Quantum Physics 2D
Key Homework Ch. 6,8,9
These problems are to be worked on in discussion. They contain most of the physics concepts needed to solve
other problems. The other problems have solutions that can be used to help you with these more basic problems.
The solutions to these problems are intentionally not given so you can learn to really solve things yourself.
For each problem, state the physics principle (in words) that you will use to solve the problem and choose the
best formulas from the formula sheet to answer the problem. Then write an algebraic expression for the answer
based on the formulas and finally compute a numerical result if that is possible
The text in blue is provided to help you get started on the problem. Don’t expect similar text on the quiz. Do be
ready to find the formulas needed on the formula sheet which is on Tritoned and will be provided with the quiz.
1. In 1923, De Broglie realized that to maintain Lorentz invariance, the wave function for a free particle should
depend on the Lorentz scalar pµxµ. Schr¨odinger postulated that the wave function should be complex, implying
that ψfree−particle = Ceipµxµ/¯h where C is just a normalization constant.
See HW 6:3
a) From this formula, derive the frequency of this free particle. Hint: The definition of the frequency is the
number of oscillations per second at a fixed location. Using your relativity dot product, write the de Broglie’s
wave moving in the x direction, then set the location to x = 0. Explain why the frequency appears like
cos(2πf t) in a wave using the definition of the frequency. Now just see what corresponds to the frequency
in the de Broglie wave.
b) Similarly, derive the wavelength (Length in x of a full oscillation at a fixed time) for the de Broglie wave.
c) We always use the kinetic energy to give the frequency. How would physics change if we used the total
energy E including the rest mass? (Hint: Factor the energy exponential and consider the relative phases of
d) Plug this free particle wavefunction into the 1D Schr¨odinger equation to find out what that equation gives
for a free particle.
e) In relativistic QM, ψfree−particle = Ce−ipµxµ/¯h
is also a solution to the Dirac equation, also with momentum
+p but going backward in time. Can you imagine what this means? This is really hard but interesting to
2. A 1D free particle (no forces) has a wavefunction ψfree(x, t) = √
i(px−Et)/¯h where p and E are constants.
See Lecture 6.3 slide 153.
a) Show that Eop = ih¯
∂t is the total Energy operator because Eopψfree(x, t) = Eψ(x, t). These
operators are derived in the lecture notes.
b) Find the differential operator for the momentum.
c) Find the NR differential operator for the Kinetic energy.
d) This free particle wavefunction is said to be an eigenstate of momentum and of energy, or a definite
momentum state. Consider the state ψ(x, t) = √
cos((px − Et)/h¯). What momenta are present in that
state? Hint: Try writing this as a linear combination of definite momentum states.
e) A free electron has a wavefunction ψ(x) = e
i(3.5×1012)x where x is measured in meters. What is the electron’s
kinetic energy in electron volts?
3. It can be proven that if the operators for two physical variables don’t commute, then there is an uncertainty
principle between those two physical variables. The most basic uncertainty principle is between a coordinate and
its “conjugate” momentum. For example, since the laws of physics are invariant under translations
in the x-direction, momentum in the x-direction is conserved. Momentum in the x-direction, px, is the
“conjugate” of the coordinate x. You have shown above that the operator for the physical variable px is
∂x . The operator for the physical variable x is just x
(op) = x (when wave functions are written in terms
of the coordinate x).
See Lecture 8.1 slide 12.
a) Compute the commutator between p
x and x
(op) − x
by carefully computing
(op) − x
Since this is your first time computing a commutator with differential operators, it is wise to keep the
function ψ(x) on the right to avoid dropping important terms containing the derivative. When you are
done, drop the wavefunction on the right to quote the correct expression for the commutator.
b) Now compute the commutator h
c) The uncertainty principle can be written as (∆p)
2 ≥ hi
. Show that your result is
consistent with this expression. This works for any pair of operators.
4. The nuclear potential that binds protons and neutrons in the nucleus of an atom is often approximated by a
square well. Imagine a particle confined in a 1D infinite square well (a box) of length 10−5 nm (a typical
See HW 6:9 6:10.
a) Calculate the energy of the photon that is emitted when a proton undergoes a transition from the 2nd
excited state to the ground state.
b) Calculate the energy of an electron in the 2nd excited state of the box.
5. In a region of space, a particle with zero energy has a wavefunction: ψ(x) = Axe−ax2
. What is the potential
energy function V (x). Hint: Use thge Schr¨odinger equation to find V (x).
See Lecture 6.1 slide 129.
