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Problems and Solutions
Exercises, Problems, and Solutions
Section 1 Exercises, Problems, and Solutions
Review Exercises
- Transform (using the coordinate system provided below) the following functions accordingly:
r
X
Z
Y
a. from cartesian to spherical polar coordinates 3x + y - 4z = 12 b. from cartesian to cylindrical coordinates y^2 + z^2 = 9 c. from spherical polar to cartesian coordinates r = 2 Sinθ Cosφ
- Perform a separation of variables and indicate the general solution for the following expressions:
a. 9x + 16y
∂y ∂x
b. 2y +
∂y ∂x
- Find the eigenvalues and corresponding eigenvectors of the following matrices:
a. (^) ^
b.
- For the hermitian matrix in review exercise 3a show that the eigenfunctions can be normalized and that they are orthogonal.
- For the hermitian matrix in review exercise 3b show that the pair of degenerate eigenvalues can be made to have orthonormal eigenfunctions.
- Solve the following second order linear differential equation subject to the specified "boundary conditions":
d^2 x dt^2
- k^2 x(t) = 0 , where x(t=0) = L, and
dx(t=0) dt
Exercises
- Replace the following classical mechanical expressions with their corresponding quantum mechanical operators.
a. K.E. =
mv^2 2 in three-dimensional space. b. p = m v , a three-dimensional cartesian vector. c. y-component of angular momentum: Ly = zpx - xpz.
- Transform the following operators into the specified coordinates:
a. L x =
h− i (^)
y
∂z
∂y
from cartesian to spherical polar coordinates.
b. L z =
h- i
∂φ
from spherical polar to cartesian coordinates.
- Match the eigenfunctions in column B to their operators in column A. What is the eigenvalue for each eigenfunction? Column A Column B
i. (1-x^2 )
d^2 dx^2
d dx 4x^4 - 12x^2 + 3
ii.
d^2 dx^2
5x^4
iii. x
d dx e3x^ + e-3x
iv.
d^2 dx^2
d dx x^2 - 4x + 2
v. x
d^2 dx^2
d dx 4x^3 - 3x
- Show that the following operators are hermitian.
b. L x, L z, with function: Y 0
0 (θ,φ) =
4 π
c. L z, L^2 , with function: Y 1
0 (θ,φ) =
4 π
Cosθ.
d. L x, L z, with function: Y 1
0 (θ,φ) =
4 π
Cosθ.
- For a "particle in a box" constrained along two axes, the wavefunction Ψ(x,y) as given in the text was :
Ψ(x,y) =
2Lx
1 2
2Ly
1 2
e
in (^) xπx Lx (^) - e
-inxπx Lx
e
in (^) yπy Ly (^) - e
-inyπy Ly (^) ,
with nx and ny = 1,2,3, .... Show that this wavefunction is normalized.
- Using the same wavefunction, Ψ(x,y), given in exercise 9 show that the expectation value of p x vanishes.
- Calculate the expectation value of the x^2 operator for the first two states of the harmonic oscillator. Use the v=0 and v=1 harmonic oscillator wavefunctions given below
which are normalized such that ⌡⌠
Ψ(x)^2 dx = 1. Remember that Ψ 0 =
α π
1/ e-αx (^2) / and Ψ 1
4 α^3 π
1/ xe-αx (^2) / .
- For each of the one-dimensional potential energy graphs shown below, determine: a. whether you expect symmetry to lead to a separation into odd and even solutions, b. whether you expect the energy will be quantized, continuous, or both, and c. the boundary conditions that apply at each boundary (merely stating that Ψ
and/or
∂x
is continuous is all that is necessary).
- Consider a particle of mass m moving in the potential:
V(x) = ∞ for x < 0 Region I V(x) = 0 for 0 ≤ x ≤ L Region II V(x) = V(V > 0) for x > L Region III a. Write the general solution to the Schrödinger equation for the regions I, II, III,
assuming a solution with energy E < V (i.e. a bound state). b. Write down the wavefunction matching conditions at the interface between regions I and II and between II and III.
c. Write down the boundary conditions on Ψ for x → ±∞. d. Use your answers to a. - c. to obtain an algebraic equation which must be satisfied for the bound state energies, E.
j
CjΨje -iEjt/h-^ , the average value of the energy does not vary with time but the
expectation values of other properties do vary with time.
- A particle is confined to a one-dimensional box of length L having infinitely high walls
and is in its lowest quantum state. Calculate: , ,
, and . Using the
definition ∆Α = ( − ^2 )1/2^ , to define the uncertainty , ∆A, calculate ∆x and ∆p.
