### Probability density, probabilities,  and the mean value of an observable

Probability density

#### Problem:

A particle confined in the region [−a, +a] has a wave function ψ(x) = N(a2 − x2).  What is the probability that a position measurement would find it located in the interval [−a/2, +a/2]?

Solution:

• Concepts:
The fundamental assumptions of quantum mechanics
• Reasoning:
The wave function ψ(r,t) is interpreted as a probability amplitude of the particles presence.  |ψ(r,t)|2 is the probability density.
• Details of the calculation:
Normalize the wave function:  ∫-aa(a2 − x2)2dx = 2∫0a (a2 − x2)2dx = 16a5/15.
N2 = 15/(16a5).
P[−a/2, +a/2] = 30/(16a5)∫0a/2 (a2 − x2)2dx = 0.793.

Probabilities

#### Problem:

A quantum system can exist in two states |a1> and |a2>, which are normalized eigenstates of the observable A with eigenvalues 0 and 1.  A second observable B is defined by B|a1> = 7|a1> - 24i|a2>,  B|a2> = 24i|a1> - 7|a2>.
(a)  Find the eigenstates of B.
(b)  The system is in the state |a1> when B is measured.  Immediately afterwards A is measured.  Find the probability that a measurement of A gives the result 0.

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 an is given by
P(an) = Σi=0gn|<uni|ψ>|2,  where {|uni>} (i=1,2,...,gn) is an orthonormal basis in the eigensubspace En associated with the eigenvalue an.
If a measurement on a system in the state |ψ> gives the result an, then the state of the system immediately after the measurement is the normalized projection of |ψ> onto the eigensubspace associated with an.
• Details of the calculation:
Bij = <ai|B|aj>b)(-7 -b)
B11 = 7,  B12 = 24i,
B21 = -24i,  B22 = -7.
det(B - bI) = 0,   (7 - b)(-7 - b) - 242 = 0, b = ±25.

b1 = +25:  (7 - 25)c1 + 24i c2 = 0,  c2 = -i¾c1
|c1|2 + |c2|2 = 1,  c1 = 4/5,  c2 = -i3/5.
|b1> = (4/5)|a1> - i(3/5)|a2>.

b2 = -25:  (7 + 25)c1 + 24i c2 = 0,  c2 = i(4/3)c1
|c1|2 + |c2|2 = 1,  c1 = 3/5,  c2 = i4/5.
|b2> = (3/5)|a1> + i(4/5)|a2>.

(b)  There are two paths.
(i)  Measure b1, then measure a1 = 0.
Pa1(b1) = |<b1|a1>|2 = (4/5)2.  Pb1(a1) = |<a1|b1>|2 = (4/5)2.
Pa1(a1 after b1) = (4/5)4.
(ii)  Measure b2, then measure a1 = 0.
Pa1(b2) = |<b2|a1>|2 = (3/5)2.  Pb2(a1) = |<a1|b2>|2 = (3/5)2.
Pa1(a1 after b2) = (3/5)4.
Pa1(a1 after measurement of B) = (4/5)4  + (3/5)4 = 337/625 = 0.54.

#### Problem:

Let A and B be two observables of a system with a two dimensional state space, and suppose measurements are made of A, B, and A again in quick succession.
Show that the probability that the second measurement of A gives the same result as the first is independent of the initial state of the system.

Hint:
Let {|a1>,|a2> } be an othonormal eigenbasis of A and let {|b1>,|b2> } be an othonormal eigenbasis of B.
A|ai> = ai|ai>,  B|bi> = bi|bi>.
|b1> = cos(θ)|a1> + sin(θ)e|a2>,  |b2> = -sin(θ)|a1> + cos(θ)e|a2>
is the most general expansion of the |bi> in terms of the |ai>.
<b1|b2> = 0,  <b1|b1> = <b2|b2> = 1.
The initial state can be written as some linear combination of |a1> and a2>.
|Ψ> = c1|a1> + c2|a2>.