6. A 1D Harmonic Oscillator with mass m and classical angular frequency ω is in a linear combination of
energy eigenstates. At t = 0, it is in the state ψ(x, 0) = √
(u0(x) + iu1(x)), where un is the n
eigenstate of the HO.
See HW 6:32, Lecture 6.2 slises 140-143, Lecture 6.3 slide 157.
a) What is the wavefunction at a later time t ? Recall that the time dependence of energy eigenstates
is quite simple, ψn(x, t) = un(x)e
b) What is the expected value of x as a function of time for the above state? This problem is not difficult
but requires the use of several important ”Algebra of Quantum Mechanics” skills which I review here in
blue. First, the particle is in the state ψ, the expectation value of x (for example) is expressed in Dirac
Notation as hψ|x
|ψi. Computationally, this expression means:
|ψi ≡ Z∞
When you plug in the time dependent ψ be careful to take the complex conjugate when you should. When
we operate on an energy eigenstate of the Harmonic Oscillator with x
(op) or p
, we can write the result
again in terms of the energy eigenstates. We have shown by example in lecture (and it can be proven) that
n + 1un+1 +
pun = i
n + 1un+1 −
This allows us to rather easily compute the expected value of x or p in any HO state. Finally remember
that the energy eigenstates are orthonormal, making all of the integral essentially trivial to perform.
|uj i = δij
c) What is the expected value of p as a function of time?
d) At t = 0, a Harmonic Oscillator with mass m and classical angular frequency ω is in the state ψ(x, 0) =
(u2(x) − u3(x))/
2. What is the expected value of x as a function of time?
7. What is the expected value of x
in the ground state of a harmonic oscillator? Hint: Use the normal formula
for expected value and apply the x operator twice.
See HW 6:32, Lecture 6.2 slises 140-143, Lecture 6.3 slide 157.
8. A particle of mass m moves in a three-dimensional box with sides L . It is found that the ground state
has an energy of 3 eV, that the first excited state has an energy of 6 eV, and that the second excited state has
an energy of 9 eV. Recall that the 3D box is separable in Cartesian coordinates. When a problems separates
like this, the total energy is the sum of energies from each coordinate and the wavefunction is the product of
wavefunctions from the three coordinates. This product means that the state in x is the one given by the x
quantum number AND the wavefunction in y is…
See HW 8:1,3,5
a) What are the quantum numbers of the ground state?
b) What are (all the possible sets of) quantum numbers of the first excited state?
c) What are (all the possible sets of) quantum numbers of the second excited state?
d) Give one set of possible quantum numbers of the third excited state
e) Write the wavefunction for this third excited state?
9. In any time independent Schr¨odinger equation problem the time independence leads to the Energy being conserved and the states being Energy eigenstates. To solve the Hydrogen atom problem, we rely on an
important additional symmetry to reduce the number of coordinates in the differential equation from 3 to one
a) Describe the important symmetry we use to help solve the Hydrogen problem.
b) In a problem without electron spin, there are 4 operators that commute with the Hamiltonian giving rise to
conserved quantities. What are they?
c) Because these 4 operators don’t all commute with each other, we choose two mutually commuting operators in addition to the Hamiltonian on which to base our solution. Which pairs of operators have zero
d) Write out the 2s wavefunction of Hydrogen in terms of (r, θ, φ).
e) Write out the 2p(m=0), ψ210, wavefunction of Hydrogen.
10. A Hydrogen atom is in the n = 3 state. Recall that an energy eigenstate ψn`m of Hydrogen has 3 quantum
numbers if we don’t consider spin. The simple Energy formula of Rydberg only depends on n and the corrections
to Rydberg are very small. Angular momentum is quantized and measurements can only return one of the
eigenvalues of the operator. Given just n = 3 we have not specified the angular quantum numbers but with 4d
we have specified ` = 2.
See HW 8:7,10,16
a) If a measurement of L
is made, what are the possible outcomes?
b) If a measurement of Lz is made, what are the possible outcomes?
c) If a measurement of Ly is made, what are the possible outcomes?
d) If a measurement of S
, the square of the electron’s spin is made, what are the possible outcomes? Why
do we not bother to label any Hydrogen states with a quantum number giving the electron’s spin?
e) For an electron in the 4d shell, what is the expected value of L
11. Many calculations of important atomic physics problems come down to expectation values (or similar matrix
elements) of some power of the radius in Hydrogen states. It turns out that these calculations are rather easy,
so we will learn to do them. We use the orthonormality of the Spherical Harmonics Y`m, and a simple
formula for a definite integral over the radial coordinate that comes up all the time in these calculations with
Hydrogenic wavefunctions. This formula comes from integration by parts but since it is used so often, it is
important for us to remember it.
