Chapter 3: Q18P (page 118)

Test the energy-time uncertainty principle for the wave function in Problem $2.5$ and the observable x, by calculating $σ_{H} σ_{X}$ and $d < x > / d t$ exactly.

Short Answer

Expert verified

The values are:

$σ_{H} = \frac{1}{2} (E_{2} - E_{1}) σ_{x}^{2} = \frac{a^{2}}{4} [\frac{1}{3} - \frac{5}{4 π^{2}} - {(\frac{32}{9 π^{2}})}^{2} \cos^{2} (3 ω t)] \frac{d <x>}{d t} = \frac{8 ħ}{3 m a} \sin (3 ω t)$

Step by step solution

Step 1: Wavefunction.

The wavefunction is given by $ψ (x, 0) = A [ψ_{1} (x) + ψ_{2} (x)]$ .

First, find the normalization constant as:

$1 = \int_{- \infty}^{\infty} {|ψ (x, 0)|}^{2} d x = \int_{- \infty}^{\infty} ψ^{*} ψ d x = {|A|}^{2} \int_{- \infty}^{\infty} {[ψ_{1} (x) + ψ_{2} (x)]}^{*} [ψ_{1} (x) + ψ_{2} (x)] d x = {|A|}^{2} [\int_{- \infty}^{\infty} ψ_{1}^{*} ψ_{1} d x + \int_{- \infty}^{\infty} ψ_{1}^{*} ψ_{2} d x + \int_{- \infty}^{\infty} ψ_{2}^{*} ψ_{1} d x + \int_{- \infty}^{\infty} ψ_{2}^{*} ψ_{2} d x +]$

Use the orthogonality condition, eliminate the second and the third integrations, whereas the first and the fourth integrations each one of them is equal to one; therefore,

$1 = {|A|}^{2} [1 + 0 + 0 + 1] = 2 {|A|}^{2} A = \sqrt{\frac{1}{2}}$

Assume $ω = π^{2} ħ / 2 m a^{2}$ this will lead equation ( ) to become $E_{n} = n^{2} ω ħ$ , and therefore:

$ψ (x, t) = A [ψ_{1} (x) e^{- i E_{1} t / ħ} + ψ_{2} (x) E^{- i E_{2} t / ħ} +] = \sqrt{\frac{1}{2}} [\sqrt{\frac{2}{a}} \sin (\frac{π}{a} x) e^{- i ω t} + \sqrt{\frac{2}{a}} s in (\frac{2 π}{a} x) e^{- 4 i ω t}] = \sqrt{\frac{1}{a}} [\sin (\frac{π}{a} x) e^{- i ω t} + s in (\frac{2 π}{a} x) e^{- 4 i ω t}]$

The modulus squared of the wavefunction is:

|ψ {(x, t)}^{2}| = ψ^{*} ψ = \frac{1}{a} {[\sin (\frac{π}{a} x) e^{- i ω t} + \sin (\frac{2 π}{a} x) e^{- 4 i ω t}]}^{*} \times [\sin (\frac{2 π}{a}) e^{- 4 i ω t}] = \frac{1}{a} {[\sin (\frac{π}{a} x) e^{- i ω t} + \sin (\frac{2 π}{a} x) e^{4 i ω t}]}^{*} \times [\sin (\frac{2 π}{a}) e^{- 4 i ω t}] = \frac{1}{a} [\sin^{2} (\frac{π}{a} x) + 2 \sin (\frac{π}{a} x) \sin (\frac{2 π}{a} x) \cos (3 ω t) + \sin^{2} (\frac{2 π}{a} x)]

Expectation value of the position.

Now find the expectation value of the position as follow:

$<x> = \int_{- \infty}^{\infty} x {|ψ (x, t)|}^{2} d x = \frac{1}{a} \int_{0}^{a} x [\sin^{2} (\frac{π}{a} x) + 2 \sin (\frac{π}{a} x) \sin (\frac{π 2}{a} x) \cos (3 ω t) + \sin^{2} (\frac{2 π}{a} x)] d x = \frac{1}{a} \int_{0}^{a} x [\sin^{2} (\frac{π}{a} x) d x + 2 \cos (3 ω t) \int_{0}^{a} \sin (\frac{π}{a} x) \sin (\frac{2 π}{a} x) d x + \int_{0}^{a} \sin^{2} (\frac{2 π}{a} x) d x]$

