Practice Questions and Problem Sheets Flashcards
Kinetic energy (in thermal equilibrium)
Ek = 3/2 kT
Ekin =
p^2/2m(p)
Energy in terms of voltage
E = qoVo
the Dirac function describes
a particle with a perfectly-defined momentum p0 : plane wave.
Hamiltonian Operator
H(hat) = T(hat) + V(hat)
H(hat) = Hamiltonian
T(hat) = Kinetic energy
V(hat) = Potential energy
potential of a simple harmonic oscillator (classically)
V = 1/2ω^2 m x^2
=> k = ω^2 m
ω = 2πv/2πr0
potential of an electron
V = -e^2/(4πεr)
kinetic spatial-picture operator
T(hat) = - ℏ^2/2m ∇^2
two operators that are the same will always
commutes with itself
Q^m Q^n operators
they commute
[p(hat),H(hat)] =
[p(hat),p(hat)^2/2m] = [p(hat),p(hat)p(hat)]/2m = 0
Probability is obtained by
mod square integral
P = ∫|Ψ|^2 dx
P = ∫ Ψ*Ψ dx
normalisation
( -∞ ∫∞ )|Ψ|^2 dx = 1
Hermitian transpose of the integrand gives
the same answer
a is the
half-width of the well
i.e a = L/2
e^(-ikx)e^(ikx) =
= 1
what are the possible forms of the wavefunction immediately after a measurement
will return the principle quantum number
if as V(x) -> 0 x -> ∞
x ≠ 0 then the type of well is
the delta well
Delta function well
potential
SE
k value
general solution
C - aδ(x)
d^2Ψ/dx^2 = -2mE/ℏ^2Ψ = Κ^2Ψ
Κ = √(-2mE/ℏ^2)
Ψ(x) = Ae^-Κx + Be^Κx
decay constant in the delta well
k = √((-2mE)/ℏ^2)
Delta well is
symmetric
calculate 1 side of the integral
tan(ka) = k/λ derivation
solving the finite square well
Ψ(x) = { A^(-)exp(kx) x < -a
Ψ(x) = { Bcos(λx)+Csin(λx) |x| < a
Ψ(x) = { A^(+)exp(-kx) x > a
outside the well a^2 = 2m(V-E)/ℏ^2
inside the well k^2 = 2mE/ℏ^2
even so C = 0
equate
take derivative
divide the two
Bcos(λa) = Aexp(-ka)
-λBsin(λa) = -kAexp(-ka)
solving the finite square well diagram
see notes
k/λ = √((V-E)/E)
as V -> 0 reduction is solution. but always at least one even solution
E(k = π/2a) < V < E(k = 3π/2a)
the expectation value of x is
zero
the expectation value of p is
zero
complex conjugate of a real
is the real
uniform energy spacing =
ℏω
T =
1 - R
the probability that an electron with energy incident on the surface will tunnel into the second piece of metal
T = 1 - R
Rotational energy
Erot = 1/2 I ω^2
L(v) = I ω(v)
=> I = ma^2
<L^2> =
l(l+1)ℏ^2
L(h)(z) =
-iℏ(x∂(y) - y∂(x))
if L(z) = 0
Ψ is an eigenstate with m = 0
Y^0_0
1/√4π
Y^0_1
√3 / √4π cosθ
Y^±1_1
∓ √3 / √8π sinθ exp(±iΦ)
derive a normalisation constant for Ψ ∝ f(x)
Ψ = 1/√Z f(x)
where Z is the normalisation constant
∫|Ψ|^2 dx = 1
when integrating θ rather than x use
dΩ
dΩ =(2π ∫ 0) dφ (π ∫ 0) sinθ dθ
u^n du =
u^(n+1)/n+1
derive the energy levels and wavefunction for a particle in a 3D, cube-shaped potential V(x,y,z)
Ψ(x,y,z) = {φ1(x)φ2(x)φ3(x) inside box
{0 otherwise
Apply TISE : divide by φ1 φ2 φ3
E = constant
d^2φ1/dx^2 = -k1^2φ1 etc..
