__Probability density__

A particle confined in the
region [−a, +a] has a wave function ψ(x) = N(a^{2} − x^{2}).
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: ∫_{-a}^{a}(a^{2}− x^{2})^{2}dx = 2∫_{0}^{a}(a^{2}− x^{2})^{2}dx = 16a^{5}/15.

N^{2}= 15/(16a^{5}).

P[−a/2, +a/2] = 30/(16a^{5})∫_{0}^{a/2}(a^{2}− x^{2})^{2}dx = 0.793.

__Probabilities__

A quantum system can exist in two states |a_{1}>
and |a_{2}>, which are normalized eigenstates of the observable A
with eigenvalues 0 and 1. A second observable B is defined by B|a_{1}> = 7|a_{1}> - 24i|a_{2}>, B|a_{2}> = 24i|a_{1}> - 7|a_{2}>.

(a) Find the eigenstates of B.

(b) The system is in the state |a_{1}> 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 a_{n}is given by

P(a_{n}) = Σ_{i=0}^{gn}|<u_{n}^{i}|ψ>|^{2}, where {|u_{n}^{i}>} (i=1,2,...,g_{n}) is an**orthonormal basis**in the eigensubspace E_{n}associated with the eigenvalue a_{n}.

If a measurement on a system in the state |ψ> gives the result a_{n}, then the state of the system immediately after the measurement is the normalized projection of |ψ> onto the eigensubspace associated with a_{n}. - Details of the calculation:

B_{ij}= <a_{i}|B|a_{j}>b)(-7 -b)

B_{11}= 7, B_{12}= 24i,

B_{21}= -24i, B_{22}= -7.

det(B - bI) = 0, (7 - b)(-7 - b) - 24^{2}= 0, b = ±25.b

_{1}= +25: (7 - 25)c_{1}+ 24i c_{2}= 0, c_{2}= -i¾c_{1},

|c_{1}|^{2}+ |c_{2}|^{2}= 1, c_{1}= 4/5, c_{2}= -i3/5.

|b_{1}> = (4/5)|a_{1}> - i(3/5)|a_{2}>.b

_{2}= -25: (7 + 25)c_{1}+ 24i c_{2}= 0, c_{2}= i(4/3)c_{1},

|c_{1}|^{2}+ |c_{2}|^{2}= 1, c_{1}= 3/5, c_{2}= i4/5.

|b_{2}> = (3/5)|a_{1}> + i(4/5)|a_{2}>.(b) There are two paths.

(i) Measure b_{1}, then measure a_{1}= 0.

P_{a1}(b1) = |<b_{1}|a_{1}>|^{2}= (4/5)^{2}. P_{b1}(a1) = |<a_{1}|b_{1}>|^{2}= (4/5)^{2}.

P_{a1}(a_{1}after b_{1}) = (4/5)^{4}.

(ii) Measure b_{2}, then measure a_{1}= 0.

P_{a1}(b_{2}) = |<b_{2}|a_{1}>|^{2}= (3/5)^{2}. P_{b2}(a1) = |<a_{1}|b_{2}>|^{2}= (3/5)^{2}.

P_{a1}(a_{1}after b_{2}) = (3/5)^{4}.

P_{a1}(a_{1}after measurement of B) = (4/5)^{4}+ (3/5)^{4}= 337/625 = 0.54.

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 {|a_{1}>,|a_{2}> } be an othonormal eigenbasis of A
and let {|b_{1}>,|b_{2}> } be an othonormal eigenbasis of
B.

A|a_{i}> = a_{i}|a_{i}>, B|b_{i}> = b_{i}|b_{i}>.

|b_{1}> = cos(θ)|a_{1}> + sin(θ)e^{iφ}|a_{2}>, |b_{2}> =
-sin(θ)|a_{1}>
+ cos(θ)e^{iφ}|a_{2}>

is the most general expansion of the |b_{i}> in terms of the |a_{i}>.

