Practice Questions and Problem Sheets Flashcards
1
Q
Kinetic energy (in thermal equilibrium)
A
Ek = 3/2 kT
2
Q
Ekin =
A
p^2/2m(p)
3
Q
Energy in terms of voltage
A
E = qoVo
4
Q
the Dirac function describes
A
a particle with a perfectly-defined momentum p0 : plane wave.
5
Q
Hamiltonian Operator
A
H(hat) = T(hat) + V(hat)
H(hat) = Hamiltonian
T(hat) = Kinetic energy
V(hat) = Potential energy
6
Q
potential of a simple harmonic oscillator (classically)
A
V = 1/2ω^2 m x^2
=> k = ω^2 m
ω = 2πv/2πr0
7
Q
potential of an electron
A
V = -e^2/(4πεr)
8
Q
kinetic spatial-picture operator
A
T(hat) = - ℏ^2/2m ∇^2
9
Q
two operators that are the same will always
A
commutes with itself
10
Q
Q^m Q^n operators
A
they commute
11
Q
[p(hat),H(hat)] =
A
[p(hat),p(hat)^2/2m] = [p(hat),p(hat)p(hat)]/2m = 0
12
Q
Probability is obtained by
A
mod square integral
P = ∫|Ψ|^2 dx
P = ∫ Ψ*Ψ dx
13
Q
normalisation
A
( -∞ ∫∞ )|Ψ|^2 dx = 1
14
Q
Hermitian transpose of the integrand gives
A
the same answer
15
Q
a is the
A
half-width of the well
i.e a = L/2
16
Q
e^(-ikx)e^(ikx) =
A
= 1
17
Q
what are the possible forms of the wavefunction immediately after a measurement
A
will return the principle quantum number
18
Q
if as V(x) -> 0 x -> ∞
A
x ≠ 0 then the type of well is
the delta well
19
Q
Delta function well
potential
SE
k value
general solution
A
C - aδ(x)
d^2Ψ/dx^2 = -2mE/ℏ^2Ψ = Κ^2Ψ
Κ = √(-2mE/ℏ^2)
Ψ(x) = Ae^-Κx + Be^Κx
20
Q
decay constant in the delta well
A
k = √((-2mE)/ℏ^2)
21
Q
Delta well is
A
symmetric
calculate 1 side of the integral
22
Q
tan(ka) = k/λ derivation
solving the finite square well
A
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)
23
Q
solving the finite square well diagram
A
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)
24
Q
the expectation value of x is
A
zero
25
the expectation value of p is
zero
26
complex conjugate of a real
is the real
27
uniform energy spacing =
ℏω
28
T =
1 - R
29
the probability that an electron with energy incident on the surface will tunnel into the second piece of metal
T = 1 - R
30
Rotational energy
Erot = 1/2 I ω^2
L(v) = I ω(v)
=> I = ma^2
31
l(l+1)ℏ^2
32
L(h)(z) =
-iℏ(x∂(y) - y∂(x))
33
if L(z) = 0
Ψ is an eigenstate with m = 0
34
Y^0_0
1/√4π
35
Y^0_1
√3 / √4π cosθ
36
Y^±1_1
∓ √3 / √8π sinθ exp(±iΦ)
37
derive a normalisation constant for Ψ ∝ f(x)
Ψ = 1/√Z f(x)
where Z is the normalisation constant
∫|Ψ|^2 dx = 1
38
when integrating θ rather than x use
dΩ
dΩ =(2π ∫ 0) dφ (π ∫ 0) sinθ dθ
39
u^n du =
u^(n+1)/n+1
40
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)
41
L± Y^m_l =
ℏ√(l(l+1)-m(m±1)) Y^m±1_l
42
Lx(h) in terms of ladder operators
1/2 (L+(h) + L-(h))
43
Ly(h) in terms of ladder operators
1/2i (L+(h) - L-(h))
44
U =
- µ(v) . B(v)
= -g e/2m(e) L(h)
45
ΔE =
±µ(B) B(v)
46
µ(v) is the
magnetic moment measured in J T^-1
47
∇B(v)
is the field gradient
48
F = m(Ag) =
- ∇U = ∓µ(B)∇B(v) µ(v)
49
Δz is the
separation of the beams
Δz = 1/2 at^2 + at . T
= F/m(Ag)v^2 [ d^2/2 + dD]
50
Heisenberg Uncertainty principle
ΔxΔp ≥ ℏ/2
where Δx can = 2R (radius)
and Δp = p(min)
51
if there are 3 separate translational degrees of freedom
multiply by 3
52
work function
hc/λ = W + E
where W is the work function
53
the Compton shift =
λ - λ' = Δλ
54
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
55
The spectrum of scattered X-Rays at 90° has two peaks
1. The Compton wavelength-shifted peak
2. 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.
56
One-dimensional time independent Schrodinger equation
H(hat) Ψ = EΨ
where H is the hamiltonian
which leads to the TISE in the formula sheet
57
To show that something is a solution to the wave equation
1. Take Ψ derivatives
2. Substitute into Schrodinger Equation
3. Solve to find H(hat) Ψ is constant and Ψ is an eigenstate and solution of SE
58
probability density graph
1. a gaussian centered around zero
2. labels |Ψ|^2 and x
3. |Ψ(Δx)|^2 is the probability density at the uncertainty value
4. uncertainty at half the gaussian
see notes
59
raising ladder operator
a+(hat)
see formula sheet
60
to apply the raising ladder operator on a wavefunction
1. take derivates of the wavefunction
2. substitute into raising ladder operator a+(hat)
3. solve
61
A perfect black body is
an object that perfectly absorbs all radiation that falls upon it regardless of wavelength
62
when a measurement is made, the wavefunction will
collapse into an eigenstate |φi> with probability |ci|^2
63
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
64
probability of an eigenfunction
Pi = |ci|^2
65
uncertainty of the eigenfunction expectation value
ΔE^2 = - ^2
66
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
67
Solving the time-independent Schrodinger equation for E < V
Write the time-independent Schrodinger equation with RHS (V-E)Ψ
68
Ehrenfest's Theorem
the recovery of classical dynamics in the expectation values of QM
example of the general correspondence principle
69
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
70
Ramsaur-Townsend effect
low energy electrons scattering from atoms, usually noble gases
cannot be explained classically
71
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.