Verify the Heisenberg uncertainty principle that ∆x∆p ≥ h−^ /2.
- It has been claimed that as the quantum number n increases, the motion of a particle in a box becomes more classical. In this problem you will have an oportunity to convince yourself of this fact. a. For a particle of mass m moving in a one-dimensional box of length L, with ends of the box located at x = 0 and x = L, the classical probability density can be shown to be
independent of x and given by P(x)dx =
dx L regardless of the energy of the particle. Using
this probability density, evaluate the probability that the particle will be found within the
interval from x = 0 to x =
L
b. Now consider the quantum mechanical particle-in-a-box system. Evaluate the
probability of finding the particle in the interval from x = 0 to x =
L
for the system in its
nth quantum state.
c. Take the limit of the result you obtained in part b as n → ∞. How does your result compare to the classical result you obtained in part a?
- According to the rules of quantum mechanics as we have developed them, if Ψ is the
state function, and φn are the eigenfunctions of a linear, Hermitian operator, A , with
eigenvalues an, A φn = anφn, then we can expand Ψ in terms of the complete set of
eigenfunctions of A according to Ψ = ∑
n
cnφn , where cn = ⌡⌠φn*Ψ d τ. Furthermore, the
probability of making a measurement of the property corresponding to A and obtaining a
value an is given by cn^2 , provided both Ψ and φn are properly normalized. Thus, P(an) =
cn^2. These rules are perfectly valid for operators which take on a discrete set of eigenvalues, but must be generalized for operators which can have a continuum of eigenvalues. An example of this latter type of operator is the momentum operator, p x,
which has eigenfunctions given by φp(x) = Aeipx/h
- where p is the eigenvalue of the p x operator and A is a normalization constant. Here p can take on any value, so we have a continuous spectrum of eigenvalues of p x. The obvious generalization to the equation for
Ψ is to convert the sum over discrete states to an integral over the continuous spectrum of states:
Ψ(x) = ⌡⌠
C(p)φp(x)dp = ⌡
C(p)Aeipx/h
The interpretation of C(p) is now the desired generalization of the equation for the
probability P(p)dp = C(p)^2 dp. This equation states that the probability of measuring the
momentum and finding it in the range from p to p+dp is given by C(p)^2 dp. Accordingly, the probability of measuring p and finding it in the range from p 1 to p 2 is given by
p 1
p 2
P(p)dp = ⌡⌠ p 1
p 2
C(p)*C(p)dp. C(p) is thus the probability amplitude for finding the particle
with momentum between p and p+dp. This is the momentum representation of the
wavefunction. Clearly we must require C(p) to be normalized, so that ⌡⌠
C(p)*C(p)dp = 1.
With this restriction we can derive the normalization constant A =
2 πh−
, giving a direct
relationship between the wavefunction in coordinate space, Ψ(x), and the wavefunction in momentum space, C(p):
Ψ(x) =
2 πh−
C(p)eipx/h
and by the fourier integral theorem:
C(p) =
2 πh−
Ψ(x)eipx/h
Lets use these ideas to solve some problems focusing our attention on the harmonic oscillator; a particle of mass m moving in a one-dimensional potential described by V(x) =
kx^2 2
a. Write down the Schrödinger equation in the coordinate representation. b. Now lets proceed by attempting to write the Schrödinger equation in the momentum representation. Identifying the kinetic energy operator T , in the momentum
representation is quite straightforward T =
p^2 2m
Error!. Writing the potential, V(x), in the momentum representation is not quite as straightforward. The relationship between position and momentum is realized in their
commutation relation [ x,p ] = ih−^ , or ( xp - px ) = ih− This commutation relation is easily verified in the coordinate representation leaving x
untouched ( x = x.) and using the above definition for p. In the momentum representation
we want to leave p untouched ( p = p.) and define the operator x in such a manner that the commutation relation is still satisfied. Write the operator x in the momentum representation. Write the full Hamiltonian in the momentum representation and hence the Schrödinger equation in the momentum representation.
c. Verify that Ψ as given below is an eigenfunction of the Hamiltonian in the coordinate representation. What is the energy of the system when it is in this state?
now, the potential V(x) in the Hamiltonian ( H = -
h− 2m
d^2 dx^2
- V(x)) for which Ψ(x) is an
eigenfunction is unknown.
a. Find a value of A which makes Ψ(x) normalized. Is this value unique? What
units does Ψ(x) have? b. Sketch the wavefunction for positive and negative values of x, being careful to show the behavior of its slope near x = 0. Recall that | |x is defined as:
| |x =
x i f x > 0 -x if x < 0 c. Show that the derivative of Ψ(x) undergoes a discontinuity of magnitude 2(a)3/ as x goes through x = 0. What does this fact tell you about the potential V(x)? d. Calculate the expectation value of | |x for the above normalized wavefunction (obtain a numerical value and give its units). What does this expectation value give a measure of?
