Q7P Consider a particle of mass m th... [FREE SOLUTION]

Chapter 6: Q7P (page 260)

Consider a particle of mass m that is free to move in a one-dimensional region of length L that closes on itself (for instance, a bead that slides frictionlessly on a circular wire of circumference L, as inProblem 2.46).
(a) Show that the stationary states can be written in the form $蠄_{n} (x) = \frac{1}{\sqrt{L}} e^{2 蟺 i n x / L}, (- L / 2 < x < L / 2),$
where $n = 0, 卤 1, 卤 2, . . . .$ and the allowed energies are $E_{n} = \frac{2}{m} {(\frac{n 蟺鈩�}{L})}^{2} .$ Notice that with the exception of the ground state (n = 0 ) 鈥� are all doubly degenerate.
(b) Now suppose we introduce the perturbation, $H^{'} = - V_{0} e^{- x^{2} / a^{2}}$ where $a 鈮� L a$ . (This puts a little 鈥渄imple鈥� in the potential at x = 0, as though we bent the wire slightly to make a 鈥渢rap鈥�.) Find the first-order correction to $E_{n}$ , using Equation 6.27. Hint: To evaluate the integrals, exploit the fact that $a 鈮� L a$ to extend the limits from $卤 L / 2 to 卤鈭�$ after all, H鈥� is essentially zero outside $- a < x < a$ .
$E_{卤}^{1} = \frac{1}{2} [W_{a a} + W_{b b} 卤 \sqrt{{(W_{a a} - W_{b b})}^{2} + 4 {|W_{a b}|}^{2}}]$ (6.27).
(c) What are the 鈥済ood鈥� linear combinations of $蠄_{n}$ and $蠄_{- n}$ , for this problem? Show that with these states you get the first-order correction using Equation 6.9.
$E_{n}^{'} = <蠄_{n}^{0} |H^{'}| 蠄_{n}^{0}>$ (6.9).
(d) Find a hermitian operator A that fits the requirements of the theorem, and show that the simultaneous Eigenstates of $H_{0}$ and A are precisely the ones you used in (c).

Short Answer

Expert verified

(a) Stationary states $E_{n} = \frac{2 n^{2} 蟺^{2} {鈩�}^{2}}{m L^{2}}$

(b)The integrals exploit $E_{卤}^{1} = \frac{V_{0} a \sqrt{蟺}}{L} [- 1 卤 e^{- {(\frac{2 蟺 n a}{L})}^{2}}]$

(c) The good linear combinations $E_{2}^{1} = \frac{伪^{2}}{2 L} (2)^{2} (- V_{0}) a \sqrt{蟺} [1 - e^{- {(2 蟺 n a / L)}^{2}}]$

(d)Hermitian operator $P |唯_{- n}> = - 1 |唯_{n}>, P |唯_{n}> = + 1 |唯_{n}>$

Step by step solution

Step 1:(a) Showing that the stationary states can be written in the formψn(x)=1Le2πinx/L, (-L/2<x<L/2).

For a particle moving freely in ID, the Schroedinger Eq. can be written as:

$\frac{- {鈩�}^{2}}{2 m} \frac{d^{2}}{d x^{2}} 唯 = E 唯$

$\frac{d^{2}}{d x^{2}} 唯 = \frac{- 2 m E}{{鈩�}^{2}} 唯$

where

$k^{2} = \frac{2 m E}{{鈩�}^{2}}$

Therefore,

$\frac{d^{2}}{d x^{2}} 唯 = - k^{2} 唯$

Solve this differential eq. as:

$\begin{array}{rcl} 唯 (x) & = & A e^{- i k x} + B e^{i k x} \\ = & A \cos k x + B \sin k x \end{array}$ 鈥�(颈)

Now, from the periodicity

$唯 (x) = 唯 (x + L)$

So using boundary condition to obtain the particular solution to this general solution:

$唯 (x + L) = A e^{- i k x} e^{- i k L} + B e^{i k x} e^{i k L}$ 鈥�(颈颈)

for $x = 0$

$唯 (0) = 唯 (L), k x = n 蟺$

By equating the two equations (i) and (ii), and applying B.C

$A + B = A e^{- i k L} + B e^{i k L}$ 鈥�(颈颈颈)

