Probability a particle is in a certain region

[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!

 

[haruspex]

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.

 

[vela]

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 $x \to \infty$.

 

[haruspex]

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?

 

[etotheipi]

$k \xi = \pi / 2$ 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...

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... )

 

[haruspex]

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 $k\xi=\pi/2$?

 

[etotheipi]

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.

- 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).

 

[haruspex]

I think that is my mistake.

Yes. I note that at 6.2.2 of http://physics.mq.edu.au/~jcresser/Phys201/LectureNotes/SchrodingerEqn.pdf the value of k is tuned to achieve the continuity of the derivative.

 

[vela]

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.