Applications of Quantum Physics Flashcards
What is the wavefunction and its conditions
Particle in the state Ψ(r,t):
P(r,t)d³r = |Ψ(r,t)|²d³r
is the probability of finding it at point r in volume d³r at time t.
Conditions:
* Continuous
Wave function Ψ(r,t) and all its first derivatives ∇Ψ must be continuous everywhere. No sudden jumps in probability density, |Ψ(r,t)|, throughout the space.
- Single valued
Wave function must be single-valued, for any given values of r and t, Ψ(r,t) must take a unique value. - Integrability
The wave function must be square-integrable. Integrating |Ψ(r,t)|² over all space must be finite and, more explicitly, equal to 1:
∫±∞ P(r,t)d³r = ∫±∞ |Ψ(r,t)|²d³r = 1
TDSE
ĤΨ = iℏ ∂/∂t Ψ(r,t)
Hamiltonian
Ĥ = p̂²/2m + V̂ = −ħ²/2m ∇² + V(r)
Kinetic + Potential energies
TISE
Ĥψ = Eψ
Time dependent solution
T(t) = T₀exp(−iEt/ħ)
TDSE general solution
Ψ(r, t) = Σₙ cₙψₙ(r) Tₙ(t)
where |cₙ|² is probability of system being in nth state.
Σₙ |cₙ|² = 1
Wave function collapse
Wave function in its general form is a superposition of an infinite number of possibilities, each with its own finite probability. Once measured, the wave function reduces to one eigenfunctions, making its probability equal to 1, regardless of what value it had before the measurement.
1D momentum operator
p̂ = -iħ ∂/∂x
Quantum tunneling solution for classically allowed regime
E > V₀
From TISE:
d²ψ(x)/dx² = −2m/ħ²(E−V₀)ψ(x) = −k²ψ(x)
This is SHO with general solution
ψ(x) = Aexp(ikx) + Bexp(−ikx)
Right + left propagation
Quantum tunneling solution for classically forbidden regime
E ≤ V₀
From TISE:
d²ψ(x)/dx² = −2m/ħ²(V₀−E)ψ(x) = β²ψ(x)
General solution is
ψ(x) = Cexp(βx) + Dexp(−βx)
Right + left propagation
Transmission rate decays exponentially.
Waves at a square barrier
Incident wave Aexp(ikx)
Reflected wave Bexp(-ikx)
Transmitted wave Fexp(ikx)
Gamow factor, and its assumption conditions
Boundary conditions of square wave reflection/transmission shows
T = |F|² / |A|²
= 16k²β² / (k²+β²)² exp(−2βa) = αexp(−2βa)
If E≪V₀, α ≈ 1 for most quantum systems. Transmission rate T can be simplified
T = exp(−2βa)
where β = −2m/ħ²(V₀−E)
This is Gamow factor.
How does Scanning Tunnelling Microscopy work?
Electron transmission between two parallel metallic plates separated infinitesimally from each other. Metallic nature allows electrons inside to move as free particles exp(±ikx)
Thin layer of vacuum acts effectively as a square potential with height V₀ that can be several orders of magnitude larger than the electrons energy.
Probability of electron tunnelling from one plate to another is
T = exp (−2βδ)
for vacuum spacing of δ.
Energy W required to release one electron, aka work function, can be regarded as W = V₀−E, for most metallic systems is ~4 eV.
Results in T ≈ e^−20 ≈ 10−9 per electron.
Looks to be a neglible transmission rate. However, each Molar unit of metal contains 6.02 × 10^23 electrons per valence state, so total transmissionis much more significant.
Scanning Tunnelling Microscopy (STM), a conducting probe with a very sharp tip scans a metallic surface. Bringing the tip close to the surface and applying a voltage between them, we can image the surface by measuring the tunnelling current passing from surface to
tip through the vacuum.
Current can vary from one point to another,
providing information about the surface topography and electronic structure of the material in question. Enables us to directly access the wave aspect of the matter and see how it can manifest itself into collective phases such as superconductivity, magnetism, optical instabilities and many more peculiar quasi-particle states such as the quantum mirage effect in solids.
Wentzel-Kramers-Brillouin
approximation and its assumptions
Potential must slowly vary with x.
Varies slowly enough that it can spread over an interval, L, much larger than the deBroglie wavelength.
λ = 2π/k ≪ L
Can treat wavefunction in forbidden region as
ψ(x) = Aexp(iu(x))
where u(x) = k(x) · x
and k(x) = √2m(E−V(x)) / ħ² ‘
Substituting into Schrodinger and taking du(x)/dx ≈ constant (because slowly varying) solution is
u(x) = ±∫ k(x)dx + u₀ = ±∫ √2m(E−V(x)) / ħ² ‘ dx + u₀
where u₀ is constant.
