Q9P A particle of mass m is in the s... [FREE SOLUTION]

Chapter 1: Q9P (page 20)

A particle of mass m is in the state:
$蠄 (x, t) = {Ae}^{鈭� a [({mx}^{2} / h) + it]}$
where A and a are positive real constants.
(a) Find A.
(b) For what potential energy function, V(x), is this a solution to the Schr枚dinger equation?
(c) Calculate the expectation values of x, $x^{2}$ , p, and $p^{2}$ .
(d) Find 蟽_x and 蟽_p. Is their product consistent with the uncertainty principle?

Short Answer

Expert verified

a. $A = {(\frac{2 am}{蟺 h})}^{1 / 4}$

b. $V_{(x, t)} = 2 {ma}^{2} x^{2}$

c. $鉄� x 鉄� = 0$ , $鉄� x^{2} 鉄� = \frac{h}{4 am}$ , $鉄� p 鉄� = 0$ , $鉄� p^{2} 鉄� = h am$

d. Yes, it鈥檚 consistent with the uncertainty principle

Step by step solution

Normalizing the wave equation (a)

Calculating for A by normalizing the wave function.

$\begin{matrix} 1 = {鈭�}_{鈭� 鈭�}^{鈭�} {| 蠄_{(x, t)} |}^{2} dx \\ 1 = {鈭�}_{鈭� 鈭�}^{鈭�} 蠄 *_{(x, t)} 蠄_{(x, t)} dx \\ 1 = {鈭�}_{鈭� 鈭�}^{鈭�} ({Ae}^{鈭� a [({mx}^{2} / h) + it]}) ({Ae}^{鈭� a [({mx}^{2} / h) 鈭� it]}) dx \\ 1 = A^{2} {鈭�}_{鈭� 鈭�}^{鈭�} e^{鈭� 2 {amx}^{2} / h} dx \end{matrix}$

Further solving above equation,

$\begin{matrix} 1 = 2 A^{2} {鈭�}_{0}^{鈭�} e^{鈭� x^{2} / {[\sqrt{h / (2 am)}]}^{2}} dx \\ 1 = 2 A^{2} . \sqrt{蟺} (\frac{\sqrt{h / 2 am}}{2}) \\ 1 = A^{2} \sqrt{\frac{蟺丑}{2 am}} \\ A^{2} = {(\frac{2 am}{蟺 h})}^{1 / 2} \\ A = {(\frac{2 am}{蟺 h})}^{1 / 4} \end{matrix}$

Thus, the value of A is ${(\frac{2 am}{蟺 h})}^{1 / 4}$ .

Finding the potential energy for which it is a solution of the Schrödinger equation (b)

Schrodinger equation is given by,

$\frac{鈭� 蠄}{鈭� t} = \frac{i h}{2 m} \frac{{鈭�}^{2} 蠄}{鈭� x^{2}} 鈭� \frac{i}{h} V_{(x, t)} 蠄_{(x, t)}$

Solving for $V_{(x, t)}$ ,

$\begin{matrix} V_{(x, t)} = \frac{h}{i} \frac{1}{蠄_{(x, t)}} (\frac{i h}{2 m} \frac{{鈭�}^{2} 蠄}{鈭� x^{2}} 鈭� \frac{鈭� 蠄}{鈭� t}) \\ V_{(x, t)} = \frac{h}{i} {(\frac{2 am}{蟺 h})}^{1 / 4} e^{a [({mx}^{2} / h) + it]} [\frac{i h}{2 m} {(\frac{2 am}{蟺 h})}^{1 / 4} \frac{{鈭�}^{2}}{鈭� x^{2}} (e^{鈭� a [({mx}^{2} / h) + it]}) 鈭� {(\frac{2 am}{蟺 h})}^{1 / 4} (e^{鈭� a [({mx}^{2} / h) + it]}) (鈭� ia)] \\ V_{(x, t)} = h e^{a [({mx}^{2} / h) + it]} [\frac{i h}{2 m} \frac{{鈭�}^{2}}{鈭� x^{2}} (e^{鈭� a [({mx}^{2} / h) + it]}) 鈭� {ae}^{鈭� a [({mx}^{2} / h) + it]}] \\ V_{(x, t)} = h e^{a [({mx}^{2} / h) + it]} \{\frac{i h}{2 m} [鈭� \frac{2 am}{h} (1 鈭� \frac{2 am}{h} x^{2}) e^{鈭� a [({mx}^{2} / h) + it]}] + {ae}^{鈭� a [({mx}^{2} / h) + it]}\} \end{matrix}$

Further solving above equation,

$\begin{matrix} V_{(x, t)} = h \{a [\frac{2 am}{h} x^{2} 鈭� 1] + a\} \\ V_{(x, t)} = 2 {ma}^{2} x^{2} \end{matrix}$

Thus, the potential energy function is $V_{(x, t)} = 2 {ma}^{2} x^{2}$ .

