Q8P A particle starts out in the gro... [FREE SOLUTION]

Chapter 10: Q8P (page 391)

A particle starts out in the ground state of the infinite square well (on the interval 0 鈮� x 鈮� a) .Now a wall is slowly erected, slightly off center:
$V (x) = f (t) 未 (x - \frac{a}{2} - 脪)$
where $f (t)$ rises gradually from $0 to 鈭�$ According to the adiabatic theorem, the particle will remain in the ground state of the evolving Hamiltonian.
(a)Find (and sketch) the ground state at $t 鈫� 鈭�$ Hint: This should be the ground state of the infinite square well with an impenetrable barrier at $a / 2 + 蔚$ . Note that the particle is confined to the (slightly) larger left 鈥渉alf鈥� of the well.
(b) Find the (transcendental) equation for the ground state energy at time t.
Answer:
$z \sin z = T [\cos z - \cos (z 未)] z \sin z, where z 鈮� k a, T 鈮� maf (t) / h^{2}, 未鈮� 2 脪 / a, and k 鈮� \sqrt{2 m E} / h .$ $z \sin z = T [\cos z - \cos (z 未)] z \sin z, where z 鈮� k a, T 鈮� maf (t) / h^{2}, 未鈮� 2 脪 / a, and k = \sqrt{2 m E} / h$
(c) Setting 未 = 0 , solve graphically for z, and show that the smallest z goes from 蟺 to 2蟺 as T goes from 0 to 鈭�. Explain this result.
(d) Now set 未 = 0.01 and solve numerically for z, using
$T = 0, 1, 5, 20, 100, and 1000$
(e) Find the probability $P_{r}$ that the particle is in the right 鈥渉alf鈥� of the well, as a function of z and 未. Answer:
$P_{r} = 1 / [1 + (I_{+} I I_{-})] P r = 1 / [1 + (I_{+} I I ?)], w h e r e I_{+} 鈮� [1 卤未 - (z (1 卤未)) s i n^{2} [z (1 鈭� 未) / 2]$
. Evaluate this expression numerically for the T鈥檚 and 未 in part (d). Comment on your results.
(f) Plot the ground state wave function for those same values of T and 未.
Note how it gets squeezed into the left half of the well, as the barrier grows.

Short Answer

Expert verified

$(a) 蠄 (x) = {\begin{cases} \sqrt{\frac{1}{2} a + 脪} s i n (\frac{\begin{matrix} 蟺虫 \end{matrix}}{\frac{1}{2} a + 脪}), 0 鈮� x 鈮� \frac{1}{2} a + 脪 \\ 0, \frac{1}{2} a + 脪鈮� x 鈮� a \end{cases} .$

(b) $s i n k a = - \frac{T}{z} (c o s 2 k 脪 - \cos k a) 鈬� z s i n z = T [c o s z - c o s (z 未)]$

(c) As $t : 0 鈫� 鈭�, T : 0 鈫� 鈭�$ , and the straight line rotates counterclockwise from 6 o鈥檆lock to 3 o鈥檆lock, so the smallest z goes from 蟺 to 2蟺, and the ground state energy goes from

$k a = 蟺鈬� E (0) = \frac{h^{2} 蟺^{2}}{2 m a^{2}} .$

(d)

1000	100	20	5	1	0	T
6.21452	6.13523	5.72036	4.76031	3.67303	3.14159	z

(e) $P_{r} = \frac{I_{r}}{I_{r} + I_{I}} = \frac{1}{1 + (I_{I} I I_{r})}$

Step by step solution

(a) Finding the ground state at t=∞

Schr枚dinger equation:

$- \frac{h^{2}}{2 m} \frac{d^{2} 蠄}{d x^{2}} = E 蠄, o r \frac{d^{2} 蠄}{d x^{2}} = - k^{2} 蠄 (k 鈮� \sqrt{2 mE} / h) \{\begin{cases} 0 < x < \frac{1}{2} a + 脪 \\ \frac{1}{2} a + 脪 < x < a \end{cases} .$

Boundary conditions: $蠄 (0) = 蠄 (\frac{1}{2} a + 脪) = 蠄 (a) .$

Solution:

