Probability a particle is in a certain region

That leaves two things to question:- does the derivative need to be continuous? I note that in the infinite well we have zero outside the well and ##\sin(nx\pi/L) ## inside. That does not have a continuous derivative at the boundaries.Infinite potential barriers are a bit weird; from the Schroedinger equation it's possible to show that solutions have continuous derivatives so long as ##V(x)## does not have any infinite discontinuities. So whilst the discontinuity in ##\varphi'## at ##x=0## can be tolerated, we still need to impose continuity in ##\varphi'## everywhere else.And I'm now seeing that this is (essentially) where I
  • #1
etotheipi
Homework Statement
Given the particle is in its ground state with ##E <0##, calculate ##\mathbb{P}(x > \xi)## for the following potential:\begin{align*} V(x) =
\begin{cases}



\infty & x < 0 \\

- \lambda & 0\leq x\leq \xi \\

0 & x > \xi

\end{cases}

\end{align*}where ##\lambda > 0##.
Relevant Equations
N/A
I think I made an error somewhere. In ##[0,a]## I let ##\varphi(x) = \varphi_1(x) := p\sin{kx}## whilst in ##(\xi, \infty)## I let ##\varphi(x) = \varphi_2 (x) := re^{-\gamma x}##, and the constraints at ##x=\xi## are \begin{align*}

\varphi_1'(\xi) = \varphi_2'(\xi) &\implies pk\cos{k\xi} = -\gamma re^{-\gamma \xi} \\
\varphi_1(\xi) = \varphi_2(\xi) &\implies p\sin{k\xi} = re^{-\gamma \xi}

\end{align*}which implies ##\gamma = -k/\tan{k\xi}##. Then we can normalise,\begin{align*}

\int_0^\xi dx \, \varphi_1^2(x) + \int_\xi^{\infty} dx \, \varphi_2^2(x) &= \frac{(2k \xi - \sin{2 k \xi})p^2}{4k} + \frac{r^2 e^{-2\gamma \xi}}{2\gamma} \overset{!}{=} 1

\end{align*}and, since ##p^2 \sin^2{k \xi} = r^2 e^{-2\gamma \xi}## it follows that\begin{align*}

\mathbb{P}(x > \xi) = \frac{r^2 e^{-2\gamma \xi}}{2\gamma} = \frac{2\sin^2{k \xi} \tan{k \xi}}{2\sin^2{k \xi} \tan{k \xi} + \sin{2k \xi} - 2k \xi}\end{align*}That the particle is in its ground state implies ##k \xi = \pi / 2##, such that the sinusoidal part joins nicely onto the decaying exponential solution in the ##x > \xi## region. But that gives a non-sense answer; so I wonder where I made my mistake(s). Thanks!
 
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  • #2
etotheipi said:
Homework Statement:: Given the particle is in its ground state with ##E <0##, calculate ##\mathbb{P}(x > \xi)## for the following potential:\begin{align*} V(x) =
\begin{cases}
\infty & x < 0 \\

- \lambda & 0\leq x\leq \xi \\

0 & x > \xi

\end{cases}

\end{align*}where ##\lambda > 0##.
Relevant Equations:: N/A

I think I made an error somewhere. In ##[0,a]## I let ##\varphi(x) = \varphi_1(x) := p\sin{kx}## whilst in ##(\xi, \infty)## I let ##\varphi(x) = \varphi_2 (x) := re^{-\gamma x}##, and the constraints at ##x=\xi## are \begin{align*}

\varphi_1'(\xi) = \varphi_2'(\xi) &\implies pk\cos{k\xi} = -\gamma re^{-\gamma \xi} \\
\varphi_1(\xi) = \varphi_2(\xi) &\implies p\sin{k\xi} = re^{-\gamma \xi}

\end{align*}which implies ##\gamma = -k/\tan{k\xi}##. Then we can normalise,\begin{align*}

\int_0^\xi dx \, \varphi_1^2(x) + \int_\xi^{\infty} dx \, \varphi_2^2(x) &= \frac{(2k \xi - \sin{2 k \xi})p^2}{4k} + \frac{r^2 e^{-2\gamma \xi}}{2\gamma} \overset{!}{=} 1

\end{align*}and, since ##p^2 \sin^2{k \xi} = r^2 e^{-2\gamma \xi}## it follows that\begin{align*}

\mathbb{P}(x > \xi) = \frac{r^2 e^{-2\gamma \xi}}{2\gamma} = \frac{2\sin^2{k \xi} \tan{k \xi}}{2\sin^2{k \xi} \tan{k \xi} + \sin{2k \xi} - 2k \xi}\end{align*}That the particle is in its ground state implies ##k \xi = \pi / 2##, such that the sinusoidal part joins nicely onto the decaying exponential solution in the ##x > \xi## region. But that gives a non-sense answer; so I wonder where I made my mistake(s). Thanks!
##k \xi = \pi / 2## implies ##\varphi_1'(\xi) =0##, no? Which is a bit of a problem for ##\varphi_2'(\xi)## with the chosen exponential form.
Not knowing anything much about QM, I do not know how you are able to assume those forms for the two regions.
 
  • #3
haruspex said:
Not knowing anything much about QM, I do not know how you are able to assume those forms for the two regions.
They're the solutions to the Schrödinger equation that satisfy the boundary conditions at ##x=0## and ##x \to \infty##.
 
