3. Gaussian wavefunction

(a) Normalize the wavefunction.
(b) Make sure $V(x)$ satisfies the Schrodinger equation.
(c) Expectation value of an operator $O$ is defined as

$$ \langle O\rangle = \int \psi(x)^* O \psi(x) dx. $$

Momentum operator $p = -i\hbar d/dx$ is a derivative operator in the position space.

(d) Uncertainty is defined as the standard deviation corresponding to an operator. For example,

$$ \sigma_O = \sqrt{\langle O^2 \rangle - \langle O \rangle^2}. $$

(a)
Let's normalize the wavefunction

$$ \int |\psi(x, t)|^2 dx = \int C^2 e^{-2a(mx^2/\hbar)}dx $$ $$ = C^2 \sqrt{\frac{\pi\hbar}{2am}} = 1 $$

Therefore,

$$ C = (\frac{2am}{\pi\hbar})^{1/4}. $$

(b)
Since the potential energy $V(x)$ is time independent, we can separate position and time dependence $\psi(x, t) = \psi(x)\psi(t)$. Therefore,

$$ i\hbar\frac{d}{dt}\psi(x, t) = E\psi(x, t) $$

The energy of the particle is $E = \hbar a$. Using the time-independent Schrodinger equation we have

$$ (-i\hbar\frac{d}{dx})^2 \psi + V(x)\psi = E \psi. $$ $$ -(-2amx)^2 \psi + V(x)\psi = E\psi $$

Finally, $$ V(x) = \hbar a + (2am)^2x^2. $$

(c)
The expectation value of position operator is 0 since the wavefunction is an even function of $x$. The expectation value of $x^2$ can be calculated using integration by parts:

$$ \langle x^2 \rangle = \int x^2 |\psi(x)|^2 dx $$ $$ = \frac{\hbar}{4am} x |\psi(x)|^2 \bigg|_{-\infty}^{\infty} - \int \frac{\hbar}{4am} |\psi(x)|^2 dx $$ $$ = \frac{\hbar}{4am}, $$

where the boundary term is zero.

Same goes with the momentum operator. The expectation value of momentum is zero since the kernel of the following integral is an odd function of position.

$$ \langle p\rangle = \int \psi(x)^*(-i\hbar\frac{d}{dx})\psi(x)dx = 0. $$

On the other hand we have

$$ \langle p^2\rangle = \int \psi(x)^*(-i\hbar\frac{d}{dx})^2\psi(x)dx $$ $$ = \int (2am)^2 x^2 |\psi(x)|^2 dx = (2am)^2 \langle x^2\rangle $$

(d)
Uncertainties in position and momentum are

$$ \sigma_x = \sqrt{\langle x^2 \rangle - \langle x\rangle^2} = \sqrt{\frac{\hbar}{4am}}, $$ $$ \sigma_p = \sqrt{\langle p^2 \rangle - \langle p\rangle^2} = \sqrt{(2am)^2 \frac{\hbar}{4am}}, $$

Let's multiply the uncertainties

$$ \sigma_x \sigma_p = (2am) \frac{\hbar}{4am} = \frac{\hbar}{2} $$

This satisfies the Heisenberg uncertainty principle. Note that the Gaussian wavefunction has the minimum uncertainty that is $\hbar/2$.