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 . . . .
Four tunnelling regimes for a semiconductor quantum dot
- eV_S < eV_D < E₀
No conduction happens. - eV_S ≤ E₀ ≤ V_D
Resonant tunnelling.
Steady transmission of two electrons with effective quantum tunnelling current IQ from the drain to source via the ground state E₀. - E₀ < eV_S < eV_D
Pauli locking.
Two electrons trapped in the E₀ state, cannot tunnel into source as source states of the same energy are already occupied. - eV_S ≤ E₁ ≤ eV_D
Resonant tunnelling / Pauli locking.
Hybrid regime, E₀ remains subject to Pauli locking while E₁ enable new 4 electron resonant tunnelling state with an effective IQ twice larger than that of (2)
Explain Coulomb blockade
Plot of V_G against V_SD shows conduction only when energy levels are between V_S and V_G. This results in diamond shaped regions of no conduction, each corresponding to a magic number, N.
Coulomb blockade results from Coulomb interaction splitting each spin-degenerate state splits into two singly occupied substates. This breacks each diamond region into N smaller regions.
See pg 58 of notes for diagram
Quantisation of energy levels in a nanotube/nanowire
Solving the TISE in a cylindrical coordinate system (r, ϕ, z)
Describe the Laplacian term in Hamiltonian
∇² = 1/r ∂/∂r (r ∂/∂r) + 1/r² ∂²/∂ϕ² + ∂²/∂z²
Can set potential energy V to zero, assuming metallic nanowire.
Separatie variables ψ(r, ϕ, z) = Φ(r, ϕ)Z(z)
E = Eϕ + E₂
E₂ = ħ²k₂²/2m∗
Boundary conditions show
k₂ = nπ/L
with n = 1, 2, 3,…
Energy states are quantised along the z-axis, similar to an infinite quantum well.
Electrons are radially confined, so can ignore terms involving ∂/∂r.
Solve
−ħ²/2m∗R² d²Φ(ϕ)/dφ² = −EϕΦ(ϕ)
by taking
ν² = 2m∗R²Eϕ/ħ²
SHO solved by
Φ(ϕ) = Ae(±iνϕ)
where A is a constant.
Boundary condition
Φ(ϕ) = Φ(ϕ + 2π)
which requires ν to be an integer number including zero.
Eϕ =ħ²ν²/2m∗R²
where ν = 0, 1, 2, …
Energy eigenvalues for a nanotube/nanowire
Eᵥₙ =ħ²/2m∗ (ν²/R² + π²n²/L²)
Equation for density of states
DOS(E) = dN/dE
Usually use dN/dk dk/dE
dN/dk can be calculated spacially/volumetrically
Orthonormal basis set equations
êᵢ · êⱼ = δᵢⱼ
A = Aₓêₓ + Aᵧêᵧ + A₂ê₂
with
A = ||A|| = √Aₓ² + Aᵧ² + A₂² ‘
For any two vectors, if A · B = 0, they are orthogonal
bra-ket notation
Eigenstate
ψₙ(r) = |ψₙ〉or |n〉
Complex conjugate
ψₙ∗ = 〈ψₙ| or〈n|
Orthonormality integrals
∫ᵣ ψₙ∗(r)ψₙ(r)
〈m|n〉= δₘₙ
Superposition
|Ψ〉= Σₙ cₙ|n〉
corresponding overlap integrals
cₖ =〈k|Ψ〉
Can show
〈Φ|Ψ〉= Σₙ dₙ∗ cₙ
(assuming |Φ〉= Σₙ dₙ|n〉)
Overlap integrals involving an operator
∫ᵣ ψₘ∗(r)Âψₙ(r)dr
=〈m|Â|n〉= Aₘₙ
First order perturbation
Eₙ⁽¹⁾ =〈ϕₙ|V̂|ϕₙ〉
For any given eigenvalue Eₙ⁽⁰⁾, the perturbation caused by potential V̂ to the first order of correction is simply the expectation value of V̂ arising from the unperturbed eigenstate |ϕₙ〉
Basic equation for angular momentum
L̂ = r̂ × p̂
Heisenberg’s uncertainty principle
∆rᵢ∆pᵢ ≥ ħ/2
[Â,B̂]
(ÂB̂ - B̂Â)
Only if = 0, commutative, can be known at the same time.
