The ground state wavefunction for a particle in a shallow potential well in 1 dimension is of the form: Ae−|x|2d. Given that the particle must be found somewhere in the range x∈[−∞,∞], the born rule then places a constraint on the modulus of A. Assuming that A is real and positive, what is the value of A?

To find the value of A, we need to normalize the wavefunction. Normalization ensures that the probability of finding the particle in the range x∈[−∞,∞] is equal to 1.

The normalization condition for a wavefunction ψ(x) is given by:
∫ |ψ(x)|^2 dx = 1, where the integral is taken over the entire range.

In this case, the wavefunction is ψ(x) = A * e^(-|x|/2d).

Let's first calculate the squared modulus of the wavefunction, |ψ(x)|^2:
|ψ(x)|^2 = (A * e^(-|x|/2d))^2
= A^2 * e^(-|x|/d)

Now, we need to evaluate the integral of |ψ(x)|^2 over the entire range:
∫ |ψ(x)|^2 dx = ∫ A^2 * e^(-|x|/d) dx

Since the potential well extends over the entire range, we can write the integral as:
∫ A^2 * e^(-|x|/d) dx = 2 ∫ A^2 * e^(-x/d) dx (integration from 0 to ∞ and then multiply by 2)

Integrating A^2 and e^(-x/d), we get:
2 ∫ A^2 * e^(-x/d) dx = 2A^2 * d * e^(-x/d) evaluated from 0 to ∞
= 2A^2 * d * (e^(-∞/d) - e^(0/d))

Since e^(-∞/d) approaches 0, the term with e^(-∞/d) becomes negligible.

2A^2 * d * (e^(-∞/d) - e^(0/d)) ≈ 2A^2 * d * (-e^(0/d))

Now, we need to set this equal to 1, as per the normalization condition:
2A^2 * d * (-e^(0/d)) = 1

Simplifying further, we have:
-2A^2 * d = 1

To solve for A, divide both sides by -2d:
A^2 = 1 / (-2d)

Since A is real and positive, we can take the square root of the right-hand side:
A = +√(1 / (-2d))

Therefore, the value of A is +√(1 / (-2d)).