Colour and Hardness Problem
Description
- imagine that we are measuring two properties of electrons: colour which is either black or white, and hardness which is either hard or soft
- we have boxes which can measure colour or hardness, send the electron into the box and the port that it comes out of is corresponds to its colour or hardness
- when white electrons are sent into a colour box they come out white again
- when white electrons are sent into a hardness box, 50% come out hard and 50% come out soft, this is the same for any combination of two boxes and properties
- but if the soft electrons are sent into a third box measuring colour, instead of 100% being white, 50% are white and 50% are black
Colour and Hardness Problem
Explanation
- the two properties are actually representative of measuring spin using two different axis
- imagine to operators for measuring the two properties colour and hardness
- when the property is measured the value is an eigenvalue of that operator, and after measurement the state of the particle becomes the eigenstate corresponding to that eigenvalue
- so when we are measuring colour for a second time, we are measuring it given that the particle is prepared in the soft state, and measuring from that state there is a 50/50 chance of measuring either colour
- this all happens because the two operators do not commute, which is equivalent to the uncertainty principle, when we are sure of one characteristic, we cannot be certain of the other
Bound States
-occur when particle motion is restricted to finite space
Particle in an Infinite Well
Description
- well with width L
- potential V(x)=0 for 0
Particle in an Infinite Well
Possible Wave Functions - Derivation
-Schrodinger’s time-independent equation
-ℏ²/2m ∂²ψ(x)/∂t² + V(x) ψ(x) = E(x) ψ(x)
-inside the well V(x)=0, rearrange:
∂²ψ(x)/∂t² + 2mE/ℏ² * ψ(x) = 0
-where k² = 2mE/ℏ²
-we can guess a solution of the form
ψ(x) = A*sin(kx)
-apply boundary conditions:
ψ(0) = ψ(L) = 0
=> sin(kL) = 0 => kL=nπ, n=1,2,3,…
k = nπ/L
-normalise to find A:
∫ dx |ψ(x)|² = 1
-where the integral is taken between x=0 and x=L
this gives A=√(2/L)
-so the allowed wave functions are:
ψ(x) = √(2/L) * sin(nπx/L) , n=1,2,3,…
Particle in an Infinite Well
Possible Wave Functions - Formula
ψ(x) = √(2/L) * sin(nπx/L) n = 1, 2, 3, ...
Particle in an Infinite Well
Possible Energies
E = ℏ²k²/2m and k=nπ/L
=>
En = n²ℏ²π²/2mL² , n = 1, 2, 3, …
-since E∝n² the higher the energy levels, the greater the difference between them
Particle in an Infinite Well
Experiments and Applications
Quantum Coral -diameter ~10nm -electrons are bound to a circular are on a surface of copper -the are is defined by 48 iron atoms around the perimeter -the electrons form 2D standing waves Quantum Dot -confinement potential in all directions -behaves like an artificial atom
Potential Step
Description
- an electron moves freely to the right and at x=0, it encounters a potential step
- at points x>0, the potential has constant value Uo>0
- region I is the area before the step, region II is the area after the step
- when the incident electron reaches the potential step it can either be reflected or transmitted
- there are two cases:
1) E>Uo
2) E
Potential Step - Case 1 : E>Uo
ψinc(x,t) = e^(ikx) ψref(x,t) = e^(-ikx) ψtrans(x,t) = e^(ik'x) -where: k²=2mE/ℏ² (k')²=2m/ℏ² * (E-Uo) -total wave function for region I: ψI = A*e^(ikx) + B*e^(-ikx) -total wave function for region II: ψII = C*e^(ik'x) -in region I there is no potential so k²=2mE/ℏ² -in region II there is a potential so the Schrodinger equation rearranged to: ∂²ψ(x)/∂t² + 2m/ℏ² *(E-Uo) ψ(x) = 0 -the coefficient of ψ(x) is equal to (k')² and (E-Uo)>0 since in this case E>Uo -sub in boundary conditions: ψI(0) = ψII(0) dψI/dx = dψII/dx evaluated at x=0 -two equations and three unknowns so the wave functions can all be expressed in terms of one unknown
Potential Step
Reflection Coefficient and Transmission Coefficient
R = no. of particles reflected per unit time / no. of particles incident per unit time T = no. of particles transmitted per unit time / no. of particles incident per unit time
