SHO

Question 1

Potential in SHO

Answer 1

v(X) = 1/2 kx^2 = 1/2 mw^2x^2 h^2

since w = \sqrt(k/m)

Question 2

first step to sovling SSE

Answer 2

change vars to dimensionless quantities

Question 3

why charcteristic length sclae for QHO but not CHO

what are the other dimensionful parameters

Answer 3

Quantum intorudes hbar (length scale) where classical no length scale

other dimensionful parameters: m and omega

Question 4

charcteristic length scale

Answer 4

[Length] = a = \sqrt(h/mw)

Question 5

Characteristic energy scale

resulting from new dimensionfull parameter

Answer 5

1/2 h w

energy if you were osciallating at characteristic length scale

Question 6

techniques to solve QHO

Answer 6

1. Solve DE (Recursion rel.)
2. algebraic method: raising and lowering ladder operators
3. shooting method where you tune E till see correct boudnary conditions where WF decays exp adn get even/odd solns oscil. (numerical methods)

Question 7

properties of WF of QHO

Answer 7

All WF are bound states
even odd ocillataing solns
all normalisable

Question 8

NBNBNB Energy of QHO in 1d

Answer 8

En = h w (n+1/2)

for n = 0, 1, 2

Question 9

Quantum number labelling comaprision

Answer 9

QHO n = 0, 1, 2 (since E still != 0 when n =0 , E = hw/2)

in ISW n = 1,2,3 to void E = 0 , since there E ~ n^2 (w/o constant factor)

Question 10

How are the energy evals of QHO spaced

Answer 10

Evenly spaced (since E ~n)

unlike ISW where E~n^2 and so gaps increased quadratically

Question 11

what are the evenly spaced gaps in QHO

Answer 11

Gaps evenly spaced by hw

since E = hw (n + 1/2)

Question 12

Eigenfunctions of QHO consis of which parts

Answer 12

3 parts

1. normalisation constnat (ugly but not significant properties)
2. Hermite polynomial (degree of n , either odd or even)
3. Gaussian

Question 13

Hermite polynomial ‘function’ in QHO solution

Answer 13

H_n(x/a)

ORGTHOGONAL polynomials (give rise eigfunc orth)

polynomial degree n
even or odd dep on whether n is even or odd

Question 14

Gaussian in QHO soln

Answer 14

width of a

exp(-x^2/2a^2)

Question 15

Hermite polynomia expression in QHO

Answer 15

H_n(x/a)

where n = 0, 1, 2

Question 16

First & second &thrid polynomial for n= 0, 1, 2

Answer 16

H_0 (x) = 1
H_1 (x) = 2x
H_2 (x) =4x^2 -2

Question 17

Hermite polynomial def

Answer 17

H_n(x) = (-1)^n e^(x^2) d^n/dx^n e^(x^2)

Question 18

GS of QHO from full soln

NBNBNB

Answer 18

WF = (norm const. ) (Hermite) (gauss)

for Ground state:
WF = (norm) H_0 (x/a) (gaussian width ~a)
= (norm) (1) (gaussian width ~a)
~ gaussian

so the GS of QHO is a pure gaussian

Question 19

GS of QHO shpae and reason

Answer 19

H_0 = 1 so GS is constn times gaussian (width ~a)

gs is also even (expected for symm potential)

nodes = 0

Question 20

WHy enegry index match wf #nodes

Answer 20

nodes = n

n starts at 0 and there are 0 nodes in GS

Question 21

Degen in QHO

Answer 21

No degen (in 1d)
Each WF has 1 E

Question 22

what happens in classical forbidden regions

Answer 22

QHO soln WF decays to 0 exponantially

Question 23

where is probability of WF maximal in QHO

Answer 23

At the classical turning points

Question 24

In the CHO the min energy at the bottom of well is E = 0, in QM qhat is the min energy and why

Answer 24

QHO E_0 = hw(0+1/2) = hw/2

in CHO = <p> = 0 sim wwen the particle stands still at the bottom of the well with lowest energy.

In QHO whe CANNOT have = = 0 sim ,since x and p DO NOT COMMUTE. so there must be some uncertainty but we do have that = </p><p> = 0 s, but this des not tell us their uncertainties are zero as in classical case

(sigma_x sigma_p >= h/2) so at minimise uncertainty E_0 = (h/2) (w) mult by w do get enrgy units
=> E_0 = hw/2</p>

Question 25

Which parts of WF are gaussian resp, and which the Hermite poly

Answer 25

for n = 0: H_0 = 1 so only gaussian width ~a inside well

for n > 0 H-n > 1 and so Hermite poly responsible for oscillating behavoiur inside well,
Gaussian resp for exp decay outside well

Question 26

WHy can tunneling resulst in -kinetic energy

Answer 26

Particle simply does not have well defined kintetic energyin QHO so -kinetic energy is okay

Question 27

expectations values of QHO

, <p></p>

Answer 27

<p> Undergo perfect SHO
e.g of correspondence principle that expectation values of observables behave classically
</p>

Question 28

What happens at large ENegies(large n)

Answer 28

The hermite polynomial term grows and thus osciallated rapidly inside well (so that it appears relatively constnat) like one big parabola

e.g. of BOhrs formualtion of corresp. principle, that for large energies the quantum proabbility (expectation val) resemble`s classical

Question 29

FTramsform frpm x to p

Answer 29

ionside integral mult by exp(ikx)

Question 30

FTramsfpr, from p to x

Answer 30

inside integral mult by exp(-ikx)

Question 31

Delta function fourier representaton

what does it revel

Answer 31

delta centred at x0 = constant factor times int over all space of (exp(ik(x-x0)))

it revelas that the fourier transform over a delta function is just a constant function