## Related questions with answers

A particle of mass m has the wave function $\Psi(x, t)=A e^{-a\left[\left(m x^{2} / \hbar\right)+i t\right]}$, where A and a are positive real constants. (a) Find A. (b) For what potential energy function, V(x), is this a solution to the Schrödinger equation? (c) Calculate the expectation values of x, $x^{2}$, p, and $p^{2}$. (d) Find $\sigma_{x}$ and $\sigma_{p}$. Is their product consistent with the uncertainty principle?

Solution

Verified$\begin{align*} \psi (x,t) = A e^{-a\left( \frac{mx^{2}}{\hbar} + it\right)}\\ \psi^{*} (x,t) = A e^{-a\left( \frac{mx^{2}}{\hbar} - it\right)} \end{align*}$

Disclamer: If $A$ was a complex number, then in $\psi^{*} (x,t)$ we would have $A^{*}$.

To find $A$ we need to $\textbf{normalize}$ the wave function $\psi (x,t)$:

$\begin{align*} 1 &= \int_{-\infty}^{\infty} \left| \psi \left( x,t \right) \right| ^{2}dx \\ &= \int_{-\infty}^{\infty} \psi^{*} \left( x,t \right)\psi \left( x,t \right) dx \\ & = \int_{-\infty}^{\infty} A e^{-a\left( \frac{mx^{2}}{\hbar} - it\right)} A e^{-a\left( \frac{mx^{2}}{\hbar} + it\right)} dx \\ & = A^{2} \int_{-\infty}^{\infty} e^{\frac{-2amx^{2}}{\hbar} } dx \end{align*}$

Integrals in form $\int_{-\infty}^{\infty} e^{-kx^{2}} dx$ for $k>0$ have solution $\sqrt{\frac{\pi}{k}}$.

In comparison to previous formula $k=\frac{2am}{\hbar}$. It follows:

$\begin{align*} A^{2}\sqrt{\dfrac{\pi \hbar}{2am}}=1\\ A^2=\sqrt{\dfrac{2am}{\pi \hbar}} \\ \boxed{A=\left( \dfrac{2am}{\pi \hbar} \right)^{\frac{1}{4}}} \end{align*}$

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