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.

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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

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