e. The potential V(x) appearing in the Schrödinger equation for which Ψ = Ae-a| |x^ is
an exact solution is given by V(x) =
h−^2 a m δ(x). Using this potential, compute the
expectation value of the Hamiltonian ( H = -
h− 2m
d^2 dx^2
- V(x)) for your normalized
wavefunction. Is V(x) an attractive or repulsive potential? Does your wavefunction
correspond to a bound state? Is negative or positive? What does the sign of tell
you? To obtain a numerical value for use
h−^2 2m = 6.06 x 10-28^ erg cm^2 and 1eV = 1.
x 10 -12^ erg.
f. Transform the wavefunction, Ψ = Ae-a| |x^ , from coordinate space to momentum space.
g. What is the ratio of the probability of observing a momentum equal to 2ah−^ to the
probability of observing a momentum equal to -ah−^?
- The π-orbitals of benzene, C 6 H6, may be modeled very crudely using the wavefunctions and energies of a particle on a ring. Lets first treat the particle on a ring problem and then extend it to the benzene system. a. Suppose that a particle of mass m is constrained to move on a circle (of radius r) in the xy plane. Further assume that the particle's potential energy is constant (zero is a good choice). Write down the Schrödinger equation in the normal cartesian coordinate representation. Transform this Schrödinger equation to cylindrical coordinates where x =
rcosφ, y = rsinφ, and z = z (z = 0 in this case).
Taking r to be held constant, write down the general solution, Φ(φ), to this Schrödinger
equation. The "boundary" conditions for this problem require that Φ(φ) = Φ(φ + 2π). Apply this boundary condition to the general solution. This results in the quantization of the energy levels of this system. Write down the final expression for the normalized wavefunction and quantized energies. What is the physical significance of these quantum
numbers which can have both positive and negative values? Draw an energy diagram representing the first five energy levels.
b. Treat the six π-electrons of benzene as particles free to move on a ring of radius 1.40 Å, and calculate the energy of the lowest electronic transition. Make sure the Pauli principle is satisfied! What wavelength does this transition correspond to? Suggest some reasons why this differs from the wavelength of the lowest observed transition in benzene, which is 2600 Å.
- A diatomic molecule constrained to rotate on a flat surface can be modeled as a planar
rigid rotor (with eigenfunctions, Φ(φ), analogous to those of the particle on a ring) with fixed bond length r. At t = 0, the rotational (orientational) probability distribution is
observed to be described by a wavefunction Ψ(φ,0) =
3 π
Cos^2 φ. What values, and with
what probabilities, of the rotational angular momentum,
-ih−
∂φ
, could be observed in this
system? Explain whether these probabilities would be time dependent as Ψ(φ,0) evolves
into Ψ(φ,t).
- A particle of mass m moves in a potential given by
V(x,y,z) =
k 2 (x^2 + y^2 + z^2 ) =
kr^2 2
a. Write down the time-independent Schrödinger equation for this system. b. Make the substitution Ψ(x,y,z) = X(x)Y(y)Z(z) and separate the variables for this system. c. What are the solutions to the resulting equations for X(x), Y(y), and Z(z)? d. What is the general expression for the quantized energy levels of this system, in terms of the quantum numbers nx, ny, and nz, which correspond to X(x), Y(y), and Z(z)? e. What is the degree of degeneracy of a state of energy
E = 5.5h−^
k m for this system?
f. An alternative solution may be found by making the substitution Ψ(r,θ,φ) =
F(r)G(θ,φ). In this substitution, what are the solutions for G(θ,φ)? g. Write down the differential equation for F(r) which is obtained when the
substitution Ψ(r,θ,φ) = F(r)G(θ,φ) is made. Do not solve this equation.
- Consider an N 2 molecule, in the ground vibrational level of the ground electronic state, which is bombarded by 100 eV electrons. This leads to ionization of the N 2 molecule to
form N 2
- (^). In this problem we will attempt to calculate the vibrational distribution of the
newly-formed N 2
- (^) ions, using a somewhat simplified approach.
a. Calculate (according to classical mechanics) the velocity (in cm/sec) of a 100 eV electron, ignoring any relativistic effects. Also calculate the amount of time required for a 100 eV electron to pass an N 2 molecule, which you may estimate as having a length of 2Å. b. The radial Schrödinger equation for a diatomic molecule treating vibration as a harmonic oscillator can be written as:
c. Under what circumstances (i.e. large or small values of k; large or small values
of μ) is the uncertainty in internuclear distance large? Can you think of any relationship between this observation and the fact that helium remains a liquid down to absolute zero?
