Quantum Well - Ramp Potential

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

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

d 2 d x 2 Ψ N ( x ) + V ( x ) Ψ N ( x ) = E Ψ N ( x ) \large{-\frac{d^2}{d x^2} \, \Psi_N (x) + V(x) \, \Psi_N (x) = E \, \Psi_N (x)}

Let us consider a case in which E = 2.5 E = 2.5 . The potential varies as follows:

V ( x ) = x < 0 V ( x ) = x 0 x x f V ( x ) = x > x f x f 3.0267 (a zero crossing of Ψ N ( x ) ) V (x) = \infty \,\,\,\,\,\,\,\, x < 0 \\ V (x) = x \,\,\,\,\,\,\,\, 0 \leq x \leq x_f \\ V (x) = \infty \,\,\,\,\,\,\,\, x > x_f \\ x_f \approx 3.0267 \, \text{(a zero crossing of } \Psi_N (x))

The values of Ψ N ( x ) \Psi_N (x) and its first spatial derivative at x = 0 x = 0 are (see graphic):

Ψ N ( 0 ) = 0 ( d d x Ψ N ) ( 0 ) = 1 \Psi_N (0) = 0 \\ \Big ( \frac{d}{dx} \, \Psi_N \Big ) (0) = 1

Note that the potential energy V ( x ) V(x) exceeds E E for x > 2.5 x > 2.5 , and yet the particle still has a finite non-zero probability of being measured anywhere inside the well.

The probability of detecting the particle within the region a x b a \leq x \leq b is:

P ( a x b ) = a b Ψ N ( x ) 2 d x 0 x f Ψ N ( x ) 2 d x \large{P(a \leq x \leq b) = \frac{\int_a^b |\Psi_N (x)|^2 \, dx }{\int_0^{x_f} |\Psi_N (x)|^2 \, dx }}

What is the probability of detecting the particle in the region ( 0 x 1 ) ( 0 \leq x \leq 1) ? If the probability is P P , give your answer as 100 P \lfloor 100 P \rfloor .

Note: This problem lends itself well to numerical solution


The answer is 29.

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