Calculating the expectation values of x (c)

Solving for the expectation values.

$\begin{matrix} 鉄� x 鉄� = \frac{{鈭�}_{鈭� 鈭�}^{鈭�} x | {蠄_{(x, t)} |}^{2} dx}{{鈭�}_{鈭� 鈭�}^{鈭�} | {蠄_{(x, t)} |}^{2} dx} \\ 鉄� x 鉄� = {鈭�}_{鈭� 鈭�}^{鈭�} 虫蠄 *_{(x, t)} 蠄_{(x, t)} dx \\ 鉄� x 鉄� = {鈭�}_{鈭� 鈭�}^{鈭�} x [{(\frac{2 am}{蟺 h})}^{1 / 4} e^{鈭� a [({mx}^{2} / h) + it]}] [{(\frac{2 am}{蟺 h})}^{1 / 4} e^{鈭� a [({mx}^{2} / h) 鈭� it]}] dx \\ 鉄� x 鉄� = \sqrt{\frac{2 am}{蟺 h}} {鈭�}_{鈭� 鈭�}^{鈭�} {xe}^{鈭� 2 {amx}^{2} / h} dx \end{matrix}$

Since, integral of an odd function over a symmetric interval is zero.

Therefore, the expectation value is zero $鉄� x 鉄� = 0$ .

Now, for $鉄� x^{2} 鉄�$

$\begin{matrix} 鉄� x^{2} 鉄� = \frac{{鈭�}_{鈭� 鈭�}^{鈭�} x^{2} | {蠄_{(x, t)} |}^{2} dx}{{鈭�}_{鈭� 鈭�}^{鈭�} | {蠄_{(x, t)} |}^{2} dx} \\ 鉄� x^{2} 鉄� = {鈭�}_{鈭� 鈭�}^{鈭�} x^{2} 蠄_{(x, t)} 蠄 *_{(x, t)} dx \\ 鉄� x^{2} 鉄� = {鈭�}_{鈭� 鈭�}^{鈭�} x^{2} [{(\frac{2 am}{蟺 h})}^{1 / 4} e^{鈭� a [({mx}^{2} / h) + it]}] [{(\frac{2 am}{蟺 h})}^{1 / 4} e^{鈭� a [({mx}^{2} / h) 鈭� it]}] dx \\ 鉄� x^{2} 鉄� = \sqrt{\frac{2 am}{蟺 h}} {鈭�}_{鈭� 鈭�}^{鈭�} x^{2} e^{鈭� 2 \max^{2} / h} dx \end{matrix}$

$\begin{array}{l} 鉄� x^{2} 鉄� = 2 \sqrt{\frac{2 am}{蟺 h}} {鈭�}_{0}^{鈭�} x^{2} e^{鈭� x^{2} / {[\sqrt{h / (2 am)}]}^{2}} dx \\ 鉄� x^{2} 鉄� = 2 \sqrt{\frac{2 am}{蟺 h}} . \sqrt{蟺} \frac{2!}{1} {(\frac{\sqrt{h / 2 am}}{2})}^{3} \\ 鉄� x^{2} 鉄� = \frac{h}{4 am} \end{array}$

Therefore, the value of $鉄� x^{2} 鉄�$ is $\frac{h}{4 am}$ .

Calculating the expectation values for p (c)

Using the Ehrenfest鈥檚 theorem,

$\begin{array}{l} 鉄� p 鉄� = m 鉄� v 鉄� \\ 鉄� p 鉄� = m \frac{d 鉄� x 鉄�}{dt} \end{array}$

Since $鉄� x 鉄� = 0$ .

Therefore, the value of $鉄� p 鉄�$ is zero.

Now, solving for $鉄� p^{2} 鉄�$ .