And from $k x = \frac{n 蟺}{2}$

$A e^{- i 蟺 / 2} e^{- i k L} + B e^{i 蟺 / 2} e^{i k L} = A e^{i 蟺 / 2} + B e^{i 蟺 / 2}$

Where

$\begin{array}{rcl} e^{- i 蟺 / 2} & = & \cos \frac{蟺}{2} - i \sin \frac{蟺}{2} \\ = & 0 - i 脳 1 \\ = & - i e^{i 蟺 / 2} \\ = & \cos \frac{蟺}{2} + i \sin \frac{蟺}{2} \\ = & 0 + i 脳 1 \\ = & i \end{array}$

which implies that:

$\begin{array}{rcl} A e^{- i 蟺 / 2} e^{- i k L} + B e^{i 蟺 / 2} e^{i k L} & = & A e^{i 蟺 / 2} + B e^{i 蟺 / 2} - A h e^{- i k L} + B l e^{i k L} \\ = & - A A + 1 B A e^{i k L} - B e^{i k L} \end{array}$

$A e^{- i 蟺 / 2} e^{- i k L} + B e^{i 蟺 / 2} e^{i k L} = A - B$ 鈥�(颈惫)

By adding (iii) and (iv), we get

$\begin{array}{rcl} 2 A & = & 2 A e^{- i k L} e^{- i k L} \\ = & 1 \end{array}$

Note that $A 鈮� 0 .$

The system has a periodicity $2 n 蟺$

In regular problem $鈬� k L = n 蟺$ but in this case $鈬� k L = 2 n 蟺 n = 0, 卤 1, 鈥�$ , then, the solution can be written as follows:

$\begin{array}{rcl} 唯 (x) & = & A e^{- 2 i n 蟺 x / L} + B e^{2 i n 蟺 x / L} \\ = & A \cos \frac{2 n 蟺 x}{L} + B \sin 2 n 蟺 x L \\ = & 唯^{+} (x) + 唯^{-} (x) \end{array}$

where A and B are not equal to zero. If $A = 0$ and $B 鈮� 0$ then,

$唯 (x) = B e^{2 n 蟺 i x / L}$

From the normalization condition

$\begin{array}{rcl} {鈭�}_{- 鈭�}^{鈭�} | 唯 (x) |^{2} d x & = & | B |^{2} {鈭�}_{- L / 2}^{L / 2} e^{2 n x x + E_{e} - 2 n x + x + L} d x \\ = & | B |^{2} {鈭�}_{- L / 2}^{L / 2} d x \\ = & | B |^{2} {鈭�}_{0}^{L} d x \\ = & | B |^{2} L \\ = & 1 \\ | {B|}^{2} & = & \frac{1}{L} \\ B & = & \frac{1}{\sqrt{L}} \end{array}$

Then,

$唯 (x) = \frac{1}{\sqrt{L}} e^{2 n 蟺 i x / L}$

where $n 鈫� 2 n$

$\begin{array}{rcl} E_{n} & = & \frac{4 n^{2} 蟺^{2} {鈩�}^{2}}{2 m L^{2}} \\ = & \frac{2 n^{2} 蟺^{2} {鈩�}^{2}}{m L^{2}} \end{array}$

(b) To Find the first order correction

To find the first-order correction, first we need to calculate $W_{a a}, W_{b b}, W_{a b e}$ where,

$a = + ve solution (+ n) b = - ve solution (- n)$

that,

$\begin{array}{rcl} E_{卤}^{1} & = & \frac{1}{2} [W_{a a} + W_{b b} 卤 \sqrt{{(W_{a a} - W_{b b})}^{2} + 4 {|W_{a b}|}^{2}}] W_{a a} \\ = & <唯_{a}^{0} |{\hat{H}}^{1}| 唯_{a}^{0}> |唯_{a}> \\ = & \frac{1}{\sqrt{L}} e^{2 in x / L} 鉄� 唯_{a}^{0} | \\ = & \frac{1}{\sqrt{L}} e^{- 2 蟺 in x / L} \end{array}$

Then,

$= \frac{- V_{0}}{L} {鈭�}_{- L / 2}^{L / 2} e^{- x^{2} / a^{2}} d x$

The wave function is periodically repeated, so we can integrate such that;

