I am struggling to answer the following question:  Let ψ₁(x) and ψ₂(x) be normalisable energy eigenfunctions for a particle of mass m in one dimension moving in a potential V(x).  Suppose that ψ₁ and ψ₂ have the same energy eigenvalue E. By considering ψ₁ψ₂' - ψ₂ψ₁', show that these wavefunctions must be proportional to one another.  I tried differentiating the Schrödinger equations and multiplying them together appropriately, but this just yielded a lot of messy algebra with no useful outcome.  I'm also wondering if there is a physical interpretation of this? Does it mean that the wavefunction for a normalisable energy eigenstate in one dimension always be chosen to be real?
I'm reading  and I'm trying to understand the proof under "Degeneracy in One Dimension".  Here's what it says:     Considering a one-dimensional quantum system in a potential $V(x)$ with degenerate states $|\psi_1\rangle$ and $|\psi_2\rangle$ corresponding to the same energy eigenvalue $E$, writing the time-independent Schrödinger equation for the system:       $$-\frac{\hbar^2}{2m}\frac{ \partial^2\psi_1}{ \partial x^2} + V\psi_1 =E\psi_1$$       $$-\frac{\hbar^2}{2m}\frac{ \partial^2\psi_2}{ \partial x^2} + V\psi_2 =E\psi_2$$   Multiplying the first equation by $\psi_2$ and the second by $\psi_1$ and subtracting one from the other, we get:       $$\psi_1\frac{d^2}{d x^2}\psi_2-\psi_2\frac{d^2}{d x^2}\psi_1=0$$   Integrating both sides       $$\psi_1\frac{\partial \psi_2}{\partial x}-\psi_2\frac{\partial \psi_1}{\partial x}=constant$$   In case of well-defined and normalizable wave functions, the above constant vanishes, provided both the wave functions vanish at at least one point, and we find: $\psi_1(x)=c\psi_2(x)$ where $c$ is, in general, a complex constant. So, the two degenerate wave functions are not linearly independent. For bound state eigenfunctions, the wave function approaches zero in the limit $x\to\infty$ or $x\to-\infty$, so that the above constant is zero and we have no degeneracy.   Here are the parts I don't understand:   How does integrating $\psi_1\frac{d^2}{d x^2}\psi_2-\psi_2\frac{d^2}{d x^2}\psi_1=0$ yield $\psi_1\frac{\partial \psi_2}{\partial x}-\psi_2\frac{\partial \psi_1}{\partial x}=constant$.  Without having explicit formulas for $\psi_1(x)$ and $\psi_2(x)$, how do we just apparently pull them both out of the integral? I don't understand how the constant in the last equation vanishes just because $\psi_1$ and $\psi_2$ vanish at infinity. Even if it does vanish, how does that imply $\psi_1=c\psi_2$?
This is a pretty simple question, at least it seems like it should be, about sudo permissions in Linux.  There are a lot of times when I just want to append something to /etc/hosts or a similar file but end up not being able to because both &gt; and &gt;&gt; are not allowed, even with root.  Is there someway to make this work without having to su or sudo su into root? Java conditional compilation: how to prevent code chunks from being compiled? I am trying to do a new environment or command (i tried both) where i can put in the name, description and code of a Method in c++ and it gives me a new section with the MethodName, the code listed and the description below. But i am not able to use the listings environment in a new command or new environment.. Consider my MWE   \documentclass[a4paper,11pt]{article} \usepackage{listings}  \newcommand{\method}[3] { \section{#1}\label{#1} \begin{lstlisting} #2 \end{lstlisting} #3 }   \begin{document}  \method{MethodName}{Method Void MethodName()}{Description of the Method blablabla}  \end{document}   How can I fix this?   I would prefer to do it using a macro since I don't need an ending. How do I include lines in resolv.conf that won't get lost on reboot? I have finally migrated to 12.04 from 7.10. I have one last part to complete but I am stumped. I am using Puppet on each server, and in the past I have included a nameserver address and a search domain name for the puppetmaster in resolv.conf.   search puppetmaster.com nameserver 192.168.1.XXX   In 12.04 resolv.conf gets overwritten when rebooted. I cannot use a static IP for these, so using the /etc/network/interfaces to help me out is a nill point.  # Dynamic resolv.conf(5) file for glibc resolver(3) generated by resolvconf(8) #     DO NOT EDIT THIS FILE BY HAND -- YOUR CHANGES WILL BE OVERWRITTEN nameserver 127.0.0.1   Is there a way to get resolvconf to handle this either in the head, tail or base? If there is, are there any examples I can use to tweak on my server.   Any help is much appreciated. Installing Kali on a USB drive - NOT Live USB or Persistence partition Greeting's, to clarify; my question is not creating a Live USB drive or a persistence partition but creating a USB with GRUB bootloader (or other) and Kali which can be used on any machine. This is similar to running other linux OS directly from USB. I am currently running Ubuntu from a pen drive. I installed the OS (Ubuntu) and the bootloader on the pen drive and I can boot off it anywhere. Here is a video explaining the same (video is not by me: )  Since, Kali's installer doesn't give option to select which drive to write the booloader to and directly writes on the HDD (sda1). This means even if I partition and install Kali on a pen drive since the bootloader will be installed on the machines fixed primary drive, the USB drive will fail to boot on a different device.  I have two questions:   Anyone has any idea how to write bootloader to a USB drive and what parameters should be given so that it boots with Kali installed on it. I am using graphic install mode for installing Kali on the USB drive, however there is no option for the install to be encrypted (Full disk encryption or at least /home folder to be encrypted). Does anyone know how can I install Kali with full disk / home folder encryption? Or enable is post install?   