Section 5

Question 1

If the energy of the state is lower than the bound energy then

Answer 1

it will be bound

Question 2

Specifically, tunnelling of wave-functions through finite-width barrier means we define a state as bound if

Answer 2

E < V(±∞)

Question 3

In general a potential can have both

Answer 3

bound and scattering states

Question 4

Particle inside a box: infinite square-well

inside the well, behaves like a free particle

Answer 4

-ℏ/2m d^2Ψ(x)/dx^2 = EΨ(x)

with k = √(2mE) / ℏ

Question 5

infinite square-wells have the general solution most conveniently express as

Answer 5

Ψ(x) = Acoskx + Bsinkx

Question 6

infinite square-wells have boundary conditions at

Answer 6

|x| = a => Ψ(±a) = 0

Question 7

Boundary conditions quantize

Answer 7

eigenstates in even/odd pairs Ψ(x)

Question 8

u(x)n+ =

Answer 8

1/√a cos nπx/2a

with n = 1,3, … ∞

Question 9

u(x)n- =

Answer 9

1/√a sin nπx/2a

with n = 2,4, … ∞

Question 10

Energies formula

Answer 10

En+ = n^2 π^2ℏ^2/8ma^2

Question 11

u(x)1+ is the

Answer 11

ground state

Question 12

Normalisation <un±|un±>

Answer 12

= 1

Question 13

Orthogonality <un±|um∓>

Answer 13

= 0

<un±|um±> = δnm

Question 14

Expectation values

Answer 14

<x> = 0
<p> = 0
<p^2> =2mEn±
</p></x>

Question 15

The spatial eigenstates have definite

Answer 15

parity, i.e. reflection symmetry x -> -x

Question 16

An asymmetric potential, V(x) ≠ V(-x), will not

Answer 16

have eigenstates of definite parity

Question 17

The delta function well takes the form

Answer 17

d^2Ψ/dx^2 = -2mE/ℏ^2 Ψ = κ^2Ψ

for κ = √(-2mE)/ℏ

Question 18

For the delta function well the general solution for real κ

Answer 18

Ψ(x) = Ae^(-κx) + Be^(κx)

Question 19

Finite square well: eigenfunctions in the well are with

Answer 19

λ = √(2mE)/ℏ

Question 20

Finite square well: in the walls exponential decay with

Answer 20

κ = √(-2m(E-V))/ℏ

Question 21

solving the finite well:

Answer 21

Even => tan(λa) = κ/λ

Odd => -cot(λa) = κ/λ

Question 22

Schrodinger Equation as a Simple Harmonic Oscillator

Quadratic well

Answer 22

-ℏ^2/2m d^2Ψ/dx^2 + 1/2mω^2x^2Ψ = EΨ

Question 23

The Simple Harmonic Oscillator Hamiltonian is

Answer 23

H(hat) = [p(hat)^2 + (mωx(hat)^2]/2m

Question 24

a(hat)± are the

Answer 24

simple harmonic oscillator ladder operators

Question 25

number operator

Answer 25

note that n(hat) = a(hat)+a(hat)- returns the principal quantum number

Question 26

In general harmonic oscillator eigenfunctions solutions are built from

Answer 26

Hermite Polynomials

Question 27

General potentials

Answer 27

general potentials can be thought of as a stack of thin square potentials

wave-function is oscillatory where net positive kinetic energy E-V.

Question 28

in an allowed region it always tends

Answer 28

to zero

Question 29

in a forbidden region it will always tend

Answer 29

away from zero

Question 30

double - well potential

Answer 30

must have at least two extrema

Question 31

periodic boundary conditions

Answer 31

possible to not have any boundary conditions that fix wave-function at a certain place