The first integral could be solve use integration by parts and assume $u = x$ , then $d u = d x$ , and $d v = \sin^{2} (\frac{2 π}{a} x)$ and by the use of integration table $v = \frac{x}{2} - \frac{a}{4 π} \sin (\frac{2 π}{a} x)$ , therefore:

localid="1656316627360" $= \int_{0}^{a} x \sin^{2} (\frac{π}{a} x) d x = \frac{x^{2}}{2} - \frac{a x}{4 π} \sin (\frac{2 π}{a} x)_{0}^{a} - [\int_{0}^{a} \frac{x}{2} - \frac{a}{4 π} \sin (\frac{2 π}{a} x) d x] = {[\frac{x^{2}}{4} - \frac{a x}{4 π} \sin (\frac{2 π}{a} x) - \frac{a^{2}}{8 π^{2}} \cos (\frac{2 π}{a} x)]}_{0}^{a} \frac{a^{2}}{4}$

Do the same thing to the third integration and get the same result again. For the second integration use identity $\sin (α) \sin (β) = 1 / 2 [\cos (α - β) - \cos (α - β)]$ then perform an integration by parts and obtain the result as:

$= \cos (3 ω t [3]) {[\frac{a^{2}}{π^{2}} \cos (\frac{π}{a} x) + x \frac{a}{π} \sin (\frac{π}{a} x) - \frac{a^{2}}{9 π^{2}} \cos (\frac{3 π}{a} x) - x \frac{a}{3 π} \sin (\frac{3 π}{a} x)]}_{0}^{a} = \cos (3 ω t) [- \frac{16 a^{2}}{9 π^{2}}]$

Find the average energy.

Therefore, the expectation value of the position will be:

$<x> = \frac{1}{a} [\frac{a^{2}}{4} - \cos (3 ω t) \frac{16 a^{2}}{9 π^{2}} + \frac{a^{2}}{4}] = a [\frac{9 π^{2} - 32 \cos (3 ω t)}{18 π^{2}}] = \frac{a}{2} [1 - \frac{32}{9 π^{2}} \cos (3 ω t)]$

The expectation value of the momentum could be found by the use of the expectation value of the position, as:

$<p> = m \frac{d <x>}{d t} = m \frac{d}{d t} [\frac{a}{2} - \frac{16 a}{9 π^{2}} \cos (3 ω t)] = m \frac{16 a}{9 π^{2}} (3 ω) \sin (3 ω t) = m \frac{16 a}{9 π^{2}} (\frac{3 π^{2} ħ}{2 m a^{2}}) \sin (3 ω t) = \frac{8 ħ}{3 a} \sin (4 ω t)$

Given below is two possible energies’ $E_{1}$ and $E_{2}$ , where,

$E_{1} = \frac{π^{2} ħ^{2}}{2 m a^{2}} E_{2} = \frac{{(2)}^{2} π^{2} ħ^{2}}{2 m a^{2}}$ and each one has a probability of one half (i.e., $p_{1} = p_{2} = 1 / 2$ ), therefore, the average energy (i.e., the expectation value of the energy) is:

$<\hat{H}> = \frac{1}{2} (E_{1} + E_{2}) = \frac{5 π^{2} ħ^{2}}{4 m a^{2}}$

Calculate the value ofσH2.

The expression for $σ_{H}^{2}$ is given by:

$σ_{H}^{2} = <H^{2}> - <H^{2}>$

The expression for $<H^{2}>$ is given by:

$H^{2} ψ = \frac{1}{\sqrt{2}} [(H^{2} ψ_{1}) e^{- E_{1} t / ħ} + (H^{2} ψ_{2}) e^{- E_{n} t / ħ}]$

Where,

$H ψ_{1} = E_{1} ψ_{1} H^{2} ψ_{1} = E_{1} H ψ_{1} = E_{1}^{2} ψ_{1}$

And $H^{2} ψ_{2} = E_{2}^{2} ψ_{2}$ .

