recap: the general solution to the hydrogen atom is given by: where three quantum numbers (discrete constants of integration) are required for the solution. n (principal quantum number) = 1, 2, 3 n`m(r,, )=R(r)Y m ` (, ) l (oribtal angular momentum quantum number) = 0, 1, 2,..., n-1 m (magentic quantum number) = -l, -l+1,..., l-1, l 10

The energy of a state depends only on n. E n = µe 4 32 2 2 0 ~2 n 2 Same as the Bohr model! But for each n, there are n values of l. And for each l, there are (2l+1) values of m. 11

n l m (2l+1) degeneracy 1 0 0 1 1 2 3 0 0 1 1 0, ±1 3 0 0 1 1 0, ±1 3 2 0, ±1, ±2 5 4 9 For each n, there are n 2 states all with the same energy. So if the energy is the same, what s different about the states? 12

The quantum number l expresses the quantization of the magnitude of the orbital angular momentum: L = ~ p`(` + 1) The magnetic quantum number m expresses the quantization of the direction of orbital angular momentum: L z = m~ Since Lz=L cosθ, then cos = L z L = m~ p`(` + 1)~ = m p`(` + 1) so there are 2l+1 possible angles of L with respect to the z axis. 13

Spectroscopic notation: l = 0 1 2 3 4 5... s p d f g h... s = sharp p = principal d = diffuse f = fundamental g =? Sober Physicists Don't Find Giraffes Hiding In Kitchens Like Mine... 14

PRS Question: For the Hydrogen atom in the n=4, l=3 state, how many possible values are there for the z-component of the angular momentum? A) 4 B) 5 C) 6 D) 7 E) 8 15

If l=3, then m=-3, -2, -1, 0, 1, 2, 3. So the answer is 7. PRS Question: For the Hydrogen atom in the n=4, l=3 state, how many possible values are there for the z-component of the angular momentum? A) 4 B) 5 C) 6 D) 7 E) 8 16

Since the electron is orbiting the proton, it s movement constitutes a current. Let s pretend it s moving in a circle. The period of one orbit ist = 2 r v So the current is I = q t = e 2 r/v = ev 2 r And any current has a magnetic moment: µ = I A = ev 2 r r2 = evr 2 Therefore µ L = evr/2 m e vr = e! ~µ = 2m e e L ~ 2m e 17

For convenience, let s introduce a constant g called the gyromagentic-factor or g-factor, where g=1. Then we have: When a magnetic moment is placed in an external magnetic field in the z direction, the potential energy of the interaction is: E = ~µ ~B = ~µ = g e 2m e ~ L g e 2m e ~L ~B = g eb 2m e L z Substituting for Lz, we get Bohr Magneton E = g eb 2m e (m~) =mgµ B B where µ B e~ 2m e 5.8 10 5 ev T 18

By applying an external magnetic field to the Hydrogen atom, the levels should be split according to their magnetic quantum number m. This is the Normal Zeeman effect. (This is not what actually happens) E = mgµ B B where m =0, ±1, ±2, +2-13.6/9 ev 3s 0 3p 0 +1 3d 0 +1-1 -1-2 -13.6/4 ev 2s 0 2p 0 +1-1 -13.6 ev 1s 0 19

Selection rules for radiative transitions: Allowed if Δl=±1 and Δm=0,±1 (Δn can be anything) Forbidden transition can occur, but much less frequently Notice that the selection rule strongly suggests that the photon has spin 1. 20

Consider a Balmer transition in a magnetic field. There are many more transitions allowed, but because of the selection rules, only three are possible: E = E Bohr + gµ B B E = E Bohr E = E Bohr gµ B B Balmer lines are now split into three: 3d 2p Δm=0 Δm=-1 Δm=1 m=2 m=1 m=0 m=-1 m=-2 m=1 m=0 m=-1 21

In an external B-field, every Bohr-model photon line should split into exactly 3 equally-spaced lines. This is the normal Zeeman effect. The size of the spacing is ΔE=gμB. B. Experimentally, the Zeeman effect does occur, but for many elements (such as in Hydrogen) it is not normal in that the spacing is not gμb. B. This is known as the anomalous Zeeman effect. Photon lines can often be split into more than three in an external B-field. Even when B=0, there can still be observed a splitting. 22