Solution:

• Concepts:
Fundamental assumptions of QM
• 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 an is given by
P(an) = ∑i=0gn|<uni|Ψ>|2,  where {|uni>} (i=1,2,...,gn) is an orthonormal basis in the eigensubspace Εn associated with the eigenvalue an.
• Details of the calculation:
The initial state can be written as some linear combination of |a1> and a2>.
|Ψ> = c1|a1> + c2|a2>,  |c1|2 + |c2|2 = 1 for a normalized state.
Branch 1:
Assume the first measurement yields a1.  The probability for this outcome is |c1|2.
What is the probability of measuring a1 again after a measurement of B?
Pa1(b1) = cos2(θ) (system is now in state |b1>), Pb1(a1) = cos2(θ),
Pa1(b1) Pb1(a1)  = cos4(θ),
Pa1(b2) = sin2(θ) (system is now in state |b2>), Pb2(a1) = sin2(θ),
Pa1(b2) Pb2(a1)  = sin4(θ).
Probability of the first path + probability of the second path = cos4(θ) + sin4(θ).
Total probability for this branch:  |c1|2(cos4(θ) + sin4(θ)).

Branch 2:
Assume the first measurement yields a2.  The probability for this outcome is |c2|2.
What is the probability of measuring a2 again after a measurement of B?
Pa2(b1) = sin2(θ) (system is now in state |b1>), Pb1(a2) = sin2(θ),
Pa2(b1) Pb1(a2)  = sin4(θ),
Pa2(b2) = cos2(θ) (system is now in state |b2>), Pb2(a2) = cos2(θ),
Pa2(b2) Pb2(a2)  = cos4(θ).
Probability of the first path + probability of the second path = sin4(θ) + cos4(θ).
Total probability for this branch:  |c2|2(cos4(θ) + sin4(θ)).
Probability of obtaining the same result =  (|c1|2 + |c2|2)(cos4(θ) + sin4(θ))
= cos4(θ) + sin4(θ), independent of the initial state.

Mean value of an observable

#### Problem:

The wave function ψ of a particle is written as a linear combination of the three orthonormal eigenfunctions {Φi} of the observable A with eigenvalues ai (i = 1,2,3).
|ψ> = (1/√6)|Φ1> + (1/√3 )|Φ2> + (1/√2) |Φ3>
Find <A>.  What is the probability that the measurement of A yields a2?   Find the wave function immediately after this measurement.

Solution:
• Concepts:
The postulates of Quantum Mechanics, the mean value
• Reasoning:
The expression for the mean value of an observable A in the normalized state |ψ> is <A> = <ψ|A|ψ>.
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 an is given by
P(an) = Σi=0gn|<uni|ψ>|2,  where {|uni>} (i=1,2,...,gn) is an orthonormal basis in the eigensubspace En associated with the eigenvalue an.
If a measurement on a system in the state |ψ> gives the result an, then the state of the system immediately after the measurement is the normalized projection of |ψ> onto the eigensubspace associated with an.
• Details of the calculation:
<ψ|ψ> = 1/6 + (1/3) + ½ = 1.  |ψ> is normalized.
<A> = <ψ|A|ψ> = a1/6 + a2/3 + a3/2.
The probability that a measurement will yield a2 is |<Φ2|ψ>|2 = (1/3).
Immediately after the measurement the wave function is Φ2.

#### Problem:

At some time t the wave function of a particle is a triangular hat wave function given by
Ψ(x,t) = Ax/a                       for 0 < x < a,
Ψ(x,t) = A(b - x)/(b - a)        for a < x < b,
Ψ(x,t) = 0                            otherwise,
where A, a, and b are constants.
(a)  Sketch Ψ and find the most probable location of the particle at time t.
(b)  Determine the normalization constant A in terms of a and b.
(c)  At time t, what are the probabilities of the particle being found left and right of a, respectively?
(d)  What is <x(t)>?

Solution:

• Concepts:
Probability density |Ψ(x,t) |2
• Reasoning:
We are given a wave function and asked to normalize it and find probabilities.
• Details of the calculation:
(a)  The most probable location is at x = a where Ψ(x,t) and therefore |Ψ(x,t) |2 has its maximum value.

(b)  ∫|Ψ(x,t)|2dx = 1,  (A/a)20ax2 dx  + [A2/(b - a)2]∫ab(b - x)2 dx = 1.
A2b/3 = 1,  A2 = 3/b.
(c)  P(left of a) = a/b.
P(right of a) = 1 - a/b.
If b = 2a then P(left of a) = P(right of a) = ½.
(d)  <x> = ∫x|Ψ(x,t)|2dx  = (A/a)20ax3 dx  + [A2/(b - a)2]∫ab(b - x)2 x dx.
<x> = ¼(b + 2a).
If b = 2a then <x> = a.