L̂ components
L̂ₓ = ŷ p̂₂ − ẑ p̂ᵧ
L̂ᵧ = ẑ p̂ₓ − x̂ p̂₂
L̂₂ = x̂ p̂ᵧ − ŷ p̂ₓ,
[L̂ₓ, L̂ᵧ] = iħL̂₂
[L̂ᵧ, L̂₂] = iħL̂ₓ
[L̂₂, L̂ₓ] = iħL̂ᵧ.
L̂² commutes with all components
Ladder operators
L̂± = L̂ₓ ± iL̂ᵧ
Commutation of the ladder operators
[L̂₊, L̂₋] = L̂₊L̂₋ − L̂₋L̂₊
L̂₊L̂₋ = L̂² − L̂₂² + ħL̂₂
L̂₋L̂₊ = L̂² − L̂₂² − ħL̂₂
[L̂₊, L̂₋] = 2ħL̂₂
What do ladder operators do?
L̂±|ψ〉are still eigenfunctions of L̂₂ with eigenvalues ħ above or below the initial L̂₂ eigenvalue of |ψ〉respectively.
L̂₊ and L̂₋ raise and lower eigenvalues of L̂₂ by equally-spaced steps ħ.
L̂±|ψ〉are still the same eigenfunctions
of L̂² without changes in eigenvalues:
L̂±L̂²|ψ〉= L̂± λ|ψ〉= λL̂±|ψ〉
Given that L₂² ≤ |L|²,
λ₂² ≤ λ.
Constraint asserts upper and lower limits for λ₂.
At the top of the ladder
L̂₊|ψ₁〉= 0
and at the bottom
L̂₋|ψ₂〉= 0
What can ladder operators show us about L̂₂ eigenvalues?
L̂₊|ψ₁〉= 0
L̂₋|ψ₂〉= 0
Apply L̂₋ to L̂₊ equation
L̂₋L̂₊|ψ₁〉= 0
(L̂² − L̂₂² − ħL̂₂)|ψ₁〉= 0
(λ − λ₂² − ħλ₂,₁)|ψ₁〉= 0
λ = λ₂,₁(λ₂,₁ + ħ)
Applying L̂₊ to L̂₋ equation gives
λ = λ₂,₂(λ₂,₂ − ħ)
These can only be satisfied if
λ₂,₁ = −λ₂,₂
Therefore, the top and bottom rungs of ladder must be symmetrically placed around zero.
λ₂,₁ − λ₂,₂ = 2λ₂,₁ = nħ
where n is a positive integer.
−nħ/2 ≤ λ₂ ≤ nħ/2
L̂₂ eigenvalues and their relationship with types of angular momenta
−nħ/2 ≤ λ₂ ≤ nħ/2
- If n is even, n=2l
−lħ ≤ λ₂ ≤ lħ
λ = l(l+1)ħ²
System carries orbital angular momentum
L₂ = λ₂ = mħ with −l ≤ m ≤ l - If n is odd, n=1
−½ħ ≤ λ₂ ≤ ½ħ
λ = 3/4 ħ²
System carries spin angular momentum
λ₂ = ħ/2 is spin-up
λ₂ = −ħ/2 is spin-down
Spin eigenvalue equations
Ŝ²|χ〉= s(s + 1)ħ²|χ〉= 1/2 · 3/2 ħ²|χ〉= 3/4ħ²|χ〉
Ŝ₂|χ〉= mₛħ|χ〉= ±ħ/2 |χ〉
where |χ〉is spin eigenstate of the system
and −1/2 ≤ mₛ ≤ 1/2
Define spinors and apply to an arbitrary eigenfunction.
|α₂〉= (¹₀) for mₛ = 1/2 (spin up)
|β₂〉= (⁰₁) for mₛ = −1/2 (spin down)
|α₂〉and |β₂〉are spin eigenvectors/”spinors” of the wave function. Subscript z stands for quantisation along the z-axis
To ensure |α₂〉and |β₂〉form an orthonormal basis, define
〈α₂|α₂〉= α₂†α₂ = (1 0)(¹₀) = 1
and
〈β₂|α₂〉= β₂†α₂ = (0 1)(¹₀) = 0
(†) denotes transposed complex conjugate.