$(1) 0 < x < \frac{1}{2} a + 脪 : 蠄 (x) = A \sin k x + B \cos k x . But 蠄 (0) = 0 鈬� B = 0 (1) 0 < x < 2 ., and$

$蠄 (\frac{1}{2} a + 脪) = 0 鈬� \{\begin{cases} k (\frac{1}{2} a + 脪) = n 蟺 (n = 1, 2, 3, . .) 鈬� E_{n} = n^{2} 蟺^{2} h^{2} / 2 m (a / 2 + 脪) \\ or else A = 0 \end{cases}$ ,

localid="1656131500084" $(2) \frac{1}{2} a + 脪 < x < a : 蠄 (x) = Fsin k (a - x) + G \cos k (a - x) . But 蠄 (a) = 0 鈬� G = 0, and 蠄 (\frac{1}{2} a + 脪) = 0 鈬� \{\begin{cases} k (\frac{1}{2} a - 脪) = n' 蟺 (n' = 1, 2, 3, . . .) 鈬� E_{n'} = {(n')}^{2} 蟺^{2} h^{2} / 2 m {(a / 2 - 脪)}^{2} \\ or else F = 0 \end{cases}$

The ground state energy is $\{\begin{cases} either E_{1} = \frac{蟺^{2} h^{2}}{2 m (\frac{1}{2} a + 脪)} (n' = 1), with F = 0 \\ or else E_{1} = \frac{蟺^{2} h^{2}}{2 m (\frac{1}{2} a - 脪)} (n' = 1), with A = 0 \end{cases}$

Both are allowed energies, but $E_{1}$ is (slightly) lower (assuming 蔚 is positive), so the ground state is

$蠄 (x) = \{\begin{cases} \sqrt{\frac{1}{2} a + 脪} \sin (\frac{\begin{matrix} 蟺虫 \end{matrix}}{\frac{1}{2} a + 脪}), 0 鈮� x 鈮� \frac{1}{2} a + 脪 \\ 0, \frac{1}{2} a + 脪鈮� x 鈮� a \end{cases}$

(b)Finding the equation for the ground state energy at time t.

$- \frac{h^{2}}{2 m} \frac{d^{2} 蠄}{d x^{2}} + f (t) 未 (x - \frac{1}{2} a - 脪) 蠄 = E 蠄鈬� 蠄 (x) = \{\begin{matrix} A \sin kx, 0 鈮� x < \frac{1}{2} a + 脪 \\ F \sin k (a - x), \frac{1}{2} a + 脪 < x 鈮� a, \end{matrix}\} where k = \frac{\sqrt{2 mE}}{h}$

Continuity in $蠄 at x = \frac{1}{2} a + 脪 :$

$A \sin k (\frac{1}{2} a + 脪) = F \sin (a - \frac{1}{2} a + 脪) = F \sin k (\frac{1}{2} a + 脪) 鈬� F = A \frac{\sin k (\frac{1}{2} a + 脪)}{\sin k (\frac{1}{2} a + 脪)} .$

Discontinuity in $蠄' at x = - \frac{2 尘伪}{h^{2}} 蠄 (0)$ (Eq. 2.128):

$- F k \cos k (a - x) - A k \cos k x = \frac{2 m f}{h^{2}} A \sin k x 鈬� F \cos k (\frac{1}{2} a - 脪) + A \cos k (\frac{1}{2} a + 脪) = - (\frac{2 m f}{h^{2} k}) A \sin k (\frac{1}{2} a - 脪) . A \frac{\sin k (\frac{1}{2} a + 脪)}{\sin k (\frac{1}{2} a + 脪)} \cos k (\frac{1}{2} a + 脪) + A \cos k (\frac{1}{2} a + 脪) = - (\frac{2 T}{z}) A \sin k (\frac{1}{2} a + 脪) . \sin k (\frac{1}{2} a + 脪) \cos k (\frac{1}{2} a - 脪) + \cos k (\frac{1}{2} a + 脪) \sin k (\frac{1}{2} a - 脪) = - (\frac{2 T}{z}) \sin k (\frac{1}{2} a + 脪) \sin k (\frac{1}{2} a - 脪) \sin k (\frac{1}{2} a + 脪 + \frac{1}{2} a - 脪) = - (\frac{2 T}{z}) \frac{1}{2} [\cos k (\frac{1}{2} a + 脪 - \frac{1}{2} a + 脪) - \cos k (\frac{1}{2} a + 脪 + \frac{1}{2} a - 脪)] . \sin k a = - \frac{T}{z} (\cos 2 k 脪 - \cos k a) 鈬� z \sin z = T [\cos z - \cos (z 未)] .$

(c)showing that the smallest z goes from π to 2π as T goes from 0 to ∞.

