Assume that the highest energy (Fermi energy) of an electron inside a block
of metal is
5 eV and that the work function of the metal, i.e. the additional energy that is
necessary to remove an electron from the metal, is 3 eV.

(a) Estimate the distance through which the wave function of an electron at
the Fermi level penetrates the barrier responsible for the work function,
assuming that the width of the 3 eV barrier is much greater than the penetration
distance. (In this estimate you can assume that the barrier is a simple step
function.)

(b) Estimate the transmission coefficient (i.e., provide a numerical value
for it) for an electron at the Fermi level if the 3 eV barrier is now assumed
have a width of 20 Å.

Note: the probability of transmission through a barrier of width a is given by

T = 4E(U_{0} - E) / [(U_{0}^{2}sinh^{2}[(2m(U_{0}
- E)/ħ^{2})^{½} a] + 4E(U_{0} - E)],

or

T = (4E/U_{0})(1 - E/U_{0}) / [sinh^{2}[(2m(U_{0}
- E)/ħ^{2})^{½} a] + (4E/U_{0})(1 - E/U_{0})].

Solution:

- Concepts:

This is a "square potential" problem. - Reasoning:

We are given a piecewise constant potential. - Details of the calculation:

Let region 1 be the region to the left of the step and region 2 the region to the right of the step. For an electron at the Fermi level with E < U_{0}we define

k_{1}^{2}= (2m/ħ^{2})E and ρ_{2}^{2}= (2m/ħ^{2})(U - E).

Then

Φ_{1}(x) = A_{1}exp(ik_{1}x) + A_{1}'exp(-ik_{1}x)

and

Φ_{2}(x) = B_{2}exp(ρ_{2}x) + B_{2}'exp(-ρ_{2}x)

are the most general solutions. We need B_{2 }= 0 for the solution to remain finite at x --> ∞.

Φ_{2}(x) = B_{2}'exp(-ρ_{2}x). The penetration depth is on the order of 1/ρ_{2}.

ρ_{2}= ((2m(3 eV)/ħ^{2})^{½}= (2*9.1*10^{-31 }kg*3*1.6*10^{-19}J/(1.054*10^{-34}Js)^{2})^{½}

= 8.87*10^{9}/m.

1/ρ_{2}= 1.13*10^{-10}m = 1.13 Å.

(b) For E < U_{0}the probability of transmission through a barrier of width a is

T = (4E/U_{0})(1 - E/U_{0}) / [sinh^{2}[(2m(U_{0}- E)/ħ^{2})^{½}a] + (4E/U_{0})(1 - E/U_{0})].

Here U_{0 }= 8 eV and E = 5 eV and a = 2*10^{-9 }m.

T = (20/8)(1 - 5/8 ) / [sinh^{2}[ρ_{2}*2*10^{-9}] + (20/8)(1 - 5/8)] = 1.4*10^{-15}.

Since ρ_{2}a = 17.74 we can approximate sinh(ρ_{2}a) = ½(exp(ρ_{2}a) - exp(-ρ_{2}a)) by ½exp(ρ_{2}a) and the expression for T by

T = (16E/U_{0})(1 - E/U_{0})exp(-2((2m/ħ^{2})(U_{0}- E))^{½}a) = 1.4*10^{-15}.

Consider the wave function in one dimension ψ(x) = C exp(-a|x|), where C and
a are positive real numbers.

(a) Normalize ψ(x).

(b) Evaluate ∂Ψ(x)/∂x|_{+ε}
- ∂Ψ(x)/∂x|_{-ε} = ∫_{-ε}^{+ε}(∂^{2}/∂x^{2})Ψ(x)dx
as ε --> 0.

(c) Write down the expression for (∂^{2}/∂x^{2})Ψ(x).

(d) Calculate p^{2}ψ(x). Does your answer give the
correct sign for <p^{2}>?

Solution:

- Concepts:

Postulates of quantum mechanics - Reasoning:

We have a function with a discontinuous derivative. - Details of the calculation:

(a) Normalization: ∫_{-∞}^{+∞}|ψ(x)|^{2}dx = 2*C^{2}∫_{0}^{+∞}exp(-2ax)dx = 1, C = √a.

(b) ∂Ψ(x)/∂x|_{+ε}= -aCexp(-aε) ~ -Ca(1 - aε).

∂Ψ(x)/∂x|_{-ε}= aCexp(aε) ~ Ca(1 + aε).

∂Ψ(x)/∂x|_{+ε}- ∂Ψ(x)/∂x|_{-ε}= -2Ca as ε --> 0.

The derivative of exp(-a|x|) is discontinuous at x = 0.

We recognize ψ(x) a the bound state wave function in a delta-function potential.

