Chapter 10: Q8P (page 391)

A particle starts out in the ground state of the infinite square well (on the interval 0 ≤ x ≤ a) .Now a wall is slowly erected, slightly off center:
$V (x) = f (t) δ (x - \frac{a}{2} - Ò)$
where $f (t)$ rises gradually from $0 to \infty$ 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 $t \to \infty$ Hint: This should be the ground state of the infinite square well with an impenetrable barrier at $a / 2 + ε$ . Note that the particle is confined to the (slightly) larger left “half” of the well.
(b) Find the (transcendental) equation for the ground state energy at time t.
Answer:
$z \sin z = T [\cos z - \cos (z δ)] z \sin z, where z \equiv k a, T \equiv maf (t) / h^{2}, δ \equiv 2 Ò / a, and k \equiv \sqrt{2 m E} / h .$ $z \sin z = T [\cos z - \cos (z δ)] z \sin z, where z \equiv k a, T \equiv maf (t) / h^{2}, δ \equiv 2 Ò / a, and k = \sqrt{2 m E} / h$
(c) Setting δ = 0 , solve graphically for z, and show that the smallest z goes from π to 2π as T goes from 0 to ∞. Explain this result.
(d) Now set δ = 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 “half” of the well, as a function of z and δ. Answer:
$P_{r} = 1 / [1 + (I_{+} I I_{-})] P r = 1 / [1 + (I_{+} I I ?)], w h e r e I_{+} \equiv [1 \pm δ - (z (1 \pm δ)) s i n^{2} [z (1 \mp δ) / 2]$
. Evaluate this expression numerically for the T’s and δ in part (d). Comment on your results.
(f) Plot the ground state wave function for those same values of T and δ.
Note how it gets squeezed into the left half of the well, as the barrier grows.

Short Answer

Expert verified

$(a) ψ (x) = {\begin{cases} \sqrt{\frac{1}{2} a + Ò} s i n (\frac{\begin{matrix} πx \end{matrix}}{\frac{1}{2} a + Ò}), 0 \leq x \leq \frac{1}{2} a + Ò \\ 0, \frac{1}{2} a + Ò \leq x \leq a \end{cases} .$

(b) $s i n k a = - \frac{T}{z} (c o s 2 k Ò - \cos k a) \Rightarrow z s i n z = T [c o s z - c o s (z δ)]$

(c) As $t : 0 \to \infty, T : 0 \to \infty$ , and the straight line rotates counterclockwise from 6 o’clock to 3 o’clock, so the smallest z goes from π to 2π, and the ground state energy goes from

$k a = π \Rightarrow E (0) = \frac{h^{2} π^{2}}{2 m a^{2}} .$

(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=∞

Schrödinger equation:

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

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

Solution:

$(1) 0 < x < \frac{1}{2} a + Ò : ψ (x) = A \sin k x + B \cos k x . But ψ (0) = 0 \Rightarrow B = 0 (1) 0 < x < 2 ., and$

$ψ (\frac{1}{2} a + Ò) = 0 \Rightarrow \{\begin{cases} k (\frac{1}{2} a + Ò) = n π (n = 1, 2, 3, . .) \Rightarrow E_{n} = n^{2} π^{2} h^{2} / 2 m (a / 2 + Ò) \\ or else A = 0 \end{cases}$ ,

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

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

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

$ψ (x) = \{\begin{cases} \sqrt{\frac{1}{2} a + Ò} \sin (\frac{\begin{matrix} πx \end{matrix}}{\frac{1}{2} a + Ò}), 0 \leq x \leq \frac{1}{2} a + Ò \\ 0, \frac{1}{2} a + Ò \leq x \leq a \end{cases}$

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

$- \frac{h^{2}}{2 m} \frac{d^{2} ψ}{d x^{2}} + f (t) δ (x - \frac{1}{2} a - Ò) ψ = E ψ \Rightarrow ψ (x) = \{\begin{matrix} A \sin kx, 0 \leq x < \frac{1}{2} a + Ò \\ F \sin k (a - x), \frac{1}{2} a + Ò < x \leq a, \end{matrix}\} where k = \frac{\sqrt{2 mE}}{h}$

Continuity in $ψ at x = \frac{1}{2} a + Ò :$

$A \sin k (\frac{1}{2} a + Ò) = F \sin (a - \frac{1}{2} a + Ò) = F \sin k (\frac{1}{2} a + Ò) \Rightarrow F = A \frac{\sin k (\frac{1}{2} a + Ò)}{\sin k (\frac{1}{2} a + Ò)} .$

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

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

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

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

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

As $t : 0 \to \infty, T : 0 \to \infty$ , and the straight line rotates counterclockwise from 6 o’clock to 3 o’clock, so the smallest z goes from π to 2π, and the ground state energy goes from

$k a = π \Rightarrow E (0) = \frac{h^{2} π^{2}}{2 m a^{2}}$ (appropriate to a well of width a) to $k a = 2 π \Rightarrow E (\infty) = \frac{h^{2} π^{2}}{2 m {(a / 2)}^{2}} .$

(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

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

$I_{r} = \int_{a / 2 + Ò}^{a} F^{2} \sin^{2} k (a - x) d x . L e t u = a - x, d u = - d x = - F^{2} \int_{a / 2 - Ò}^{a} F^{2} \sin^{2} k u d u = F^{2} \int_{a}^{a / 2 - Ò} F^{2} \sin^{2} k u d u = \frac{a}{4} F^{2} [1 - δ - \frac{1}{z} \sin (z - z δ)] . \frac{I_{I}}{I_{r}} = \frac{A^{2} [1 + δ - (1 / z) \sin (z + z δ)]}{F^{2} [1 - δ - (1 / z) \sin (z - z δ)]} . But (from (b)) \frac{A^{2}}{F^{2}} = \frac{\sin^{2} k (a / 2 - Ò)}{\sin^{2} k (a / 2 + Ò)} = \frac{\sin^{2} [z (1 - δ) / 2]}{\sin^{2} [z (1 + δ) / 2]} . = \frac{I_{+}}{I_{-}} = I ? I +, w h e r e I_{+} = [1 \pm δ - \frac{1}{2} \sin (1 \pm δ)] \sin^{2} [z (1 \mp δ) / 2], P_{r} = \frac{1}{1 + (I_{+} I I_{-})}$

Using δ = 0.01 and the z’s from (d), Mathematical gives

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

As $t : 0 \to \infty (so T : 0 \to \infty)$ 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).