Thank You! How to convert from one grading scheme to another? Upvoting a downvoted answer will do +2 upvote? I have first downvoted an answer. Then, I decided to leave it the way it was before, meaning to undo my downvote just like it never existed, so I upvote to remove my downvote. Suprise! SO seems to consider it as two upvotes!  Let's say I have an answer with -3, then I decide to downvote it making it -4. After deeper thoughts, I prefer to remove my downvote to let it the way it was before I downvoted, meaning -3, but SO makes it -2 automatically!     Wouldn't it be supposed -3, and not -2? If someone else would have upvoted, it would have made it an upvote +1, not a +2!   Am I missing something here?  In fact, is there a way to tell SO that I only wish to undo my downvote, and not upvote the answer? I may understand why it does behave this way. I only wish to undo my downvote, though. probability that there is at least one defective part I study economics and I am trying to solve this problem - without success, unfortunately.     A machine has a defect rate of 6%.       a) Chosen at random 4 pieces (with replacement) from the production flow, compute the probability that none is defective.   b) In the event that extractions are performed 60, compute the probability that there is at least one defective part.   I have no idea how to answer the first question. What logical processes can I use?  Regarding the second question, I thought of using the binomial random variable with $n = 60$ and $p = 0.06$. Then:  $$\begin{multline} P (X \geq 1) = 1 - P (X &lt; 1) = 1 - P (X = 0) = \\ = 1 - \binom{60}{0}  \cdot (0.06)^ 0 \cdot (0.94)^{60} = 1 - 0.02446 = 0.97558.\end{multline}$$  Can somebody please explain me how to solve this problem? Thank you very much. Seeking Postcode Shapefiles for UK? Cron fails with exit status 127 Behaviour of increment and decrement operators in Python Background: One of my co-authors faced problem of compiling the paper that I had built using TexStudio and MikteX 2.9 on his Mac Machine running TexLive.   ShareLaTeX Role: So, one solution is to use ShareLaTeX. But the first problem pops in mind is how safe is uploading unpublished article to a publicly accessible portal?  What I did: I googled security and data breach of ShareLaTex and found nothing except some words from the ShareLaTex community about their word to ensure that our data are safe; perhaps this would not be enough.  My 1st Question: To what extent can I trust ShareLaTeX and put my ready-to-publish content there?   My 2nd Question: Is there any alternate way of sharing LaTeX with others around the globe without changing the underlying platform? Repeatedly rolling a six sided die four times and summing the highest three results gives you a distribution with what mean and standard deviation?  I've only taken AP statistics, but I would like to learn how to do this. Some questions about Hilbert matrix This Exercise $12$ page $27$ from Hoffman and Kunze's book Linear Algebra.     The result o Example $16$ suggests that perhaps the matrix      $$A = \begin{bmatrix} 1 &amp; \frac{1}{2} &amp; \ldots &amp; \frac{1}{n}\\  \frac{1}{2} &amp; \frac{1}{3} &amp; \ldots &amp; \frac{1}{n+1}\\ \vdots &amp; \vdots &amp;  &amp; \vdots \\ \frac{1}{n} &amp;\frac{1}{n+1} &amp;\ldots &amp;\frac{1}{2n-1} \end{bmatrix}$$      is invertible and $A^{-1}$ has integer entries. Can you prove that?   I will copy their words before that example:      Some people find it less awkward to carry along two sequences of   matrices, one describing the reduction of $A$ to the identity and the   other recording the effect of the same sequence of operations starting   from the identity.   In this , Adrián Barquero says:      If your linear algebra textbook is Kenneth Hoffmann and Ray Kunze's   Linear Algebra book then I was in the same situation you're right now   a few years ago, but I was adviced at the moment that this particular   exercise wasn't as easy as one might expect at first.      The matrix you have is called the Hilbert matrix and the question you   have was already asked a couple of times in math overflow  and   . They have excellent answers so I will just point you to them.   My question: Is it possible answer the question made by the authors without using high techniques?  I am not allowed to use determinants here.  PS: Those answers in MO don't answer the question above. I apologise if this question is unappropriated or if I was unable to understand their answers. I have a custom script and I want to send a desktop notification (the one that appears in the top right corner of the screen) with a custom message. How do I do that? How to Hide sections of InfoPath form based on Group membership I need to Hide "Sections" of InfoPath form using SharePoint or AD group membership (either is fine). How can I achieve this?  InfoPath 2010 in SharePoint 2010 How do open files behave on linux systems? In a ring, I was trying to prove that for all $a$, $a0 = 0$.  But I found that this depended on a lemma, that is, for all $a$ and $b$, $a(-b) = -ab = (-a)b$.  I am wondering how to prove these directly from the definition of a ring.  Many thanks! In the following sentence:     The city is populated by many people.   "many" can be replaced by something like "a great number of" (let us leave the perceived differences between the meanings of these two modifiers out of discussion):     The city is populated by a great number of people.   I understand that this replacement does not change the object "populating the city" from the original "people" to a "number", but rather shows the ability of expressions in the form "[numeral] of" to provide information about the quantity of nouns they modify.  Is this ability due to some special qualities of nouns such as "number", "quantity", "fraction", &amp;c., or is it just one usage of the preposition "of"? More Generating Functions problems