$\begin{matrix} 鉄� p^{2} 鉄� = {鈭�}_{鈭� 鈭�}^{鈭�} 蠄 *_{(x, t)} {(鈭� i h \frac{鈭�}{鈭� x})}^{2} 蠄_{(x, t)} dx \\ 鉄� p^{2} 鉄� = 鈭� h^{2} {鈭�}_{鈭� 鈭�}^{鈭�} 蠄 *_{(x, t)} \frac{{鈭�}^{2} 蠄}{鈭� x^{2}} dx \\ 鉄� p^{2} 鉄� = 鈭� h^{2} {鈭�}_{鈭� 鈭�}^{鈭�} [{(\frac{2 am}{蟺 h})}^{1 / 4} e^{鈭� a [({mx}^{2} / h) 鈭� it]}] \frac{{鈭�}^{2}}{鈭� x^{2}} [{(\frac{2 am}{蟺 h})}^{1 / 4} e^{鈭� a [({mx}^{2} / h) + it]}] dx \end{matrix}$

Further solving above equation,

$\begin{matrix} 鉄� p^{2} 鉄� = 鈭� h^{2} \sqrt{\frac{2 am}{蟺 h}} (\frac{2 am}{h}) {鈭�}_{鈭� 鈭�}^{鈭�} (1 鈭� \frac{2 am}{h} x^{2}) e^{鈭� 2 {amx}^{2} / h} dx \\ 鉄� p^{2} 鉄� = 鈭� h^{2} \sqrt{\frac{2 am}{蟺 h}} (\frac{2 am}{h}) ({鈭�}_{鈭� 鈭�}^{鈭�} e^{鈭� 2 {amx}^{2} / h} dx 鈭� \frac{2 am}{h} {鈭�}_{鈭� 鈭�}^{鈭�} x^{2} e^{鈭� 2 {amx}^{2} / h} dx) \\ 鉄� p^{2} 鉄� = 鈭� h^{2} \sqrt{\frac{2 am}{蟺 h}} (\frac{4 am}{h}) ({鈭�}_{0}^{鈭�} e^{鈭� x^{2} / {[\sqrt{h / 2 am}]}^{2}} dx 鈭� \frac{2 am}{h} {鈭�}_{0}^{鈭�} x^{2} e^{鈭� x^{2} / {[\sqrt{h / 2 am}]}^{2}} dx) \\ 鉄� p^{2} 鉄� = 鈭� h^{2} \sqrt{\frac{2 am}{蟺 h}} (\frac{4 am}{h}) \{\sqrt{蟺} (\frac{\sqrt{h / 2 am}}{2}) 鈭� \frac{2 am}{h} \sqrt{蟺} \frac{2!}{1!} {(\frac{\sqrt{h / 2 am}}{2})}^{3}\} \end{matrix}$

Further solving above equation,

$\begin{array}{l} 鉄� p^{2} 鉄� = h^{2} (\frac{4 am}{h}) (\frac{1}{2} 鈭� \frac{1}{4}) \\ 鉄� p^{2} 鉄� = h am \end{array}$

Therefore, the value of $鉄� p^{2} 鉄�$ is $h am$ .

Checking the consistency with Heisenberg Uncertainty Principle

The standard variation of an observable gives the uncertainty, hence,

$\begin{array}{l} 蟽_{x} = 螖虫 \\ 螖虫 = \sqrt{鉄� x^{2} 鉄� 鈭� {鉄� x 鉄�}^{2}} \\ 螖虫 = \sqrt{\frac{h}{4 am} 鈭� 0} \\ 螖虫 = \frac{1}{2} \sqrt{\frac{h}{am}} \end{array}$

The calculation of standard deviation $蟽_{p}$ is,

$\begin{matrix} 蟽_{p} = 螖辫 \\ 螖辫 = \sqrt{鉄� p^{2} 鉄� 鈭� {鉄� p 鉄�}^{2}} \\ 螖辫 = \sqrt{h am 鈭� 0} \\ 螖辫 = \sqrt{h am} \end{matrix}$

The uncertainty can be written as,

$\begin{array}{l} 螖虫螖辫 = \frac{1}{2} \sqrt{\frac{h}{am}} \sqrt{ham} \\ 螖虫螖辫 = \frac{h}{2} \end{array}$

And Heisenberg Uncertainty Principle states that

$螖虫螖辫鈮� \frac{h}{2}$

Therefore, it is consistent with the Heisenberg Uncertainty Principle.

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Short Answer

Step by step solution

Normalizing the wave equation (a)

Finding the potential energy for which it is a solution of the Schrödinger equation (b)

Calculating the expectation values of x (c)

Calculating the expectation values for p (c)

Checking the consistency with Heisenberg Uncertainty Principle

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