QM in 3D

Question 1

What is the 3D version of the TISE?

Answer 1

-ћ/2m *∇^2 u(r) + V(r)u(r) = Eu(r), where ∇^2 is the laplacian operator and r is the position vector

Question 2

Which 3 ODEs do we get when we separate the TISE with central potential u(r)?

Answer 2

• Equation for ф(Ф) with new constant m(l)
• Equation for Θ(θ) with new constant l(l+1)
• Equation for R(r) involving the potential U(r)

Question 3

What does m(l) represent? What does l represent?

Answer 3

m(l) = magnetic quantum number
l = orbital quantum number

Question 4

What is the first ODE?

Answer 4

d^2ф/dФ^2 + m(l)^2 *ф = 0

Question 5

How can we solve the first ODE?

Answer 5

Try ф(Ф) = Aexp(im(l)Ф), and sub into ODE: find it works

Question 6

What do we find from solving the first ODE?

Answer 6

That ф must be single-valued: ф(Ф) = ф(Ф+2kπ), so Aexp(im(l)2π) = 1, so m(l) = 0, +/-1, +/- 2 etc

Question 7

How is l related to m(l)?

Answer 7

l >= |m(l)|

Question 8

What do we get from combined Θ and ф solutions?

Answer 8

Spherical harmonics.

Question 9

What do we need to solve the third ODE?

Answer 9

The potential U(r). Most important is coulomb potential U(r) = -e^2/4πε0r

Question 10

What equation do we obtain for the enrgy En?

Answer 10

En = -me^4/(2(4πε0)^2*ћ^2) * 1/n^2, where n = 1,2,3… and n>l+1 (n = principle quantum number)

Question 11

What can the 3 quantum numbers equal?

Answer 11

n = 1,2,3…, l = 0,1, … (n-1), m(l) = 0, +-1, +-2, … +-l
Both energy and angular momentum are quantised.

Question 12

What happens if the energy E in the third ODE is >0?

Answer 12

The electron is free and the H atom is ionised.

Question 13

What is the equation for the total energy E?

Answer 13

E = Kr + K0 - e^2/4πε0r, where Kr is purely radial motion and K0 is orbital motion

Question 14

When is the square bracket in the radial equation (third OD) purely radial?

Answer 14

Only when K0 cancels with the l(l+1) term, so K0 must equal this term.

Question 15

What is the third ODE equal to?

Answer 15

1/r^2 * d/dr(r^2 * dR/dr) + [2m/ћ^2 * (E-U(r)) - l(l+1)/r^2]*R = 0

Question 16

What does K0 therefore have to equal?

Answer 16

K0 = ћ^2 *l(l+1)/2mr^2

Question 17

How can we relate K0 to classical angular momentum?

Answer 17

L = mv0r, so K0 = 1/2mv0^2

Question 18

How can we find L from K0?

Answer 18

ћ^2 l(l+1)/2mr^2 = 1/2mv0^2 = 1/2L^2/mr^2, where L^2 = ћ^2 *l(l+1) -> L quantized in units of ћ

Question 19

What does the quantum number l mean?

Answer 19

l determines the magnitude of the angular momentum, but the angular momentum is a vector.

Question 20

What direction does the vector L point in for a plane rotating?

Answer 20

Points perpendicular to the plane of rotation.

Question 21

What does the quantum number m(l) mean?

Answer 21

Lz = ћ*m(l)

Question 22

If, for example, l = 2, what does L and Lz equal?

Answer 22

L = ~ 2.45ћ, Lz = +-2ћ, +-ћ, 0 -> L cannot point along z-axis so Lz < |L|

Question 23

What would be the problem if L pointed along the z-axis exactly?

Answer 23

The linear momentum p would have no z-component, so HUP would equal infinity which is not possible for a localised atomic state.

Question 24

Which components of L can we know?

Answer 24

Can only know one component in one direction, so if we know Lz, we can’t know Lx and Ly.

Question 25

Why do we ‘choose’ a z-axis?

Answer 25

Example of external field e.g. magnetic field with orbiting electron as current loop, magnetic moment of electron m = πr^2 * I z(hat)
-Component of L parallel to B-field is Lz which determines strength of magnetic interaction.