QM of a particle in a 1D infinite square well

A particle in the infinite square well potential

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

has stationary states in position basis for $𝑛 \in ℕ$

𝜓 𝑛 (𝑥) = ⎧ { { {⎨ { { {⎩ \sqrt 1 𝑎 s i n 𝑛 𝜋 𝑥 2 𝑎 𝑛 e v e n, 𝑥 \in [- 𝑎, 𝑎] \sqrt 1 𝑎 c o s 𝑛 𝜋 𝑥 2 𝑎 𝑛 o d d, 𝑥 \in [- 𝑎, 𝑎] 0 o t h e r w i s e

with energies

𝐸 𝑛 = 𝑛 2 𝜋 2 ℏ 2 8 𝑚 𝑎

General solutions to the full Schrödinger equation therefore have the form $\sum 𝑐 𝑛 𝜓 𝑛 (𝑥) 𝑒 - 𝑖 𝐸 𝑛 𝑡 / ℏ$ .

Proof

Inside $[- 𝑎, 𝑎]$ the TISE reads
$- ℏ 2 2 𝑚 𝜕 2 𝜕 𝑥 2 𝜓 = 𝐸 𝜓$
or
$𝜕 2 𝜕 𝑥 2 𝜓 = - 𝑘 2 𝜓, 𝑘 = \sqrt 2 𝑚 𝐸 ℏ$
noting that $𝐸 < 0$ states are forbidden (which would come up in the solutions anyway). The general solution is then
$𝜓 (𝑥) = ˜ 𝐴 s i n 𝑘 𝑥 + ˜ 𝐵 c o s 𝑘 𝑥$
we take boundary conditions $𝜓 (- 𝑎) = 𝜓 (𝑎) = 0$ . Now
$𝜓 (- 𝑎) = 𝐴 s i n (- 𝑘 𝑎) + 𝐵 c o s (- 𝑘 𝑎) = - 𝐴 s i n 𝑘 𝑎 + 𝐵 c o s 𝑘 𝑎$
giving solutions giving solutions for $𝑘 𝑛 = 𝑛 𝜋 2 𝑎$ with $𝐴 = 0$ for odd $𝑛$ and $𝐵 = 0$ for even $𝑛$ . The $𝑛 = 0$ solution is not normalizable and hence is rejected as unphysical, and negative $𝑛$ gives a rescaling of a positive $𝑛$ solution. Thus the energies are
$𝐸 𝑛 = ℏ 2 𝑘 2 𝑛 2 𝑚 = 𝑛 2 𝜋 2 ℏ 2 8 𝑚 𝑎$
Normalisation for odd $𝑛$ gives
$1 = ⟨ 𝜓 𝑛 | 𝜓 𝑛 ⟩ = | 𝐵 | 2 \int 𝑎 - 𝑎 c o s 2 𝑛 𝜋 𝑥 2 𝑎 𝑑 𝑥 = | 𝐵 | 2 \int 𝑎 0 (1 + c o s 𝑛 𝜋 𝑥 𝑎) 𝑑 𝑥 = | 𝐵 | 2 [𝑥 + 𝑎 𝜋 𝑛 s i n 𝑛 𝜋 𝑥 𝑎] 𝑥 = 𝑎 𝑥 = 0 = | 𝐵 | 2 𝑎$
Likewise for even $𝑛$ we have
$1 = ⟨ 𝜓 𝑛 | 𝜓 𝑛 ⟩ = | 𝐴 | 2 \int 𝑎 - 𝑎 s i n 2 𝑛 𝜋 𝑥 2 𝑎 𝑑 𝑥 = | 𝐴 | 2 \int 𝑎 0 (1 - c o s 𝑛 𝜋 𝑥 𝑎) 𝑑 𝑥 = | 𝐴 | 2 [𝑥 - 𝑎 𝜋 𝑛 s i n 𝑛 𝜋 𝑥 𝑎] 𝑥 = 𝑎 𝑥 = 0 = | 𝐴 | 2 𝑎$
So $𝐴 = 𝐵 = \sqrt 1 / 𝑎$ .

tidy | en | SemBr

QM of a particle in a 1D infinite square well

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