Wave function inside barrier becomes
ψ(x) = Aexp( i (±∫ √2m(E−V(x)) / ħ² ‘ dx+u₀))
or
ψ(x) = Aexp(∓∫ √2m(V(x)−E) / ħ² ‘ dx+u₀)
right or left wave
Note: can also derive by taking arbitrary potential as many square barriers.
Why does the sun shine?
Short answer: the sun shines because of thermonuclear fusion caused by the
immense gravitational pressures at its core. When two protons are roughly 1fm from each other, attractive nuclear forces can surpass repulsive Coulombic forces, fusing them together and releasing a massive amount of energy.
How do they get so close?
Classical distance of closest approach in Sun is 1106 fm.
Thermal nuclear energy alone cannot cause nuclear fusion.
Protons tunnel through the Coulombic barrier
Can use WKB approximation and regard proton pairs as single particles with reduced mass
µₚ = mₚmₚ / mₚ+mₚ = mₚ / 2
Eventually end up with
T = exp (√−2µₚπ²e²r꜀ / 4πε₀ħ² ‘)
= exp(−√r꜀/r_G’)
This ends up being
T ≈ exp(−20) ≈ 2 × 10^−9
Very unlikely, but sun so huge that it happens loads.
In reality, protons fuse to deuterium, then helium 3, then release protons and alpha, so fusion is self driven.
DeBroglie equations
λ = 2π/k
p = h/λ = ħk
Wavefunction inside an infinite square well
ψ(x) = C sin(kₙx)
where kₙ = nπ/a
ψ(x) = C sin(nπx/a)
n = 1, 2, 3…
Normalised, so C = √2/a’
Wavelength and momentum in infinite square well
λₙ = 2π/kₙ = 2a/n
pₙ = h/λₙ = ħkₙ = ħnπ/a
Energy in infinite square well
V (x) = 0 inside the well, so
Eₙ = pₙ²/2m
= ħ²kₙ²/2m
= ħ²n²π² / 2ma²
= n²E₁ where n = 1, 2, 3… and E₁ is the lowest level.
Bound states in a finite square well
Barrier has a finite height. ψₙ(x) retains sinusoidal form inside the well to a large extent, but gains exponential tails extending into the barrier due to quantum tunnelling. It remains a bound state as it has no way to escape the barrier as long as Eₙ < V₀
Particle inside a double barrier
Turning square quantum well into a double barrier.
Assume at time t = t₀ we have abruptly dropped potential to zero everywhere except two small areas on the right and left sides of the well.
ψₙ(x) can now leak through quantum tunnelling, allowing particle to gradually escape the well. Wave function undergoes continuous decay with lifetime τ dependent on its energy and the tunnelling properties of the surrounding barriers.
Particle cannot be in a bound state, but decay process may take such a long time that we can assume it is in a quasi-bound state.
Energy-time uncertainty principle
∆t∆E ≥ ħ
Take ∆t = τ and ∆E as the change in the particle’s energy, extreme case
∆E ∼ ħ/τ
Particle effectively stationary, aka bound to the interior of the double barrier. Small ∆E, however, requires E≪V₀, a condition only satisfied if the particle is massively heavy and resides in its lowest quantisation states, i.e. small n values.
Particle incident on a double barrier
Incident on the double barrier from the left side. Split into 5 regions.
ψ₁(x) = exp(ikx) + B/A exp(−ikx)
ψ₃(x) = F/A exp(ikx)
where B/A and F/A are reflection and transmission amplitude. Complex, so can write as
B/A = rexp(iφᵣ)
F/A = texp(iφₜ)
Also define reflection and transmission probabilities
R = |rexp(iφᵣ)|² = r²
T = |texp(iφₜ)|² = t²
R + T = 1
Can derive wavefunction and transmission probability for ψ₅ by considering ψ₃ as the incident wave. Must consider initial ψ₃, and all subsequent reflections between the barriers.
Gives geometric series.
ψ₅ = texp(iφₜ) [1 + rexp(iφᵣ) rexp(iφᵣ) exp(2ika) +
( rexp(iφᵣ) rexp(iφᵣ) exp(2ika) )² + … ] texp(iφₜ) exp(ikx)
Simplifies to
ψ₅ = t² exp(2iφₜ) exp(ikx) / 1−r² exp(2i(φᵣ+ka))
Double tunnelling probability is
T_D = |T / 1−Rexp(2i(φᵣ+ka))|²
If exp(2i(φᵣ+ka)) = 1
T_D will reach its maximum value of 1
If (φᵣ+ka) = nπ, overall transmission rate through a double barrier will be 100% regardless of the tunnelling properties of each barrier.
Phase shift φᵣ is usually much smaller than ka and thus can be neglected.
