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

Physics 228 Today: Ch 41: 1-3:

3D quantum mechanics, hydrogen atomWebsite: Sakai 01:750:228 or www.physics.rutgers.edu/ugrad/228

Happy April Fools Day

Monday, April 1, 2013

http://www.physics.rutgers.edu/ugrad/228

http://www.physics.rutgers.edu/ugrad/228

Example / Worked ProblemsWhat is the ratio of the energy of the 3rd to the 2nd excited states of a harmonic oscillator?

The harmonic oscillator energy levels are En = (n+½)ħω, with ω2 = 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 = E3/E2 = (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 En = n2ħ2π2/2mL2. The ground state is n = 1, and the second excited state is n = 3.

Thus the photon or transition energy is

Eγ = En=3 - En=1 = (32-12)ħ2π2/2mL2 = 8ħ2π2/2mL2 = 4ħ2π2/mL2.


Example / Worked ProblemsWhat 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.


Example / Worked Problems IIWhat is the momentum of an electron in the 1st excited state of an infinite square well potential?

The infinite square well energy levels are En = n2ħ2π2/2mL2. 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 E2 = 4ħ2π2/2mL2 = 2ħ2π2/mL2. 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!Monday, April 1, 2013

3D Quantum Mechanics

The time-independent S.E. now becomes

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:

�∇ =∂

∂xx +

∂

∂yy +

∂

∂zz

−�2

2m∇2ψ(�x) + U(x)ψ(�x) = Eψ(�x)

or, for a 3D square well potential we can write−�2

2m∇2ψ(�x) =

�E − U(�x)

�ψ(�x)

where inside the box U0 = 0, and outside the box U0 = ∞.Monday, April 1, 2013

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.

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)

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∇2ψ(�x) = Eψ(�x)

Note the new indices: q in x, r in y, and s in z directions.


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 Eq = q2ħ2π2/2mL2, 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 kx = qπ/L, ψx(x) = (2/L)1/2sin(qπx/L), and Eqx = q2ħ2π2/2mL2, with q = 1, 2, 3 ... and similar in the y and z directions. Then

Eqrs = Eq + Er + Es = (q2+r2+s2)ħ2π2/2mL2.

The g.s. will have q=r=s=1, and E111 = 3ħ2π2/2mL2.


iClicker

Eqrs = Eq + Er + Es = (q2+r2+s2)ħ2π2/2mL2.

The g.s. will have q=r=s=1, and E111 = 3ħ2π2/2mL2.

What is the energy of the 1st excited state? How many of them are there?

a) E2 = Egs, 1

b) E2 = 2Egs, 1

c) E2 = 4Egs, 1

d) E2 = 2Egs, 3

e) E2 = 4Egs, 3


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

Eqrs = Eqx + Ery + Esz = (q2+r2+s2)ħ2π2/2mL2 →

Eqrs = Eqx + Ery + Esz = (q2/Lx2 +r2/Ly2 +s2/Lz2) ħ2π2/2m.


Probability Distributions

Again in 3D QM, as in the 1D case, the probability of finding a particle at some point x0, y0, z0 in space is given by ψ*ψ = |ψ|2 at that point.

Note an odd feature of QM: the particle in, e.g., the ψ2,1,1 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!


A 3D spherical boxA 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:

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 = r0, but not at the center, r=0. The angular dependences are usually written as "spherical harmonics" Ylm(θ,φ), which you might have encountered elsewhere. The are of the form of Legendre polynomials times e±imφ.

∇2ψ(�x) =1r2

∂

∂r

�r2 ∂ψ

∂r

�+

1r2 sin θ

∂

∂θ

�sin θ

∂ψ

∂θ

�+

1r2 sin2 θ

∂2ψ

∂φ2=−2mE

�2ψ

(Actually solving the equations is too much math for us.)Monday, April 1, 2013

The Hydrogen AtomFor hydrogen, like the spherical box, we have spherical symmetry, but the potential energy is the Coulomb potential energy between opposite charges:

U(r) = -e2/4πε0r.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 ...


The Principal Quantum Numbern = 1, 2, 3, ...: The wave function is of the form of a polynomial series in r (∑i=0,...n-1 ciri) (some ci are 0) times an exponential exp(-r/na0). Here a0 is the Bohr radius, 0.053 nm. The resulting energies are the same as in the Bohr model, En = -13.6 eV / n2.

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nction

s

n = 1 n = 2 n = 3

We refer to the n=1, 2, 3... sets of states as the K, L, M, N, ... shells.


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)]½ħ. But we commonly refer to:

l = 0 1 2 3 4 5 ...

as an s p d f g h .... state

(or L/ħ = 0 √2 √6 √12 √20 √30)

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.


The Magnetic Quantum Number

ml = 0, ±1, ... ±l: "magnetic" or "z-component of angular momentum" quantum number. The z-projection of the orbital angular momentum has magnitude mlħ.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 ml = +2 makes an angle of cosθ = 2/√6 = 0.816 ➭ θ = 35o with the z-axis. For ml = +1, cosθ = 1/√6 = 0.408 ➭ θ = 66o.


Summary of Hydrogen StatesStates within the same shell are degenerate.

Shell n l ml notationK 1 0 0 1sL 2 0 0 2s" 2 1 0, ±1 2pM 3 0 0 3s" 3 1 0, ±1 3p" 3 2 0, ±1, ±2 3dN 4 0 0 4s

etc...


iClickerHow many degenerate states are there in the n=3 shell? (If you know about the spin of the electron, ignore it.)

Shell n l ml notationK 1 0 0 1sL 2 0 0 2s" 2 1 0, ±1 2pM 3 0 0 3s" 3 1 0, ±1 3p" 3 2 0, ±1, ±2 3dN 4 0 0 4s

etc...

a) 1

b) 5

c) 9

d) 12

e) 14


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 = r2dr dcosθ dφso the probability distribution uses r2ψ*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?


More probability distributions

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 sphere + shell donut dumbbell


iClickerWhat angle does the angular momentum make with the z-axis in the semi-classical picture for the ml = 0 2p state?

a) There is no 2p state, you need at least 3p

b) |cosθ| = 1/√2

c) |sinθ| = 1/√2

d) 0o

e) 90o


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

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