Complete basis set, can use to expand any arbitrary eigenfunction
|χ〉= c₁|α₂〉+ c₂|β₂〉= c₁(¹₀) + c₂(⁰₁) = (c₁ // c₂)
where c₁ and c₂ are complex numbers that satisfy the normalisation requirement:
〈χ|χ〉= (c₁∗ c₂∗) (c₁ // c₂) = c₁∗c₁ + c₂∗c₂ = |c₁|² + |c₂|² = 1
Ŝ commutations
[Ŝₓ, Ŝᵧ] = iħŜ₂
[Ŝᵧ, Ŝ₂] = iħŜₓ
[Ŝ₂, Ŝₓ] = iħŜᵧ
[Ŝ₂, Ŝ²] = 0
Pauli spin matrices and their commutations
σₓ =
(0 1
1 0)
σᵧ =
(0 −i
i 0)
σ₂ =
(1 0
0 −1)
Ŝₓ = ħ/2 σₓ
Ŝᵧ = ħ/2 σᵧ
Ŝ₂ = ħ/2 σ₂
[σₓ, σᵧ] = 2iσ₂
[σᵧ, σ₂] = 2iσₓ
[σ₂, σₓ] = 2iσᵧ
Spin quantisation along x-axis
Solve
σₓ|χ〉= λ|χ〉
using Pauli matrices.
Get λ = ±1, so Ŝₓ eigenvalues are ±ħ/2
For λ = 1
|αx〉= 1/√2’ (¹₁)
For λ = −1
|βx〉= 1/√2’ (¹₋₁)
Using the completeness property of spinors:
αₓ = 1/√2’ (¹₀) + 1/√2’ (⁰₁) = 1/√2’ (|α₂〉+ |β₂〉)
and
βₓ = 1/√2’ (¹₀) − 1/√2’ (⁰₁) = 1/√2’ (|α₂〉− |β₂〉)
Magnetic moment due to orbital angular momentum
µ = I · A = qv/2πr · πr² = q/2 vr = q/2m pr = q/2m Lz
(Using L = r × p, Lz = rp)
µ = q/2m L
Change in energy in external B field
∆E = −µ · B
Orbital energy levels when under influence of an external magnetic field
µ̂_L = q/2m L̂
Ĥ = −µ̂_L · B
Under application of perpendicular magnetic field B = B e_z
Ĥ = e/2m L̂ · B
= eB/2m L̂₂
= µ_B B /ħ L̂₂
where µ_B = eħ/2m is Bohr magneton.
Eigenvalues of L̂₂ are mₗħ with −l ≤ mₗ ≤ l
Each energy level in presence of B splits into (2l + 1) equally-spaced sub-levels
Emₗ = mₗµ_B B
with − l ≤ mₗ ≤ l
Energy spacing between the adjacent sub-levels is µ_B B, of order < 1 µeV.
Change in energy levels in external magnetic field due to spin
Electrons have intrinsic spin angular momentum Ŝ.
Spin magnetic moment
µ̂ₛ = −gₛ e/2m Ŝ
gₛ = 2.002318 for an electron
Under influence of an external magnetic field
Ĥ = −µ̂ₛ · B = egₛ/2m Ŝ · B
= egₛB/2m Ŝz = gₛµ_B B/ħ Ŝz
Ŝz has two eigenvalues ±ħ/2.
Each energy level is spin-split into two branches
Eₘₛ = mₛgₛµ_B B
= ±gₛµ_B B/2
mₛ = 1/2 is higher than mₛ = −1/2
Spinor with time dependence
Using Ψ(t) = Ψ(0) exp(−iEt/ħ)
|χ(t)〉=
(c₁exp(−iωt)
c₂exp(iωt) )
Energy eigenvalue E is taken as
gₛµ_B B/2 for |αz〉
-gₛµ_B B/2 for |βz〉
implying ω = egₛB/4m
What happens to an electron initially quantised along the z-axis when influenced by a magnetic field also along the z-axis?