$鈭� \frac{- V_{0}}{L} {鈭�}_{- L / 2}^{L / 2} e^{- x^{2} / a^{2}} d x = \frac{- V_{0}}{L} {鈭�}_{- 鈭�}^{鈭�} e^{- x^{2} / a^{2}} d x$

This integral has the form of Gaussian integral, so we can evaluate it as Gaussian integral, where,

$\begin{array}{rcl} {鈭�}_{- 鈭�}^{鈭�} e^{- 伪 x^{2}} d x & = & \sqrt{\frac{蟺}{伪}} \frac{- V_{0}}{L} {鈭�}_{- 鈭�}^{鈭�} e^{- 伪 x^{2}} d x \\ = & \frac{- V_{0}}{L} \sqrt{\frac{蟺}{伪}} \frac{- V_{0}}{L} {鈭�}_{- 鈭�}^{鈭�} e^{- x^{2} / a^{2}} d x \\ = & \frac{- V_{0}}{L} \sqrt{\frac{蟺}{1 / a^{2}}} n \\ = & \frac{- V_{0}}{L} \sqrt{蟺 a^{2}} n \\ = & W_{a a} 鈭� W_{a a} \\ = & \frac{- V_{0}}{L} \sqrt{蟺 a^{2}} \\ = & \frac{- V_{0} a}{L} \sqrt{蟺} \end{array}$

For $W_{b b}$

$\begin{array}{rcl} W_{b b} & = & <唯_{b}^{0} |{\hat{H}}^{1}| 唯_{b}^{0}> \\ = & \frac{- V_{0}}{L} {鈭�}_{- L / 2}^{L / 2} e^{- x^{2} / a^{2}} d x \\ = & \frac{- V_{0} a}{L} \sqrt{蟺} \\ = & W_{a a} \end{array}$

For $W_{a b}$

$\begin{array}{rcl} W_{a b} & = & \frac{- V_{0}}{L} {鈭�}_{- L / 2}^{L / 2} e^{- 2 蟺 i n x / L} e^{- x^{2} / a^{2}} e^{- 2 蟺 i n x / L} d x \\ = & \frac{- V_{0}}{L} {鈭�}_{- L / 2}^{L / 2} e^{- 4 蟺 i n x / L} e^{- x^{2} / a^{2}} d x \end{array}$

To evaluate this integral, first we should complete the square,

$\begin{array}{rcl} \frac{- x^{2}}{a^{2}} - 4 蟺 i n \frac{x}{L} & = & \frac{- 1}{a^{2}} [x^{2} + \frac{4 a^{2} 蟺 i n x}{L}] \\ = & \frac{- 1}{a^{2}} [{[x + \frac{2 蟺 i n a^{2}}{L}]}^{2} - {[\frac{2 蟺 i n a^{2}}{L}]}^{2}] \\ W_{a b} & = & \frac{- V_{0}}{L} {鈭�}_{- 鈭�}^{鈭�} e^{(- 1 / a^{2}) {(x + \frac{2 a^{2} 蟺 i n}{L})}^{2}} e^{{(\frac{2 蟺 i n a}{L})}^{2}} d x \end{array}$

By using the Gaussian integral, take

$y = x + \frac{2 a^{2} 蟺 i n}{L} d y = d x$

Then,

$\begin{array}{rcl} (\frac{- V_{0}}{L}) {鈭�}_{- 鈭�}^{鈭�} e^{- y^{2} / a^{2}} e^{- {(\frac{2 蟺 n a}{L})}^{2}} d y & = & (\frac{- V_{0}}{L}) e^{- {(\frac{2 蟺 n a}{L})}^{2}} 路 a \sqrt{蟺} \\ W_{a b} & = & \frac{- V_{0} a \sqrt{蟺}}{L} 路 e^{- {(\frac{2 蟺 n a}{L})}^{2}} \end{array}$

Now, we can calculate the first-order correction.

