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Quantum Mechanics > Practice Questions and Problem Sheets > Flashcards

Practice Questions and Problem Sheets Flashcards

1
Q

Kinetic energy (in thermal equilibrium)

A

Ek = 3/2 kT

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2
Q

Ekin =

A

p^2/2m(p)

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3
Q

Energy in terms of voltage

A

E = qoVo

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4
Q

the Dirac function describes

A

a particle with a perfectly-defined momentum p0 : plane wave.

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5
Q

Hamiltonian Operator

A

H(hat) = T(hat) + V(hat)

H(hat) = Hamiltonian

T(hat) = Kinetic energy

V(hat) = Potential energy

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6
Q

potential of a simple harmonic oscillator (classically)

A

V = 1/2ω^2 m x^2

=> k = ω^2 m

ω = 2πv/2πr0

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7
Q

potential of an electron

A

V = -e^2/(4πεr)

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8
Q

kinetic spatial-picture operator

A

T(hat) = - ℏ^2/2m ∇^2

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9
Q

two operators that are the same will always

A

commutes with itself

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10
Q

Q^m Q^n operators

A

they commute

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11
Q

[p(hat),H(hat)] =

A

[p(hat),p(hat)^2/2m] = [p(hat),p(hat)p(hat)]/2m = 0

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12
Q

Probability is obtained by

A

mod square integral

P = ∫|Ψ|^2 dx
P = ∫ Ψ*Ψ dx

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13
Q

normalisation

A

( -∞ ∫∞ )|Ψ|^2 dx = 1

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14
Q

Hermitian transpose of the integrand gives

A

the same answer

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15
Q

a is the

A

half-width of the well

i.e a = L/2

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16
Q

e^(-ikx)e^(ikx) =

A

= 1

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17
Q

what are the possible forms of the wavefunction immediately after a measurement

A

will return the principle quantum number

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18
Q

if as V(x) -> 0 x -> ∞

A

x ≠ 0 then the type of well is

the delta well

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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

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20
Q

decay constant in the delta well

A

k = √((-2mE)/ℏ^2)

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21
Q

Delta well is

A

symmetric

calculate 1 side of the integral

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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)

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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)

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24
Q

the expectation value of x is

A

zero

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25
Q

the expectation value of p is

A

zero

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26
Q

complex conjugate of a real

A

is the real

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27
Q

uniform energy spacing =

A

ℏω

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28
Q

T =

A

1 - R

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29
Q

the probability that an electron with energy incident on the surface will tunnel into the second piece of metal

A

T = 1 - R

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30
Q

Rotational energy

A

Erot = 1/2 I ω^2

L(v) = I ω(v)

=> I = ma^2

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31
Q

<L^2> =

A

l(l+1)ℏ^2

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32
Q

L(h)(z) =

A

-iℏ(x∂(y) - y∂(x))

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33
Q

if L(z) = 0

A

Ψ is an eigenstate with m = 0

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34
Q

Y^0_0

A

1/√4π

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35
Q

Y^0_1

A

√3 / √4π cosθ

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36
Q

Y^±1_1

A

∓ √3 / √8π sinθ exp(±iΦ)

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37
Q

derive a normalisation constant for Ψ ∝ f(x)

A

Ψ = 1/√Z f(x)

where Z is the normalisation constant

∫|Ψ|^2 dx = 1

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38
Q

when integrating θ rather than x use

A

dΩ

dΩ =(2π ∫ 0) dφ (π ∫ 0) sinθ dθ

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39
Q

u^n du =

A

u^(n+1)/n+1

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40
Q

derive the energy levels and wavefunction for a particle in a 3D, cube-shaped potential V(x,y,z)

A

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)

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41
Q

L± Y^m_l =

A

ℏ√(l(l+1)-m(m±1)) Y^m±1_l

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42
Q

Lx(h) in terms of ladder operators

A

1/2 (L+(h) + L-(h))

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43
Q

Ly(h) in terms of ladder operators

A

1/2i (L+(h) - L-(h))

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44
Q

U =

A
  • µ(v) . B(v)
    = -g e/2m(e) L(h)
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45
Q

ΔE =

A

±µ(B) B(v)

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46
Q

µ(v) is the

A

magnetic moment measured in J T^-1

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47
Q

∇B(v)

A

is the field gradient

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48
Q

F = m(Ag) =

A
  • ∇U = ∓µ(B)∇B(v) µ(v)
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49
Q

Δz is the

A

separation of the beams

Δz = 1/2 at^2 + at . T

= F/m(Ag)v^2 [ d^2/2 + dD]

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50
Q

Heisenberg Uncertainty principle

A

ΔxΔp ≥ ℏ/2

where Δx can = 2R (radius)

and Δp = p(min)

