# Is the given wave function a stationary state?

• Marioweee
In summary, the wave function of a stationary state is:$$\Psi(x,t)=f(x)e^{-iEt/\hbar}$$ with E being the energy of the 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

Delta2
Sorry is my fisrt post and i post it in the wrong thread

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.

Last edited:
vanhees71 and 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.

vanhees71 and hilbert2

## 1. What is a stationary state in quantum mechanics?

A stationary state in quantum mechanics is a state in which a system's wave function does not change over time. This means that the probability of finding the system in a particular state remains constant. In other words, the system is in a stable, unchanging state.

## 2. How can I determine if a given wave function is a stationary state?

To determine if a given wave function is a stationary state, you need to solve the time-independent Schrödinger equation for the system. If the resulting wave function does not depend on time, then it is a stationary state.

## 3. What is the significance of a system being in a stationary state?

A system being in a stationary state is significant because it allows us to make predictions about the system's behavior. Since the wave function is not changing, we know that the system will remain in the same state and the probabilities of finding it in different states will not change over time.

## 4. Can a system be in a superposition of stationary states?

Yes, a system can be in a superposition of stationary states. This means that the wave function of the system is a combination of two or more stationary states, and the system can exist in multiple states simultaneously.

## 5. What happens if a system is not in a stationary state?

If a system is not in a stationary state, then its wave function will change over time. This means that the probabilities of finding the system in different states will also change over time, and the system's behavior will be unpredictable.

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