Modern Physics. Unit 6: Hydrogen Atom - Radiation Lecture 6.3: Vector Model of Angular Momentum

Transcription

1 Modern Physics Unit 6: Hydrogen Atom - Radiation ecture 6.3: Vector Model of Angular Momentum Ron Reifenberger Professor of Physics Purdue University 1

2 Summary of Important Points from ast ecture The magnitude of the angular momentum is related to the quantum number l according to = ( + 1) ; = 0, 1,... l= Allowed m 0-1,0,1 ±, ±1,0 ±3, ±, ±1, 0 ±4, ±3, ±, ±1, 0 = The z-component of is quantized to certain values specified by the allowed values of m.

3 Example I: What are the eigenvalues for the operators and z when an electron is in the quantum state l=1, m=1? Ψ n,1,1( r, θφ, ) = Rn, 1( r) Y1, 1( θφ, ) Y1,1( θφ, ) = sinθ + sin θe sinθ θ θ sin θ φ 3 1 ( sinθ ) ( θ ) ( sinθ cosθ) ( i) = + sinθ θ sin θ = + = = e ( sinθ) ( sin θ cos θ) e sinθ sinθ 3 3 sin e e 3

4 z Y 1,1 3 ( θφ, ) = i sinθe φ 3 i = i ( i) sinθ e φ 3 = + 1 sinθ e If you understand the theory, you can simply write down the answers = ( + 1) = z and = 1 = m = m= 1 4

5 * * 1 1 = Ψn. 1.1 op Ψn,1,1 dv = Ψn.1.1 sinθ + Ψn,1,1 dv sinθ θ θ sin θ φ = Example II: What is the expectation value for when an electron is in the quantum state l=1, m=1? sinθ θ φ dv r dr d d Ψ ( r, θφ, ) = R ( ry ) ( θφ, ) n,1,1 n,1 1,1 * * 1 1 = R n,1 () r Y 1, 1( θ, φ) sin θ + R n,1( r) Y1, 1( θ, φ) r sinθ drdθ dφ sinθ θ θ sin θ φ = ( ) ( ) sin sin r R r θ θ e sinθ dr dθ dφ sinθ θ θ sin θ φ * * rr n,1 n,1 Y 1,1( θφ, ) + * * 3 1 = r R n,1 ( r) Rn, 1( r) Y 1,1( θφ, ) ( sinθcosθ) + ( sinθ) e sinθ dr dθ dφ sinθ θ sin θ = * * rr n, 1() r Rn,1( r) Y 1, 1( θφ, ) ( ) ( i) sin θ + cos θ e sinθ dr dθ dφ sinθ sinθ * * 3 = r R n,1 () r Rn, 1() r Y 1,1( θ, φ) ( sinθ ) e sinθ dr dθ dφ 5

6 Repeat last line on previous slide * * 3 = r R n,1 ( r) Rn,1( r) Y 1,1( θφ, ) ( sinθ) e sinθ dr dθ dφ = = = 3 r R r R r Y θφ, ) ( sin θ ) e θ dr dθ dφ * * n,1 () n,1() 1,1( sin * () ( 3 r R r R r dr sin d dφ n,1 n,1 ) ( θ ) sinθ θ π * 3 r R n,1 () r Rn,1() r dr ( sin θ ) sin 0 0 π θ dθ π * 3 1 r R n,1 ( r) Rn,1( r) dr ( cos3θ 9cosθ) = * r R n,1 ( r) Rn,1( r) dr ( 1 (1) ) (( 1 (1) )) * * n,1 n,1 π n,1 n,1 π = r R () r R () r dr r R () r R () r dr 8 + = 6 π 6 π π 0 dφ = = π = 3 as expected π π normalized! 6

7 Note: to save time, use on-line integrator; it s free See: 7

8 The vector model of orbital angular momentum for l= a sphere of radius = ( + 1) z z m cosθ = = ( + 1) θ x y z z z z = ; m= = ; m= 1 = 0; m= 0 = ; m= 1 z = ; m= x, y are not defined! Note: While is called a "vector", the quantum angular momentum is a special kind of vector because it's projection along a direction in space is quantized to specific values. 8

9 Visualizing the vector model of angular momentum l+σ = 5/ l= Spin Family (009) by physicist-turned-sculptor Julian Voss-Andreae. 9

10 Example III: The square of the orbital angular momentum of an electron in a H atom is measured in a quantum lab and found to have a value of 30. If the z-component of the angular momentum is measured, what are the possible results for this measurement? ( ) = + 1 = 30 = 5 m + = 5, 4, 3,, 1,0, + 1, +, + 3, + 4, + 5, z 10

11 How to select (specify) the z-axis? According to quantum physics, we can only know one of three components of. By convention, we always align the known quantized component along the z-axis, usually by an external magnetic field. There is an equivalent uncertainty principal for angular momentum that can be stated: φ / z It follows if we know z exactly (which we do), then Δ z =0 and we have no knowledge of the angle φ. This just means x and y are not known 11

12 SUMMARY: Space quantization in quantum physics Only certain orientations of angular momentum are allowed: Suppose we could prepare hydrogen atoms in l= state Perform experiment to measure z-component of We would measure Now perform another experiment with completely different z-axis We would still measure =,1,0, 1, z =,1,0, 1, z We get the same answer, no matter what direction we choose for z! This behavior is completely different from classical vectors These results for angular momentum are critical to understanding the magnetic moments of atoms. Stay tuned! 1

13 The Effective Potential that appears in the Radial Wave equation for H d e 1 ( + 1) me dr 4πε o r mer ( rr() r ) = + E ( rr() r ) Kinetic Energy (radial direction) Electrostatic Potential Energy Rotational Energy Effective Potential Energy Total Energy l=0 l=1 Energy (ev) Effective Potential l=0 U(r) Rotational Energy (ev) Effective Potential l=1 Effective Potential (l=1) U(r) Rotational r/a o r/a o 13

14 Up next Radiative Transitions 14

15 Appendix: Why we can t know x and y? x z x and y not specified z quantized y Complete Knowledge z m e v x û x and y specified y If we knew x, y and z, then we would know precisely (both it s magnitude and the direction of its orientation specified by the vector û ) If we know the orientation of, then we know the electron motion must be confined to a plane perpendicular to. If the motion is confined to a plane, then we know precisely that the electron s position is also confined to that plane; in other words, we know precisely that the electron s position along û must be zero. If the motion is confined to a plane, then we know precisely that the electron s momentum must also be confined to that plane; in other words, we also know precisely that the electron s momentum along û must be zero. But the Heisenberg Uncertainty Principle tells us that we cannot know precisely both position and momentum in the same direction. This implies that we cannot know x, y AND z simultaneously with high precision. 15

16 Summary of results in spherical polar coordinates In spherical polar coordinates, the gradient operator is written as 1 ˆ 1 = rˆ + θ + ˆ φ r r θ rsinθ φ The momentum operator is given by 1 ˆ 1 ˆ ˆ i = i r + θ + φ sin r r θ r θ φ The angular momentum operator is given by ˆ 1 ˆ 1 ˆ ˆ op = i r = i r r + θ + φ r r θ rsinθ φ The expectation value for the angular momentum is given by = Ψ Ψ dv = i Ψ r ΨdV * * ˆop Is Ψ AWAYS an eigenvalue of op? ( ) 16