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51
Q

if there are 3 separate translational degrees of freedom

A

multiply by 3

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52
Q

work function

A

hc/λ = W + E

where W is the work function

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53
Q

the Compton shift =

A

λ - λ’ = Δλ

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54
Q

Compton scattering is evidence for the particle nature of light

A

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

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55
Q

The spectrum of scattered X-Rays at 90° has two peaks

A
  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.
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56
Q

One-dimensional time independent Schrodinger equation

A

H(hat) Ψ = EΨ

where H is the hamiltonian

which leads to the TISE in the formula sheet

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57
Q

To show that something is a solution to the wave equation

A
  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
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58
Q

probability density graph

A
  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

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59
Q

raising ladder operator

A

a+(hat)

see formula sheet

60
Q

to apply the raising ladder operator on a wavefunction

A
  1. take derivates of the wavefunction
  2. substitute into raising ladder operator a+(hat)
  3. solve
61
Q

A perfect black body is

A

an object that perfectly absorbs all radiation that falls upon it regardless of wavelength

62
Q

when a measurement is made, the wavefunction will

A

collapse into an eigenstate |φi> with probability |ci|^2

63
Q

what does the expectation value of the energy represent

A

the average of many measurements, assuming that the measurements over a large number of identical systems

64
Q

probability of an eigenfunction

A

Pi = |ci|^2

65
Q

uncertainty of the eigenfunction expectation value

A

ΔE^2 = <E^2> - <E>^2</E>

66
Q

Show that

∂/∂t ( ∫ V) Pd^3r = - ( ∫ S) J.dS

^ Green’s theorem

A

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
Q

Solving the time-independent Schrodinger equation for E < V

A

Write the time-independent Schrodinger equation with RHS (V-E)Ψ

68
Q

Ehrenfest’s Theorem

A

the recovery of classical dynamics in the expectation values of QM

example of the general correspondence principle

69
Q

the current is identically zero if either

A

the wavefunction is real valued

or

it has a complex phase which applies uniformly to all space positions

70
Q

Ramsaur-Townsend effect

A

low energy electrons scattering from atoms, usually noble gases

cannot be explained classically

71
Q

quantum tunelling

A

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
Q

constraints on the wavefunction at the interface between two regions of different potential energy

A

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
Q

energy from the lifetime

A

ΔE = ℏ/τ

74
Q

Writing the Schrodinger equation in the time-independent form

A

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
Q

minimum transmission when

A

sin^2(ka) = 1

76
Q

an example of physical process that demonstrates a variation in transmission in a 1D potential

A

Ramsaur effects

show a variation in transmission with kinetic energy, with a minimum of about one eV for Xenon.

77
Q

prove the angular momentum operators, Lx, Ly, Lz are given by

A

L(v hat) = r(v hat) x p(v hat)

L(v hat) = -iℏ | i j k
|x y z
|∂(x) ∂(y) ∂(z)

78
Q

the physical significance of [Lx,Ly] = iℏLz

A

shows that a system cannot simultaneously be in an eigenstate of Lx and Ly and therefore the two components cannot be measured simultaneously

79
Q

the physical significance of [L^2,Lx] = 0

A

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
Q

inner-atomic distance are order

A

10^-10m

81
Q

Ψ = sin(kx)

A

Ψ = 1/√a sin(kx)
Ψ = 1/√a 1/2i (exp{ikx}-exp{-ikx})
Ψ = Φ(+) - Φ(-)

82
Q

p(hat)Φ(+) =

A

-iℏikΦ(+)

83
Q

Conservation of probability implies

A

dP/dt = 0

84
Q

cos^2(x) =

sin^2(x) =

A

1/2 + cos(2x)/2

1/2 - cos(2x)/2

85
Q

P =

A

(-∞ ∫ ∞) |Ψ|^2 dx

86
Q

expression for mass in terms of h,c and the oscillation period of a photon

A

E = mc^2
E = hv

m = h/(c^2 T)

87
Q

the rest mass energy of an electron

A

E(e) = m(e)c^2

88
Q

number of transition photons corresponding to this is

A

N = m(e) c^2 T/h

89
Q

l = 0,1,2,3…

A

s , p , d , f atomic orbitals

90
Q

J =

A

ℏ√(j(j+1))

91
Q

mismatch in the angular momentum due to

A

uncertainty principle

92
Q

general gaussian function

A

exp{-ax^2} = √(π/a)

93
Q

T(t) is proportional to

A

exp(iEt/ℏ)

94
Q

U(-)(n) is proportional to

A

sin(±nπ) = 0

95
Q

U(+)(n) is proportional to

A

cos(±(2n-1)nπ/2) = 0

96
Q

H(hat) U(-)(n) =

A

E(-)(n) U(-)(n)