apply boundary conditions
Ψ(x,y,z) = Asin(n1π(x)/a)sin(n2π(y)/a)sin(n3π(z)/a)
E = Σi ℏ^2 ki^2 /2m = (ℏ^2π^2)/(2ma^2) (n1^2+n2^2+n3^2)
L± Y^m_l =
ℏ√(l(l+1)-m(m±1)) Y^m±1_l
Lx(h) in terms of ladder operators
1/2 (L+(h) + L-(h))
Ly(h) in terms of ladder operators
1/2i (L+(h) - L-(h))
U =
- µ(v) . B(v)
= -g e/2m(e) L(h)
ΔE =
±µ(B) B(v)
µ(v) is the
magnetic moment measured in J T^-1
∇B(v)
is the field gradient
F = m(Ag) =
- ∇U = ∓µ(B)∇B(v) µ(v)
Δz is the
separation of the beams
Δz = 1/2 at^2 + at . T
= F/m(Ag)v^2 [ d^2/2 + dD]
Heisenberg Uncertainty principle
ΔxΔp ≥ ℏ/2
where Δx can = 2R (radius)
and Δp = p(min)
if there are 3 separate translational degrees of freedom
multiply by 3
work function
hc/λ = W + E
where W is the work function
the Compton shift =
λ - λ’ = Δλ
Compton scattering is evidence for the particle nature of light
the observed shift in the wavelength of the scattered photons cannot be explained classically.
Classically the incoming radiation should be absorbed by the electrons and radiation emitted at the same wavelength.
Compton scattering can be described by considering the X-Rays to be a stream of photons with momentum p = h/λ, and the process to be scattering of those photons off the orbital electron
The spectrum of scattered X-Rays at 90° has two peaks
- The Compton wavelength-shifted peak
- photons scattering off tightly bound electrons: the photon is then effectively scattering off the nucleus. Hence the mass of the nucleus should be used in the Compton scattering equation, resulting in a very small and effectively undetectable shift in wavelength.
One-dimensional time independent Schrodinger equation
H(hat) Ψ = EΨ
where H is the hamiltonian
which leads to the TISE in the formula sheet
To show that something is a solution to the wave equation
- Take Ψ derivatives
- Substitute into Schrodinger Equation
- Solve to find H(hat) Ψ is constant and Ψ is an eigenstate and solution of SE
probability density graph
- a gaussian centered around zero
- labels |Ψ|^2 and x
- |Ψ(Δx)|^2 is the probability density at the uncertainty value
- uncertainty at half the gaussian
see notes
raising ladder operator
a+(hat)
see formula sheet
to apply the raising ladder operator on a wavefunction
- take derivates of the wavefunction
- substitute into raising ladder operator a+(hat)
- solve
A perfect black body is
an object that perfectly absorbs all radiation that falls upon it regardless of wavelength
when a measurement is made, the wavefunction will
collapse into an eigenstate |φi> with probability |ci|^2
what does the expectation value of the energy represent
the average of many measurements, assuming that the measurements over a large number of identical systems
probability of an eigenfunction
Pi = |ci|^2
uncertainty of the eigenfunction expectation value
ΔE^2 = <E^2> - <E>^2</E>
Show that
∂/∂t ( ∫ V) Pd^3r = - ( ∫ S) J.dS
^ Green’s theorem
Take the time derivative of P(r,t) = |psi (r,t)|^2
int V dP/dt = int V nabla . J dr = 0
Identify J
use green’s theorem
J is the probability currrent
Solving the time-independent Schrodinger equation for E < V
Write the time-independent Schrodinger equation with RHS (V-E)Ψ
Ehrenfest’s Theorem
the recovery of classical dynamics in the expectation values of QM
example of the general correspondence principle
the current is identically zero if either
the wavefunction is real valued
or
it has a complex phase which applies uniformly to all space positions
Ramsaur-Townsend effect
low energy electrons scattering from atoms, usually noble gases
cannot be explained classically
quantum tunelling
finite wavefunction manages to leak through the classically forbidden region and resume its oscillatory behaviour on reaching the classically allowed region on the other side.
