The figure below shows one of the possible energy eigenfunctions ψ(x) for a particle bouncing freely back and forth between impenetrable walls located at x = -a and x = +a. The potential energy equals zero for |x| < a. If the energy of the particle is 2 eV when it is in the quantum state associated with this eigenfunction, find the energy when it is in quantum state of lowest possible energy.

Solution:

- Concepts:

The infinite square well - Reasoning:

We are asked to identify which eigenfunction of the infinite square well is shown in the figure. - Details of the calculation:

The eigenfunctions of H are ψ_{n}(x) = (1/a)^{½}sin(nπ(x+a)/(2a)) with eigenvalues

E_{n}= n^{2}π^{2}ħ^{2}/(8ma^{2}). ψ_{2}(x) is shown in the figure.

E_{2}= 2 eV, E_{1}= ½ eV is the lowest possible energy eigenvalue.

A particle in a 1-D box has a minimum allowed energy of 2.5 eV.

(a) What is the next higher energy it can have? And the next higher after
that? Does it have a maximum allowed energy?

(b) If the particle is an electron, how wide is the box?

(c) The fact that particles in a 1-D box have a minimum energy is not completely
unrelated to the uncertainty principle. Using the minimum energy, find
the minimum magnitude of the momentum of a particle, with mass m, trapped in a
1-D box of size L in classical mechanics. How does this compare with the
momentum uncertainty required by the uncertainty principle, if we assume Δx = L?

Solution:

- Concepts:

The 1-D infinite square well, the uncertainty principle. - Reasoning:

We are asked to find the energy levels of a particle in a one-dimensional infinite square well. - Details of the calculation:

(a) E_{n}= n^{2}π^{2}ħ^{2}/(2mL^{2})

E_{1}= 2.5 eV, E_{2}= 4*2.5 eV = 10 eV, E_{3}= 9*2.5 eV = 22.5 eV. There is no upper limit

(b) 2.5 eV*1.6*10^{-19}J/eV = π^{2}(1.05*10^{-34 }Js)^{2}/(2*9.109*10^{-31 }kg * L^{2})

L = 3.86*10^{-10}m.

The box has a width that is comparable to the typical size of an atom.

(c) E_{1}= p_{min}^{2}/(2m), p_{min}^{2}= (2*9.109*10^{-31 }kg*2.5 eV*1.6*10^{-19}J/eV), p_{min}= 8.5*10^{-25}.

Δp = ħ/Δx = (1.05*10^{-34 }Js)/(3.86*10^{-10}m) = 2.7*10^{-25}kgm/s. The minimum momentum required by the uncertainty principle is p = Δp.

Δp and p_{min}have the same order of magnitude.

An electron (m = 9.11*10^{-31} kg)^{ }moves
with a speed v = 3.8*10^{6} m/s (non-relativistic) back and forth inside
a one-dimensional box (U = 0) of length L. The potential is infinite elsewhere,
hence the electron may not escape the box.

(a) If the electron were a classical particle, what would
be its energy?

(b) Now treat the electron quantum-mechanically but assume
it has the energy you found in part (a). If the quantum number associated with
the state of the electron is n = 2, what is the length of the box?

(c) What is the energy of the ground state?

(d) Write down the wave function for the first excited state.

Solution:

- Concepts:

The infinite square well - Reasoning:

The electron moves in an infinite square well potential in one dimension. - Details of the calculation:

(a) We are told that we can approach the problem non-relativistically.

K = ½mv^{2}= 0.5*(9.11*10^{-31}kg)*(3.8*10^{6}m/s) = 6.577*10^{-18}J = 41.1 eV.

(b) The energy for a particle in a box is given by E_{n}= ħ^{2}n^{2}π^{2}/(2mL^{2}).

Solving the above equation for L gives L = (ħ^{2}n^{2}π^{2}/(2mE))^{½}= 0.191 nm.

(c) E_{1}= E_{n}/n^{2}= 41.1 eV/4 = 10.3 eV.

(d) The general expression for the wave function for a particle in a box is given by

ψ_{n}(x) = (2/L)^{½}sin(nπx/L).

With n = 2 this expression becomes ψ_{2}(x) = (2/L)^{½}sin(2πx/L).

We can substitute in the value for L that we found in part (b).

For the infinite well shown, the wave function for a particle of mass m,
at t = 0, is

ψ(x, t = 0) = (2/a)^{½}sin(3πx/a).

