Problem 1:

An electron is moving freely in the x-direction.  At t = 0 the electron is described by the wave function (neglect spin)
Ψ(x,0) = Aexp{-x2/2b2}exp{ip0x/ħ}.
(a)  Compute the constant A such that ∫-∞+∞|ψ(x,0)|2 dx = 1.
(b)  Compute ∆x at t = 0.
(c)  Compute ∆p at t = 0 and show that for this electron ∆x∆p = ħ/2.
(d)  Assume that the electron has a position uncertainty of ∆x = 10-10 m.
Compute its velocity uncertainty compared to the speed of light.
(me = 9.1*10-31 kg,  ħ = 1.05*10-34 J s,  c = 3*108 m/s).
Hint:  ∫-∞+∞dx exp(-(ax2 + bx + c)) = (π/a)1/2 exp((b2 - 4ac)/(4a)).
To obtain, for example, ∫-∞+∞dx x exp(-ax2), differentiate with respect to b and then set b = c = 0.

Solution:

• Concepts::
The mean value and the root mean square deviation of an observable
• Reasoning:
The expression for the mean value of an observable A in the normalized state |Ψ> is <A> = <Ψ|A|Ψ>.  If |Ψ> is not normalized then  <A> = <Ψ|A|Ψ>/<Ψ|Ψ>.
The root mean square deviation ΔA characterizes the dispersion of the measurement around <A>.
ΔA = (<(A - <A>)2>)1/2 = (<A2> - <A>2)1/2.
• Details of the calculation:
(a)  ∫-∞+∞dx|Ψ(x)|2 = |A|2 -∞+∞dx exp{-x2/b2} = |A|21/2.
|A|2 = 1/(bπ1/2).
(b)  Δx = (<x2> - <x>2)1/2.  <x> = 0 from symmetry.
<x2> = ∫-∞+∞x2|Ψ(x)|2dx
= |A|2-∞+∞x2exp{-x2/b2} dx = b21/2-∞+∞x'2exp{-x'2} dx' = b2/2.
Δx = b/21/2.
(c)   Δp = (<p2> - <p>2)1/2.
pΨ(x) = (ħ/i)(∂/∂x)Ψ(x) = (p0 + iħx/b2)Ψ(x).
p2Ψ(x) = [(p0 + iħx/b2)2 + ħ2/b2]Ψ(x).
<p> = ∫-∞+∞ dx(p0 + iħx/b2)|Ψ(x)|2 = p0.
<p2> = ∫-∞+∞ dx[(p0 + iħx/b2)2 + ħ2/b2]|Ψ(x)|2
= p02 + ħ2/b2 -  ħ2/b4<x2> = p02 + ħ2/b2 - ħ2/(2b2) = p02 + ħ2/(2b2).
Δp = ħ/(21/2b),  Δx Δp = (b/21/2) ħ/(21/2b) = ħ/2.
(d)  Δv/c = Δp/(mec) = ħ/(2mecΔx) = 1.05*10-34/[2*9.1*10-31*3*108*10-10] = 1.9*10-3.

Problem 2:

The wave function of a particle at t = 0 is
Ψ(x) = 1/L1/2,  |x| < L/2,  Ψ(x) = 0 otherwise.
At t = 0, what possible values of the momentum of the particle can be found, and with what probability?

Solution:

• Concepts:
The postulates of quantum mechanics
• Reasoning:
P is a Hermitian operator.  The possible results of a measurement are the eigenvalues of the P, -∞ < p < ∞.
When a physical quantity described by the operator P is measured on a system in a normalized state |Ψ>, the probability of measuring the eigenvalue p is given by dP(p) = |<p|Ψ>|2dp = |Ψ(p)|2dp, where |p> is the eigenvector corresponding to the eigenvalue p; we assume p is a non-degenerate continuous eigenvalue of P.
• Details of the calculation:
Ψ(p) = (2πħ)-1/2∫Ψ(x,0) exp(-ipx/ħ)dx,
Ψ(p) = (2πħL)-1/2-L/2+L/2 exp(-ipx/ħ)dx
= (2πħL)-1/2(iħ/p) [exp(-ipL/(2ħ)) - exp(-pL/(2ħ))]
= (2πħL)-1/2(2ħ/p) sin(pL/(2ħ)).
The probability of finding the particle with momentum between p and p + dp  is |Ψ(p)|2dp.
|Ψ(p)|2  = (2ħ/(πLp2) )sin2(pL/(2ħ)).
Problem 3:

A particle of mass m is confined to an infinite one-dimensional potential well of width L, i.e. U(x) = 0,  0 < x < L,  U(x) = ∞ everywhere else.  At t = 0 the particle is equally likely to be found in the ground state or the firsts excited state.

(a)  What is the expectation value of the energy of the system?
(b)  Write down a properly normalized wave function to describe the system at subsequent times.
(c)  Find <px> for times  t > 0.

Solution:

• Concepts:
Postulates of quantum mechanics, the infinite square well.
• Reasoning:
The particle is in a linear superposition of two eigenstates of the infinite square well.  The mean value of any observable A is given by <ψ|A|ψ>.
• Details of the calculation:
The eigenfunctions of the infinite well are
ψn(x) = (2/L)1/2sin(nπx/L), En = n2π2ħ2/(2mL2).
(a)
<H> = ½(E1 + E2) = (5/2)π2ħ2/(2mL2).
(b)  ψ(x,t) = (1/L)1/2[sin(πx/L)
exp(-iE1t/ħ) + sin(2πx/L) exp(-iE2t/ħ)]
E1 = π2ħ2/(2mL2),  E2 = 4π2ħ2/(2ma2).
We can multiply the whole expression by e, where Χ is an arbitrary phase factor, and multiply one of the terms in the superposition by e, where φ is an arbitrary phase factor.