  • #4
vela said:
They're the solutions to the Schrödinger equation that satisfy the boundary conditions at ##x=0## and ##x \to \infty##.
Ok, I've done a little reading..
The sign of E-V(x) is critical, yes? Where positive, we should get a trig solution and where negative an exponential one.
But V is given as ##-\lambda## in the central range, and we are told E<0 so it should be exponential solutions for all x>0?
 
  • #5
haruspex said:
##k \xi = \pi / 2## implies ##\varphi_1'(\xi) =0##, no? Which is a bit of a problem for ##\varphi_2'(\xi)## with the chosen exponential form.
Yeah, that's the non-sense bit that I can't seem to figure out 😄. That seems to be the false assumption (i.e. I believe we are instead only allowed to write ##k\xi > \pi/2##), but the question remains to then find the actual value of ##k \xi## in the ground state...

haruspex said:
Ok, I've done a little reading..
The sign of E-V(x) is critical, yes? Where positive, we should get a trig solution and where negative an exponential one.
But V is given as ##-\lambda## in the central range, and we are told E<0 so it should be exponential solutions for all x>0?
How I read it was that ##- \lambda < E < 0##, so ##E > - \lambda## means you have sinusoidal solutions in the central region and ##E < 0## means you have a decaying exponential in the ##x > \xi## region. The question doesn't explicitly state that, but it's the only way the situation seems to make sense in the first place (as far as I can tell... ☺️)
 
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  • #6
etotheipi said:
How I read it was that ##- \lambda < E < 0##, so ##E > - \lambda## means you have sinusoidal solutions in the central region and ##E < 0## means you have a decaying exponential in the ##x > \xi## region. The question doesn't explicitly state that, but it's the only way the situation seems to make sense in the first place (as far as I can tell... ☺️)
Just realized I tripped over the double negative.
##E-V=E+\lambda##, so it is unclear which sign it will have.
You are saying we should take it as positive... ok.
That leaves two things to question:
- does the derivative need to be continuous? I note that in the infinite well we have zero outside the well and ##\sin(nx\pi/L) ## inside. That does not have a continuous derivative at the boundaries.
- why should ##k\xi=\pi/2##?
 
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  • #7
haruspex said:
That leaves two things to question:
- does the derivative need to be continuous? I note that in the infinite well we have zero outside the well and ##\sin(nx\pi/L) ## inside. That does not have a continuous derivative at the boundaries.
Infinite potential barriers are a bit weird; from the Schroedinger equation it's possible to show that solutions have continuous derivatives so long as ##V(x)## does not have any infinite discontinuities. So whilst the discontinuity in ##\varphi'## at ##x=0## can be tolerated, we still need to impose continuity in ##\varphi'## everywhere else.

haruspex said:
- why should ##k\xi=\pi/2##?
I think that is my mistake. I edited my previous post #5 just a short while ago to reflect that the actual statement should be ##k \xi > \pi/2##; that's because it's only possible to satisfy the continuity conditions at ##x = \xi## if the gradient of the wave-function is negative. And without loss of generality we can take ##k## to be positive (because the wave-function itself is not physically meaningful, only it's [modulus] square).
 
  • #9
etotheipi said:
Yeah, that's the non-sense bit that I can't seem to figure out 😄. That seems to be the false assumption (i.e. I believe we are instead only allowed to write ##k\xi > \pi/2##), but the question remains to then find the actual value of ##k \xi## in the ground state...
You have to solve it numerically.

You can also look at the system graphically. Let ##k_0^2 = 2m\lambda/\hbar^2##. Then you have
$$k^2 = \frac{2m(E+\lambda)}{\hbar^2} = -\gamma^2 + k_0^2.$$ So the solutions to the system correspond to the intersection of ##\gamma = -k \cot k\xi## with the circle ##\gamma^2 + k^2 = k_0^2## in the ##k\gamma## plane.
 
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FAQ: Probability a particle is in a certain region

1. What is the meaning of "probability a particle is in a certain region"?

"Probability a particle is in a certain region" refers to the likelihood or chance that a particle will be found within a specific area or region in space. This probability is based on various factors such as the particle's energy, position, and the potential energy of the region.

2. How is the probability of a particle being in a certain region calculated?

The probability of a particle being in a certain region is calculated using the wave function of the particle, which describes the probability amplitude of finding the particle at a specific position. The square of the wave function gives the probability density, which is used to calculate the probability of finding the particle in a given region.

3. Can the probability of a particle being in a certain region ever be 100%?

No, according to the principles of quantum mechanics, the probability of a particle being in a certain region can never be 100%. This is because the position of a particle is described by a wave function, which allows for the possibility of the particle being found in multiple locations simultaneously.

4. How does the uncertainty principle relate to the probability of a particle being in a certain region?

The uncertainty principle states that it is impossible to know both the exact position and momentum of a particle at the same time. This means that the more accurately we know the position of a particle, the less certain we are about its momentum, and vice versa. Therefore, the probability of a particle being in a certain region is affected by the uncertainty principle.

5. Can the probability of a particle being in a certain region change over time?

Yes, the probability of a particle being in a certain region can change over time. This is because the wave function of a particle can evolve and change over time, which affects the probability of finding the particle in a specific region. This is known as wave function collapse and is a fundamental concept in quantum mechanics.

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