(Given: $\int_{-\infty}^{+\infty} x^{2n} e^{-x^2} dx = \frac{1 \cdot 3 \cdot 5 \dots (2n-1)}{2^n} \sqrt{\pi}$)
To find the normalization constant $N$, we use the normalization condition for a wavefunction $\psi(x)$: $ \int_{-\infty}^{+\infty} |\psi(x)|^2 dx = 1 $ Here, the wavefunction is $\psi(x) = N(2x^2 -1)e^{-x^2/2}$.
Wavefunction Squared Calculation
First, calculate the square of the wavefunction's magnitude:
$ |\psi(x)|^2 = \left| N(2x^2 -1)e^{-x^2/2} \right|^2 $ $ |\psi(x)|^2 = N^2 (2x^2 -1)^2 (e^{-x^2/2})^2 $ $ |\psi(x)|^2 = N^2 (4x^4 - 4x^2 + 1) e^{-x^2} $Normalization Integral Setup
Now, substitute this into the normalization condition:
$ \int_{-\infty}^{+\infty} N^2 (4x^4 - 4x^2 + 1) e^{-x^2} dx = 1 $Factor out $N^2$ and split the integral:
$ N^2 \left( 4\int_{-\infty}^{+\infty} x^4 e^{-x^2} dx - 4\int_{-\infty}^{+\infty} x^2 e^{-x^2} dx + \int_{-\infty}^{+\infty} e^{-x^2} dx \right) = 1 $Integral Evaluation
We use the given integral formula: $\int_{-\infty}^{+\infty} x^{2n} e^{-x^2} dx = \frac{1 \cdot 3 \cdot 5 \dots (2n-1)}{2^n} \sqrt{\pi}$.
- For $n=0$: $\int_{-\infty}^{+\infty} e^{-x^2} dx = \frac{1}{2^0} \sqrt{\pi} = \sqrt{\pi}$.
- For $n=1$: $\int_{-\infty}^{+\infty} x^2 e^{-x^2} dx = \frac{1}{2^1} \sqrt{\pi} = \frac{\sqrt{\pi}}{2}$.
- For $n=2$: $\int_{-\infty}^{+\infty} x^4 e^{-x^2} dx = \frac{1 \cdot 3}{2^2} \sqrt{\pi} = \frac{3\sqrt{\pi}}{4}$.
Substitute these values back into the equation:
$ N^2 \left( 4 \left(\frac{3\sqrt{\pi}}{4}\right) - 4 \left(\frac{\sqrt{\pi}}{2}\right) + \sqrt{\pi} \right) = 1 $ $ N^2 (3\sqrt{\pi} - 2\sqrt{\pi} + \sqrt{\pi}) = 1 $ $ N^2 (2\sqrt{\pi}) = 1 $Solving for Normalization Constant N
Solve for $N^2$:
$ N^2 = \frac{1}{2\sqrt{\pi}} $Take the square root to find $N$. We typically choose the positive root for normalization constants:
$ N = \sqrt{\frac{1}{2\sqrt{\pi}}} = \left(\frac{1}{2\sqrt{\pi}}\right)^{\frac{1}{2}} $This matches Option C.
Important Questions from Harmonic Oscillator
- The correct option for the average value of kinetic energy and the average value of potential energy of a one-dimensional harmonic oscillator with frequency $\nu$ in its ground state is
The lowest energy of a quantum mechanical one-dimensional simple harmonic oscillator is $300$ cm$^{-1}$. The energy (in cm$^{-1}$) of the next higher level is _________________
- The wave function for a Harmonic oscillator described by $Nxexp(-ax^2/2)$ has
