Classic Quantum Well

Classical Mechanics Level 3

A particle in a one-dimensional quantum well is governed by a variant of the time-independent Schrodinger equation, expressed in terms of wave function Ψ ( x ) \Psi (x) .

The quantities V V and E E are the potential energy and total energy, respectively.

− d 2 d x 2 Ψ ( x ) + V ( x ) Ψ ( x ) = E Ψ ( x ) -\frac{d^2}{d x^2} \, \Psi (x) + V(x) \, \Psi (x) = E \, \Psi (x)

The potential varies as follows:

V ( x ) = ∞ x < 0 V ( x ) = 0 0 ≤ x ≤ 2 V ( x ) = ∞ x > 2 V (x) = \infty \,\,\,\,\,\,\,\, x < 0 \\ V (x) = 0 \,\,\,\,\,\,\,\, 0 \leq x \leq 2 \\ V (x) = \infty \,\,\,\,\,\,\,\, x > 2

The boundary conditions on Ψ ( x ) \Psi (x) are:

Ψ ( x ) = 0 x ≤ 0 Ψ ( x ) = 0 x ≥ 2 \Psi (x) = 0 \,\,\,\,\,\,\,\, x \leq 0 \\ \Psi (x) = 0 \,\,\,\,\,\,\,\, x \geq 2

Let E 1 E_1 and E 2 E_2 be the lowest and second lowest allowable values of E E , subject to the constraint E > 0 E > 0 . What is the sum of E 1 E_1 and E 2 E_2 ?

Note: This problem is easy to solve analytically


The answer is 12.34.

Steven Chase by Steven Chase , 37, USA

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1 solution

Karan Chatrath
Jun 21, 2020

We are asked to solve:

d 2 Ψ ( x ) d x 2 + E Ψ ( x ) = 0 \frac{d^2 \Psi(x)}{dx^2}+E\Psi(x) = 0 Ψ ( 0 ) = Ψ ( 2 ) = 0 \Psi(0)=\Psi(2)=0

Solution after applying the condition that x ( 0 ) = 0 x(0)=0 is:

Ψ ( x ) = A sin ⁡ ( x E ) \Psi(x) = A \sin\left(x\sqrt{E}\right)

The constant A A can be found by solving the following equation which normalises the square of the wave function to ensure that the probablitity of finding the particle anywhere between x = 0 x=0 and x = 2 x=2 is unity:

∫ 0 2 A 2 sin ⁡ 2 ( x E ) d x = 1 \int_{0}^{2} A^2 \sin^2\left(x\sqrt{E}\right) \ dx = 1

Applying the second boundary condition:

Ψ ( 2 ) = 0 = A sin ⁡ ( 2 E ) \Psi(2) = 0 = A \sin\left(2\sqrt{E}\right)

∴ n π = 2 E \therefore n\pi = 2\sqrt{E}

Where n = 1 , 2 , 3 … n = 1,2,3 \dots . So the lowest allowable energy value occurs at n = 1 n=1 and the second lowest occurs at n = 2 n=2 . This gives:

E 1 = π 2 4 ; E 2 = π 2 E_1 = \frac{\pi^2}{4} \ ; \ E_2 = \pi^2 ⟹ E 1 + E 2 = 5 π 2 4 \implies \boxed{ E_1 + E_2 = \frac{5 \pi^2}{4}}

4 Helpful 4 Interesting 2 Brilliant 0 Confused

@Karan Chatrath sir I am not able to understand the meaning of standard deviation in your recent problem does it mean that λ \lambda can vary from 90 90 to 110 110 ??

A Former Brilliant Member - 11 months, 3 weeks ago

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λ o \lambda_o can take any value in the interval ( − ∞ , ∞ ) (-\infty,\infty) . I suggest you read about random variables and probability distributions.

Karan Chatrath - 11 months, 3 weeks ago

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