72
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
73
energy from the lifetime
ΔE = ℏ/τ
74
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
75
minimum transmission when
sin^2(ka) = 1
76
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.
77
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)
78
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
79
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
80
inner-atomic distance are order
10^-10m
81
Ψ = sin(kx)
Ψ = 1/√a sin(kx)
Ψ = 1/√a 1/2i (exp{ikx}-exp{-ikx})
Ψ = Φ(+) - Φ(-)
82
p(hat)Φ(+) =
-iℏikΦ(+)
83
Conservation of probability implies
dP/dt = 0
84
cos^2(x) =
sin^2(x) =
1/2 + cos(2x)/2
1/2 - cos(2x)/2
85
P =
(-∞ ∫ ∞) |Ψ|^2 dx
86
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)
87
the rest mass energy of an electron
E(e) = m(e)c^2
88
number of transition photons corresponding to this is
N = m(e) c^2 T/h
89
l = 0,1,2,3...
s , p , d , f atomic orbitals
90
J =
ℏ√(j(j+1))
91
mismatch in the angular momentum due to
uncertainty principle
92
general gaussian function
exp{-ax^2} = √(π/a)
93
T(t) is proportional to
exp(iEt/ℏ)
94
U(-)(n) is proportional to
sin(±nπ) = 0
95
U(+)(n) is proportional to
cos(±(2n-1)nπ/2) = 0
96
H(hat) U(-)(n) =
E(-)(n) U(-)(n)
97
the parity eigenvalue of U(-)(n) is
-1 for all n
98
the parity eigenvalue of U(+)(n) is
is +1 for all n
99
0 for all n and m
100
since L(x) and L(y) commute with L^2
the ladder operators commute with the total angular momentum
101
ladder operators change
the magnetic quantum number, m
102
µ(z) =
-µ(B) L(z) / ℏ
103
number of beams
corresponds to the number of m(j) values
104
x(hat) in terms of the ladder operators
x(hat) = √(ℏ/2mω) (a(+)(hat) + a(-)(hat))
105
p(hat) in terms of the ladder operators
p(hat) = i√(ℏmω/2) (a(+)(hat) - a(-)(hat))
106
a(+)(hat) a(-)(hat)
√n √n
107
a(-)(hat) a(+)(hat)
√n+1 √n+1
108
the final term in Planck's model
is a modification of equipartition coming from discretisation of the available energy levels.
109
The ultraviolet catastrophe
ever-increasing modes with same energy: frequency doesn't scale as expected.
110
the peak of the spectrum
is found by taking the derivative of frequency or wavelength and = 0
dy/dv = 0
111
mathematically conditions applying to wavefunctions
square-integrable
have continuous values and derivatives at all points
112
A discontinuous wavefunction value
would produce an infinite probability
flux at that point
113
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.
114
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
115
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
116
the hamiltonian in terms of ladder opperators
solve a(-)(hat)a(+)(hat)
insert H
117
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
118
stationary states
'standing wave' solutions of time-invariant probability density
119
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'
120
scattering from forbidden potential step (E
Ψ1(x) = Aexp(ik1x) + Bexp(-ik1x)
k1 = √(2mE/ℏ^2)
Ψ2(x) = Dexp(-k2x)
k2 = √(2m(V-E)/ℏ^2)
Ψ1 = Ψ2
Ψ1' = Ψ2'
121
Rayleigh's method was
to count modes of electromagnetic standing waves in an idealised box of sides L
122
the energy spectrum is not continuous, but
quantized into discrete units of epsilon = nhv
123
classical equipartition comes from
computing the expectation value of a Boltzmann distribution
124
stopping voltage
for a given frequency of light, the voltage that sends the current to zero.
125
heavier objects recoil
less than light ones
126
the canonical experiment demonstrating wavelike nature of light is
Young's double-slit
127
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.
128
the FT can be thought of as
an integral map between the momentum/k-space and the spatial x-space.
129
the tool we use to derive the wavepacket motion is the
principle of stationary phase
130
the fourier-transform of a single-frequency plane wave will return
the definition of the dirac-delta function
131
the uncertainty principle is found through
the operator algebra
132
hilbert space
vectors in an infinite-dimensional space
133
the overlap integral tells us
how much shared information there is between two states
134
the commutator measures the
compatibility between the eigenstate bases of the two operators
135
repeatedly applying the lowering operator will eventually result in
the ground state, on which a further lowering just returns 0
136
the QSHO always has a finite
zero point energy E(ground) > 0
137
l is the
orbital angular momentum quantum number
138
orbital g factor reflects
the efficiency of conversion of angular momentum into magnetic momentum
139
Zeeman effect
splitting of energy-degenerate spectral lines
140
s is the
spin quantum number
141
ms is
the spin magnetic quantum number
142
total angular momentum
J = L + S
143
j is
the total angular momentum quantum number
144
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:
145
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.
146
the density matrix is a key object for
understanding entanglement and decoherence.