Resonance condition reduces to
ka =2π/λ a = nπ, where n = 1, 2, 3 …
or
a = nλ/2
Resonant tunnelling happens if the distance between the barriers is an integer multiple of half of the particle’s wavelength.
Has applications in semiconductor devices, including high-frequency oscillators, high-speed switching and multi-level logic.
Resonance condition
a = nλ/2
Difference between a quantum dot, a nanowire/nanotube, and a quantum well
How many dimensions they are trapped in. Dot all three, wire only two, well only one.
Applications of quantum dots
Photoluminescence to quantum computing. They can absorb light and emit a rainbow of colours with tunable brightness and frequencies.
Silver nanocubes used in plasmonic sensing, catalysis and bionanotechnology.
TISE in a nanocube
−ħ²/2m [∂²/∂x²+∂²/∂y²+∂²/∂z²] ψ(x,y,z) = Eψ(x,y,z)
Can be simplified to
[∂²/∂x²+∂²/∂y²+∂²/∂z²] ψ(x,y,z) = −k² ψ(x,y,z)
where
k² = 2mE/ħ²
Solution of wavefunction in nanocube
Use separation of variables, get 3 eqs of form
1/X(x) d²X(x)/dx² = −kₓ²
with
kₓ² + kᵧ² + k₂² = k²
Each eq solved by
X(x) = √2/a’ sin(nₓπx/a)
with nₓ = 1, 2, 3…
Overall solution
ψ(x, y, z) = (2/a)³’² sin(nₓπx/a) sin(nᵧπy/a) sin(n₂πz/a)
with corresponding eigenvalues
Eₙₓ,ₙᵧ,ₙ₂ = ħ²π²/2ma² [nₓ² + nᵧ² + n₂²]
Three rules for energy level occupation
- Aufbau principle:
Electrons occupy levels in ascending energy order, with the lowest energy level being filled first. - Pauli’s exclusion principle:
No more than two electrons can occupy the same energy level, and the two electrons must have opposite spin orientation. - Hund’s principle:
Degenerate energy levels (eigenvalues with the same energy) must all be singly occupied before any are doubly occupied.
What’s in a semiconductor quantum dot
Drain, insulator, quantum dot with surrounding gate, insulator, source.
How does a semiconductor quantum dot work?
Drain/source and gate potentials can be varied to move energy levels, allowing or disallowing quantum tunnelling between different parts of the dot.
TISE in a semiconductor quantum dot
−ħ²/2m∗ [∂²/∂x²+∂²/∂y²+∂²/∂z²] ψ(x,y,z) + 1/2 k(x²+y²) = Eψ(x,y,z)
where m∗ is reduced mass
Potential energy due to gate voltage is
V(x,y) = 1/2 k(x²+y²)
where k is a constant proportional to Vg.
Along z-axis, bias voltage Vsd can slightly tilt the V(z) inside the quantum dot. This can be safely ignored as a perturbation with no effect on the energy states, at least to the first order of approximation.
Separation of variables leads to two harmonic oscillators in X and Y, and an infinite square well in Z.
SHO energy eigenvalues
Eₙ = (n + ½)ħω
with n = 0,1,2…
Energy levels in a semiconductor quantum dot
Eₙₓ = (nₓ + ½)ħω
Eₙᵧ = (nᵧ + ½)ħω
with nₓ,nᵧ = 0,1,2…
Eₙ₂ = n₂²π²ħ²/2m∗a²
with n₂ = 1, 2, 3…
Eₙₓ,ₙᵧ,ₙ₂ = (nₓ + nᵧ + 1)ħω + n₂²π²ħ²/2m∗a²
NOTE difference in range of n for x,y, and z
Eₙ₂, in general, is much larger than its x and y counterparts, so the lowest z level is much higher than low x and y levels, meaning n₂ can be taken as 1 for low occupation.
How do magic numbers arise from semiconductor quantum dots?
Ground state:
nₓ = nᵧ = 0
Occupation number N = 2.
First excited shell:
{nₓ = 1, nᵧ = 0} or {nₓ = 0, nᵧ = 1}
These combinations accommodate four electrons between them.
Total occupation N = 6.
Second excited shell:
{nₓ = 2, nᵧ = 0}, {nₓ = 0, nᵧ = 2} or {nₓ = 1, nᵧ = 1}
Two electrons each.
Total occupation N = 12.
Third excited shell:
{nₓ = 3, nᵧ = 0}, {nₓ = 0, nᵧ = 3}, {nₓ = 2, nᵧ = 1} or {nₓ = 1, nᵧ = 2}
Total occupation N = 20.
Each shell is again ħω higher than the last. Abrupt jumps in energy upon completion of an energy shell are similar to that of atomic
and nuclear structures.
The magic numbers are N = 2, 6, 12, 20 . . . .
