Document 23217546

Document technical information

Format pdf
Size 2.6 MB
First found May 22, 2018

Document content analysis

Category Also themed
Language
English
Type
not defined
Concepts
no text concepts found

Persons

Organizations

Places

Transcript

Quantum Physics Lecture 4
The Uncertainty Principle - continued
“Thought-experiments”
- microscope, single slit and 2-slit diffraction
Some applications
- propagation of wave group, minimum confinement energy
Alternative E-t form Thermal Properties
Specific Heat of solids
- Classical model, failure at low temperature, Einstein model
Black Body radiation (introduction)
Uncertainty and Error?
Uncertainty looks like some sort of “experimental error” - It is not!
- Measurement of one of x or p alters the value of the other. Experimental error can be arbitrarily reduced by better experiment.
But uncertainty is a fundamental limit, and property of the wave
nature of matter! ΔxΔp ≥  2
Note central role of Planck’s constant h Commonly use “h-bar”  or h/2π = 1.054 x 10-34 J s
(later) will see that

is basic unit of angular momentum!
Uncertainty in ‘wave’ experiments? - Microscope and Diffraction
Thought-Experiment (microscope)
Use optical microscope to find particle (e.g. electron) position.
“see” the electron by scattering a photon into the lens…..
…..anywhere within the lens angle 2α.
Photon momentum (p=h/λ) change causes recoil of electron!
Along horizontal, change ranges from -psinα to +psinα
i.e. range in photon momentum Δp = 2psinα = 2(h/λ)sinα
…….which becomes the uncertainty in particle momentum
Thought-Experiment (microscope)
Uncertainty in particle position associated with “diffraction limit” : minimum separation of points is
Δx = λ/sinα
Δx.Δp = (λ/sinα)(2(h/λ)sinα) = 2h Δx.Δp ≥ 
2
How could we “improve” microscope? by decreasing λ: decreasing Δx but increasing Δp?
by decreasing α, decreasing Δp but increasing Δx?
Quantum concept of photon is intrinsic: Classically, could decrease Δx without increasing Δp (lower intensity and wait?)
Thought-Experiment (single-slit)
Remove “complication” of photon and electron by
single-slit diffraction (of either)
Slit width (s) is the
uncertainty in position: Δx = s s
θ
λ
At 1st diffraction minimum: s sinθ = λ
Therefore, Δx = s = λ/sinθ = h/psinθ
Thought-Experiment (single-slit)
Electron (or photon) arriving within central maximum must be deflected through angle range 0 to θ : this means uncertainty in tranverse momentum :
p
Δp
θ
If momentum is p, then, Δp = psinθ
Δx.Δp = (h/psinθ)(psinθ) = h
“showing” that
Δx.Δp ≥ 
2
Equivalent analysis of Young’s (Two) Slits using 1st maximum,
Where slit separation is the uncertainty in position (exercise) Q: “which slit does the particle (or photon) go through?” !!
Two-slit experiment
Observe:
- Close one slit (i.e. the particle must go through the other) ⇒lose the 2-slit diffraction pattern!
-  Single particle causes single point of scintillation
⇒ pattern results from addition of many particles!
- Pattern gives probability of any single particle location G.I.Taylor
Low intensity beam
Two slit experiment - summary
(1)  Both slits required to give pattern, even for single particle
(2) Single particle arrives at single point. i.e. “explores” all regions available (see 1), but occupies only one point when actually
“measured”
(3) Arrival of individual particle conforms to statistical pattern of diffraction (complementarity). (4) Average over many particles gives standard diffraction pattern.
(complementarity)
Key features of quantum mechanics!
“Practical” applications of UP:
propagation of a wave group?
Establish particle position to an uncertainty Δxo at time zero:
what is uncertainty Δxt at later time t?
UP implies Δp ≥ 2Δx 0
and p = mv
So Δv = Δp m ≥  2mΔx 0
uncertainty in velocity implies uncertainty in position at time t
Δx t = Δv.t ≥ t 2mΔx
o
Δxt ∝ t: uncertainty in position increases with time (dispersion)
Δxt ∝ 1/Δxo: “more you know now, less you know later”
Application of UP :
minimum energy of confinement
Rough estimate KE of electron in hydrogen atom (in full, later lecture)
Δx ~ radius of H atom = 5.3 x 10-11 m
Δp ≥ h/4πΔx = 1 x 10-24 kg m s-1
Treat electron as non-relativistic, KE = p2/2mo
where p ~ Δp at least:
⎛ Δp ⎞
1x10 )
(
−1 9
⎜
⎟
KE ≥
=
=
5.4x10
J = 3.4 eV
−3 1
⎝ 2mo ⎠ (2)(9.1x10 )
2
−2 4 2
(see later lecture: KE=13.6 eV so correct order of magnitude)
Alternative form of UP:
Δx.Δp ≥ h/4π related to spatial extent needed to measure λ
What about the temporal extent needed to measure ω (or f)?
- at least one period?
Estimate: Δf.Δt ≥ 1
E=hf
ΔE=hΔf
So
ΔEΔt ≥ h
(correct maths gives)
ΔEΔt ≥  2
Eg. ΔE is the spectral “width” of optical emission lines, where Δt is “lifetime” of transition (see atomic transitions, later)
Specific heat Cv of solids
Solid is N atoms coupled, each having 3 degrees of freedom
Classically
Energy = kbT per oscillator
(equipartition principle – kbT/2 per deg of freedom, oscillator has 2, PE & KE)
Total energy U = 3NkbT
Now Cv= dU/dT
So
(N is number of atoms, kb Boltzmann const)
Cv = 3Nkb
(or 3R/mole, Dulong & Petit) i.e. a constant, independent of temperature T…
Experimental observation:
OK at high T
BUT…
Cv falls (towards zero) at low temperatures…
Why is classical result wrong?
3Nk
3Nkbb
Specific heat of solids (cont.)
Planck - Assumed energy of oscillators is quantised! E = nω where n is a positive integer
Probability of an energy E is P( E ) = e −E / kb T = e −nω / kb T
Mean (expectation) energy is
So total Energy U is 3N
E =
3N ∑ nωe −nω / kb T
n
∑e
−nω / kb T
E
∑ E P( E )
=
∑ P( E )
n
n
⎡ ω / k T ⎤
= 3NkbT ⎢ ω / kb Tb ⎥
−1 ⎦
⎣e
n
i.e. Quantum term on R.H.S. freezes out energy exchange at low temperature. Happens because the finite gap between states, ω becomes greater than kbT
Similar ‘quenching’ effect for molecule modes
⎛ ω ⎞
⎛ ∂U ⎞
C
=
=
Nk
⎜
⎟
⎟
b⎜
⇒ V ⎝ ∂T ⎠
k
T
⎝
V
b ⎠
2
(e
e ω / kb T
ω / kb T
)
−1
2
Einstein formula for specific heat
Blackbody Radiation
At finite temperature matter “glows” i.e. emits radiation with a continuous spectrum. e.g Infrared imaging of people, planet etc.
Surface dependent (emissivity, silvery, black, etc.)
Blackbody = ideal 100% emitter/absorber in thermal equilibrium with its surroundings. Practical realisation is a thermal cavity.
Measure: spectrum energy density u(ω)
Observe that increasing temperature
(1)  increases u overall
(2) shifts peak emission to higher frequencies
i.e. colour and intensity of hot objects vary with T Examples – Bar fire, molten iron, stars, universe µwave background…..
×

Report this document