Applications of Quantum Physics Flashcards

Question

Applications of quantum dots

Answer 1

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.

Answer 2

−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

Answer 3

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]

Answer 4

1. Aufbau principle: Electrons occupy levels in ascending energy order, with the lowest energy level being filled first. 2. Pauli’s exclusion principle: No more than two electrons can occupy the same energy level, and the two electrons must have opposite spin orientation. 3. Hund’s principle: Degenerate energy levels (eigenvalues with the same energy) must all be singly occupied before any are doubly occupied.

Answer 5

Drain, insulator, quantum dot with surrounding gate, insulator, source.

Answer 6

Drain/source and gate potentials can be varied to move energy levels, allowing or disallowing quantum tunnelling between different parts of the dot.

Answer 7

−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.

Answer 8

Eₙ = (n + ½)ħω with n = 0,1,2...

Answer 9

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.

Answer 10

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 . . . .

Answer 11

1. eV_S < eV_D < E₀ No conduction happens. 2. 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. 3. 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. 4. 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)

Answer 12

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

Answer 13

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)

Answer 14

DOS(E) = dN/dE Usually use dN/dk dk/dE dN/dk can be calculated spacially/volumetrically

Answer 15

êᵢ · êⱼ = δᵢⱼ **A** = Aₓêₓ + Aᵧêᵧ + A₂ê₂ with A = ||**A**|| = √Aₓ² + Aᵧ² + A₂² ' For any two vectors, if **A** · **B** = 0, they are orthogonal

Answer 16

Answer 17

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〉

Answer 18

L̂ = r̂ × p̂

Answer 19

∆rᵢ∆pᵢ ≥ ħ/2

Answer 20

(ÂB̂ - B̂Â) Only if = 0, commutative, can be known at the same time.

Answer 21

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

Answer 22

L̂± = L̂ₓ ± iL̂ᵧ

Answer 23

[L̂₊, L̂₋] = L̂₊L̂₋ − L̂₋L̂₊ L̂₊L̂₋ = L̂² − L̂₂² + ħL̂₂ L̂₋L̂₊ = L̂² − L̂₂² − ħL̂₂ [L̂₊, L̂₋] = 2ħL̂₂

Answer 24

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

Answer 25

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

Answer 26

−nħ/2 ≤ λ₂ ≤ nħ/2 * If n is even, n=2*l* −*l*ħ ≤ λ₂ ≤ *l*ħ 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

Answer 27

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

Answer 28

Answer 29

[Ŝₓ, Ŝᵧ] = iħŜ₂ [Ŝᵧ, Ŝ₂] = iħŜₓ [Ŝ₂, Ŝₓ] = iħŜᵧ [Ŝ₂, Ŝ²] = 0

Answer 30

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

Answer 31

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〉)

Answer 32

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**

Answer 33

∆E = −**µ** · **B**

Answer 34

**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 (2*l* + 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.

Answer 35

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

Answer 36

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

Answer 37

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.

Answer 38

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.

Answer 39

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.

Answer 40

Answer 41

* 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

Answer 42

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.

Answer 43

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.

Answer 44

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

Answer 45

Eₙ = −E_R/n² where E_R = 13.6 eV is the Rydberg energy

Answer 46

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

Answer 47

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.

Answer 48

Answer 49

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** = e**Bₑₓₜ**/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

Answer 50

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.

Answer 51

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.

Answer 52

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)

Answer 53

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

Answer 54

Answer 55

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.

Answer 56

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.

Answer 57

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.

Answer 58

Answer 59

**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.

Answer 60

|α₂〉= 1/√2' [|αₓ〉+ |βₓ〉] |β₂〉= 1/√2' [|αₓ〉− |βₓ〉]

Answer 61

µ̂ₛ = −egₛ/2m Ŝ