Physics 228 Today: Ch 41: 1-3: 3D quantum mechanics, hydrogen atom

Transcription

1 Physics 228 Today: Ch 41: 1-3: 3D quantum mechanics, hydrogen atom Website: Sakai 01:750:228 or Happy April Fools Day

2 Example / Worked Problems What is the ratio of the energy of the 3rd to the 2nd excited states of a harmonic oscillator? The harmonic oscillator energy levels are E n = (n+½)ħω, with ω 2 = k'/m. The ground state is n = 0, the second excited state is n = 2, and the third excited state is n = 3. Thus: r = E 3 /E 2 = (3+½)/(2+½) = 7/5. What energy photon is given off when an electron in an infinite 1-D square well transitions from the 2nd excited state to the ground state? The infinite square well energy levels are E n = n 2 ħ 2 π 2 /2mL 2. The ground state is n = 1, and the second excited state is n = 3. Thus the photon or transition energy is E γ = E n=3 - E n=1 = ( )ħ 2 π 2 /2mL 2 = 8ħ 2 π 2 /2mL 2 = 4ħ 2 π 2 /ml 2.

3 Example / Worked Problems What is the ratio of the energy of the 3rd to the 2nd excited states of a harmonic oscillator? The ground state is n = 0. What energy photon is given off when an electron in an infinite 1-D square well transitions from the 2nd excited state to the ground state? The ground state is n = 1. While we use n to denote the levels, note that for some potentials we use n = 0 for the ground state, while for other potentials we use n = 1 for the ground state. On a test, we will expect you to know n of the ground state.

4 Example / Worked Problems II What is the momentum of an electron in the 1 st excited state of an infinite square well potential? The infinite square well energy levels are E n = n 2 ħ 2 π 2 /2mL 2. The first excited state is n = 2. The only way we know how to get the momentum is from the energy. The kinetic energy in this state is E 2 = 4ħ 2 π 2 /2mL 2 = 2ħ 2 π 2 / ml 2. So we expect the magnitude of the momentum is p = (2mE) ½ = 2πħ/L = h/l. But is this the answer to the question? You need to be careful about how it is asked. If the "expectation value" of the momentum is requested, this is the signed momentum, not the magnitude, and the answer is 0. Since the particle in a box isn't going anywhere, its average momentum must be 0, although the average magnitude of the momentum is non-0. Be careful about whether the average momentum or the magnitude of the momentum is requested. If you have this on a test, and are not sure, ASK!

5 3D Quantum Mechanics When we go from 1D to 3D, we need to take the 3D derivative. We need vector calculus since the derivative can be different in different directions. We use the gradient operator: = x ˆx + yŷ + zẑ The time-independent S.E. now becomes 2 2m 2 ψ(x )+U(x)ψ(x )=Eψ(x ) or, for a 3D square well potential we can write 2 2m 2 ψ(x )= E U(x ) ψ(x ) where inside the box U 0 = 0, and outside the box U 0 =.

6 Infinite 3D Square Well Let's consider a simple cubical box, each side of length L, and the box extending from 0 to L in x, y, and z directions. Again, as in the 1D case, outside the box U =, so the differential equation is solved if the wave function ψ(x) = 0. Thus the particle is in the box and we require ψ(x) = 0 at x (or y or z) = 0 or x (or y or z) = L. In the box U(x) = 0, so S.E. is simply: 2 2m 2 ψ(x )=Eψ(x ) We again need a wave function that has the same form as its 2nd derivative, and the right boundary conditions. It is similar to the 1D case, and we can guess the result: ψ(x) = (2/L) 3/2 sin(qπx/l) sin(rπy/l) sin(sπz/l) Note the new indices: q in x, r in y, and s in z directions.

7 Infinite 3D Square Well Energy Levels In the 1D case, we had k = qπ/l, ψ(x) = (2/L) 1/2 sin(qπx/l), and E q = q 2 ħ 2 π 2 /2mL 2, with q = 1, 2, 3... What happens now with the 3D case where ψ(x) = (2/L) 3/2 sin(qπx/l) sin(rπy/l) sin(sπz/l)? The three directions are independent, so we will have k x = qπ/l, ψ x (x) = (2/L) 1/2 sin(qπx/l), and E qx = q 2 ħ 2 π 2 /2mL 2, with q = 1, 2, 3... and similar in the y and z directions. Then E qrs = E q + E r + E s = (q 2 +r 2 +s 2 )ħ 2 π 2 /2mL 2. The g.s. will have q=r=s=1, and E 111 = 3ħ 2 π 2 /2mL 2.

8 iclicker E qrs = E q + E r + E s = (q 2 +r 2 +s 2 )ħ 2 π 2 /2mL 2. The g.s. will have q=r=s=1, and E 111 = 3ħ 2 π 2 /2mL 2. What is the energy of the 1st excited state? How many of them are there? a) E 2 = E gs, 1 b) E 2 = 2E gs, 1 c) E 2 = 4E gs, 1 d) E 2 = 2E gs, 3 e) E 2 = 4E gs, 3

9 Degenerate States States of the same energy are called "degenerate states". Is is possible to break the degeneracy by having all 3 sides of the box different lengths, so that E qrs = E qx + E ry + E sz = (q 2 +r 2 +s 2 )ħ 2 π 2 /2mL 2 E qrs = E qx + E ry + E sz = (q 2 /L x 2 +r 2 /L y 2 +s 2 /L z2 ) ħ 2 π 2 /2m.