(c) We have to take care of the discontinuity at the origin.

∫_{-ε}^{+ε}(∂^{2}/∂x^{2})Ψ(x) dx = ∫_{-ε}^{+ε}(∂/∂x)(∂Ψ(x)/∂x) dx

= ∂Ψ(x)/∂x|_{+ε}- ∂Ψ(x)/∂x|_{-ε}= -2Ca as ε --> 0.

If the integral of (∂^{2}/∂x^{2})Ψ(x) is a constant as ε --> 0,

then (∂^{2}/∂x^{2})Ψ(x) must be the constant times a delta function.

(∂^{2}/∂x^{2})Ψ(x) = a^{2}Ψ(x) - 2aδ(x)C = a^{2}Ψ(x) - 2aδ(x)Ψ(x).

Or use the Schroedinger equation for a delta-function potential.

∂Ψ(x)/∂x|_{+ε}- ∂Ψ(x)/∂x|_{-ε}= -(2m/ħ^{2})∫_{-ε}^{+ε}((E - Aδ(x))Ψ(x)dx

= (2m/ħ^{2})AΨ(0) = -2Ca --> A = -aħ^{2}/m, U(x) = -(aħ^{2}/m)δ(x).

At x ≠ 0 (∂^{2}/∂x^{2})Ψ(x) = Ca^{2}exp(-a|x|) = a^{2}Ψ(x) = -(2m/ħ^{2})EΨ(x) --> E = a^{2}ħ^{2}/(2m).

Therefore (∂^{2}/∂x^{2})Ψ(x) = -(2m/ħ^{2})(E - U(x))Ψ(x) --> (∂^{2}/∂x^{2})Ψ(x) = a^{2}Ψ(x) - 2aδ(x)Ψ(x).

(d) p^{2}ψ(x) = -ħ^{2}(∂^{2}/∂x^{2})Ψ(x) = -ħ^{2}a^{2}ψ(x) + 2ħ^{2}aδ(x)Ψ(x).

<p^{2}> = ħ^{2}a^{2}∫_{-∞}^{+∞}|ψ(x)|^{2}dx + 2ħ^{2}a∫_{-∞}^{+∞}δ(x)|ψ(x)|^{2}dx = ħ^{2}a^{2}+ 2ħ^{2}a|ψ(0)|^{2}

= -ħ^{2}a^{2}+ 2ħ^{2}a^{2}= ħ^{2}a^{2}> 0 as required.

The wave function ψ(**r**) of a spinless particle is ψ(**r**) = Nz^{2}exp(-r^{2}/b^{2}),
where b is a real constant and N is a normalization constant.

(a) If L^{2} is measured, what results can be obtained and
with what probabilities?

(b) If L_{z} is measured, what results can be obtained and with
what probabilities?

(c) Is ψ(**r**) an eigenfunction of L^{2} or L_{z}?

Solution:

- Concepts:

The eigenfunctions of the orbital angular momentum operator, the spherical harmonics - Reasoning:

The common eigenfunctions of L^{2}and L_{z}are the spherical harmonics. We have to write the given wave functions in terms of the spherical harmonics. - Details of the calculation:

ψ(**r**) = Nz^{2}exp(-r^{2}/b^{2}) = Nr^{2}exp(-r^{2}/b^{2})cos^{2}(θ) in spherical coordinates.

We have

Y_{00}= (4π)^{-½}, Y_{1±1}= ∓(3/8π)^{½}sinθ exp(±iφ), Y_{10}= (3/4π)^{½}cosθ,

Y_{2±2}= (15/32π)^{½}sin^{2}θ exp(±i2φ), Y_{2±1}= ∓(15/8π)^{½}sinθ cosθ exp(±iφ),

Y_{20}= (5/16π)^{½}(3cos^{2}θ - 1).

Therefore cos^{2}θ = [(4π)^{½}/3][2Y_{20}/√5 + Y_{00}].

Normalize: N^{2}∫dΩ [2Y_{20}/√5 + Y_{00}] [2Y_{20}/√5 + Y_{00}]* = 1. N = √5/3.

cos^{2}θ = (4π/5)^{½}[2Y_{20}/3 + √5 Y_{00}/3].

ψ(**r**) = f(r)χ(θ), χ(θ) = 2Y_{20}/3 + √5 Y_{00}/3. χ(θ) is a normalized function of angle on the unit sphere.

(a) If L^{2}is measure we can obtain 2(2 + 1)ħ^{2}with probability 4/9 and 0 with probability 5/9.

(b) If L_{z}is measured we obtain 0 with probability 1.

(c) ψ(**r**) is not an eigenfunction of L^{2}. ψ(**r**) is an eigenfunction of L_{z}.