The expectation value of $<H^{2}>$ will be:

localid="1656347269478" $<H^{2}> = \frac{1}{2} <(ψ_{1} e^{- i E_{1} t / ħ} + ψ_{2} e^{- i E_{2} t / ħ}) I (E_{1}^{2} ψ_{1} e^{- {iE}_{1} t / ħ} + E_{2}^{2} ψ_{2} e^{- {iE}_{2} t / ħ})> \{\begin{matrix} \frac{1}{2} ((ψ_{1} ψ_{1}) e^{- i E_{1} t / ħ} E_{1}^{2} e^{- i E_{2} t / ħ} + (ψ_{1} | ψ_{2}) e^{- i E_{1} t / ħ} E_{2}^{2} e^{- i E_{2} t / ħ}) \\ + (ψ_{2} | ψ_{1}) e^{- i E_{1} t / ħ} E_{1}^{2} e^{- i E_{2} t / ħ} + (ψ_{2} | ψ_{2}) e^{- i E_{1} t / ħ} E_{2}^{2} e^{- i E_{2} t / ħ} \end{matrix}\} = \frac{1}{2} (E_{1}^{2} + E_{2}^{2}) \{H^{2} = \frac{1}{2} (E_{1}^{2} + E_{2}^{2})\}$

Substitute all the known values in the expression of $σ_{H}^{2}$ .

$σ_{H}^{2} = \frac{1}{2} (E_{1}^{2} + E_{2}^{2}) - \frac{1}{4} {(E_{1} + E_{2})}^{2} = \frac{1}{4} (2 E_{1}^{2} + 2 E_{2}^{2} - E_{2}^{2} - E_{1}^{2} - 2 E_{1} E_{2} - E_{2}^{2}) = \frac{1}{4} (E_{1}^{2} - 2 E_{1} E_{2} + E_{2}^{2}) = \frac{1}{4} {(E_{2} - E_{1})}^{2} T h u s : σ_{H} = \frac{1}{2} (E_{2} - E_{1})$

Find the expectation value of x2

Find the expectation value of $x^{2}$ , which is given by:

$<x^{2}> = \frac{1}{2} [<ψ_{1} | x^{2} ψ_{1}> + <ψ_{2} | x^{2} ψ_{2}> + <ψ_{1} | x^{2} ψ_{2}> e^{i (E_{1} - E_{2}) t / ħ} + <ψ_{2} | x^{2} ψ_{1}> e^{i (E_{2} - E_{1}) t / ħ}]$

Solve the steps as given below,

$<ψ_{n} | x^{2} | ψ_{m}> = \frac{2}{a} \int_{0}^{a} x^{2} \sin (\frac{n π}{a} x) \sin (\frac{m π}{a} x) d x = \frac{2 a^{2}}{π^{2}} [\frac{{(- 1)}^{n - m}}{{(n - m)}^{2}} - \frac{{(- 1)}^{n + m}}{{(n + m)}^{2}}] = \frac{2 a^{2}}{π^{2}} {(- 1)}^{n + m} \frac{4 n m}{{(n^{2} - m^{2})}^{2}}$

And from problem 2.4 obtain the result as :

$<ψ_{n} | x^{2} ψ_{n}> = a^{2} [\frac{1}{3} - \frac{1}{2 {(n π)}^{2}}]$

From the calculated values the expectation value of $x^{2}$ will be,

$<x^{2}> = \frac{1}{2} \{a^{2} [\frac{1}{3} - \frac{1}{2 π^{2}}] + a^{2} [\frac{1}{3} - \frac{1}{8 π^{2}}] - \frac{16 a^{2}}{9 π^{2}} [e^{3 i ω} + e^{- 3 i ω}]\} Here : \frac{E_{2} - E_{1}}{ħ} = \frac{(4 - 1) π^{2} ħ^{2}}{2 m a^{2} ħ} = \frac{3 π^{2} ħ}{3 ω} b u t 2 \cos (3 ω t) = e^{3 i ω} + e^{- 3 i ω}; Thus, we can write : <x^{2}> = \frac{a^{2}}{2} [\frac{2}{3} - \frac{5}{8 π^{2}} - \frac{32}{9 π^{2}} \cos (3 ωt)]$

Calculate the value of σx2 .