constraints on the wavefunction at the interface between two regions of different potential energy
Ψ1(x=0) = Ψ2(x=0) wave function must be continuous conservation of mass
(dΨ1/dx)(x=0) = (dΨ2/dx)(x=0) first derivative of wave function must be continuous, conservation of momentum
energy from the lifetime
ΔE = ℏ/τ
Writing the Schrodinger equation in the time-independent form
H = T + V
use separation of variables Ψ(r,t) = Υ(r)φ(t) in the schrodinger equation
divide both sides by Ψ(r,t) = Υ(r)φ(t)
since RHS depends on r and LHS depends only on t, both sides must be equal to a constant which we donate as E
minimum transmission when
sin^2(ka) = 1
an example of physical process that demonstrates a variation in transmission in a 1D potential
Ramsaur effects
show a variation in transmission with kinetic energy, with a minimum of about one eV for Xenon.
prove the angular momentum operators, Lx, Ly, Lz are given by
L(v hat) = r(v hat) x p(v hat)
L(v hat) = -iℏ | i j k
|x y z
|∂(x) ∂(y) ∂(z)
the physical significance of [Lx,Ly] = iℏLz
shows that a system cannot simultaneously be in an eigenstate of Lx and Ly and therefore the two components cannot be measured simultaneously
the physical significance of [L^2,Lx] = 0
shows that the system can simultaneously be in an eigenstate of L^2 and Lz and therefore the total angular momentum squared and one of the components can be measured simultaneously
inner-atomic distance are order
10^-10m
Ψ = sin(kx)
Ψ = 1/√a sin(kx)
Ψ = 1/√a 1/2i (exp{ikx}-exp{-ikx})
Ψ = Φ(+) - Φ(-)
p(hat)Φ(+) =
-iℏikΦ(+)
Conservation of probability implies
dP/dt = 0
cos^2(x) =
sin^2(x) =
1/2 + cos(2x)/2
1/2 - cos(2x)/2
P =
(-∞ ∫ ∞) |Ψ|^2 dx
expression for mass in terms of h,c and the oscillation period of a photon
E = mc^2
E = hv
m = h/(c^2 T)
the rest mass energy of an electron
E(e) = m(e)c^2
number of transition photons corresponding to this is
N = m(e) c^2 T/h
l = 0,1,2,3…
s , p , d , f atomic orbitals
J =
ℏ√(j(j+1))
mismatch in the angular momentum due to
uncertainty principle
general gaussian function
exp{-ax^2} = √(π/a)
T(t) is proportional to
exp(iEt/ℏ)
U(-)(n) is proportional to
sin(±nπ) = 0
U(+)(n) is proportional to
cos(±(2n-1)nπ/2) = 0
H(hat) U(-)(n) =
E(-)(n) U(-)(n)
the parity eigenvalue of U(-)(n) is
-1 for all n
the parity eigenvalue of U(+)(n) is
is +1 for all n
<U(+)(n)|U(+)(m)> =
0 for all n and m
since L(x) and L(y) commute with L^2
the ladder operators commute with the total angular momentum
ladder operators change
the magnetic quantum number, m
µ(z) =
-µ(B) L(z) / ℏ
number of beams
corresponds to the number of m(j) values
x(hat) in terms of the ladder operators
x(hat) = √(ℏ/2mω) (a(+)(hat) + a(-)(hat))
p(hat) in terms of the ladder operators
p(hat) = i√(ℏmω/2) (a(+)(hat) - a(-)(hat))
a(+)(hat) a(-)(hat)
√n √n
a(-)(hat) a(+)(hat)
√n+1 √n+1
the final term in Planck’s model
is a modification of equipartition coming from discretisation of the available energy levels.