Consider a spin ½ particle in the presence of a uniform static
magnetic field **B
**= B_{0}**i**. Suppose that at t = 0 the spin state of the
particle is the |->_{z} eigenket of S_{z}.

(a) State briefly how you would prepare such a state.

(b) Suppose at time t > 0 we measure the z-component of the spin.
What values can be obtained and with what probabilities?

(c) Evaluate the mean value of that measurement and comment on the physics
of your result.

Solution:

- Concepts:

The two dimensional state space of a spin ½ particle, the evolution operator - Reasoning:

The Hamiltonian of the system is H = ω_{0}S_{x}. The evolution operator is exp(-iHt/ħ). We use the evolution operator to find the state of the system at time t. - Details of the calculation:

(a) Example: A beam of unpolarized neutral atoms in the ground state with a single electron outside a closed shell is passed passes through a Stern-Gerlach (SG) devices whose magnetic field is directed along the z-axis.

The force on a magnetic dipole in that field is given by F_{z}= μ_{z}dB_{z}/dz. For the electron**μ**= γ**S**with γ < 0. The |->_{z}state is deflected downward. We block off the upward deflected beam.(b) In a magnetic field

**B**the Hamiltonian of an atom is H = -**μ**∙**B**= -γ**S**∙**B**. Here H = -γS_{x}B_{0}= ω_{0}S_{x}, ω_{0}> 0. The evolution operator is U(t,0) = exp(-iω_{0}S_{x},t/ħ).

The matrix of S_{x}in the eigenbasis of S_{z}is

S

_{x}= (ħ/2)0 1 1 0 .

The eigenvectors of S

_{x}are |+>_{x}and |->_{x}.

|+> = 2^{-½}(|+>_{x}+ |->_{x}), |-> = 2^{-½}(|+>_{x}- |->_{x}).

|ψ(0)> = |->.

|ψ(t)> = U(t,0)|-> = 2^{-½}U(t,0)(|+>_{x}- |->_{x})

= 2^{-½}(exp(-iω_{0}t/2)|+>_{x}- exp(iω_{0}t/2)|->_{x})

= ½[exp(-iω_{0}t/2)(|+> + |->) - exp(iω_{0}t/2)(|+> - |->)]

= -i sin(ω_{0}t/2)|+> + cos(ω_{0}t/2)|->.

The only values we can obtain from a measurement of S_{z}are ±ħ/2.

We want P(S_{z}=-ħ/2,t) = cos^{2}(ω_{0}t/2), P(S_{z}=ħ/2,t) = sin^{2}(ω_{0}t/2).

(c) <Sz(t)> = (ħ/2)(sin^{2}(ω_{0}t/2) - cos^{2}(ω_{0}t/2)) = -(ħ/2)cos(ω_{0}t).

This is spin precession.

An electron is contained in a one dimensional potential
well, having a potential energy of 0 when between x = 0 and x = 8 nm, and a
potential energy of ∞ for all other values of x.

(a) Write Schroedinger's equation for this problem, obtain
well-behaved solutions, and determine the energy eigenvalues.

(b) Obtain normalized wave functions, which will give unit
probability of the electron existing in all of space.

(c) Find the probability that the electron in its lowest
energy state will exist in the space between x = 2 nm and x = 4 nm.

Solution:

- Concepts:

The infinite square well, postulates of Quantum Mechanics - Reasoning:

We are supposed to find the eigenfunctions and eigenvalues of H for a particle in an infinite square well. - Details of the calculation:

(a) Let L = 8 nm, m = m_{e}. The time-independent Schroedinger equation is

(∂^{2}/∂x^{2})ψ(x) + (2m/ħ^{2})E ψ(x) = 0 for x < L.

(∂^{2}/∂x^{2}) + k^{2}ψ(x) = 0. E = ħ^{2}k^{2}/(2m).

The most general solution is ψ(x) = Aexp(ikx) + A'exp(-ikx), with A and A' complex constants. The boundary conditions require that ψ(x) = 0 at x = 0 and x = L.

This yields ψ(x) = Nsin(kx), kL = nπ/2, E_{n}= ħ^{2}n^{2}π^{2}/(2mL^{2}).

(b) For a particle in an infinite well we have ψ(x) = (2/L)^{½}sin(nπx/L).

(c) The probability that the electron in its lowest energy state will exist in the space between x_{1}= 2 nm and x_{2}= 4 nm is

P = ∫_{x1}^{x2}|ψ(x)|^{2}dx = (2/L)∫_{x1}^{x2 }sin^{2}(πx/L)dx

= (2/π)∫_{π/4}^{π/2 }sin^{2}x dx

= 1/4 + 1/(2π) = 0.41.