Calculate $σ_{x}^{2}$ as:

$σ_{x}^{2} = <x^{2}> - <x^{2}> = \frac{a^{2}}{4} [\frac{4}{3} - \frac{5}{4 π^{2}} - \frac{64}{9 π^{2}} \cos (3 ω t) - 1 + \frac{64}{9 π^{2}} \cos (3 ω t) - {(\frac{32}{9 π^{2}})}^{2} \cos^{2} (3 ω t)] = \frac{a^{2}}{4} [\frac{1}{3} - \frac{5}{4 π^{2}} - {(\frac{32}{9 π^{2}})}^{2} \cos^{2} (3 ω t)] = \frac{a^{2}}{4} [\frac{1}{3} - \frac{5}{4 π^{2}} - {(\frac{32}{9 π^{2}})}^{2} \cos^{2} (3 ω t)] T h e r e f o r e, σ_{x}^{2} = \frac{a^{2}}{4} [\frac{1}{3} - \frac{5}{4 π^{2}} - {(\frac{32}{9 π^{2}})}^{2} \cos^{2} (3 ω t)]$

It is given that:

$\frac{d <x>}{d t} = \frac{8 ħ}{3 m a} \sin (3 ω t) Now check the energy - time uncertainty principle : σ_{H}^{2} σ_{H}^{2} \geq \frac{ħ^{2}}{4} {(\frac{d <x>}{dt})}^{2} The LHS of the equation is : σ_{H}^{2} σ_{x}^{2} = \frac{1}{4} {(3 ħω)}^{2} \frac{a^{2}}{4} [\frac{1}{3} - \frac{5}{4 π^{2}} - {(\frac{32}{9 π^{2}})}^{2} \cos^{2} (3 ωt)] = {(ħωa)}^{2} {(\frac{3}{4})}^{2} [\frac{1}{3} - \frac{5}{4 π^{2}} - {(\frac{32}{9 π^{2}})}^{2} \cos^{2} (3 ωt)] And the RHS is : \frac{ħ^{2}}{4} {(\frac{d <x>}{dt})}^{2} = {(\frac{ħ}{2} \frac{8 ħ}{3 {ma}^{2}})}^{2} \sin^{2} (3 ωt) = {(\frac{8}{3 π^{2}})}^{2} ({ħωa}^{2}) \sin^{2} (3 ωt)$

Test the energy-time uncertainty principle.

$The uncertainty principle holds if : {(\frac{3}{4})}^{2} [\frac{1}{3} - \frac{5}{4 π^{2}} - {(\frac{32}{9 π^{2}})}^{2} \cos^{2} (3 ω t)] \geq {(\frac{8}{3 π^{2}})}^{2} \sin^{2} (3 ω t) Further solving above equation as, \frac{1}{3} - \frac{5}{4 π^{2}} \geq {(\frac{32}{9 π^{2}})}^{2} \cos^{2} (3 ωt) + {(\frac{4}{3} \frac{8}{3 π^{2}})}^{2} \sin^{2} (3 ωt) = {(\frac{32}{9 π^{2}})}^{2} \frac{1}{3} - \frac{5}{4 π^{2}} \geq {(\frac{32}{9 π^{2}})}^{2} (\cos^{2} (3 ωt) + \sin^{2} (3 ωt)) \frac{1}{3} - \frac{5}{4 π^{2}} \geq {(\frac{32}{9 π^{2}})}^{2} Solving for the given values : \frac{1}{3} - \frac{5}{4 π^{2}} = 0.20668 {(\frac{32}{9 π^{2}})}^{2} = 0.12978 Thus, the uncertainty principle holds the condition .$

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Test the energy-time uncertainty principle for the wave function in Problem $2.5$ and the observable x, by calculating $σ_{H} σ_{X}$ and $d < x > / d t$ exactly.

Short Answer

Step by step solution

Step 1: Wavefunction.

Expectation value of the position.

Find the average energy.

Calculate the value ofσH2.

Find the expectation value of x2

Calculate the value of σx2 .

Test the energy-time uncertainty principle.

One App. One Place for Learning.

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Test the energy-time uncertainty principle for the wave function in Problem2.5and the observable x, by calculatingσHσXandd<x>/dtexactly.

Short Answer

Step by step solution

Step 1: Wavefunction.

Expectation value of the position.

Find the average energy.

Calculate the value ofσH2.

Find the expectation value of x2

Calculate the value of σx2 .

Test the energy-time uncertainty principle.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Geometrical and Physical Optics

Thermodynamics

Famous Physicists

Engineering Physics

Magnetism and Electromagnetic Induction

Torque and Rotational Motion

Study anywhere. Anytime. Across all devices.

Test the energy-time uncertainty principle for the wave function in Problem $2.5$ and the observable x, by calculating $σ_{H} σ_{X}$ and $d < x > / d t$ exactly.