Initial state
|χ〉= |αz〉= (¹₀)
Apply uniform magnetic field along the z-direction, must evolve as
|χ(t)〉= (exp(−iωt) ₀) = exp(−iωt)|αz〉
〈Ŝz〉=〈χ(t)|Ŝz|χ(t)〉
= ħ/2 ( exp(iωt) 0 ) (¹₀⁰₋₁) ( exp(−iωt) ₀ )
= ħ/2
Can show〈Ŝx〉=〈Ŝy〉= 0
External magnetic field does not affect the expectation value of Ŝ for an electron initially quantised along the z-direction, allowing it to remain stationary.
What happens to an electron initially quantised along the x-axis when influenced by a magnetic field along the z-axis?
Initial state
|χ〉= |αx〉= 1/√2’ (¹₁)
Under influence of field
|χ(t)〉= 1/√2’
(exp(−iωt)
exp(iωt) )
Can get Ŝ expectation values
〈Ŝz〉=〈χ(t)|Ŝz|χ(t)〉
= ħ/4 ( exp(iωt) exp(−iωt) ) (¹₀⁰₋₁) (exp(−iωt) // exp(iωt) )
= 0
〈Ŝx〉=〈χ(t)|Ŝx|χ(t)〉
= ħ/4 [exp(2iωt) + exp(−2iωt)]
= ħ/2 cos(2ωt)
〈Ŝy〉=〈χ(t)|Ŝy|χ(t)〉
= −iħ/4 [exp(2iωt) − exp(−2iωt)]
= ħ/2 sin(2ωt)
System not in a stationary state.
It precesses about the z-axis with
ω_L = 2ω = egₛB/2m
ω_L is the Larmor frequency.
Consequence of Larmor precession
ω_L = 2ω
Spinors subject to continuous precession will only return to their original state after 4π rotation.
At ω_L t = 2π, ωt = π
|χ(t=2π/ω_L)〉= 1/√2’ (exp(−iπ) // exp(iπ))
= 1/√2’ (⁻¹₋₁)
= −|αx〉
At ω_L t = 2π, all spin expectation values return to their initial values, but spinors undergo a sign reversal.
System must undergo an additional 2π precession to return to the original spinor states.
Spin ladder operators
Ŝ₊ = Ŝₓ + iŜᵧ
= ħ/2 (⁰₁¹₀) + iħ/2 (⁰ᵢ⁻ᶦ₀) = ħ (⁰₀¹₀)
Ŝ₋ = Ŝₓ − iŜᵧ
= ħ/2 (⁰₁¹₀) − iħ/2 (⁰ᵢ⁻ᶦ₀) = ħ (⁰₁⁰₀)
When applied to spinors
Ŝ₊|αz〉= ħ (⁰₀¹₀) (¹₀) = 0
Ŝ₋|βz〉= ħ (⁰₁⁰₀) (⁰₁) = 0
Ŝ₋|αz〉= ħ (⁰₁⁰₀) (¹₀) = ħ (⁰₁) = ħ|βz〉
Ŝ₊|βz〉= ħ (⁰₀¹₀) (⁰₁) = ħ (¹₀) = ħ|αz〉
Apply spin ladder operations to a two spin-1/2 particle system
- Ŝ₊|α⁽¹⁾α⁽²⁾〉= Ŝ₊⁽¹⁾|α⁽¹⁾α⁽²⁾〉+ Ŝ₊⁽²⁾|α⁽¹⁾α⁽²⁾〉
=|α⁽²⁾〉Ŝ₊⁽¹⁾|α⁽¹⁾〉+|α⁽¹⁾〉Ŝ₊⁽²⁾|α⁽²⁾〉
= 0 + 0
= 0 - Ŝ₋|α⁽¹⁾α⁽²⁾〉= ħ [|β⁽¹⁾α⁽²⁾〉+|α⁽¹⁾β⁽²⁾〉]
Normalised as
= 1/√2’ [|β⁽¹⁾α⁽²⁾〉+|α⁽¹⁾β⁽²⁾〉]
Linear combination of mₛ = 0 states - Ŝ₋ [|β⁽¹⁾α⁽²⁾〉+|α⁽¹⁾β⁽²⁾〉] = 2ħ|β⁽¹⁾β⁽²⁾〉
- Ŝ₋|β⁽¹⁾β⁽²⁾〉=0
Eigenvectors of spin-triplet states
Triplet states have total spin angular momentum s = 1 and projections 1 ≤ mₛ ≤ 1
Eigenvectors are
* |α⁽¹⁾α⁽²⁾〉for mₛ = +1
* 1/√2’ [|β⁽¹⁾α⁽²⁾〉+|α⁽¹⁾β⁽²⁾〉] for mₛ = 0
* |β⁽¹⁾β⁽²⁾〉for mₛ = −1
Characteristic feature of this state is its invariance under swapping the label of the two particles.