The ultraviolet catastrophe
ever-increasing modes with same energy: frequency doesn’t scale as expected.
the peak of the spectrum
is found by taking the derivative of frequency or wavelength and = 0
dy/dv = 0
mathematically conditions applying to wavefunctions
square-integrable
have continuous values and derivatives at all points
A discontinuous wavefunction value
would produce an infinite probability
flux at that point
A discontinuous gradient
produces a step-change in
flux at a boundary, hence in
general an infinite rate of change of probability density at that point.
The quadratic potential well
ground state
spread
solutions
the ground-state wavefunction is a Gaussian
spread in x increases with energy
solutions are Hermite polynomials
solving the harmonic oscillator: ground state
a(-)(hat) = 0
dΨ(0)/dx = -mω/ℏ x Ψ(0)
gives the solution in the formula sheet
with energy = E(n) = (n+1/2)ℏω
the hamiltonian in terms of ladder opperators
solve a(-)(hat)a(+)(hat)
insert H
infinite square well
SE
k value
general solution
eigenstates
energies
-ℏ^2/2m d^2Ψ/dx^2 = EΨ
k = √(2mE/ℏ^2)
Ψ(x) = Acos(kx) + Bsin(kx)
u(x)+_n
U(x)-_n
En = n^2 π^2ℏ^2/8ma^2
stationary states
‘standing wave’ solutions of time-invariant probability density
Scattering from allowed potential step (E>V)
Ψ1(x) = Aexp(ik1x) + Bexp(-ik1x)
k1 = √(2mE/ℏ^2)
Ψ2(x) = Cexp(ik2x)
k2 = √(2m(E-V)/ℏ^2)
Ψ1 = Ψ2
Ψ1’ = Ψ2’
scattering from forbidden potential step (E<V)
Ψ1(x) = Aexp(ik1x) + Bexp(-ik1x)
k1 = √(2mE/ℏ^2)
Ψ2(x) = Dexp(-k2x)
k2 = √(2m(V-E)/ℏ^2)
Ψ1 = Ψ2
Ψ1’ = Ψ2’
Rayleigh’s method was
to count modes of electromagnetic standing waves in an idealised box of sides L
the energy spectrum is not continuous, but
quantized into discrete units of epsilon = nhv
classical equipartition comes from
computing the expectation value of a Boltzmann distribution
stopping voltage
for a given frequency of light, the voltage that sends the current to zero.
heavier objects recoil
less than light ones
the canonical experiment demonstrating wavelike nature of light is
Young’s double-slit
path integral formalism of QM
in which an infinite set of amplitudes contribute to all processes, with the dominant contributions coming from constructive interference near the classical solution, but not-quite-cancellations between non-classical paths give what are known as quantum corrections.
the FT can be thought of as
an integral map between the momentum/k-space and the spatial x-space.
the tool we use to derive the wavepacket motion is the
principle of stationary phase
the fourier-transform of a single-frequency plane wave will return
the definition of the dirac-delta function
the uncertainty principle is found through
the operator algebra
hilbert space
vectors in an infinite-dimensional space
the overlap integral tells us
how much shared information there is between two states
the commutator measures the
compatibility between the eigenstate bases of the two operators
repeatedly applying the lowering operator will eventually result in
the ground state, on which a further lowering just returns 0
the QSHO always has a finite
zero point energy E(ground) > 0
l is the
orbital angular momentum quantum number
orbital g factor reflects
the efficiency of conversion of angular momentum into magnetic momentum
Zeeman effect
splitting of energy-degenerate spectral lines
s is the
spin quantum number
ms is
the spin magnetic quantum number
total angular momentum
J = L + S
j is
the total angular momentum quantum number
is the wavefunction real, or just a mathematically artifact?
the aharonov-bohm effect confirms it is real.
Interference seen via two paths with different EM potentials A:
Quantum measurement problem
how is a measurement sharp, non-unitary and what is a measurement?
largely understood via the Copenhagen interpretation as decoherence from coupling.
the density matrix is a key object for
understanding entanglement and decoherence.
