Answer :
Explanation:
To find the normalization constant \( A \) for the wave function \( \Psi(x) = Ax e^{-x^2/2} \), we need to ensure that the total probability of finding the particle over all space is equal to 1. Mathematically, this is expressed as:
\[ \int_{-\infty}^{\infty} |\Psi(x)|^2 \, dx = 1 \]
First, let's square the wave function:
\[ |\Psi(x)|^2 = |Ax e^{-x^2/2}|^2 = |A|^2 x^2 e^{-x^2} \]
Now, we integrate this squared wave function over the entire real line:
\[ \int_{-\infty}^{\infty} |A|^2 x^2 e^{-x^2} \, dx = 1 \]
We can simplify this integral using the fact that the integral of the Gaussian function \( e^{-x^2} \) over the entire real line is \( \sqrt{\pi} \). So, we have:
\[ |A|^2 \int_{-\infty}^{\infty} x^2 e^{-x^2} \, dx = 1 \]
The integral \( \int_{-\infty}^{\infty} x^2 e^{-x^2} \, dx \) is a standard integral that evaluates to \( \frac{\sqrt{\pi}}{2} \). Substituting this back into our equation:
\[ |A|^2 \cdot \frac{\sqrt{\pi}}{2} = 1 \]
Now, solve for \( A \):
\[ |A|^2 = \frac{2}{\sqrt{\pi}} \]
\[ |A| = \sqrt{\frac{2}{\sqrt{\pi}}} \]
\[ A = \pm \sqrt[4]{\frac{2}{\pi}} \]
Therefore, the normalization constant \( A \) for the given wave function is \( A = \pm \sqrt[4]{\frac{2}{\pi}} \). Usually, the positive sign is chosen for the normalization constant, making \( A = \sqrt[4]{\frac{2}{\pi}} \).
Answer:
To find the value of the normalization constant
A for the wave function
(
)
=
−
2
2
ψ(x)=Axe
−
2
x
2
, we need to ensure that the wave function is properly normalized. This means that the integral of the absolute square of the wave function over all space must equal 1:
∫
−
∞
∞
∣
(
)
∣
2
=
1
∫
−∞
∞
∣ψ(x)∣
2
dx=1
Given the wave function
(
)
=
−
2
2
ψ(x)=Axe
−
2
x
2
, we first find
∣
(
)
∣
2
∣ψ(x)∣
2
:
∣
(
)
∣
2
=
(
−
2
2
)
2
=
2
2
−
2
∣ψ(x)∣
2
=(Axe
−
2
x
2
)
2
=A
2
x
2
e
−x
2
Next, we set up the normalization condition:
∫
−
∞
∞
2
2
−
2
=
1
∫
−∞
∞
A
2
x
2
e
−x
2
dx=1
We can factor out
2
A
2
since it is a constant:
2
∫
−
∞
∞
2
−
2
=
1
A
2
∫
−∞
∞
x
2
e
−x
2
dx=1
Now, we need to evaluate the integral:
∫
−
∞
∞
2
−
2
∫
−∞
∞
x
2
e
−x
2
dx
This is a standard integral in mathematical physics and can be evaluated using known results. The result of this integral is:
∫
−
∞
∞
2
−
2
=
2
∫
−∞
∞
x
2
e
−x
2
dx=
2
π
Substituting this result back into the normalization condition:
2
⋅
2
=
1
A
2
⋅
2
π
=1
Solving for
2
A
2
:
2
=
2
A
2
=
π
2
Taking the square root of both sides, we find
A:
=
2
=
(
2
)
1
/
2
A=
π
2
=(
π
2
)
1/2
Therefore, the normalization constant
A is:
=
(
2
)
1
/
2
=
(
2
1
/
2
)
1
/
2
=
(
2
1
/
2
)
1
/
2
A=(
π
2
)
1/2
=(
π
1/2
2
)
1/2
=(
π
1/2
2
)
1/2
So,
=
(
2
1
/
2
)
1
/
2
=
(
2
1
/
2
)
1
/
2
=
(
2
)
1
/
2
A=(
π
1/2
2
)
1/2
=(
π
1/2
2
)
1/2
=(
π
2
)
1/2
Hence, the value of the normalization constant
A is
(
2
)
1
/
2
(
π
2
)
1/2
.
m
Explanation:
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