0m0 dΩ = Z
0m0 d cos θ dφ = δ“0δmm0
ψn`m(r, θ, φ) = Rn`(r)Y`m(θ, φ)
Just use the orthonormality to do the angular integral, then do the radial integral.
SeeHW 8:22,23,24,29. Lecture 8.3 slides 36-37.
a) What is the expected value of the radius squared, hψ21m|r
|ψ21mi, for the 2p state of hydrogen?
b) What is the expected value of the radius, for the ground state of hydrogen? How does this
compare to the Bohr radius?
c) What is the expected value of 1
, for the ground state of hydrogen?
12. The Hydrogen states can be written in terms of their radial wavefunctions and the spherical harmonics.
a) Write out the 2p state ψ211 of Hydrogen.
b) Write out the 2s state ψ200 of Hydrogen.
c) Why does the 2p state go to zero at r = 0? Why doesn’t the 2s state?
d) Which state had more ”nodes”, that is more zeros of the wavefunction excluding those at infinity and r = 0.
13. Electrons with orbital angular momentum are very similar to a current loop which generates a magnetic dipole
field. The calculated magnetic dipole moment is proportional to the angular momentum: ~µ =
L~ . When a
magnetic field is applied, these magnetic moments have an energy that depends on thier orientation relative to
the field: E = −~µ · B~ . Choosing the field to be along the z direction, we have:
E = −~µ · B~ =
m`B = µBBm`
The Zeeman Effect is the splitting of atomic states in a magnetic field.
See HW 9:1, Lecture 9.1 slide 50.
a) An atom has a single valence electron in the 2p state. Since all the other electrons add up to have net
zero orbital angular momentum, the magnetic moment is entirely due to this valence electron. What energy
shifts do we expect to see in a 0.5 Tesla magnetic field? What element would have this single 2p valence
b) What Energy shifts would we see for a single valence electron in 3d state? What element would have this
single 3d electron.
c) Why do we expect the magnetic field to split the state into an odd number of energies?
14. The expected Zeeman effect was not really observed in atoms. Rather than an odd number of lines with equal
spacing as expected for orbital angular momentum, an even number of lines was usually observed and the splitting
between them was not the same. This has been called the Anomalous Zeeman Effect because it was very
hard to understand. Stern and Gerlach separated a beam of Silver atoms into two beams in a B field with a large
gradient. Silver has a single valence electron in the 5s state which would not split according to the normal Zeeman
Effect. The Silver beam has a magnetic moment because the electron has internal angular momentum
that we call spin. Spin one-half electrons have a magnetic moment of one Bohr Magneton, the same size as
the magnetic moment due to ` = 1 orbital angular momentum. That is, they have a gyromagnetic ratio of g = 2.
These magnetic moments give an energy in a magnetic field of E = −~µ · B~ . The magnetic moment operator is
proportional to S~ and the resulting energies are thus quantized. Choosing the field to be along the z direction,
E = −~µ · B~ =
mshB¯ = g
msB = µBBgms
For electrons with orbital angular momentum, the splitting due to L~ must be added to those due to S~. Addition
of angular momentum J~ = L~ + S~, helps but because the gyromagnetic ratio for spin is 2 while the gyromagnetic
ratio for orbital is 1, its still complicated.
a) Consider the original Stern-Gerlach experiment in which the magnetic moment of Silver atoms is entirely
due to the spin of a single valence electron. What is the difference in energy between the two spin states of
the Silver atom in a 1 T magnetic field?
b) An electron is placed in a 2T magnetic field. The magnetic moment of the electron is entirely due to its
spin. What frequency of EM radiation will cause transitions between the two spin states?
15. An electron is in the +
eigenstate of Sx at t = 0. That is its spin is up along the x direction. The electron is in
a B-field Bz in the z direction. The state will be given as a linear combination of spin-up and spin-down (along
the z direction). This can be expressed as a 2-component vector χ =
where a is the amplitude to have
spin-up and b is the amplitude to have spin-down. The state should be normalized to 1 so that |a|
2 + |b|
2 = 1.
Lets determine what that state is. The Sx operator is given by Sx =
. We want Sxχ = +¯h
χ to be
in that eigenstate. So we have ¯h
1 0 a
, implying χ =
to keep χ
See Lecture 9.2 slide 64.
a) What is the electron’s state as a function of time?
b) What is the expected value of Sx as a function of time? The expected value is computed in this vector-state,
matrix-operator notation as hSxi = χ
c) What is the expected value of Sy =
as a function of time?
