Q19P Test the energy-time uncertainty... [FREE SOLUTION]

Chapter 3: Q19P (page 118)

Test the energy-time uncertainty principle for the free particle wave packet in Problem 2.43and the observable x , by calculating $蟽_{H} 蟽_{x}$ , and $d < x > / d t$ exactly.

Short Answer

Expert verified

The result obtained are

$蟽_{x}^{2} = \frac{1}{4 a} [1 + {(\frac{2 魔 a t}{m})}^{2}] 蟽_{H}^{2} = \frac{魔^{4} a}{2 m^{2}} (a + 2 l^{2}) \frac{d (x)}{d t} \frac{魔 l}{m}$

Step by step solution

The wave function and find |ψ|2

From problem2.43, the wave function is given by:

$蠄 (x, t) = {(\frac{2 a}{蟺})}^{1 / 4} \frac{\frac{- a x^{2} + i (l x - 魔 l^{2} t / 2 m}{\sqrt{1 + 2 i 魔 a t / m}}}{\sqrt{1 + 2 i 魔 a t / m}}$ 鈥︹€� (1)

Now find ${|蠄|}^{2}$ ,

${|蠄|}^{2} = \sqrt{\frac{2 a}{蟺}} \frac{1}{\sqrt{1 + \frac{4 魔^{2} a^{2} t^{2}}{m^{2}}}} e^{- l^{2} / 2 a} e x p \{a [\frac{{(i x + I / 2 a)}^{2}}{(1 + 2 i 魔 a t / m)} + \frac{{(i x + I / 2 a)}^{2}}{(1 + 2 i 魔 a t / m)}]\} Let 胃 = 2 魔补迟 / m, so result is : {|蠄|}^{2} = \sqrt{\frac{2 a}{蟺}} \frac{1}{\sqrt{1 + 胃^{2}}} e^{- I / 2 a} \exp \{a [\frac{{(ix + I / 2 a)}^{2}}{(1 + 2 i魔补迟 / m)} + \frac{{(ix + I / 2 a)}^{2}}{(1 - 颈胃)}]\} Expand the term in square brackets as : {|蠄|}^{2} = \sqrt{\frac{2 a}{蟺}} \frac{1}{\sqrt{1 + 胃^{2}}} e^{- I / 2 a} \exp \{\frac{- 2 a}{1 + 胃^{2}} {(x - \frac{胃濒}{2 a})}^{2} + \frac{l^{2}}{2 a}\} {|蠄|}^{2} = \sqrt{\frac{2 a}{蟺}} \frac{1}{\sqrt{1 + 胃^{2}}} \exp \{\frac{- 2 a}{1 + 胃^{2}} {(x - \frac{胃濒}{2 a})}^{2}\}$

Find the expectation value of x .

Considering the following function:

$w = {(\frac{a}{1 + {(2 魔 a t / m)}^{2}})}^{1 / 2} = {(\frac{a}{1 + 胃^{2}})}^{1 / 2} The wave function will be : {|蠄 (x, t)|}^{2} = \sqrt{\frac{2}{蟺}} {we}^{- 2 w^{2} (x - 胃 / 2 a)}^{2} Find the expectation value of x, as : (x) = {鈭�}_{- 鈭�}^{鈭�} x {|蠄 (x, t)|}^{2} dx Substituting the value of wavefunction : (x) = {鈭�}_{- 鈭�}^{鈭�} \sqrt{\frac{2}{蟺}} wx {e^{- 2 w^{2} (x - 胃 / 2 a)}}^{2} dx Let y = x - 胃滨 / 2 a = x - vt, and thus x = y + vt where v = 魔滨 / m, substitute these and get : (x) = {鈭�}_{- 鈭�}^{鈭�} (y + vt) \sqrt{\frac{2}{蟺}} w {e^{- 2 w^{2} (x - 胃 / 2 a)}}^{2} dy In two intergrals obtained the first one is zero, since the funtion id odd, and the integration of an odd function from - 鈭� to 鈭� is zero, and the second integral is 1 by normalization, so : (x) = vt = \frac{魔濒}{m} t . . . . . . (2) On differentiating above equation : \frac{d (x)}{dt} \frac{魔濒}{m}$

Find the expectation value of x2.

