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$\Psi$ in function of 'position'

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The general exercise:

Let's focus on c).

The value of $\sigma$:

$$\sigma =\sqrt{ < x^2 > -< x >^2 }= \frac{1}{\lambda \sqrt{2}}$$

We are asked to plug in $\sigma$:

$$| \Psi (\sigma ) | = \lambda e^{\frac{-2}{\sqrt{2}}}$$

Where $\lambda$ is a positive constant.

I calculated < x > and its value is 0. Check this link to obtain further details:

https://math.stackexchange.com/questions/2893056/misleading-expected-value

About the probability

I have difficulties solving this probability as this time I got:

$P_i = \int_{-\sigma}^{\sigma} \lambda e^{-2 \lambda |x|} dx =$

$= \lambda \int_{-\sigma}^{0} e^{-2 \lambda |x|} dx$

$+ \lambda \int_{0}^{+\sigma} e^{-2 \lambda |x|} dx = 0$

Where did I get wrong? I have been trying to get this probability but none outcome seems to make sense.

$$P_o = 1 - P_i $$

Where $P_i$ and $P_o$ mean inside and outside respectively

Thank you

asked Aug 25 in Physics Problems by JD_PM (490 points)
edited Aug 31 by JD_PM
Sorry I don't understand your difficulty. The question asks you to sketch the graph of $|\Psi|^2$ vs $x$ and mark the lines $x= \pm \sigma$. Where is the difficulty?

For the probability, you have the right idea but you don't need $\lambda$ in the integral. You are just finding a probability, not an average value within a range. $\lambda$ does not vary anyway so there is no need to integrate to find an average.
But we are not working with discrete distributions, so I thought we had to integrate. How would you do it?

I uploaded the graph I got.
That is not the graph of $|\Phi|^2$, which is symmetrical about the $y$ axis.

Sorry I should have explained better : you don't need to integrate to find an average value of $\lambda$ within a fixed range (if that is the thinking behind your expression for $P_i$), because $\lambda$ is a constant. You do need to integrate to find the area under the graph. This is $\int_{-\sigma}^{+\sigma} |\Phi|^2 dx$.  You don't need the factor of $\lambda$ to calculate $P_i$.
So there has to be an exponential function symmetrically going up. What I do not understand is why. I mean, we have the function |x| in the exponential and to obtain the symmetrical graph do not we have to have both negative and positive exponential functions? |x|function does not transform negative exponential into positive?

OK so no $\lambda$. But we agree with the fact that first we have to obtain $P_i$ by $\int_{-\sigma}^{+\sigma} |\Phi|^2 dx$ and after that $P_o = 1 - P_i$ I found that the given solution to this probability is 0.368 while I got 0.632. It appears they just computed the internal probability. I do not understand why to be honest.
1. The argument of $e^{-2\lambda |x|}$ is never +ve, so the maximum value is $e^0=1$.  On your graph there is no maximum.
2. The value of $|x|$ is the same whether $x \gt 0$ or $x \lt 0$ so the same can be said for  $e^{-2\lambda |x|}$.
3. You calculated $< x > = 0$ so this means the graph cannot be as you plotted it.

I will look into the probability later.
Okey I fixed the graph. Do you agree with it?

Fine, I will wait for probability issue.
Graph is ok except there should be nothing above $y=1$.

Probability : I agree with $\sigma=\frac{1}{\lambda\sqrt2}$. My apologies you DO need $\lambda$ in the integral for $P_o$ - I forgot it is there to normalise $|\Psi|^2$.  Then I get $P_o=e^{-\sqrt2}=0.243$.
OK graph fixed.

Probability: I got this time that $P_i = 0$, which has to be wrong (check my question, I updated it). Could you illustrate your point? And do you really need to integrate for getting $P_o$ ?
You can see from your graph that $P_i \ne 0$ and that $\int_{-\sigma}^0 |\Psi|^2dx = \int_0^{+\sigma} |\Psi|^2dx$. You have assumed that one integral is the -ve of the other without any justification. $e^{-2\lambda |x|}$ is an even function, not an odd function. See your graph.

$P_i=1-P_o=\int_{-\sigma}^{+\sigma} |\Psi|^2dx=2\lambda \int_0^{+\sigma} e^{-2\lambda x}dx=[e^{-2\lambda x}]_{+\sigma}^0=1-e^{\sqrt2}$ therefore $P_o=e^{-\sqrt2}=0.243$.

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