Is the given wave function a stationary state?

[Marioweee]

Homework Statement

psb

Relevant Equations

$$H\Psi=E\Psi$$

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.

1. Is it a stationary state? What can you say about energy?
2. It is possible to find stationary states with lower energy?
3. Find V(x)

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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

 

[Marioweee]

Sorry is my fisrt post and i post it in the wrong thread

 

[hilbert2]

1. You have already noted that $E=c$, by comparing the phase factor $e^{-ict/\hbar}$ with $e^{-iEt/\hbar}$. No more solution is needed for the first question. Just keep in mind that the "c" here can't mean the speed of light, because that wouldn't be dimensionally correct.

2. The lowest energy bound state of a 1D quantum system has 0 nodes in its wave function. What's the number of nodes in $\psi (x) = xe^{-bx}$ if the boundary condition node $\psi (0) = 0$ isn't counted?

3. If the time independent Schrödinger equation is

$\displaystyle -\frac{\hbar^2}{2m}\frac{d^2 \psi (x)}{dx^2} + V(x)\psi (x) = E\psi (x)$,

then what's the $V(x)$ when solved as a function of $\psi (x)$ and $E$? The solution should contain a term proportional to $\psi'' (x)/\psi (x)$, with the double prime denoting the 2nd derivative. This is one of the so called "quantum inverse problems". Also note that the $V(x)$ has to be $\infty$ when $x\leq 0$, because otherwise the wave function wouldn't be zero for all negative $x$ values.

 

[Marioweee]

Thanks so much for the help.
Regarding question 2, the number of nodes would be zero since:
$$e^{-bx} \neq 0$$
And we would have only de boundary condicition node which means that the state is already the one with the lowest energy.
Again, thank you very much, everything is clear now.