Find the expectation value of $x^{2}$ as:

$(x^{2}) = {鈭�}_{- 鈭�}^{鈭�} {(y + v t)}^{2} \sqrt{\frac{2}{蟺}} w e^{- 2 w^{2} y^{2}} d y$

Since ${(y + v t)}^{2} = y^{2} + 2 y v t + v^{2} t^{2}$ , to find the values of the first integral, use the integral calculator, where the second integral is zero since it is an odd function, and the third integral is 1 by the normalization, so the value of the integral will be:

$(x^{2}) = \frac{1}{4 w^{2}} + 0 + {(v t)}^{2} Substitute back with and get : (x^{2}) = \frac{1}{4 w^{2}} + {(\frac{魔濒t}{m})}^{2} Substitute back with w, and get : (x^{2}) = \frac{1 + {(2 补魔濒t / m)}^{2} + a {(2 补魔濒t / m)}^{2}}{4 a} . . . . . . . (3) The 蟽_{x}^{2} can be calculated as : 蟽_{x}^{2} = (x^{2}) - {(x)}^{2} = \frac{1}{4 w^{2}} + (\frac{魔濒t}{m}) - (\frac{魔濒t}{m}) = \frac{1}{4 w^{2}} 蟽_{x}^{2} = \frac{1}{4 a} [1 + {(\frac{2 魔补迟}{m})}^{2}] Write (1) in terms of 胃 as : 蠄 (x, t) = {(\frac{2 a}{蟺})}^{1 / 4} \frac{1}{\sqrt{1 + 颈胃}} e^{- \frac{I^{2}}{4 a}} e^{a (ix + \frac{I}{2 a}) / (1 + 颈胃)} Thus : 桅 (p, t) = \frac{1}{\sqrt{2 蟺魔}} {(\frac{2 a}{蟺})}^{1 / 4} \frac{1}{\sqrt{1 + 颈胃}} e^{- \frac{I^{2}}{4 a}} {鈭�}_{- 鈭�}^{鈭�} e^{a (ix + \frac{I}{2 a}) / (1 + 颈胃)} dx let y = x - \frac{il}{2 a}, so : 桅 (p, t) = \frac{1}{\sqrt{2 蟺魔}} {(\frac{2 a}{蟺})}^{1 / 4} \frac{1}{\sqrt{1 + 颈胃}} e^{- I^{2} / 4 a} e^{pI / 2 补魔} {鈭�}_{- 鈭�}^{鈭�} e^{- ipy / 魔} e^{- {ey}^{2} I (1 + 颈胃)} dy$

Find the expectation value of p4.

By the use of the integral from part (a) of problem 2.22:

$桅 (p, t) = \frac{1}{\sqrt{2 蟺魔}} {(\frac{2 a}{蟺})}^{1 / 4} \frac{1}{\sqrt{1 + i 胃}} e^{- I^{2} / 4 a} e^{p I / 2 a 魔} \sqrt{\frac{蟺 (1 + 颈胃)}{a}} e^{- \frac{p^{2} (1 + i 胃)}{4 a 魔^{2}}} = \frac{1}{\sqrt{魔}} {(\frac{1}{2 补蟺})}^{1 / 4} e^{- \frac{t^{2}}{4 a}} e^{- \frac{p I}{2 补魔}} e^{- \frac{p^{2} (1 + i 胃)}{4 a 魔^{2}}} Thus :$