16. Ten identical, non-interacting, spin 1
particles with mass m are placed into a cubical box of side L = 0.25 nm.
What is the ground state energy of this system? Hint: Because of the two spin states, we can put 2 electrons
into each spatial state of the cubic box. The spin state of the 2 electrons will be the antisymmetric state.
See Lecture 9.3 slide77.
17. A Hydrogen atom is in the 3d state. If a measurement of J
is made, what are the possible outcomes? In the
calculation of the Hydrogen Fine Structure, we find that the energy depends on the total angular momentum
quantum number j and not on ` or s. The formulas in the Atomic physics section should be used. The limits
on j are given and the eigenvalues of J
are given in terms of j.What are the Fine-Structure corrections to the
energy for these two states?
See Lecture 9.2 slide 68.
18. Every electron in the universe is the same as any other electron. It is a symmetry of physics, that if you interchange
any pair of electrons, the Schr¨odinger equation is invariant. Consider one pair of electrons which we call electron-1
and electron-2. Define the interchange operator P12 to simple interchange the two electrons.
Hψ1ψ2 = Eψ1ψ2
HP12ψ1ψ2 = EP12ψ1ψ2
See Lecture 9.2 slide 69.
a) Use the above equations to show that the P12 operator commutes with the Hamiltonian.
b) Use that fact that if we operate twice with P12 we get back to the same state, to show what the eigenvalues
of the P12 operator are.
c) The overall state of two electrons must be antisymmetric under interchange. Usually, we can
factorize the spin state of two electrons from the spatial state: ψ = ψspinψspace. Since the overall state
of two electrons has to be antisymmetric under interchange, we can have a symmetric spin state with an
antisymmetric space state, or vice versa. Lets delve into the spin states of two electrons. Its very common
to use arrows to designate whether the spin of an electron is up or down since these are the only two states
of an electron and to use the first arrow for electron-1 and the second for electron-2. So, for example, the
state ↑↓ says that electron-1 has spin up and electron two has spin down. Since there are only 2 spin states
for any electron, there are just 4 states of two electrons: ↑↑, ↑↓, ↓↑, and ↓↓. Of these four states, which
are symmetric under interchange and which are antisymmetric.
d) From the two states that are neither symmetric or antisymmetric, make one linear combination that is
antisymmetric and one that is symmetric. Multiply by a constant to normalize these states to one pair of
e) Write the properly-normalized, symmetric-under-interchange spin state(s) of two electrons?
f) What is the properly-normalized, antisymmetric-under-interchange spin state of two electrons?
19. The Helium ground state has two electrons in the 1s state and in the antisymmetric spin state. The Helium first
excited state has one electron in the 1s state and the other electron in the 2s state. Recall that the big effect on
the energies of the states is the Coulomb repulsion between the two electrons. We have found that the spatial
antisymmetric state has significantly lower energy than the symmetric one because of this repulsion.
See Lecture 9.3 slide 90.
a) Which of the possible (1s)(2s) states will have the lowest energy?
b) What is the symmetry of the spin state of the two electrons?
c) Write an overall antisymmetric state that is (one of) the degenerate first excited state(s) of Helium.
20. In Hydrogen, the energies of states depend on n and j but not on `, even with the fine structure corrections
included. This degeneracy in ` is strongly broken in atoms with multiple electrons.
a) What is the explanation for the fact that states with higher ` have higher energy in atoms?
b) The order of states as the periodic table is filled up is still unclear with this big change that lower-` states
have lower energy. We can get the order of orbitals right for the atoms we find in nature by following two
simple rules. First pick the state with the lowest n + ` and if that is equal, pick the state with the lowest n.
With this, write down the order of n` orbitals in atoms up to Rutherfordium (Z=104). (Not for each atom,
just 1s, 2s, 2p, …).
c) Finally, in choosing the ground state of an atom, we should avoid ”paired electrons”. That is avoid two
electrons in the same spatial state that therefore must be in the symmetric spatial state and have maximum
Coulomb repulsion. This gives rise to the Chemist’s rather crude chart that shows that we put one spin
up electron ↑ into each m` state until they are all taken, then start pairing the electrons designated by ↑↓.
Draw the Charts for Nitrogen and for Iron.
d) With all the Coulomb repulsion between electrons, it seems surprising that the atomic shell model works
well, particularly for large Z. Why does the atomic shell model work well?
Quantum Physics 2D