- Suppose you are given a trial wavefunction of the form:
φ =
Ze^3 πa 03
exp
-Zer (^1) a 0 exp
-Zer (^2) a 0
to represent the electronic structure of a two-electron ion of nuclear charge Z and suppose that you were also lucky enough to be given the variational integral, W, (instead of asking you to derive it!):
W =
Z
e^2 - 2ZZe +
Z (^) e
e^2 a 0
a. Find the optimum value of the variational parameter Ze for an arbitrary nuclear
charge Z by setting
dW dZe = 0. Find both the optimal value of Ze and the resulting value of
W. b. The total energies of some two-electron atoms and ions have been experimentally determined to be:
Z = 1 (^) H- -14.35 eV Z = 2 He -78.98 eV Z = 3 (^) Li+ -198.02 eV Z = 4 (^) Be+2 -371.5 eV Z = 5 (^) B+3 -599.3 eV Z = 6 (^) C+4 -881.6 eV Z = 7 (^) N+5 -1218.3 eV Z = 8 (^) O+6 -1609.5 eV
Using your optimized expression for W, calculate the estimated total energy of each of these atoms and ions. Also calculate the percent error in your estimate for each ion. What physical reason explains the decrease in percentage error as Z increases? c. In 1928, when quantum mechanics was quite young, it was not known whether
the isolated, gas-phase hydride ion, H-, was stable with respect to dissociation into a
hydrogen atom and an electron. Compare your estimated total energy for H-^ to the ground state energy of a hydrogen atom and an isolated electron (system energy = -13.60 eV), and
show that this simple variational calculation erroneously predicts H-^ to be unstable. (More
complicated variational treatments give a ground state energy of H-^ of -14.35 eV, in agreement with experiment.)
- A particle of mass m moves in a one-dimensional potential given by H = -
h−^2 2m
d^2 dx^2
a|x| , where the absolute value function is defined by |x| = x if x ≥ 0 and |x| = -x if x ≤ 0.
a. Use the normalized trial wavefunction φ =
2b π
1 (^4) e -bx^2 to estimate the energy of
the ground state of this system, using the variational principle to evaluate W(b).
b. Optimize b to obtain the best approximation to the ground state energy of this
system, using a trial function of the form of φ, as given above. The numerically calculated
exact ground state energy is 0.808616 h−
2 (^3) m
1 (^3) a
2 (^3). What is the percent error in your
value?
- The harmonic oscillator is specified by the Hamiltonian:
H = -
h−^2 2m
d^2 dx^2
kx^2.
Suppose the ground state solution to this problem were unknown, and that you wish to approximate it using the variational theorem. Choose as your trial wavefunction,
φ =
a
5 (^2) (a (^2) - x (^2) ) for -a < x < a
φ = 0 for |x| ≥ a where a is an arbitrary parameter which specifies the range of the wavefunction. Note that
φ is properly normalized as given.
a. Calculate ⌡⌠
φ* H φdx and show it to be given by:
φ* H φdx =
h−^2 ma^2
ka^2 14
b. Calculate ⌡⌠
φ* H φdx for a = b
h−^2 km
1 (^4) with b = 0.2, 0.4, 0.6, 0.8, 1.0, 1.5, 2.0,
2.5, 3.0, 4.0, and 5.0, and plot the result. c. To find the best approximation to the true wavefunction and its energy, find the
minimum of ⌡⌠
φ* H φdx by setting
d da
φ* H φdx = 0 and solving for a. Substitute this value
into the expression for
φ* H φdx given in part a. to obtain the best approximation for the energy of the ground
state of the harmonic oscillator. d. What is the percent error in your calculated energy of part c.?
- Einstein told us that the (relativistic) expression for the energy of a particle having rest
mass m and momentum p is E^2 = m^2 c^4 + p^2 c^2. a. Derive an expression for the relativistic kinetic energy operator which contains
terms correct through one higher order than the "ordinary" E = mc^2 +
p^2 2m
c. Compute the polarizability, α, of the electron in the n=1 state of the box, and
explain physically why α should depend as it does upon the length of the box L.
