A particle of mass m has the wave function
Ψ (x,t) = A e^{-a[(mx^2/\hbar )+it]} ,
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 σ_x and σ_p . Is their product consistent with the uncertainty principle?
(a)
1=2|A|^{2} \int_{0}^{\infty} e^{-2 a m x^{2} / \hbar} d x=2|A|^{2} \frac{1}{2} \sqrt{\frac{\pi}{(2 a m / \hbar)}}=|A|^{2} \sqrt{\frac{\pi \hbar}{2 a m}} ; A = \biggl(\frac{2 a m}{\pi \hbar }\biggr)^{1/4} .
(b)
\frac{\partial \Psi}{\partial t}=-i a \Psi ; \quad \frac{\partial \Psi}{\partial x}=-\frac{2 a m x}{\hbar} \Psi ; \quad \frac{\partial^{2} \Psi}{\partial x^{2}}=-\frac{2 a m}{\hbar}\left(\Psi+x \frac{\partial \Psi}{\partial x}\right)=-\frac{2 a m}{\hbar}\left(1-\frac{2 a m x^{2}}{\hbar}\right) \Psi .
Plug these into the Schrödinger equation, i \hbar \frac{\partial \Psi}{\partial t}=-\frac{\hbar^{2}}{2 m} \frac{\partial^{2} \Psi}{\partial x^{2}}+V \Psi :
V \Psi=i \hbar(-i a) \Psi+\frac{\hbar^{2}}{2 m}\left(-\frac{2 a m}{\hbar}\right)\left(1-\frac{2 a m x^{2}}{\hbar}\right) \Psi= \left[\hbar a-\hbar a\left(1-\frac{2 a m x^{2}}{\hbar}\right)\right] \Psi=2 a^{2} m x^{2} \Psi , so V (x) = 2ma^2x^2 .
(c)
\langle x\rangle =\int_{-\infty}^{\infty} x|\Psi|^{2} d x = 0. [Odd integrand.]
\left\langle x^{2}\right\rangle=2|A|^{2} \int_{0}^{\infty} x^{2} e^{-2 a m x^{2} / \hbar} d x=2|A|^{2} \frac{1}{2^{2}(2 a m / \hbar)} \sqrt{\frac{\pi \hbar}{2 a m}} = \frac{\hbar}{4 a m } .
\langle p\rangle = m \frac{d\langle x\rangle }{dt} = 0 .
\left\langle p^{2}\right\rangle=\int \Psi^{*}\left(\frac{\hbar}{i} \frac{\partial}{\partial x}\right)^{2} \Psi d x=-\hbar^{2} \int \Psi^{*} \frac{\partial^{2} \Psi}{\partial x^{2}} d x .
=-\hbar^{2} \int \Psi^{*}\left[-\frac{2 a m}{\hbar}\left(1-\frac{2 a m x^{2}}{\hbar}\right) \Psi\right] d x=2 a m \hbar\left\{\int|\Psi|^{2} d x-\frac{2 a m}{\hbar} \int x^{2}|\Psi|^{2} d x\right\} .
=2 a m \hbar\left(1-\frac{2 a m \mid}{\hbar}\left\langle x^{2}\right\rangle\right)=2 a m \hbar\left(1-\frac{2 a m}{\hbar} \frac{\hbar}{4 a m}\right)=2 a m \hbar\left(\frac{1}{2}\right)= a m \hbar .
(d)
σ^2_x = \langle x^2\rangle – \langle x\rangle^2 = \frac{\hbar}{a m \hbar} ⇒ σ_x = \sqrt{\frac{\hbar }{4 a m} } ;
σ^2_p = \langle p^2\rangle – \langle p\rangle^2 = a m \hbar ⇒ σ_p = \sqrt{a m \hbar} .
σ_x σ_p = \sqrt{\frac{\hbar }{4 a m} } \sqrt{a m \hbar} = \frac{\hbar}{2} . This is (just barely) consistent with the uncertainty principle.