(a) Is ψ(x, t = 0) an eigenfunction
of the Hamiltonian?

(b) Calculate <X>, <P_{x}>, and <H> at t = 0. U

Solution:

- Concepts:

The eigenfunctions of the infinite square well, the postulates of quantum mechanics. - Reasoning:

We recognize that the given wave function is an eigenfunction of the infinite square well. The mean value of any observable A is <ψ|A|ψ>. - Details of the calculation:

(a) ψ(x, t = 0) = (2/a)^{½}sin(3πx/a) is not an eigenfunction of the Hamiltonian.

(b) <x> = (2/a)∫_{0}^{a}dx x sin^{2}(3πx/a) = a/2.

<p_{x}> = (2/a)∫_{0}^{a}dx sin(3πx/a) (ħ/i) ∂sin(3πx/a)/∂x = C ∫_{0}^{a}dx sin(3πx/a) cos(3πx/a) = 0.

<H> = <T> + <U>. ψ is an eigenfunction of the infinite square well with U = 0 for 0 < x < a.

For the infinite square well we have <H> = E = <T>.

Therefore <T> = n^{2}π^{2}ħ^{2}/(2ma^{2}) = 9π^{2}ħ^{2}/(2ma^{2}).

<U> = ∫_{0}^{a }ψUψ dx = ∫_{0}^{2a/3 }ψU_{0}ψ dx = (2/a)∫_{0}^{2a/3}dx U_{0 }sin^{2}(3πx/a) = 2U_{0}/3.

<H> = 9π^{2}ħ^{2}/(2ma^{2}) + 2U_{0}/3.

A particle of mass m moves in one dimension in a square well with walls of infinite height a distance L apart. The particle is known to be in a state consisting of an equal admixture of the two lowest energy eigenstates of the system. Find the probability as a function of time that the particle will be found in the right-hand half of the well.

Solution:

- Concepts:

The postulates of quantum mechanics - Reasoning:

When a physical quantity described by the operator A is measured on a system in a normalized state |ψ>, the probability of measuring the eigenvalue a_{α}is given by dP(a_{α}) = |<u_{α}|ψ>|^{2}dα, where |u_{α}> is the eigenvector corresponding to the eigenvalue a_{α}; we assume a_{α}is a non-degenerate continuous eigenvalue of A.

Therefore dP(x,t) = |<x|ψ>|^{2}dx = |ψ(x,t)|^{2}dx. - Details of the calculation:

ψ(x,t) = 2^{-½}[ψ_{1}(x,t) + ψ_{2}(x,t)], where

ψ_{1}(x,t) = (2/L)^{½}sin(πx/L)exp(-(2πi/h)E_{1}t),

ψ_{2}(x,t) = (2/L)^{½}sin(2πx/L)exp(-(2πi/h)E_{2}t), with

E_{1 }= h^{2}/(8mL^{2}), E_{2 }= h^{2}/(2mL^{2}) are solutions to the one-dimensional infinite square-well problem.

P(x,t) = |ψ(x,t)|^{2}is the probability per unit length of finding the particle at position x.

P(x,t) = (1/L)[sin^{2}(πx/L) + sin^{2}(2πx/L) + 2sin(πx/L)sin(2πx/L)cos(2π(E_{2}-E_{1})t/h)].

∫_{L/2}^{L}P(x,t)dx = (1/L)[(L/4) + (L/4) + (4L/3π)cos(2π(E_{2}-E_{1})t/h)]

= ½ + (4/3π)cos(3h^{2}t/(8mL^{2})).

Consider a particle of mass m in an one-dimension infinite square well of
length L. Assume that the particle is in the nth eigenstate (n = 1, 2, 3, … ).

(a) The momentum is measured. Show that the probability distribution P_{n}(k)
for measuring a momentum p = ħk is P_{n}(k) = [2πLn^{2}(k^{2}L^{2}
– n^{2}π^{2})^{2}] [1 + (-1)^{n+1}cos(kL)].

(b) What outcome of a momentum measurement is most likely? Does your result
agree with your intuition?

(c) Which momenta cannot be the result of a momentum measurement? Why is that
so?

Solution:

- Concepts:

The square well potential, postulates of QM, the Fourier transform - Reasoning:

Let U(x) = 0, 0 < x < L, U(x) = infinite everywhere else.