Remember that α =
∂μ ∂ε (^)
ε=
Solutions
Review Exercises
- The general relationships are as follows:
r
X
Z
Y
x = r Sinθ Cosφ r^2 = x^2 + y^2 + z^2
y = r Sinθ Sinφ Sinθ =
x^2 + y 2
x^2 + y 2 + z 2
z = r Cosθ Cosθ =
z
x^2 + y 2 + z 2
Tanφ =
y x a. 3x + y - 4z = 12 3(rSinθCosφ) + rSinθSinφ - 4(rCosθ) = 12 r(3SinθCosφ + SinθSinφ - 4Cosθ) = 12 b. x = rCosφ r^2 = x^2 +y^2
y = rSinφ Tanφ =
y x z = z
y^2 + z^2 = 9 r^2 Sin^2 φ + z^2 = 9
c. r = 2SinθCosφ
r = 2
x r
r^2 = 2x x^2 +y^2 + z^2 = 2x x^2 - 2x +y^2 + z^2 = 0 x^2 - 2x +1 + y^2 + z^2 = 1 (x - 1)^2 + y^2 + z^2 = 1
- a. 9x + 16y
∂y ∂x
16ydy = -9xdx 16 2 y^2 = -
x^2 + c
16y^2 = -9x^2 + c' y^2 9
x^2 16 = c'' (general equation for an ellipse)
b. 2y +
∂y ∂x
2y + 6 = -
dy dx
y + 3 = -
dy 2dx
-2dx =
dy y + 3 -2x = ln(y + 3) + c c'e-2x^ = y + 3 y = c'e-2x^ - 3
- a. First determine the eigenvalues:
det
-1 - λ 2 2 2 - λ
(-1 - λ)(2 - λ) - 2^2 = 0 -2 + λ - 2λ + λ^2 - 4 = 0 λ^2 - λ - 6 = 0 (λ - 3)(λ + 2) = 0 λ = 3 or λ = -2. Next, determine the eigenvectors. First, the eigenvector associated with eigenvalue -2:
C 11
C 21 = -2^
C 11
C 21
5C 232 = 1
C 23 = 0.2 , and therefore C 33 = 2 0.. Next, find the pair of eigenvectors associated with the degenerate eigenvalue of -2. First, root one eigenvector one: -2C 11 = -2C 11 (no new information from row one) -C 21 + 2C 31 = -2C 21 (row two) C 21 = -2C 31 (again the third row offers no new information) C 112 + C 212 + C 312 = 1 (from normalization) C 112 + (-2C 31 )^2 + C 312 = 1 C 112 + 5C 312 = 1 C 11 =
1 - 5C 312 (Note: There are now two equations with three unknowns.) Second, root two eigenvector two: -2C 12 = -2C 12 (no new information from row one) -C 22 + 2C 32 = -2C 22 (row two) C 22 = -2C 32 (again the third row offers no new information) C 122 + C 222 + C 322 = 1 (from normalization) C 122 + (-2C 32 )^2 + C 322 = 1 C 122 + 5C 322 = 1 C 12 =
1 - 5C 322 (Note: Again there are now two equations with three unknowns) C 11 C 12 + C 21 C 22 + C 31 C 32 = 0 (from orthogonalization) Now there are five equations with six unknowns. Arbitrarily choose C 11 = 0
C 11 = 0 = 1 - 5C 312 5C 312 = 1 C 31 = 0. C 21 = -2 0. C 11 C 12 + C 21 C 22 + C 31 C 32 = 0 (from orthogonalization) 0 + -2 0.2(-2C 32 ) + 0.2 C 32 = 0 5C 32 = 0 C 32 = 0, C 22 = 0, and C 12 = 1 Therefore the eigenvector matrix becomes:
- Show: <φ 1 |φ 1 > = 1, <φ 2 |φ 2 > = 1, and <φ 1 |φ 2 > = 0
<φ 1 |φ 1 > =
(-2 0.2 )^2 + ( 0.2 )^2 =
<φ 2 |φ 2 > =
( 0.2 )^2 + (2 0.2 )^2 =
<φ 1 |φ 2 > = <φ 2 |φ 1 > =
- Show (for the degenerate eigenvalue; λ = -2): <φ 1 |φ 1 > = 1, <φ 2 |φ 2 > = 1, and <φ 1 |φ 2 > = 0
<φ 1 |φ 1 > =
0 + (-2 0.2 )^2 + ( 0.2 )^2 =
<φ 2 |φ 2 > =
<φ 1 |φ 2 > = <φ 2 |φ 1 > =
- Suppose the solution is of the form x(t) = eαt, with α unknown. Inserting this trial solution into the differential equation results in the following:
d^2 dt^2
eαt^ + k^2 eαt^ = 0
α^2 eαt^ + k^2 eαt^ = 0 (α^2 + k^2 ) x(t) = 0 (α^2 + k^2 ) = 0 α^2 = -k^2
