SHO Flashcards
Potential in SHO
v(X) = 1/2 kx^2 = 1/2 mw^2x^2 h^2
since w = \sqrt(k/m)
first step to sovling SSE
change vars to dimensionless quantities
why charcteristic length sclae for QHO but not CHO
what are the other dimensionful parameters
Quantum intorudes hbar (length scale) where classical no length scale
other dimensionful parameters: m and omega
charcteristic length scale
[Length] = a = \sqrt(h/mw)
Characteristic energy scale
resulting from new dimensionfull parameter
1/2 h w
energy if you were osciallating at characteristic length scale
techniques to solve QHO
- Solve DE (Recursion rel.)
- algebraic method: raising and lowering ladder operators
- shooting method where you tune E till see correct boudnary conditions where WF decays exp adn get even/odd solns oscil. (numerical methods)
properties of WF of QHO
All WF are bound states
even odd ocillataing solns
all normalisable
NBNBNB Energy of QHO in 1d
En = h w (n+1/2)
for n = 0, 1, 2
Quantum number labelling comaprision
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)
How are the energy evals of QHO spaced
Evenly spaced (since E ~n)
unlike ISW where E~n^2 and so gaps increased quadratically
what are the evenly spaced gaps in QHO
Gaps evenly spaced by hw
since E = hw (n + 1/2)
Eigenfunctions of QHO consis of which parts
3 parts
- normalisation constnat (ugly but not significant properties)
- Hermite polynomial (degree of n , either odd or even)
- Gaussian
Hermite polynomial ‘function’ in QHO solution
H_n(x/a)
ORGTHOGONAL polynomials (give rise eigfunc orth)
polynomial degree n
even or odd dep on whether n is even or odd
Gaussian in QHO soln
width of a
exp(-x^2/2a^2)
Hermite polynomia expression in QHO
H_n(x/a)
where n = 0, 1, 2
First & second &thrid polynomial for n= 0, 1, 2
H_0 (x) = 1
H_1 (x) = 2x
H_2 (x) =4x^2 -2
Hermite polynomial def
H_n(x) = (-1)^n e^(x^2) d^n/dx^n e^(x^2)
GS of QHO from full soln
NBNBNB
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
GS of QHO shpae and reason
H_0 = 1 so GS is constn times gaussian (width ~a)
gs is also even (expected for symm potential)
nodes = 0
WHy enegry index match wf #nodes
nodes = n
n starts at 0 and there are 0 nodes in GS
Degen in QHO
No degen (in 1d) Each WF has 1 E
what happens in classical forbidden regions
QHO soln WF decays to 0 exponantially
where is probability of WF maximal in QHO
At the classical turning points
In the CHO the min energy at the bottom of well is E = 0, in QM qhat is the min energy and why
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>
Which parts of WF are gaussian resp, and which the Hermite poly
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
WHy can tunneling resulst in -kinetic energy
Particle simply does not have well defined kintetic energyin QHO so -kinetic energy is okay
expectations values of QHO
, <p></p>
<p> Undergo perfect SHO
e.g of correspondence principle that expectation values of observables behave classically
</p>
What happens at large ENegies(large n)
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
FTramsform frpm x to p
ionside integral mult by exp(ikx)
FTramsfpr, from p to x
inside integral mult by exp(-ikx)
Delta function fourier representaton
what does it revel
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