<b_{1}|b_{2}> = 0, <b_{1}|b_{1}> = <b_{2}|b_{2}>
= 1.

The initial state can be written as some linear combination of |a_{1}>
and a_{2}>.

|Ψ> = c_{1}|a_{1}> + c_{2}|a_{2}>.

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 a_{n}is given by

P(a_{n}) = ∑_{i=0}^{gn}|<u_{n}^{i}|Ψ>|^{2}, where {|u_{n}^{i}>} (i=1,2,...,g_{n}) is an orthonormal basis in the eigensubspace Ε_{n}associated with the eigenvalue a_{n}. - Details of the calculation:

The initial state can be written as some linear combination of |a_{1}> and a_{2}>.

|Ψ> = c_{1}|a_{1}> + c_{2}|a_{2}>, |c_{1}|^{2}+ |c_{2}|^{2}= 1 for a normalized state.

Branch 1:

Assume the first measurement yields a_{1}. The probability for this outcome is |c_{1}|^{2}.

What is the probability of measuring a_{1}again after a measurement of B?

P_{a1}(b_{1}) = cos^{2}(θ) (system is now in state |b_{1}>), P_{b1}(a_{1}) = cos^{2}(θ),

P_{a1}(b_{1}) P_{b1}(a_{1}) = cos^{4}(θ),

P_{a1}(b_{2}) = sin^{2}(θ) (system is now in state |b_{2}>), P_{b2}(a_{1}) = sin^{2}(θ),

P_{a1}(b_{2}) P_{b2}(a_{1}) = sin^{4}(θ).

Probability of the first path + probability of the second path = cos^{4}(θ) + sin^{4}(θ).

Total probability for this branch: |c_{1}|^{2}(cos^{4}(θ) + sin^{4}(θ)).Branch 2:

Assume the first measurement yields a_{2}. The probability for this outcome is |c_{2}|^{2}.

What is the probability of measuring a_{2}again after a measurement of B?

P_{a2}(b_{1}) = sin^{2}(θ) (system is now in state |b_{1}>), P_{b1}(a_{2}) = sin^{2}(θ),

P_{a2}(b_{1}) P_{b1}(a_{2}) = sin^{4}(θ),

P_{a2}(b_{2}) = cos^{2}(θ) (system is now in state |b_{2}>), P_{b2}(a_{2}) = cos^{2}(θ),

P_{a2}(b_{2}) P_{b2}(a_{2}) = cos^{4}(θ).

Probability of the first path + probability of the second path = sin^{4}(θ) + cos^{4}(θ).

Total probability for this branch: |c_{2}|^{2}(cos^{4}(θ) + sin^{4}(θ)).

Probability of obtaining the same result = (|c_{1}|^{2}+ |c_{2}|^{2})(cos^{4}(θ) + sin^{4}(θ))

= cos^{4}(θ) + sin^{4}(θ), independent of the initial state.

__Mean value of an observable__

The wave function ψ
of a particle is written as a linear combination of the three orthonormal
eigenfunctions {Φ_{i}} of
the observable **A** with eigenvalues a_{i} (i =
1,2,3).

|ψ> = (1/√6)|Φ_{1}> + (1/√3 )|Φ_{2}> + (1/√2) |Φ_{3}>

Find <A>. What
is the probability that the measurement of A yields a_{2}? Find
the wave function immediately after this measurement.

- 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 a_{n}is given by

P(a_{n}) = Σ_{i=0}^{gn}|<u_{n}^{i}|ψ>|^{2}, where {|u_{n}^{i}>} (i=1,2,...,g_{n}) is an**orthonormal basis**in the eigensubspace E_{n}associated with the eigenvalue a_{n}.

If a measurement on a system in the state |ψ> gives the result a_{n}, then the state of the system immediately after the measurement is the normalized projection of |ψ> onto the eigensubspace associated with a_{n}. - Details of the calculation:

<ψ|ψ> = 1/6 + (1/3) + ½ = 1. |ψ> is normalized.