${|桅 (p, t)|}^{2} = \frac{1}{\sqrt{2 a 蟺}} \frac{1}{魔} e^{- I^{2} / 2 a} e^{p I / a 魔} e^{- p^{2} / 2 a 魔^{2}} = \frac{1}{魔 \sqrt{2 a 蟺}} e^{\frac{1}{2 a} (I^{2} - \frac{2 p I}{魔} + \frac{p^{2}}{魔^{2}})} = \frac{1}{魔 \sqrt{2 a 蟺}} {e^{- (I - p I 魔)}}^{2} / 2 a . . . . . . . (3) The expectation value of p^{4} is : <p^{4}> = \frac{1}{魔 \sqrt{2 补蟺}} {鈭�}_{- 鈭�}^{鈭�} p^{4} {e^{- (I - 辫滨魔)}}^{2} / 2 a dp Let \frac{p}{魔} - I = z so p = 魔 (z + I), so : <p^{4}> = \frac{1}{魔 \sqrt{2 补蟺}} 魔^{5} {鈭�}_{- 鈭�}^{鈭�} {(z + I)}^{4} e^{- z^{2} / 2 a} dz Only even powers of z survive, where the odd powers of z vanishes, since it 鈥� s an odd function and the integration of an odd function from 鈭� to 鈭� is zero . So :$

$<p^{4}> = \frac{魔^{4}}{\sqrt{2 a 蟺}} {鈭�}_{- 鈭�}^{鈭�} (z^{4} + 6 z^{2} I^{2} + I^{4}) e^{- z^{2} / 2 a} d z = \frac{魔^{4}}{\sqrt{2 a 蟺}} [\frac{3 {(2 a)}^{2}}{4} \sqrt{2 a 蟺} + 6 I^{2} \frac{(2 a)}{2} \sqrt{2 a 蟺} + I^{4} \sqrt{2 a 蟺}] = 魔^{4} (3 a^{2} + 6 a I^{2} + I^{4}) We know the following relation : (H^{2}) = \frac{(p^{4})}{4 m^{2}} The expectation value will be : <H^{2}> = \frac{魔^{4}}{4 m^{2}} (3 a^{2} + 6 {aI}^{2} I^{4}) . . . . . . (5)$

Find the expectation value of (p2)

Now find the expectation value of $(p^{2})$

$(p^{2}) = - 魔^{2} {鈭�}_{- 鈭�}^{鈭�} 唯^{*} \frac{d^{2} 唯}{d x^{2}} d x From (l), write this : \frac{诲唯}{dx} = \frac{2 ia (x + \frac{I}{2 a})}{(1 + 颈胃)} 唯; thus : \frac{d^{2} 唯}{{dx}^{2}} = [\frac{2 ia (ix + I / 2 a)}{1 + 颈胃}] \frac{诲唯}{dx} + \frac{2 i^{2} a}{1 + 颈胃} 唯 = [\frac{- 4 a^{2} (ix + I / 2 a)}{1 + 颈胃} - \frac{2 a}{1 + 颈胃}] 唯$

So the expectation value is:

