QM of particle in a 1D finite square well

A particle in the finite square well potential

𝑉 (𝑥) = {- 𝑉 0 𝑥 \in [- 𝑎, 𝑎] 0 e l s e w h e r e

where

𝜅 = \sqrt - 2 𝑚 𝐸 ℏ = \sqrt 𝑧 20 + 𝑧 2 𝑎 𝑙 = \sqrt - 2 𝑚 (𝐸 + 𝑉 0) ℏ = 𝑧 𝑎

has odd bound states

𝜓 (𝑥) = ⎧ { {⎨ { {⎩ 𝐴 𝑒 𝜅 𝑥 𝑥 < - 𝑎 𝐹 s i n 𝑙 𝑥 𝑥 \in [- 𝑎, 𝑎] - 𝐴 𝑒 - 𝜅 𝑥 𝑧 = 𝑥 > 𝑎

where

𝐴 = - 𝐹 𝑒 𝜅 𝑎 s i n 𝑙 𝑎 c o t 𝑧 = \sqrt 𝑧 20 𝑧 2 - 1

and even bound states

𝜓 (𝑥) = ⎧ { {⎨ { {⎩ 𝐴 𝑒 𝜅 𝑥 𝑥 < - 𝑎 𝐹 c o s 𝑙 𝑥 𝑥 \in [- 𝑎, 𝑎] 𝐴 𝑒 - 𝜅 𝑥 𝑧 = 𝑥 > 𝑎

where

𝐴 = 𝐹 𝑒 𝜅 𝑎 c o s 𝑙 𝑎 t a n 𝑧 = \sqrt 𝑧 20 𝑧 2 - 1

Proof

Bound states correspond to $- 𝑉 0 \leq 𝐸 < 0$ . It follows that $𝜓 (- 𝑥) = \pm 𝜓 (𝑥)$ . For $𝑥 \in (- \infty, 𝑎)$ the Schrödinger equation becomes

𝐸 𝜓 (𝑥) = - ℏ 2 2 𝑚 𝑑 2 𝑑 𝑥 2 𝜓 (𝑥) 𝑑 2 𝑑 𝑥 2 𝜓 (𝑥) = - 2 𝑚 𝐸 ℏ 2 𝜓 (𝑥) = 𝜅 2 𝜓 (𝑥)

where $𝜅 = \sqrt - 2 𝑚 𝐸 / ℏ$ . Thus $𝜓 (𝑥) = 𝐴 𝑒 - 𝜅 𝑥 + 𝐵 𝑒 𝜅 𝑥$ for $𝑥 \in (- \infty, 𝑎)$ , and applying $l i m 𝑥 \to - \infty 𝜓 (𝑥) = 0$ we conclude $𝐴 = 0$ . For $𝑥 \in [- 𝑎, 𝑎]$ the Schrödinger equation is

𝐸 𝜓 (𝑥) = - ℏ 2 2 𝑚 𝑑 2 𝑑 𝑥 2 𝜓 (𝑥) - 𝑉 0 𝜓 (𝑥) 𝑑 2 𝑑 𝑥 2 𝜓 (𝑥) = - 2 𝑚 (𝐸 + 𝑉 0) ℏ 2 𝜓 (𝑥) = - 𝑙 2 𝜓 (𝑥)

where $𝑙 = \sqrt - 2 𝑚 (𝐸 + 𝑉 0) ℏ$ . Thus $𝜓 (𝑥) = 𝐹 s i n 𝑙 𝑥 + 𝐺 c o s 𝑙 𝑥$ for $𝑥 \in [- 𝑎, 𝑎]$ . For odd solutions, $𝜓 (- 𝑥) = - 𝜓 (𝑥)$ , hence

𝜓 (𝑥) = ⎧ { {⎨ { {⎩ 𝐴 𝑒 𝜅 𝑥 𝑥 < - 𝑎 𝐹 s i n 𝑙 𝑥 𝑥 \in [- 𝑎, 𝑎] - 𝐴 𝑒 - 𝜅 𝑥 𝑧 = 𝑥 > 𝑎 𝜓 (- 𝑥) = ⎧ { {⎨ { {⎩ 𝐴 𝜅 𝑒 𝜅 𝑥 𝑥 < - 𝑎 𝐹 𝑙 c o s 𝑙 𝑥 𝑥 \in [- 𝑎, 𝑎] 𝐴 𝜅 𝑒 𝜅 𝑥 𝑥 > 𝑎

thus, by continuity we have $𝐴 𝑒 - 𝜅 𝑎 = - 𝐹 s i n 𝑙 𝑎$ and by smoothness we have $𝐴 𝜅 𝑒 - 𝜅 𝑎 = 𝐹 𝑙 c o s 𝑙 𝑎$ . Thus $𝜅 = - 𝑙 c o t 𝑙 𝑎$ Let $𝑧 = 𝑙 𝑎$ and $𝑧 0 = 𝑎 ℏ \sqrt 2 𝑚 𝑉 0$ . Since $𝜅 2 + 𝑙 2 = 2 𝑚 𝑉 0 ℏ 2$ , it follows $𝜅 𝑎 = \sqrt 𝑧 20 - 𝑧 2$ , hence

c o t 𝑧 = - 𝜅 𝑎 𝑙 𝑎 = \sqrt 𝑧 20 - 𝑧 2 𝑧 = \sqrt 𝑧 20 𝑧 2 - 1

which may be solved numerically. A similar treatment for the even case¹ gives the result stated above.

#state/tidy | #lang/en | #SemBr

2018. Introduction to quantum mechanics, §2.6, p. 72 ↩

QM of particle in a 1D finite square well

Footnotes