#### Problem:

Assume the wave function of a particle is Ψ(x) = N exp(ip0x/ħ)/(x2 + a2)½.
Here a and p0 are real constants and N is a normalization constant.
(a)  Find N so that ψ(x) is normalized.
(b)  If the position of the particle is measured, what is the probability of finding the particle between  -a/√3  and  +a/√3?
(c)  Calculate the mean value of the momentum of the particle.

Solution:
• Concepts:
The postulates of Quantum Mechanics, the mean value
• 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 aα is given by dP(aα) = |<uα|ψ>|2dα, where |uα> is the eigenvector corresponding to the eigenvalue aα; we assume aα is a non-degenerate continuous eigenvalue of A.
The expression for the mean value of an observable A in the normalized state |ψ> is <A> = <ψ|A|ψ>.
• Details of the calculation:
(a) <Ψ|Ψ> = 1.  ∫-∞+∞Ψ*(x)Ψ(x) dx = 1.
N2-∞+∞exp(-ip0x/ħ)/(x2 + a2)½ exp(ip0x/ħ)/(x2 + a2)½
= N2-∞+∞dx/(x2 + a2)  = 1.
-∞+∞dx/(x2 + a2)  = (1/a)tan-1(x/a)|-∞+∞  = (1/a)[π/2 - (-π/2)] = π/a.
N2 = a/π,  N = (a/π)½.

(b)  Let P denote the probability of finding the particle between -a/√3 and +a/√3.
Let y = a/√3.
P = ∫-y+y |Ψ(x)|2 dx = (a/π)∫-y+ydx/(x2 + a2)  = (1/π)tan-1(x/a)|-y+y

= (1/π)[tan-1(1/√3) - tan-1(-1/√3)]
= (1/π)[π/6 + π/6] = (1/3).

(c)  Let <P> denote the mean value of the momentum of the particle.
<P> = ∫-∞+∞Ψ*(x) ((ħ/i)∂/∂x) Ψ(x) dx
= (a/π)∫-∞+∞exp(-ip0x/ħ)/(x2 + a2)½ (ħ/i)[(ip0/ħ)exp(ip0x/ħ)/(x2 + a2)½ - x exp(-ip0x/ħ)/(x2 + a2)3/2]
= (a/π)p0-∞+∞dx/(x2 + a2)  - (a/π)(ħ/i)∫-∞+∞ xdx/(x2 + a2)2 = p0.
(π/a)                                 0, odd function

#### Problem:

From measurements of the differential cross section for scattering of electrons off protons (in atomic hydrogen) it was found that the proton had a charge density given by ρ(r) = αexp(-βr) where α and β are constants.
(a)  Find α and β such that the proton charge equals e, the charge on the electron.
(b)  Show that the protons mean square radius <r2> is 12/β2.
(c)  Assuming a reasonable value for <r2>½ calculate a in esu/cm3.

Solution:

• Concepts:
Fundamental assumptions of Quantum Mechanics
• Reasoning:
We interpret |ψ(r)|2 as the probability density and assume that the probability density is proportional to the charge density.  We then normalize the wave function ψ(r) and find the average value of the observable r2.
• Details of the calculation:
(a)  4π∫0ρ(r)r2dr = α4π∫0exp(-βr)r2dr = (α/β3)4π∫0exp(-r')r'2dr' = (α/β3)8π = e.
(b)  Assume ρ(r) = A2|ψ(r)|2
For a normalized wave function 4π∫0|ψ(r)|2r2dr = 1.
Therefore A2 = β3/(8πα),  |ψ(r)|2 = [β3/(8π)]exp(-βr)
<r2> = 4π∫0|ψ(r)|2r4dr = [β3/2]∫0exp(-βr)r4dr = (1/(2β2))∫0exp(-r')r'4dr'
= 4!/(2β2) = 12/β2.
(c)  Assume <r2>½ = 10-15 m = 10-13 cm.
Then β2 = 12/<r2> = 1.2*1027 cm-2.
(α/β3)8π = 4.8*10-10 esu,  α = 7.9*1029 esu cm-3.