The triplet states are symmetric under any exchange of the particles.
Eigenvector of spin-singlet states
Singlet states have total spin angular momentum s = 0 and projection mₛ = 0
Eigenvector is
1/√2’ [|β⁽¹⁾α⁽²⁾〉−|α⁽¹⁾β⁽²⁾〉]
This is normalised and orthogonal to all three triplet state components.
It cannot be transformed by ladder operations.
Characteristic feature of this state is its asymmetric behaviour under the interchange of particles; it gains a negative sign after swapping labels.
Total angular momentum equations
Ĵ = L̂ + Ŝ
|l − s| ≤ j ≤ l + s
− j ≤ mⱼ ≤ j
Ĵ²|ψ〉= j(j+1)ħ² |ψ〉
Ĵz|ψ〉= mⱼħ|ψ〉
Ĵ²= |L̂ + Ŝ|²
Ĵz = L̂z + Ŝz
Energy levels in hydrogen
Eₙ = −E_R/n²
where E_R = 13.6 eV is the Rydberg energy
Spin-orbital field experienced by an electron
Current loop I = eω/2π
In the lab (nucleus) frame,
B = µ₀I/2r ê_z
Electron angular momentum, assuming point mass,
L = mωr² ê_z
Combined
B = µ₀eω/4πr ê_z
= µ₀e/4πmr³ L
Electron frame, multiply by 1/2 for Thomas precession factor (relativistic corrections)
B̂ = µ₀e/8πmr³ L̂
Spin-orbit coupling energy correction
Internal field experienced by electron
B̂ = µ₀e/8πmr³ L̂
Electron intrinsic spin magnetic moment
µ̂ₛ = −egₛ/2m Ŝ
Field-spin interaction
Ĥₛₒ = −µ̂ₛ · B̂ = f(r) L̂ · Ŝ
where f(r) = µ₀gₛe²/16πm²r³
Most atomic systems, B relatively weak, so Ĥₛₒ can be treated as first order perturbation.
∆Eₛₒ =〈Ĥₛₒ〉
=〈n, l, s, j, mⱼ|f(r) L̂ · Ŝ|n, l, s, j, mⱼ〉
Cannot know L̂ · Ŝ as |n, l, s, j, mⱼ〉NOT eigenstates. Will need to rewrite.
Rewrite L̂ · Ŝ independent of L̂z and Ŝz, and use to rewrite spin-orbit energy correction.
Ĵ² = (L̂ + Ŝ)² = L̂² + Ŝ² + 2L̂ · Ŝ
L̂ · Ŝ = 1/2 [Ĵ² − L̂² − Ŝ²]
Rewrite
∆Eₛₒ =〈Ĥₛₒ〉=〈n, l, s, j, mⱼ|f(r) L̂ · Ŝ|n, l, s, j, mⱼ〉
as
∆Eₛₒ = 1/2〈f(r)〉〈n, l, s, j, mⱼ|Ĵ² − L̂² − Ŝ²|n, l, s, j, mⱼ〉
Use eigenvalue equations
Ĵ²|n, l, s, j, mⱼ〉= ħ² j(j+1)|n, l, s, j, mⱼ〉
L̂²|n, l, s, j, mⱼ〉= ħ² l(l+1)|n, l, s, j, mⱼ〉
Ŝ²|n, l, s, j, mⱼ〉= ħ² s(s+1)|n, l, s, j, mⱼ〉
∆Eₛₒ = Eₙₗ/2 [j(j + 1) − l(l + 1) − s(s + 1)]
with
Eₙₗ =〈f(r)〉ħ² > 0
Zeeman effect Hamiltonian and change in energy (inc. Lande factor)
Adding external magnetic field
Ĥ = Ĥ₀ + Ĥₛₒ + Ĥₘₐ₉
Assume Ĥₘₐ₉ is much weaker than Ĥₛₒ.
