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?
Why did the particle go to therapy? Because it had some negative potential! But in all seriousness, let's calculate the value of A.
According to the Born rule, the probability density for finding a particle in a specific location is given by the squared modulus of the wavefunction, |Ψ(x)|^2. So, we'll square our wavefunction Ae^(−|x|/2d):
|Ψ(x)|^2 = (Ae^(−|x|/2d))^2
= A^2e^(-|x|/d)
Now, since the particle must be found somewhere in the entire range x∈[−∞,∞], the total probability of finding the particle must be equal to 1. In other words, we need to integrate the probability density over the entire range and set it equal to 1:
∫[−∞,∞] |Ψ(x)|^2 dx = 1
∫[−∞,∞] A^2e^(-|x|/d) dx = 1
Since the integrand is an even function, we can simplify the integral as follows:
2∫[0,∞] A^2e^(-x/d) dx = 1
To solve this integral, we can use basic integration techniques:
2A^2 ∫[0,∞] e^(-x/d) dx = 1
2A^2 (-d) e^(-x/d) |[0,∞] = 1
2A^2 (-d) (0 - e^(-∞/d)) = 1
2A^2 (-d) (-e^(-∞/d)) = 1
2A^2d e^(-∞/d) = 1
Now, e^(-∞/d) is essentially zero, since the exponential function rapidly approaches zero as its argument approaches negative infinity. Therefore:
2A^2d e^(-∞/d) ≈ 0
0 = 1
Uh oh! Looks like we've hit an error here. The wavefunction does not satisfy the normalization condition, which means something must be wrong. It seems we made a mistake somewhere, and unfortunately, I can't provide you with a specific value for A in this case. My apologies for any confusion caused. Let's hope the quantum mechanics police don't come after me for this!
The Born rule states that the probability density of finding a particle in a specific state is given by the square of the absolute value of its wavefunction. Therefore, the probability of finding the particle somewhere in the range x∈[−∞,∞] should be equal to 1.
To find the value of A, we need to normalize the wavefunction such that the integral of the probability density over all space is equal to 1.
The probability density is given by |Ψ(x)|^2 = |Ae^(-|x|/2d)|^2 = A^2e^(-|x|/d).
To normalize the wavefunction, we need to compute the integral of the squared wavefunction over all space:
∫{|Ae^(-|x|/2d)|^2} dx = ∫ A^2e^(-|x|/d) dx = A^2∫e^(-|x|/d) dx.
To simplify the integral, we can split it into two parts based on the different sign of |x|:
∫e^(-|x|/d) dx = ∫e^(-x/d) dx + ∫e^(x/d) dx.
Integrating both parts separately, we obtain:
∫e^(-x/d) dx = -de^(-x/d) + C1,
∫e^(x/d) dx = de^(x/d) + C2.
where C1 and C2 are integration constants.
Combining the results, we have:
A^2(-de^(-x/d) + de^(x/d) + C1 + C2).
In order to find the value of A, we set the integral equal to 1 (since the probability of finding the particle in the range x∈[−∞,∞] should be equal to 1):
A^2(-de^(-x/d) + de^(x/d) + C1 + C2) = 1.
Since the integral should be equal to 1 for all values of x, the terms involving exponents should cancel out. Therefore, the constant terms C1 and C2 must be equal to zero.
This simplifies the equation to:
A^2(-de^(-x/d) + de^(x/d)) = 1.
Now, let's evaluate the exponential terms:
-e^(-x/d) + e^(x/d) = 1 / (Ad).
Next, we multiply both sides of the equation by Ad:
-Ad e^(-x/d) + Ad e^(x/d) = 1.
Since Ad is a positive quantity, we can see that the left side of the equation represents the difference of two exponential functions. In order for this difference to equal 1, the two exponential functions must be equal to each other:
Ad e^(-x/d) = Ad e^(x/d).
We can cancel out the common factor of Ad:
e^(-x/d) = e^(x/d).
Taking the natural logarithm of both sides of the equation:
(-x/d) = (x/d).
Multiplying both sides by d:
-x = x.
Simplifying the equation, we find that x = 0.
Since x = 0, the exponential terms in the integral vanish, leaving:
-Ad + Ad = 1.
This reduces to:
0 = 1.
Since this equation is not satisfied, it implies that there is no solution for A that would make the integral equal to 1.
Therefore, the given wavefunction cannot be normalized, and there is no valid value of A that satisfies the constraint of the Born rule for this particular wavefunction.
To find the value of A, we need to use the normalization condition for the wavefunction, which states that the integral of the probability density over all space must be equal to 1.
The probability density is given by |Ψ(x)|^2, where Ψ(x) is the wavefunction.
In this case, the wavefunction is Ae^(-|x|/2d). So, the probability density is |Ae^(-|x|/2d)|^2 = |A|^2e^(-|x|/d).
To calculate the integral of the probability density over all space, we integrate |A|^2e^(-|x|/d) from -∞ to ∞. This integral can be split into two integrals, one from -∞ to 0 and the other from 0 to ∞.
∫[-∞,∞] |A|^2e^(-|x|/d) dx = ∫[-∞,0] |A|^2e^(x/d) dx + ∫[0,∞] |A|^2e^(-x/d) dx
Since the integrals are symmetric, the two parts are equal, so we can write:
2∫[0,∞] |A|^2e^(-x/d) dx = 1
Let's evaluate this integral:
2∫[0,∞] |A|^2e^(-x/d) dx = [-dA^2e^(-x/d)] [0,∞]
= -dA^2lim(x→∞) e^(-x/d) + dA^2e^(0/d)
Since the exponential function decays to zero as x approaches infinity, the first term in the above expression is zero.
So we are left with:
2dA^2e^(0/d) = 1
e^(0/d) = 1
Therefore, the value of A that satisfies the normalization condition is:
2dA^2 = 1
A^2 = 1/(2d)
Taking the square root of both sides, we find:
A = √(1/(2d))
Since A is real and positive, the value of A is given by:
A = √(1/(2d))