10 Probability Distributions Again in 3D QM, as in the 1D case, the probability of finding a particle at some point x 0, y 0, z 0 in space is given by ψ*ψ = ψ 2 at that point. Note an odd feature of QM: the particle in, e.g., the ψ 2,1,1 state can never be found in the x=l/2 plane, but it can be on one side or the other, and one expects moves between the two sides. It crosses the plane, but it is never found there!

11 A 3D spherical box A complication, not covered in the textbook, is to have a 3D spherical box. Now using x, y, z makes the problem more difficult since the, e.g., x limits depend on the y and z values. Instead we use spherical coordinates, r, θ, φ. The 3D Laplacian operator operating on ψ makes S.E., slightly rearranged, of the form: 2 ψ(x )= 1 r 2 r r 2 ψ r + 1 r 2 sin θ θ sin θ ψ θ + 1 r 2 sin 2 θ 2 ψ φ 2 = 2mE 2 ψ It is natural to guess the solution is of the form ψ(r,θ,φ) = R(r)Θ(θ)Φ(φ). Since each term differentiates only one of the three functions, we come up with 3 independent equations, like in the 3D cubical box case. The radial function R needs to go to 0 at the edge of the box, r = r 0, but not at the center, r=0. The angular dependences are usually written as "spherical harmonics" Y lm (θ,φ), which you might have encountered elsewhere. The are of the form of Legendre polynomials times e ±imφ. (Actually solving the equations is too much math for us.)

12 The Hydrogen Atom For hydrogen, like the spherical box, we have spherical symmetry, but the potential energy is the Coulomb potential energy between opposite charges: U(r) = -e 2 /4πε 0 r. The radial wave function R(r) must go to 0 at large r, since the electrons are bound to the protons, but does not necessarily go to 0 at the origin. The angular functions must be finite and periodic in φ, with a period of 2π. The wave functions are characterized by 3 quantum numbers...

13 The Principal Quantum Number n = 1, 2, 3,...: The wave function is of the form of a polynomial series in r ( i=0,...n-1 c i r i ) (some c i are 0) times an exponential exp(-r/na 0 ). Here a 0 is the Bohr radius, nm. The resulting energies are the same as in the Bohr model, E n = ev / n 2. n = 1 n = 2 n = 3 examples of wave functions We refer to the n=1, 2, 3... sets of states as the K, L, M, N,... shells.

14 The Orbital Angular Momentum Quantum Number l = 0, 1,... n-1: (Or if you fix l, then n = l, l+1, l+2,...) Formally, the orbital angular momentum in quantum mechanics is not actually lħ, but is rather L = [l(l+1)] ½ ħ. But we commonly refer to: l = as an s p d f g h... state (or L/ħ = ) and say the orbital angular momentum is quantized to integral multiples of ħ. In the Bohr model each value of n describes a circular orbit with a particle speed and radius and orbital angular momentum. With the S.E., each orbital (each particular value of n) has n different possible degenerate orbital angular momenta states.

15 The Magnetic Quantum Number m l = 0, ±1,... ±l: "magnetic" or "zcomponent of angular momentum" quantum number. The z-projection of the orbital angular momentum has magnitude m l ħ. The azimuthal direction of the angular momentum L is arbitrary, and if, e.g., there is a magnetic field in the z direction the angular momentum will precess about the z direction. In the semiclassical picture shown, the angular momentum of an l = 2 state with m l = +2 makes an angle of cosθ = 2/ 6 = θ = 35 o with the z-axis. For m l = +1, cosθ = 1/ 6 = θ = 66 o.

16 Summary of Hydrogen States States within the same shell are degenerate. Shell n l m l notation K s L s " 2 1 0, ±1 2p M s " 3 1 0, ±1 3p " 3 2 0, ±1, ±2 3d N s etc...

17 iclicker How many degenerate states are there in the n=3 shell? (If you know about the spin of the electron, ignore it.) Shell n l m l notation K s L s " 2 1 0, ±1 2p M s " 3 1 0, ±1 3p " 3 2 0, ±1, ±2 3d N s etc... a) 1 b) 5 c) 9 d) 12 e) 14

18 More probability distributions In the case of xyz coordinate systems, the probability is ψ*ψ because the volume element is constant. But for spherical coordinates systems, the volume integral uses: dv = r 2 dr dcosθ dφ so the probability distribution uses r 2 ψ*ψ in the plots to the right. Thus all the distributions go to 0 at the origin. If we just showed ψ or ψ*ψ, you would see the s orbits do not go to 0 at the origin, but the p, d, f... orbits do. Why is this?

19 More probability distributions sphere To indicate the 3D nature of the distributions, we can show them in the xz plane. Rotate the image around the z axis for the 3D view. sphere + shell donut dumbbell

20 iclicker What angle does the angular momentum make with the z- axis in the semi-classical picture for the m l = 0 2p state? a) There is no 2p state, you need at least 3p b) cosθ = 1/ 2 c) sinθ = 1/ 2 d) 0 o e) 90 o