$\sin z = \frac{T}{z} (\cos z - 1) 鈬� \frac{z}{T} = \frac{\cos z - 1}{\sin z} = - \tan (z / 2) = - \frac{z}{T}$

$Plot \tan (z / 2) and - z / 7$ on the same graph, and look for intersections:

As $t : 0 鈫� 鈭�, T : 0 鈫� 鈭�$ , and the straight line rotates counterclockwise from 6 o鈥檆lock to 3 o鈥檆lock, so the smallest z goes from 蟺 to 2蟺, and the ground state energy goes from

$k a = 蟺鈬� E (0) = \frac{h^{2} 蟺^{2}}{2 m a^{2}}$ (appropriate to a well of width a) to $k a = 2 蟺鈬� E (鈭�) = \frac{h^{2} 蟺^{2}}{2 m {(a / 2)}^{2}} .$

(appropriate for a well of width a/2.)

(d)solving numerically for z.

Mathematical yields the following table:

1000	100	20	5	1	0	T
6.21452	6.13523	5.72036	4.76031	3.67303	3.14159	z

(e)Find the probability Pr

$P_{r} = \frac{I_{r}}{I_{r} + I_{I}} = \frac{1}{1 + (I_{I} I I_{r})}, where I_{I} = {鈭�}_{0}^{a / 2 + 脪} A^{2} \sin^{2} k x d x = A^{2} {[\frac{1}{2} x - \frac{1}{4 k} \sin (2 k x)]}_{0}^{a / 2 + 脪} = A^{2} \{\frac{1}{2} (\frac{a}{2} + 脪) - \frac{1}{4 k} s i n [2 k (\frac{a}{2} + 脪)]\} = \frac{a}{4} A^{2} [1 + \frac{2 脪}{a} - \frac{1}{k a} s i n (k a + \frac{2 脪}{a} k a)] = \frac{a}{4} A^{2} [1 + 未 - \frac{1}{z} s i n (z + z 未)]$

$I_{r} = {鈭�}_{a / 2 + 脪}^{a} F^{2} \sin^{2} k (a - x) d x . L e t u = a - x, d u = - d x = - F^{2} {鈭�}_{a / 2 - 脪}^{a} F^{2} \sin^{2} k u d u = F^{2} {鈭�}_{a}^{a / 2 - 脪} F^{2} \sin^{2} k u d u = \frac{a}{4} F^{2} [1 - 未 - \frac{1}{z} \sin (z - z 未)] . \frac{I_{I}}{I_{r}} = \frac{A^{2} [1 + 未 - (1 / z) \sin (z + z 未)]}{F^{2} [1 - 未 - (1 / z) \sin (z - z 未)]} . But (from (b)) \frac{A^{2}}{F^{2}} = \frac{\sin^{2} k (a / 2 - 脪)}{\sin^{2} k (a / 2 + 脪)} = \frac{\sin^{2} [z (1 - 未) / 2]}{\sin^{2} [z (1 + 未) / 2]} . = \frac{I_{+}}{I_{-}} = I ? I +, w h e r e I_{+} = [1 卤未 - \frac{1}{2} \sin (1 卤未)] \sin^{2} [z (1 鈭� 未) / 2], P_{r} = \frac{1}{1 + (I_{+} I I_{-})}$

Using 未 = 0.01 and the z鈥檚 from (d), Mathematical gives

1000	100	20	5	1	0	T
0.00248443	0.146529	0.401313	0.471116	0.486822	0.490001

As $t : 0 鈫� 鈭� (so T : 0 鈫� 鈭�)$ the probability of being in the right half drops from almost 1/2 to zero-the particle gets sucked out of the slightly smaller side, as it heads for the ground state in (a).