|Bₑₓₜ| ≪ |B̂|
Assume Bₑₓₜ is along the z-axis, rewrite Ĥₘₐ₉ as
Ĥₘₐ₉ = −µ̂_L · Bₑₓₜ − µ̂ₛ · Bₑₓₜ
= e/2m (L̂ + gₛŜ) · Bₑₓₜ
= eBₑₓₜ/2m (L̂z + gₛŜz)
Recall Ĵz = L̂z + Ŝz and take gₛ ≈ 2
Ĥₘₐ₉ = eBₑₓₜ/2m (Ĵz + Ŝz)
First-order perturbation,
∆Eₘₐ₉ =〈Ĥₘₐ₉〉= eBₑₓₜ/2m [mⱼħ +〈Ŝz〉]
sz not a good quantum number, derive
〈Ŝz〉= Ĵ²+Ŝ²−L̂²/2j(j+1)ħ〈Ĵz〉
= j(j+1) + s(s+1) − l(l+1) / 2j(j+1) mⱼħ
∆Eₘₐ₉ = eBₑₓₜ/2m mⱼħ [1 + j(j+1) + s(s+1) − l(l+1) / 2j(j+1)]
= gⱼmⱼµ_B Bₑₓₜ
with
gⱼ = 1 + j(j+1) + s(s+1) − l(l+1) / 2j(j+1)
This is the Lande g-factor
Paschen-Back effect
If external field effect is stronger than spin-orbit coupling term,
∆ₛₒ/gⱼmⱼµ_B ≪ Bₑₓₜ
system becomes effectively spin-orbit decoupled.
Ĥₘₐ₉ = eBₑₓₜ/2m (L̂z + gₛŜz)
〈Ĥₘₐ₉〉= µ_B Bₑₓₜ (ml + 2mₛ)
ml and mₛ can be treated as good quantum numbers, as in the case of the unperturbed Hamiltonian.
Physical setup of a Q-bit
A double quantum dot, set up such that only one state per dot contributes to tunnelling.
Denote wavefunctions as
|L〉= (¹₀)
and
|R〉= (⁰₁)
Wavefunctions can tunnel through the central barrier, requiring correction energy −∆/2
Energy difference between states is ε, which is proportional to the difference in the two gate voltages.
Hamiltonian of a Q-bit
Without mixing of the two states
Ĥ = 1/2
(ε 0
0 −ε)
= 1/2 ε (¹₀⁰₋₁)
= 1/2 ε σ₂
with |L〉and |R〉being its eigenstates. Eigenvalues are ±ε/2.
Couple |L〉and |R〉by adding the mixing parameter −∆/2 to the off-diagonal terms
Ĥ = 1/2
(ε −∆
−∆ −ε)
= 1/2 [ε (¹₀⁰₋₁) − ∆ (⁰₁¹₀)]
= 1/2 (εσ₂ − ∆σₓ)
Q-bit initialisation
ε ≫ 0 and ∆ = 0
Q-bit can be initialised by setting Vᵍᴸ ≫ Vᵍᴿ and adjusting drain and source so one electron tunnels to L channel.
ε ≫ 0, so electron tunnelling from |L〉to |R〉is negligibe. Electron is effectively residing in
|χ(t = 0)〉= |L〉= (¹₀)
with energy eigenvalue ε/2
Q-bit manipulation
ε = 0 and ∆ ≠ 0
Raise Vᵍᴿ so that Vᵍᴸ = Vᵍᴿ, and ε = 0
|L〉and |R〉are at the same energy, leading to strong resonant tunnelling.
Electron is effectively shared between the L and R channels, ∆ is no longer negligible.
Time evolution can be described using the TDSE
iħ ∂|χ(t)〉/∂t = −∆/2 σₓ|χ(t)〉= −∆/2 (⁰₁¹₀)|χ(t)〉
Solve and apply initialisation constraint|χ(t=0)〉=(¹₀)
|χ(t)〉= exp(i∆t/2ħ)/2 (¹₁) + exp(−i∆t/2ħ)/2 (¹₋₁)
= exp(i∆t/2ħ)/√2’ |E 〉+ exp(−i∆t/2ħ)/√2’ |O〉
with eigenvectors |E〉= |αₓ〉and |O〉= |βₓ〉corresponding to eigenvalues −∆/2 and ∆/2, respectively.