The normalized stationary states <x|ψ > = ψ(x) in this potential are

ψ_{n}(x) = (2/L)^{½}sin(nπx/L) = (2/L)^{½}sin(k_{n}L), k_{n}= nπ/L, in coordinate representation.

To change from the {|x>} basis to the {|p>} basis we use

<p|ψ > = ∫dx<p|x><x|ψ > = (2πħ)^{-½}∫dx exp(-ipx/ħ) ψ(x) = ψ(p).

|ψ(p)|^{2}is the probability density distribution P_{n}(p). P_{n}(p)dp = P_{n}(k)dk, P_{n}(k) = ħ P_{n}(p). - Details of the calculation:

(a) ψ_{n}(p) = (πħL)^{-½}∫_{0}^{L}dx exp(-ipx/ħ) sin(nπx/L)

∫_{0}^{L}dx exp(-ipx/ħ) sin(nπx/L) = ∫_{0}^{L}dx exp(-ikx) sin(k_{n}x)

= ∫_{0}^{L}dx cos(kx)sin(k_{n}x) – i∫_{0}^{L}dx sin(kx)sin(k_{n}x)

= (cos((k-k_{n})L) – 1)/(2(k-k_{n})) - (cos((k+k_{n})L) – 1)/(2(k+k_{n}))

-i[sin((k-k_{n})L)/(2(k-k_{n})) - sin((k+k_{n})L)/(2(k+k_{n}))

= (-1)^{n}exp(ikL)/(2(k-k_{n})) - (-1)^{n}exp(ikL)/(2(k+k_{n})) - 1/(2(k-k_{n})) + 1/(2(k+k_{n}))

= [(-1)^{n}exp(ikL) – 1]k_{n}/(k^{2}– k_{n}^{2}) = [(-1)^{n}exp(ipL/ħ) – 1](nπħ^{2}L)/(p^{2}L^{2}– n^{2}π^{2}ħ^{2}).

ψ_{n}(p) = [nħ^{3/2}(πL)^{½}/(p^{2}L^{2}– n^{2}π^{2}ħ^{2})] [(-1)^{n}exp(ipL/ħ) – 1].

P_{n}(p) = |ψ_{n}(p)|^{2}= [2πLn^{2}ħ^{3}/(p^{2}L^{2}– n^{2}π^{2}ħ^{2})^{2}] [1 + (-1)^{n+1}cos(pL/ħ)].

P_{n}(k) = ħ P_{n}(p) = [2πLn^{2}(k^{2}L^{2}– n^{2}π^{2})^{2}] [1 + (-1)^{n+1}cos(kL)].(b) If the particle is in the nth eigenstate then the probability of measuring a momentum p is

P_{n}(p) = (2πLn^{2}/ħ)[1 + (-1)^{n+1}cos(pL/ħ)]/(p^{2}L^{2}/ħ^{2}– n^{2}π^{2})^{2}

= (2πLn^{2}/ħ)[ [1 + (-1)^{n+1}cos(pL/ħ)]/(pL/ħ + nπ)^{2}(pL/ħ – nπ)^{2}.

(Note n is a given integer.)

If pL = ±n'πħ, then (-1)^{n'+1}cos(pL/ħ) = (-1)^{n'+1}cos(±n'π) = (-1)^{n'+1}(-1)^{n' }= -1, for any n'.

If n' = n, both the numerator and the denominator of P_{n}(p) are zero.

If we expand the cosine function about these points we have

1 + (-1)^{n+1}cos(pL/ħ)] = 1 + (-1)^{n+1}(-1)^{n}(1 – (pL/ħ – nπ)^{2}/2) + (pL/ħ – nπ)^{4}/4 … )

= (pL/ħ – nπ)^{2}/2 - (pL/ħ – nπ)^{4}/4

and

P_{n}(p) = (πLn^{2}/ħ)/(pL + nπħ)^{2}(1 - (pL/ħ – nπ)^{2}/2).

P_{n}(p) --> L/(4πħ)( 1 - (pL/ħ – nπ)^{2}/2) as pL --> ±nπħ.

The maxima of P_{n}(p) are at p = ±nπħ/L.

The most likely outcome of a momentum measurement is ħk = ±nπħ/L.

(c) We need [1 + (-1)^{n'+1}cos(pL/ħ)] = 0, pL = ±n'πħ, and n' ≠ n.

Momenta p = ±n’πħ/L, n’ ≠ n cannot be the result of a momentum measurement.