Consider a particle whose wave function is given by ψ(x) = Ae^(-ax) (A) What is the value of A if this wave function is normalized? (B) What is the expectation value of x for this particle? Let’s apply the normalization condition: ∫|ψ(x)|^2 dx = ∫|Ae^(-ax)|^2 dx = ∫A^2e^(-2ax) dx ∫|ψ(x)|^2 dx = A^2 ∫e^(-2ax) dx ∫|ψ(x)|^2 dx = A^2 * (-1/2a) * e^(-2ax) + C To normalize the wave function, the integral of |ψ(x)|^2 over all space must equal 1. Therefore: A^2 * (-1/2a) * e^(-2ax) + C = 1 Solving for A: A^2 * (-1/2a) * e^(-2ax) = 1 – C A^2 = (1 – C) * (-2a) * e^(2ax) A = √(1 – C) * (-2a) * e^(2ax) To find the expectation value of x, we use the formula: ⟨x⟩ = ∫x|ψ(x)|^2 dx ⟨x⟩ = ∫x * A^2e^(-2ax) dx ⟨x⟩ = A^2 * ∫xe^(-2ax) dx ⟨x⟩ = A^2 * (-1/2a) * xe^(-2ax) + A^2 * (1/4a^2) * e^(-2ax) + D ⟨x⟩ = A^2 * (-1/2a) * xe^(-2ax) + A^2 * (1/4a^2) * e^(-2ax) + D
The Correct Answer and Explanation is:
Let’s go through the problem step by step.
Part A: Normalizing the Wave Function
The wave function is given by:ψ(x)=Ae−ax\psi(x) = A e^{-ax}ψ(x)=Ae−ax
To normalize the wave function, we need to apply the normalization condition, which requires that the integral of ∣ψ(x)∣2|\psi(x)|^2∣ψ(x)∣2 over all space equals 1. This ensures that the total probability is 1.
The integral of ∣ψ(x)∣2|\psi(x)|^2∣ψ(x)∣2 is:∫−∞∞∣ψ(x)∣2dx=1\int_{-\infty}^{\infty} |\psi(x)|^2 dx = 1∫−∞∞∣ψ(x)∣2dx=1
First, let’s compute ∣ψ(x)∣2|\psi(x)|^2∣ψ(x)∣2:∣ψ(x)∣2=∣Ae−ax∣2=A2e−2ax|\psi(x)|^2 = |A e^{-ax}|^2 = A^2 e^{-2ax}∣ψ(x)∣2=∣Ae−ax∣2=A2e−2ax
Now, we integrate:∫0∞A2e−2axdx=1\int_{0}^{\infty} A^2 e^{-2ax} dx = 1∫0∞A2e−2axdx=1
The integral of e−2axe^{-2ax}e−2ax is:∫e−2axdx=−12ae−2ax\int e^{-2ax} dx = \frac{-1}{2a} e^{-2ax}∫e−2axdx=2a−1e−2ax
Now substitute this into the integral:A2∫0∞e−2axdx=A2[−12ae−2ax]0∞A^2 \int_0^\infty e^{-2ax} dx = A^2 \left[\frac{-1}{2a} e^{-2ax}\right]_{0}^{\infty}A2∫0∞e−2axdx=A2[2a−1e−2ax]0∞
At x=∞x = \inftyx=∞, e−2axe^{-2ax}e−2ax goes to zero, and at x=0x = 0x=0, e−2ax=1e^{-2ax} = 1e−2ax=1. So, the integral becomes:A2(0−(−12a))=A212aA^2 \left( 0 – \left( \frac{-1}{2a} \right) \right) = A^2 \frac{1}{2a}A2(0−(2a−1))=A22a1
For normalization, this must equal 1:A212a=1A^2 \frac{1}{2a} = 1A22a1=1
Solving for AAA:A2=2aA^2 = 2aA2=2aA=2aA = \sqrt{2a}A=2a
So, the normalization constant is:A=2aA = \sqrt{2a}A=2a
Part B: Expectation Value of xxx
The expectation value of xxx, denoted ⟨x⟩\langle x \rangle⟨x⟩, is given by the formula:⟨x⟩=∫0∞x∣ψ(x)∣2dx\langle x \rangle = \int_{0}^{\infty} x |\psi(x)|^2 dx⟨x⟩=∫0∞x∣ψ(x)∣2dx
Substitute ∣ψ(x)∣2=A2e−2ax|\psi(x)|^2 = A^2 e^{-2ax}∣ψ(x)∣2=A2e−2ax:⟨x⟩=∫0∞xA2e−2axdx\langle x \rangle = \int_{0}^{\infty} x A^2 e^{-2ax} dx⟨x⟩=∫0∞xA2e−2axdx
Now, we need to integrate xe−2axx e^{-2ax}xe−2ax. We can solve this using integration by parts. Let:
- u=xu = xu=x, so du=dxdu = dxdu=dx
- dv=e−2axdxdv = e^{-2ax} dxdv=e−2axdx, so v=−12ae−2axv = \frac{-1}{2a} e^{-2ax}v=2a−1e−2ax
Using integration by parts:∫xe−2axdx=−x2ae−2ax+∫12ae−2axdx\int x e^{-2ax} dx = \frac{-x}{2a} e^{-2ax} + \int \frac{1}{2a} e^{-2ax} dx∫xe−2axdx=2a−xe−2ax+∫2a1e−2axdx
The second integral is:∫e−2axdx=−12ae−2ax\int e^{-2ax} dx = \frac{-1}{2a} e^{-2ax}∫e−2axdx=2a−1e−2ax
So, putting everything together:∫xe−2axdx=−x2ae−2ax+14a2e−2ax\int x e^{-2ax} dx = \frac{-x}{2a} e^{-2ax} + \frac{1}{4a^2} e^{-2ax}∫xe−2axdx=2a−xe−2ax+4a21e−2ax
Evaluating this at x=0x = 0x=0 and x=∞x = \inftyx=∞, we find:[−x2ae−2ax+14a2e−2ax]0∞\left[\frac{-x}{2a} e^{-2ax} + \frac{1}{4a^2} e^{-2ax}\right]_0^\infty[2a−xe−2ax+4a21e−2ax]0∞
At x=∞x = \inftyx=∞, the terms go to 0, and at x=0x = 0x=0, they also vanish, except for the second term:14a2\frac{1}{4a^2}4a21
Thus, the expectation value becomes:⟨x⟩=A2⋅14a2\langle x \rangle = A^2 \cdot \frac{1}{4a^2}⟨x⟩=A2⋅4a21
Now, substitute A2=2aA^2 = 2aA2=2a into this:⟨x⟩=(2a)⋅14a2\langle x \rangle = (2a) \cdot \frac{1}{4a^2}⟨x⟩=(2a)⋅4a21
Simplifying:⟨x⟩=12a\langle x \rangle = \frac{1}{2a}⟨x⟩=2a1
So, the expectation value of xxx is:⟨x⟩=12a\langle x \rangle = \frac{1}{2a}⟨x⟩=2a1
Final Answers:
- The normalization constant A=2aA = \sqrt{2a}A=2a.
- The expectation value of xxx is ⟨x⟩=12a\langle x \rangle = \frac{1}{2a}⟨x⟩=2a1.