-since R and T are probsbilities and reflection and transmission are the only possible outcomes:
R + T = 1
Potential Step - Case 1 : E>Uo
Reflection Coefficient and Transmission Coefficient
R = (k-k'/k+k')² T = 4kk'/(k+k')²
Potential Step - Case 2 : E
ψinc(x,t) = e^(ikx) ψref(x,t) = e^(-ikx) ψtrans(x,t) = e^(-k'x) (or e^(k'x)) -where: k²=2mE/ℏ² -(k')²=2m/ℏ² * (E-Uo) -note the negative sign since E
Potential Barrier
Description
- particle moves to the right
- at x=0 there is a potential step up Uo
- at x=L there is a potential step back down to 0
- region I is the region before the first step, region II is the region between the two steps and region III is the region between after the second step
- like with the potential step there are two cases:
1) E>Uo
2) E
Potential Barrier - Case 1 : E>Uo
ψI = A*e^(ikx) + B*e^(-ikx) ψII = C*e^(ik'x) + D*e^(-ik'x) ψIII = F*e^(ikx) -where: k²=2mE/ℏ² (k')²=2m/ℏ² (E-Uo) -to find the constants apply the boundary conditions: ψI(0) = ψII(0) dψI/dx = dψII/dx evaluated at x=0 ψII(L) = ψIII(L) dψII.dx = dψIII/dx evaluated at x=L
Potential Barrier - Case 2 : E
ψI = A*e^(ikx) + B*e^(-ikx) ψII = C*e^(-k'x) + D*e^(k'x) ψIII = F*e^(ikx) -where: k²=2mE/ℏ² -(k')²=2m/ℏ² (E-Uo) -in this case the exponentially growing solution (D) is allowed as the solution only covers a finite area 0
Quantum Tunnelling
-for a potential barrier where the reflection coefficient is:
R = sinh²(k’L) / [ sinh²(k’L) + 4*(k²(k’)²)/(k²-k’²)²]
-since T=1-R, in general T is non-zero meaning a particle can be found in region III even though it has lower energy than the barrier
Resonant Transmission
-for a particle and a potential barrier where E>Uo, the reflection coefficient is:
R = sin²(k’L) / [ sin²(k’L) + 4*(k²(k’)²)/(k²-k’²)² ]
-since R ∝ sin² , we see that R=0 for certain values of k’:
k’L = nπ , n=1,2,3,…
OR
E = Uo + n²ℏ²π²/2mL²
-for these values of energy, particles do not reflect, R=0, instead everything goes through, T=0, this is called resonant transmission
Alpha Decay
-from different alpha emitting nuclei, emitted alpha particles had roughly the same energy yet mean time of decay differed by orders of magnitude
-this is explained by quantum tunnelling
-the alpha particle needs to quantum tunnel to escape the nucleus but the probability of that happening exponentially depends on its energy:
T ∝ e^(-c√E)
Scanning Tunneling Microscope (STM)
- metal tip positioned near sample
- free electrons in tip/sample can tunnel across because the tip is close enough to the sample
- measure the tunnelling current across the surface of the sample to produce an effective image of the surface
Linear Harmonic Oscillator
-a spring attached to a barrier and a mass horizontally on a table top
-classically:
H = p²/2m + 1/2 kx²
-total energy is conserved, it just transfers between kinetic and potential
-the motion of the classical oscillator is bounded by classical turning points
Harmonic Oscillator in Physics
-the harmonic oscillator is important in all areas of physics
-it describes the deviations of a system around points of stable equilibrium
V(x) = V(xo) + dV/dx (x-xo) + 1/2 d²V/dx² (x-xo)² + ….
-expanded using a taylor series, all derivatives evaluated at x=xo
-can take V(xo)=0 and at an equilibrium point dV/dx = 0
=>
V(x) ≈ 1/2 d²V/dx² (x-xo)² = 1/2 k (x-xo)²
-any system behaves as a simple harmonic oscillator in the vicinity of a stable equilibrium
ω = √(k/m) = √[ 1/m d²V/dx² ]
Quantum Harmonic Oscillator
^H = ^p²/2m + 1/2 mω² ^x²
-eigenvalues and eigenvectors difficult to compute since ^x and ^p don’t commute
-using Schrodingers notation to write as a differential equation:
- ℏ²/2m ∂²ψ/∂x² + 1/2 mω²x²ψ = Eψ
-Eigenenergies:
En = ℏω (n+1/2) , n=0,1,2,3,….
-eigenvectors:
ψn(x) = 1/√[√(π)2^n n! xo] * Hn(x/xo) e^(x²/2xo²)
-where Hn are the Hermite Polynomials
Applications of Harmonic Oscillators
- describes the vibrations of crystal lattices: phonons
- electrons in magnetic fields behave like harmonic oscillators