Q-bit measurement
Switch system back to initial configuration,
ε ≫ 0 and ∆ = 0
|χ(t)〉= cᴸ(t)|L〉+ cᴿ(t)|R〉
with
Pᴸ(t) = cᴸ∗(t)cᴸ(t)
and
Pᴿ(t) = cᴿ∗(t)cᴿ(t)
being the probability amplitude of finding the electron in the L or R channel.
One can find
Pᴸ(t) = cos²(∆t/2ħ)
Pᴿ(t) = sin²(∆t/2ħ)
The electron periodically oscillates between the two channels with time.
What is spin resonance?
Magnetic field, B₁, rotating in the x-y plane (around z-axis) with frequency 2ω₀.
Electron spin appears to precess around B₁ with frequency 2ω₁.
If oscillating frequency of external magnetic field reaches the Larmor frequency of the electron, ω = 2ω₀,
〈Ŝz〉= ħ/2 cos(2ω₁t).
〈Ŝz〉oscillates between ±ħ/2 with frequency 2ω₁
〈Ŝz〉absorbs energy from the field while increasing, and radiates energy while decreasing. This is spin resonance, and it is important for imaging devices such as MRIs.
What is spin blockade in a Q-bit?
Double quantum dot with field B₀ along z-axis and oscillating field B₁ cos(ωt) in the x − y plane.
Adjust gate voltages such that only one electron can tunnel into each quantum dot.
Static field B₀, so electrons are spin-up.
Next sub-level empty due to the Coulomb blockade.
Can fine-tune Vᵍᴿ so the right empty sublevel is energetically lined up with the left occupied level.
No magnetic field would lead to resonant tunnelling between the dots, but this is forbidden by Pauli exclusion as the sublevel can only host spin-down.
This phenomenon is “spin blockade”, and is vital for manipulating Q-bits via the electron’s spin degree of freedom.
Through the oscillating field, it is possible to control the spin orientation in each channel.
How does a Q-bit store more information than a classical bit?
Superposition.
|αz〉= spin-up = 1
|βz〉= spin-down = 0
1/√2’ (|αz〉+ |βz〉) is superposition and would behave like noise in a classical bit, but actually contains well defined information of |αx〉. It represents the Ŝₓ spin-up eigenstate.
Wavefunction of an entangled vs non-entangled state
Entangled:
|χ(1, 2)〉= 1/√2’ [|α₂⁽ᴬ⁾ β₂⁽ᴮ⁾〉− |β₂⁽ᴬ⁾ α₂⁽ᴮ⁾〉]
Measuring Ŝ₂⁽¹⁾ returns ±ħ/2.
* If +ħ/2,
|χ(1, 2)〉collapses into |α₂⁽ᴬ⁾ β₂⁽ᴮ⁾〉
any subsequent measurement of Ŝ₂⁽²⁾ must return −ħ/2 with 100% probability.
- If −ħ/2,
|χ(1, 2)〉collapses into |β₂⁽ᴬ⁾ α₂⁽ᴮ⁾〉
any subsequent measurement of Ŝ₂⁽²⁾ must return +ħ/2 with 100% probability.
Measurement of one determines value of the other, therefore the system is entangled.
Non-entangled:
|χ(1, 2)〉= 1/√2’ [|α₂⁽ᴬ⁾ α₂⁽ᴮ⁾〉− |β₂⁽ᴬ⁾ α₂⁽ᴮ⁾〉]
Regardless of the outcome of Ŝ₂⁽¹⁾, Ŝ₂⁽²⁾ will always give ħ/2 with 100% probability.
Any system whose spinor wavefunction can be described as a product of individual particles’ spinors is regarded as a non-entangled system.
Z spinors in x basis set.
|α₂〉= 1/√2’ [|αₓ〉+ |βₓ〉]
|β₂〉= 1/√2’ [|αₓ〉− |βₓ〉]
Electron intrinsic spin magnetic moment
µ̂ₛ = −egₛ/2m Ŝ
