Problem 1:
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(d) What is the lowest value that can possibly be obtained in any one measurement of the energy? How does it compare with the expectation value of the energy?
Solution:
The possible values that a single measurement of the energy can give are the eigenvalues of the Hamiltonian operator (the operator corresponding to total energy) and for a particle of mass m in a box of length b they are En = h2n2
8mb2 , where n = 1, 2, .... The lowest value is the one for n = 1,
E1 = h2
8mb2 . This is 4 times lower than the expectation value.
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(e) What will be the state of the system after the measurement described in part (d)?
Solution:
After a measurement of the energy where the value E1 was obtained, the state of the system is the stationary state corresponding to that value of the energy, φ1(x) =
√ 2 b
Problem 2:
Note: This problem serves as an introduction to the next chapter, chapter 5 in the text book. You do not need to have studied chapter 5 to do this problem.
(a) Show that the functions ψ0(x) = N0e −αx2/2 and ψ1(x) = N1xe
−αx2/2 are eigenfunctions of the Hamiltonian operator for the harmonic oscillator with spring constant k where α =
√ km/~.
(b) What are the eigenvalues, E0 and E1?
(c) Find the normalization constants, N0 and N1.
(d) Sketch the probability distribution for the location of the particle as a function of x for each one of the two states.
(e) Show that the two functions are orthogonal (as any two eigenfunctions of the Hamiltonian must be, unless they correspond to the same eigenvalue). (Hint: you can use symmetry arguments).
Solution:
Need to operate with the Hamiltonian operator on ψ0(x) = N0e −αx2/2
and show that the outcome is a constant, the eigenvalue, times ψ0(x). For the harmonic oscillator, the Hamiltonian operator is
H = − ~2
2/2
] 3
In order for ψ0(x) to be an eigenfunction, the x2 term needs to vanish. This means that α has to have a value such that
~2α2
Hψ0(x) = N0
2/2
] (b) The result in part (a) shows that the eigenvalue corresponding to ψ0(x) is
E0 = − ~2
2m (−α) =
~ 2
√ k
m
(c) Find the normalization constant, N0: Must have∫ ∞ −∞ |ψ0(x)|2 dx = 1
Plugging in the form of the function gives,
N2 0
e−αx 2
(e) Show that ψ0(x) and ψ1(x) are orthogonal, that is∫ ∞ −∞
ψ0(x)ψ1(x) dx = N0N1
dx.
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This can be seen most easily from a symmetry argument. The ground state wavefunction ψ0(x) is symmetric about x = 0, that is ψ0(−x) = ψ0(x). It is an even function. The first excited state wavefunction is antisymmetric, that is ψ1(−x) = −ψ1(x). It is an odd function. The product of an even function and an odd function is an odd function. The integral over an odd function is necessarily zero if the range of integration is symmetric about x = 0.
Problem 3:
A simple function that is frequently used to describe the potential energy of rare gas dimers (such as Ar2) is the Lennard-Jones potential
U(r) = 4ε
((σ r
)12 − (σ r
)6) where r is the distance between the two atoms and the two parameters, ε and σ depend on which atoms are involved. In each one of the questions (a-e) below you should give an expression that contains the parameters of the potential function, ε and σ, and possibly also the mass of the two atoms, m1 and m2.
(a) Sketch the potential energy as a function of distance and mark the axes with ε and σ.
(b) What is the distance rb at which the potential energy is smallest (the "bond length") and find the value of the potential energy at that distance (the binding energy of the dimer). Mark this point on the graph from part (a).
(c) Expand U(r) in a Taylor series about r = rb up to second order. By truncating the Taylor expansion at second order a harmonic oscillator app- roximation to the potential is obtained. Give an expression for the force constant, k, of the harmonic oscillator approximation to U(r). Sketch the harmonic oscillator approximation to the potential on the graph from part (a).
(c) Assuming the Harmonic Oscillator approximation is good enough for describing the ground state of the rare gas dimer, what is the dissociati- on energy of the dimer (that is, what is the minimum energy required to dissociate the dimer into two atoms separated by a large distance)?
(d) A dimer that is initially in the ground state can become excited to the first excited state when it collides with a surface (for example, the walls of a container). What energy is required to excite the dimer from the ground vi- brational state to the first vibrationally excited state assuming the harmonic
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oscillator approximation is valid? Sketch the two energy levels on the graph from part (a).
Solution:
(b) At the minimum, the first derivative is zero. Find dU dr
by differentiating the expression for the Lennard-Jones potential, and then set it to zero to find rb.
dU
)12 + (σ r
)6) This must be zero when r = rb, so rb must satisfy
2 (σ r
)12 = (σ r
)6 That is
rb = 21/6 σ
U(rb) = 4ε
2
) = −ε
(c) Let x = r − rb and V (x) = U(r). Need to expand V (x) up to and including the second order term in a Taylor series
V (x) = V (0) + x dV
dx (0) +
where k = d2V dx2
(0). The first derivative term vanishes because x = 0 cor- responds to a minimum of the V (x) function. This is a potential energy function for a harmonic oscillator with a spring constant k. In oder to get an expression for the spring constant in terms of the potential parameters ε and σ and the atomic masses m1 and m2, find the second derivative
k = d2V
dx2 (0) =
72ε
21/3σ2 =
36ε
22/3σ2
(d) The dissociation energy is the energy needed to bring the dimer up from the ground state to the state of two separated atoms.
Ed = E2Ar − EAr2
The energy of two separated atoms is E2Ar = limr→∞ U(r) = 0 and the energy of the ground state is EAr2 = −ε + ~
2
√ k µ where k is given by the
expression obtained in (b) and µ is the reduced mass of the dimer µ = m1m2/(m1 +m2). It is important to add to −ε the zero point energy of the harmonic oscillator, which is the energy of the ground state with respect to the potential energy minimum.
(e) The energy required to excite a harmonic oscillator from a stationary state to the next higher state is
E = ~ω = ~
√ k
µ
where the expression for k is given in part (b) and the expression for µ is given in part (c).
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