Related questions with answers

Consider the wave function $\Psi(x, t)=A e^{-\lambda|x|} e^{-i \omega t}$ , where A, $\lambda$ , and $\omega$ are positive real constants. (a) Normalize $\Psi$ . (b) Determine the expectation values of x and $x^{2}$ . (c) Find the standard deviation of x. Sketch the graph of $|\Psi|^{2}$ , as a function of x, and mark the points $(\langle x\rangle+\sigma)$ and $(\langle x\rangle-\sigma)$ , to illustrate the sense in which $\sigma$ represents the “spread” in x. What is the probability that the particle would be found outside this range?

The wave function of a quantum particle of mass m is

\psi(x)=A \cos (k x)+B \sin (k x)

where A, B, and k are constants. Assuming the particle is free (U = 0), show that $\psi(x)$ is a solution of the Schrodinger equation.

Which one of the following functions, and why, qualifies to be a wave function of a particle that can move along the entire real axis? (a) $\psi(x) = A ~ e^{-x^2}$ (b) $\psi(x) = A ~ e^{-x}$ (c) $\psi(x) = A ~ \tan{(x)}$ (d) $\psi(x) = A ~ \dfrac{\sin{(x)}}{x}$ (e) $\psi(x) = A ~ e^{-\left| x \right|}$

Question

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?

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\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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