(c)  <px> = ∫0L dx Ψ*(x) (ħ/i)(∂/∂x) Ψ*(x)
<px> = (ħ/(iL))∫0L dx [sin(πx/L) exp(iE1t/ħ) + sin(2πx/L) exp(iE2t/ħ)]
*
(∂/∂x) [sin(πx/L) exp(-iE1t/ħ) + sin(2πx/L) exp(-iE2t/ħ)]
=
(-iħπ/L2)[∫0L dx [sin(2πx/L) cos(πx/L) exp(i(E2 - E1)t/ħ)
+ 2
0L dx [sin(πx/L) cos(2πx/L) exp(-i(E2 - E1)t/ħ)]
=
(-iħ/L) exp(i(E2 - E1)t/ħ) ∫0πdx cos(x) sin(2x)
+
(2iħ/L) exp(-i(E2 - E1)t/ħ) ∫0πdx cos(2x) sin(x)
=
(-iħ/L) exp(i(E2 - E1)t/ħ)(4/3) + (2iħ/L) exp(-i(E2 - E1)t/ħ)(-2/3)
=
(-4iħ/(3L))[exp(i(E2 - E1)t/ħ) - exp(i(E2 - E1)t/ħ)] = (8ħ/(3L)sin((E2 - E1)t/ħ).

Problem 4:

Consider a quantum system for which the exact Hamiltonian is H.  Assume the quantum system is of bounded spatial extend, so that it is known rigorously that the eigenstates of H, {|Ψn>}, are complete.
(a) Show that if |Ψn> and |Ψm> are two eigenstates of H with eigenvalues En and Em
with En ≠ Em, then <Ψnm> = 0.
(b) Suppose En = Em, with n ≠ m.  Can we still have <Ψnm> = 0 ?
(c) The problem H|Ψ> = E|Ψ> is very complicated but it is suggested that we use a trial function |Ψtrial> for |Ψ> and approximate E by E = <Ψtrial|H|Ψtrial>/<Ψtrialtrial>.
Show that E > E0, where E0 is the lowest eigenvalue of H.

Solution:

• Concepts::
Hermitian operators
• Reasoning:
H is a Hermitian operator.  The eigenvalues of a Hermitian operator are real.  Every Hermitian operator has at least one basis of orthonormal eigenvectors.
• Details of the calculation:
(a)  H = H, H is a Hermitian operator
n|H|Ψm> = Emnm> =  <Ψn|Hm> = Ennm>.
Emnm> - Ennm> = 0.  Since En ≠ Em, <Ψnm> must be zero.
(b)  Yes, an orthonormal basis for the state space can be formed by the eigenfunctions of any Hermitian operator.
(c)  Any wavefunction can be expanded in terms of the eigenfunctions of H.
trial> = ∑nann>,  {|Ψn>} = orthonormal basis for the state space.
H|Ψtrial> = ∑nanEnn>.
trial|H|Ψtrial> = ∑nanEntrialn> = ∑mnam*anEnmn>
= ∑mnam*anEnδmnnn> = ∑n|an|2Ennn>  ≥  ∑n|an|2E0nn>
since E0 is the lowest eigenvalue.
trial|H|Ψtrial> ≥ E0n|an|2nn> = E0trialtrial>

Problem 5:

Assume that the Hamiltonian H for a quantum system is Hermitian.
(a) Show that its eigenvalues E are real.
(b) Show that the eigenvectors |E> and |E’> corresponding to different eigenvalues E ≠ E’ are orthogonal.
(c) If the square of the angular momentum operator L2 and its z component Lz have eigenvectors |Elm> that are simultaneous eigenvectors of H, i.e.
H|Elm> = E|Elm>,  L2|Elm> = ħ2l(l + 1)|Elm>,   Lz|Elm> = ħm|Elm>,
and these eigenvectors form a complete set of states, show that [H,L2 ]= [H,Lz] = 0.

Solution:

• Concepts:
Hermitian operators, commuting observables
• Reasoning:
The Hamiltonian is a Hermitian operator.  The eigenvalues of a Hermitian operator are real.  Every Hermitian operator has at least one basis of orthonormal eigenvectors.  Eigenvectors corresponding to different eigenvalues are orthogonal
• Details of the calculation:
(a)  H|E> = E|E>;  |E> is an eigenvector with eigenvalue E.
<E|H|E> = <E|HE> = E<E|E> = <HE|E> = E*<E|E>.
since H = H
(E* - E)<E|E> = 0,  <E|E> ≠ 0 --> E = E*.
(b)  H|E'> = E'|E'>;  |E'> is an eigenvector with eigenvalue E'.
<E'|H|E> = <E'|HE> = E'<E'|E> =<HE'|E> = E'<E'|E>.
(E' - E)<E'|E> = 0,  E' ≠ E --> <E'|E> = 0.
(c)  [H,L2] = HL2 - L2H.
HL2|Elm> = ħ2l(l + 1)H|Elm> = ħ2l(l + 1)E|Elm>.
L2H|Elm> = EL2|Elm> = Eħ2l(l+1)|Elm>.
[H,L2]|Elm> = 0.
Consider an arbitrary vector |Ψ>.
|Ψ> = ∑clmn|Elm> since {|Elm>} is a basis for the state space.
[H,L2]|Ψ> = ∑clmn[H,L2]|Elm> = 0 for any |Ψ>.
Therefore  [H,L2] = 0.
Similarly:
HLz|Elm> = mħH|Elm> = mħE|Elm>.
LzH|Elm> = ELz|Elm> = Emħ|Elm>.
[H,Lz]|Elm> = 0.
[H,Lz]Ψ> = ∑clmn[H,Lz]|Elm> = 0 for any |Ψ>.
Therefore  [H,Lz] = 0.