$\begin{array}{rcl} E_{卤}^{1} & = & \frac{1}{2} [W_{a a} + W_{b b} 卤 \sqrt{{(W_{a a} - W_{b b})}^{2} + 4 {|W_{a b}|}^{2}}] \\ = & \frac{1}{2} [\frac{- 2 V_{0} a \sqrt{蟺}}{L} 卤 \sqrt{4 {|\frac{- V_{0} a \sqrt{蟺}}{L} 路 e^{- {(\frac{2 m a}{L})}^{2}}|}^{2}}] \\ = & \frac{1}{\ not 2} [\frac{- 2 V_{0} a \sqrt{蟺}}{L} 卤 \ not 2 (\frac{- V_{0} a \sqrt{蟺}}{L} 路 e^{- {(\frac{2 m a}{L})}^{2}})] \\ = & \frac{V_{0} a \sqrt{蟺}}{L} [- 1 卤 (e^{- {(\frac{2 蟺 a}{L})}^{2}})] \\ E_{卤}^{1} & = & \frac{V_{0} a \sqrt{蟺}}{L} [- 1 卤 e^{- {(\frac{2 m a}{L})}^{2}}] \end{array}$

(c) To Find the linear combination

We need to find linear combination of $唯_{n}$ and $唯_{- n}$ By using the following:

$\begin{array}{rcl} 伪 W_{a a} + 尾 W_{a b} & = & 伪 E_{卤}^{1} E_{卤}^{1} \\ = & \frac{V_{0} a \sqrt{蟺}}{L} [- 1 卤 e^{- {(\frac{2 蟺 a a}{L})}^{2}}] W_{a a} \\ = & \frac{- V_{0} a 蟺}{L} W_{a b} \\ = & \frac{- V_{0} a 蟺}{L} 路 e^{- {(\frac{2 蟺 n a}{L})}^{2}} \end{array}$

Then,

$\begin{array}{rcl} - 伪 \frac{V_{0} 伪 \sqrt{蟺}}{L} - 尾 \frac{V_{0} 伪 \sqrt{蟺}}{L} 路 e^{- {(\frac{2 蟺 n a}{L})}^{2}} & = & 伪 \frac{V_{0} 伪 \sqrt{蟺}}{L} [- 1 卤 e^{- {(\frac{2 蟺 n a}{L})}^{2}}] - 伪 - 尾 e^{- {(\frac{2 蟺 n a}{L})}^{2}} \\ = & 伪 [- 1 卤 e^{- {(\frac{2 蟺 m a}{L})}^{2}}] 尾 e^{- {(\frac{2 蟺 n a}{L})}^{2}} \\ = & - 伪 [1 - 1 卤 e^{- {(\frac{2 蟺 m a}{L})}^{2}}] 尾 e^{- {(\frac{2 蟺 n a}{L})}^{2}} \\ = & - 伪 [卤 e^{- {(\frac{2 蟺 n a}{L})}^{2}}] 尾 \\ = & 鈭� 伪 \end{array}$

We know that

$唯 = 伪唯_{n} + 尾唯_{- n} and 尾 = 鈭� 伪$

Then,

$\begin{array}{rcl} 唯_{1} & = & 伪唯_{n} - 伪唯_{- n} \\ = & 伪 [唯_{n} - 唯_{- n}] \\ = & \frac{伪}{\sqrt{L}} [e^{2 蟺 i n x / L} - e^{- 2 蟺 i n x / L}] 唯_{2} \\ = & 伪唯_{n} + 伪唯_{- n} \\ = & 伪 [唯_{n} + 唯_{- n}] \\ = & \frac{伪}{\sqrt{L}} [e^{2 蟺 i n x / L} + e^{- 2 蟺 i n x / L}] \end{array}$

and since,

$\sin x = \frac{e^{- i x} - e^{i x}}{2 i} and \cos x = \frac{e^{- i x} + e^{i x}}{2}$ then,

$唯_{1} = \frac{- 伪 (2 i)}{\sqrt{L}} \sin (\frac{2 蟺 n x}{L}) 唯_{2} = \frac{伪 (2)}{\sqrt{L}} \cos (\frac{2 蟺 n x}{L})$

Now, we calculate the first-order correction;

