Consider an electron trapped in a box of side L, in one dimension. After making a measurement of the particle position, what is the probability of finding it within L/4 of either side of the box, if the particle is initially in:
a) the ground stationary state?
b) the first excited stationary state?
To determine the probability of finding the electron within a certain range in the box, we need to calculate the wave function for each state and then integrate the absolute square of the wave function over the desired range.
a) For the ground stationary state, the wave function can be written as:
ψ(x) = √(2/L) * sin(πx/L)
To find the probability of finding the electron within L/4 of either side of the box, we need to integrate the absolute square of the wave function from -L/4 to L/4. The probability density function (PDF) is given by |ψ(x)|^2.
P(a) = ∫ |ψ(x)|^2 dx from -L/4 to L/4
Replacing the wave function, we get:
P(a) = ∫ [√(2/L) * sin(πx/L)]^2 dx from -L/4 to L/4
P(a) = ∫ [2/L * sin^2(πx/L)] dx from -L/4 to L/4
This integral simplifies to:
P(a) = 1/2
Hence, the probability of finding the electron within L/4 of either side of the box in the ground stationary state is 1/2.
b) For the first excited stationary state, the wave function can be written as:
ψ(x) = √(2/L) * sin(2πx/L)
Using the same steps as in part a, we can calculate the probability:
P(b) = ∫ |ψ(x)|^2 dx from -L/4 to L/4
P(b) = ∫ [√(2/L) * sin(2πx/L)]^2 dx from -L/4 to L/4
P(b) = ∫ [2/L * sin^2(2πx/L)] dx from -L/4 to L/4
This integral simplifies to:
P(b) = 1/2
Thus, the probability of finding the electron within L/4 of either side of the box in the first excited stationary state is also 1/2.