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Is the given wave function a stationary state?
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[QUOTE="Marioweee, post: 6469202, member: 688512"] [B]Homework Statement:[/B] psb [B]Relevant Equations:[/B] $$H\Psi=E\Psi$$ A particle of mass m that is under the effect of a one-dimensional potential V (x) is described by the wave function: \begin{array}{c} xe^{-bx}e^{-ict/\hbar }, x \geq 0\\ 0 , x \leq 0\end{array} where $$b\geq 0,c\in R$$ and the wave function is normalized. [LIST=1] [*]Is it a stationary state? What can you say about energy? [*]It is possible to find stationary states with lower energy? [*]Find V(x) [/LIST] ----------- My solution: First of all, I am new to quantum mechanics so i may have elementary errors. (Also I am not a native english speaker so i may have some grammar errors too). As far as i know, the wave function of a stationary state is: $$\Psi(x,t)=f(x)e^{-iEt/\hbar}$$ with E being the energy of the state. In the problem the wave function given has the same form with c being E (i think here is my problem). Then I have use the time independent Schröringer equation to obteein the value of E: $$H\Psi=(-\dfrac{\hbar^2}{2m}\dfrac{\partial^2}{\partial x^2}+V(x))xe^{-bx}e^{-ict/\hbar }=e^{-ict/\hbar }(\dfrac{\hbar^2}{2m}be^{-bx}(2-bx)+V(x)xe^{-bx})=xe^{-bx}e^{-ict/\hbar }(\dfrac{\hbar^2}{m}\dfrac{b}{x}-\dfrac{\hbar^2}{2m}b^2+V(x))=Exe^{-bx}e^{-ict/\hbar }$$ so $$E=(\dfrac{\hbar^2}{m}\dfrac{b}{x}-\dfrac{\hbar^2}{2m}b^2+V(x))$$ And if c=E, the form of the wave function would be: $$\Psi(x,t)=f(x)e^{-i(\dfrac{\hbar^2}{m}\dfrac{b}{x}-\dfrac{\hbar^2}{2m}b^2+V(x))t/\hbar }$$ And now it has not the form of a stationary state because of the dependence on x and t. So, would it be a stationary state? As I said before i think the problem is that c is not equal E necessarily. I dindt try nº 2 and 3 yet because I want to understand stationary states first. Thanks for the help :smile: [/QUOTE]
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Is the given wave function a stationary state?
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