$\begin{array}{rcl} E_{1}^{1} & = & <唯_{1}^{0} |{\hat{H}}^{1}| 唯_{1}^{0}> \ hfill \\ = & \frac{- 伪^{2}}{L} (2 i)^{2} (- V_{0}) {鈭�}_{- L / 2}^{L / 2} \sin^{2} (\frac{2 蟺 n x}{L}) e^{- x^{2} / a^{2}} d x \ hfill \\ = & \frac{- 伪^{2}}{L} (2 i)^{2} (- V_{0}) {鈭�}_{- L / 2}^{L / 2} [\frac{1}{2} - \frac{1}{2} \cos (\frac{4 蟺 n x}{L})] e^{- x^{2} / a^{2}} d x \ hfill \\ = & \frac{- 伪^{2}}{L} (2 i)^{2} (- V_{0}) {鈭�}_{- L / 2}^{L / 2} [\frac{1}{2} - \frac{1}{2} (\frac{e^{- 4 蟺 i n x / L} + e^{4 蟺 i n x / L}}{2})] e^{- x^{2} / a^{2}} d x \ hfill \\ = & \frac{- 伪^{2}}{L} (2 i)^{2} (- V_{0}) {鈭�}_{- L / 2}^{L / 2} \frac{1}{2} e^{- x^{2} / a^{2}} - \frac{1}{4} e^{- 4 蟺 i n x / L} e^{- x^{2} / a^{2}} - \frac{1}{4} e^{4 蟺 i n x / L} e^{- x^{2} / a^{2}} d x \ hfill \end{array}$

From the previous point (b), we have

${鈭�}_{- 鈭�}^{鈭�} e^{- x^{2} / a^{2}} d x = a \sqrt{蟺} {鈭�}_{- 鈭�}^{鈭�} e^{- x^{2} / a^{2}} e^{- 4 蟺 i n x / L} d x = a \sqrt{蟺} e^{- {(2 蟺 n a / L)}^{2}} {鈭�}_{- 鈭�}^{鈭�} e^{- x^{2} / a^{2}} e^{- 4 蟺 i n x / L} d x = a \sqrt{蟺} e^{- {(2 蟺 n a / L)}^{2}}$

Therefore,

$\begin{array}{rcl} E_{1}^{1} & = & \frac{- 伪^{2}}{L} (2 i)^{2} (- V_{0}) [\frac{1}{2} a \sqrt{蟺} - \frac{1}{2} a \sqrt{蟺} e^{- {(2 蟺 n a / L)}^{2}}] \\ = & \frac{- 伪^{2}}{2 L} (2 i)^{2} (- V_{0}) a \sqrt{蟺} [1 - e^{- {(2 蟺 n a / L)}^{2}}] \end{array}$

The same for $E_{2}^{1} = <唯_{2} |{\hat{H}}^{1}| 唯_{2}>$

$E_{2}^{1} = \frac{伪^{2}}{L} (2 i)^{2} (- V_{0}) {鈭�}_{- L / 2}^{L / 2} \cos^{2} (\frac{2 n 蟺 x}{L}) e^{- x^{2} / a^{2}} d x$

$\begin{array}{rcl} \cos 2 x & = & 2 \cos^{2} x - 1 \cos^{2} x \\ = & \frac{1}{2} \cos 2 x + \frac{1}{2} \end{array}$

We do the same steps, we get

$E_{2}^{1} = \frac{伪^{2}}{2 L} (2)^{2} (- V_{0}) a \sqrt{蟺} [1 - e^{- {(2 蟺 n a / L)}^{2}}]$

(d) To Find the hermitian operator

The hermitian operator which commutes with $H^{0} and H^{1}$ with the even eigenstate $唯_{n}$ and odd eigenstate $唯_{- n}$ is the parity operator(operator changes the direction of coordinate)

$\begin{array}{rcl} [P, H^{0}] & = & 0 P |唯_{- n}> \\ = & - 1 |唯_{n}> P |唯_{n}> \\ = & + 1 |唯_{n}> \end{array}$

$+ 1$ is the eigenvalue for even states.

Unlock Step-by-Step Solutions & Ace Your Exams!

Full Textbook Solutions
Get detailed explanations and key concepts
Unlimited Al creation
Al flashcards, explanations, exams and more...
Ads-free access
To over 500 millions flashcards
Money-back guarantee
We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

91影视

Short Answer

Step by step solution

Step 1:(a) Showing that the stationary states can be written in the formψn(x)=1Le2πinx/L, (-L/2<x<L/2).

(b) To Find the first order correction

(c) To Find the linear combination

(d) To Find the hermitian operator

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Geometrical and Physical Optics

Thermodynamics

Circular Motion and Gravitation

Electricity and Magnetism

Dynamics

Atoms and Radioactivity

Study anywhere. Anytime. Across all devices.