97
Q

the parity eigenvalue of U(-)(n) is

A

-1 for all n

98
Q

the parity eigenvalue of U(+)(n) is

A

is +1 for all n

99
Q

<U(+)(n)|U(+)(m)> =

A

0 for all n and m

100
Q

since L(x) and L(y) commute with L^2

A

the ladder operators commute with the total angular momentum

101
Q

ladder operators change

A

the magnetic quantum number, m

102
Q

µ(z) =

A

-µ(B) L(z) / ℏ

103
Q

number of beams

A

corresponds to the number of m(j) values

104
Q

x(hat) in terms of the ladder operators

A

x(hat) = √(ℏ/2mω) (a(+)(hat) + a(-)(hat))

105
Q

p(hat) in terms of the ladder operators

A

p(hat) = i√(ℏmω/2) (a(+)(hat) - a(-)(hat))

106
Q

a(+)(hat) a(-)(hat)

A

√n √n

107
Q

a(-)(hat) a(+)(hat)

A

√n+1 √n+1

108
Q

the final term in Planck’s model

A

is a modification of equipartition coming from discretisation of the available energy levels.

109
Q

The ultraviolet catastrophe

A

ever-increasing modes with same energy: frequency doesn’t scale as expected.

110
Q

the peak of the spectrum

A

is found by taking the derivative of frequency or wavelength and = 0

dy/dv = 0

111
Q

mathematically conditions applying to wavefunctions

A

square-integrable
have continuous values and derivatives at all points

112
Q

A discontinuous wavefunction value

A

would produce an infinite probability
flux at that point

113
Q

A discontinuous gradient

A

produces a step-change in
flux at a boundary, hence in
general an infinite rate of change of probability density at that point.

114
Q

The quadratic potential well

ground state

spread

solutions

A

the ground-state wavefunction is a Gaussian

spread in x increases with energy

solutions are Hermite polynomials

115
Q

solving the harmonic oscillator: ground state

A

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
Q

the hamiltonian in terms of ladder opperators

A

solve a(-)(hat)a(+)(hat)

insert H

117
Q

infinite square well

SE

k value

general solution

eigenstates

energies

A

-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
Q

stationary states

A

‘standing wave’ solutions of time-invariant probability density

119
Q

Scattering from allowed potential step (E>V)

A

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
Q

scattering from forbidden potential step (E<V)

A

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
Q

Rayleigh’s method was

A

to count modes of electromagnetic standing waves in an idealised box of sides L

122
Q

the energy spectrum is not continuous, but

A

quantized into discrete units of epsilon = nhv

123
Q

classical equipartition comes from

A

computing the expectation value of a Boltzmann distribution

124
Q

stopping voltage

A

for a given frequency of light, the voltage that sends the current to zero.

125
Q

heavier objects recoil

A

less than light ones

126
Q

the canonical experiment demonstrating wavelike nature of light is

A

Young’s double-slit

127
Q

path integral formalism of QM

A

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
Q

the FT can be thought of as

A

an integral map between the momentum/k-space and the spatial x-space.

129
Q

the tool we use to derive the wavepacket motion is the

A

principle of stationary phase

130
Q

the fourier-transform of a single-frequency plane wave will return

A

the definition of the dirac-delta function

131
Q

the uncertainty principle is found through

A

the operator algebra

132
Q

hilbert space

A

vectors in an infinite-dimensional space

133
Q

the overlap integral tells us

A

how much shared information there is between two states

134
Q

the commutator measures the

A

compatibility between the eigenstate bases of the two operators

135
Q

repeatedly applying the lowering operator will eventually result in

A

the ground state, on which a further lowering just returns 0

136
Q

the QSHO always has a finite

A

zero point energy E(ground) > 0

137
Q

l is the

A

orbital angular momentum quantum number

138
Q

orbital g factor reflects

A

the efficiency of conversion of angular momentum into magnetic momentum

139
Q

Zeeman effect

A

splitting of energy-degenerate spectral lines

140
Q

s is the

A

spin quantum number

141
Q

ms is

A

the spin magnetic quantum number

142
Q

total angular momentum

A

J = L + S

143
Q

j is

A

the total angular momentum quantum number

144
Q

is the wavefunction real, or just a mathematically artifact?

A

the aharonov-bohm effect confirms it is real.

Interference seen via two paths with different EM potentials A:

145
Q

Quantum measurement problem

A

how is a measurement sharp, non-unitary and what is a measurement?

largely understood via the Copenhagen interpretation as decoherence from coupling.

146
Q

the density matrix is a key object for

A

understanding entanglement and decoherence.

Quantum Mechanics (11 decks)

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