$(p^{2}) = \frac{4 a^{2} 魔^{2}}{{(1 + i 胃)}^{2}} {鈭�}_{- 鈭�}^{鈭�} [{(i x + \frac{I}{2 a})}^{2} + \frac{(1 + i 胃)}{2 a}] {|唯|}^{2} d x = \frac{4 a^{2} 魔^{2}}{{(1 + i 胃)}^{2}} {鈭�}_{- 鈭�}^{鈭�} [- {(y + v t \frac{i I}{2 a})}^{2} + \frac{(1 + i 胃)}{2 a}] {|唯|}^{2} d y n = \frac{4 a^{2} 魔^{2}}{{(1 + i 胃)}^{2}} \{- {鈭�}_{- 鈭�}^{鈭�} y^{2} {|唯|}^{2} d y - 2 (v t - \frac{i I}{2 a}) {鈭�}_{- 鈭�}^{鈭�} y {|唯|}^{2} d y n + [- {(v t - \frac{i I}{2 a})}^{2} + \frac{(1 + i 胃)}{2 a}] {鈭�}_{- 鈭�}^{鈭�} {|唯|}^{2} d y\} = \frac{4 a^{2} 魔^{2}}{{(1 + i 胃)}^{2}} [- \frac{1}{4 w^{2}} + 0 {(v t - \frac{i I}{2 a})}^{2} + \frac{(1 + i 胃)}{2 a}] Further solving above equation, = \frac{4 a^{2} 魔^{2}}{{(1 + 颈胃)}^{2}} \{- \frac{1 + 胃^{2}}{4 a} - {[(\frac{- il}{2 a}) (1 + 颈胃)]}^{2} + \frac{(1 + 颈胃)}{2 a}\} = \frac{补魔}{1 + 颈胃} [- (- 1 - 颈胃) + \frac{l^{2}}{a} (- 1 + 颈胃) + 2] = \frac{{补魔}^{2}}{1 + 颈胃} [(1 + 颈胃) (1 + \frac{l^{2}}{a})] = 魔^{2} (a + l^{2})$

Find the value of σH2 and test the energy-time uncertainty principle.

The expectation value is $(H)$ :

$(H) = \frac{(p^{2})}{2 m}$

The above equation will be:

$(H) = \frac{魔^{2} (a + l^{2})}{2 m}$ (6)

Now the value of $蟽_{H}^{2}$ could be found as:

$蟽_{H}^{2} = (H^{2}) - {(H)}^{2} = \frac{魔^{4}}{4 m^{2}} (3 a^{2} + 6 a l^{2} + l^{4} - a^{2} - 2 a l^{2} - l^{4}) = \frac{魔^{4}}{4 m^{2}} (2 a^{2} + 4 a l^{2}) = \frac{魔^{4} a}{2 m^{2}} (a + 2 l^{2}) 蟽_{H}^{2} = \frac{魔^{4} a}{2 m^{2}} (a + 2 l^{2}) Finally test the energy - time uncertainty principle as : 蟽_{H}^{2} 蟽_{x}^{2} = \frac{魔^{4} a}{2 m^{2}} (a + 2 l^{2}) \frac{1}{4 a} [1 + {(\frac{2 魔补迟}{m})}^{2}] = \frac{魔^{4} a}{2 m^{2}} (1 + \frac{a}{{al}^{2}}) [1 + {(\frac{2 魔补迟}{m})}^{2}] 鈮� \frac{魔^{4} l^{2}}{4 m^{2}} = \frac{魔^{4}}{4} {(\frac{魔濒}{m})}^{2} = \frac{魔^{2}}{4} {(\frac{d (x)}{dt})}^{2}$

So, the uncertainty principle holds.

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Test the energy-time uncertainty principle for the free particle wave packet in Problem 2.43and the observable x , by calculating $蟽_{H} 蟽_{x}$ , and $d < x > / d t$ exactly.

Short Answer

Step by step solution

The wave function and find |ψ|2

Find the expectation value of x .

Find the expectation value of x2.

Find the expectation value of p4.

Find the expectation value of (p2)

Find the value of σH2 and test the energy-time uncertainty principle.

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91影视

Test the energy-time uncertainty principle for the free particle wave packet in Problem 2.43and the observable x , by calculating 蟽H蟽x , and d<x>/dtexactly.

Short Answer

Step by step solution

The wave function and find |ψ|2

Find the expectation value of x .

Find the expectation value of x2.

Find the expectation value of p4.

Find the expectation value of (p2)

Find the value of σH2 and test the energy-time uncertainty principle.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Particle Model of Matter

Fluids

Circular Motion and Gravitation

Astrophysics

Atoms and Radioactivity

Electricity

Study anywhere. Anytime. Across all devices.

Test the energy-time uncertainty principle for the free particle wave packet in Problem 2.43and the observable x , by calculating $蟽_{H} 蟽_{x}$ , and $d < x > / d t$ exactly.