<A> = <ψ|A|ψ> = a_{1}/6 + a_{2}/3 + a_{3}/2.

The probability that a measurement will yield a_{2}is |<Φ_{2}|ψ>|^{2}= (1/3).

Immediately after the measurement the wave function is Φ_{2}.

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)|^{2}dx = 1, (A/a)^{2}∫_{0}^{a}x^{2}dx + [A^{2}/(b - a)^{2}]∫_{a}^{b}(b - x)^{2}dx = 1.

A^{2}b/3 = 1, A^{2}= 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)|^{2}dx = (A/a)^{2}∫_{0}^{a}x^{3}dx + [A^{2}/(b - a)^{2}]∫_{a}^{b}(b - x)^{2}x dx.

<x> = ¼(b + 2a).

If b = 2a then <x> = a.

Assume the wave function of a particle is Ψ(x) = N exp(ip_{0}x/ħ)/(x^{2}
+ a^{2})^{½}.

Here a and p_{0} 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.

- 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 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.

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.

N^{2}∫_{-∞}^{+∞}exp(-ip_{0}x/ħ)/(x^{2}+ a^{2})^{½}exp(ip_{0}x/ħ)/(x^{2}+ a^{2})^{½}

= N^{2}∫_{-∞}^{+∞}dx/(x^{2}+ a^{2})^{ }= 1.

∫_{-∞}^{+∞}dx/(x^{2}+ a^{2})^{ }= (1/a)tan^{-1}(x/a)|_{-∞}^{+∞ }= (1/a)[π/2 - (-π/2)] = π/a.

N^{2}= 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}^{+y}dx/(x^{2}+ a^{2})^{ }= (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(-ip_{0}x/ħ)/(x^{2}+ a^{2})^{½}(ħ/i)[(ip_{0}/ħ)exp(ip_{0}x/ħ)/(x^{2}+ a^{2})^{½}- x exp(-ip_{0}x/ħ)/(x^{2}+ a^{2})^{3/2}]

= (a/π)p_{0}∫_{-∞}^{+∞}dx/(x^{2}+ a^{2}) - (a/π)(ħ/i)∫_{-∞}^{+∞ }xdx/(x^{2}+ a^{2})^{2}= p_{0}.

(π/a) 0, odd function

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 <r^{2}> is 12/β^{2}.

(c) Assuming a reasonable value for <r^{2}>^{½} calculate
a in esu/cm^{3}.

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 r^{2}. - Details of the calculation:

(a) 4π∫_{0}^{∞}ρ(r)r^{2}dr = α4π∫_{0}^{∞}exp(-βr)r^{2}dr = (α/β^{3})4π∫_{0}^{∞}exp(-r')r'^{2}dr' = (α/β^{3})8π = e.

(b) Assume ρ(r) = A^{2}|ψ(r)|^{2}.

For a normalized wave function 4π∫_{0}^{∞}|ψ(r)|^{2}r^{2}dr = 1.

Therefore A^{2}= β^{3}/(8πα), |ψ(r)|^{2}= [β^{3}/(8π)]exp(-βr)

<r^{2}> = 4π∫_{0}^{∞}|ψ(r)|^{2}r^{4}dr = [β^{3}/2]∫_{0}^{∞}exp(-βr)r^{4}dr = (1/(2β^{2}))∫_{0}^{∞}exp(-r')r'^{4}dr'

= 4!/(2β^{2}) = 12/β^{2}.

(c) Assume <r^{2}>^{½}= 10^{-15 }m = 10^{-13 }cm.

Then β^{2}= 12/<r^{2}> = 1.2*10^{27 }cm^{-2}.

(α/β^{3})8π = 4.8*10^{-10 }esu, α = 7.9*10^{29}esu cm^{-3}.